

The Mobius meter across digit designs
Fix a base base >= 3 and a digit set F inside {0..base-1} with fill = |F| >= 2. The digit-restricted set S_F holds the positive integers n whose digits at that base all lie in F, with no leading zero: a one-dimensional digit design, the same restriction rule that carves every fractal in this tree, read on the integer line instead of the square. This page measures how much the Mobius function cancels along each design, against the design's own size - and proves that the columns are not independent: digit sets that are scalar multiples of each other carry exactly transferred meters, including one family whose meter vanishes identically and one base-4 pair locked in exact anti-symmetry. Every number is printed by lab/rs/mobius-designs, the divisor section's census numbers by lab/rs/rho-decoupling, the GRH section's by lab/rs/mertens-numerology and the pair route's by lab/py/mobius-region.
Tags as everywhere in this tree: Proved means derived here from definitions, Verified means recomputed exactly and checked against an independent path, Conjecture is labelled belief, Refuted means killed here with the witness that kills it.
The zeros are the other face and they are a different page. The design's own Dirichlet series zeta_F(s) = sum_(n in S_F) n^(-s) has zeros inside its own half-plane of absolute convergence, a comb of them along the pole lattice, three products where the integers have one Euler product, and a Mertens function of its own that runs the wrong way; none of it reaches the meter measured here, and the decoupling is why this page's question is about mu restricted to S_F and nothing else: zeta.
The meter and its yardstick
A_F(x)countsS_Fup tox. The count is exact at every checkpoint:A_F(base^level) = fill^level - 1when0 in F(plus 1 when1 in Ftoo, for the boundary elementbase^levelitself), andA_F(base^level) = (fill^(level+1) - fill)/(fill - 1)when0is not inF, by counting digit strings of each length. Proved; the lane's tests pin it against direct enumeration. Between checkpointsA_F(x)/x^(log(fill) / log(base))carries the log-periodic ripple every design in this tree carries - the classical fluctuation of digital sums (Flajolet, Grabner, Kirschenhofer, Prodinger and Tichy 1994) - so a checkpoint value is a grid value, never a constant.- The meter is
M_F(x) = sum of mu(n)overn in S_F,n <= x, and the exponent istheta(F) = limsup of log|M_F(x)| / log A_F(x). A single cut of|M_F|is a bad estimator - the meter crosses zero freely - so the census prints two readings per level:M_F(base^level)itself, and the running maximummax of |M_F(x)|overx <= base^level, whose exponentthetamaxis monotone in the numerator and is the estimator the slope tables use. - The yardstick matters.
A_F(x)grows likex^(log(fill) / log(base)), soS_Fis sparse, and a bound of shapeo(x)is weaker than the trivial|M_F(x)| <= A_F(x). The indicator ofS_Fis automatic in that base, so Mullner 2017 (automatic sequences fulfill the Sarnak conjecture) givesM_F(x) = o(x)for everyF: orthogonality holds and the question is well-posed, but against the set's own mass that bound says nothing at all. The same shape repeats in base 2 through circuits: the indicator is computable in bounded depth from the binary digits, so Green 2012 also giveso(x), again below the trivial bound. Verified against the literature. The honest question istheta, and it is open at every2 <= fill <= base - 1. - The Dirichlet series over
S_Fis built territory, and this page claims nothing about it: the abscissa islog(fill) / log(base)(Kohler and Spilker 2009, with position-varying digit rules in Nathanson 2021); the series continues meromorphically toCwith simple poles amongs = log(fill) / log(base) - m + 2 pi i j / log base(the automatic-series mechanism of Allouche, Mendes France and Peyriere 2000, carried out for missing digits in Burnol 2026 and unified in Allouche, Shallit and Stipulanti 2025); a pole lattice of period2 pi i / log basereads as log-periodic oscillation through the Mellin dictionary of Flajolet, Gourdon and Dumas 1994, and the oscillation is visible in the series' own numerical moments (Burnol 2026 oscillations); the Mobius function itself is automatic in no base, so the Mobius-weighted series inherits none of that continuation (Coons 2010); and no Mobius or Mertens sum appears anywhere in that literature. Verified against the sources in REFS.md. The series does not carry the meter the way zeta carries Mertens:S_Fis not multiplicatively closed - atbase 3,F = {0,1}, both4 = 11and13 = 111lie inS_Fwhile4 x 13 = 52 = 1221does not - so there is no Euler product andM_Fis not the coefficient sum of an inverse series. Proved by that witness. - The full digit set is the classical boundary.
S_Fis then every integer,M_Fis the Mertens function of Mertens 1897, andM(x) = O(x^(1/2 + eps))for everyeps > 0is equivalent to the Riemann hypothesis (Titchmarsh 1986, Theorem 14.25 (C)), whilelimsup |M(x)|/sqrt(x) >= 1.06unconditionally by Odlyzko and te Riele 1985, so the exponent over allxequals1/2exactly when RH holds. Verified against the literature. This page claims nothing about RH: the full-set column below is a control rendered for scale, and the tree's own claims live in the restricted columns. - The even moments of the digit transform are additive energies. Proved.
sum_{a mod base^level} |hat F_level(a/base^level)|^(2r) = base^level E_r(level)withE_r(level)the number of2r-tuples of digit strings of lengthlevelwithn_1 + ... + n_r = n_(r+1) + ... + n_(2r) mod base^level, by orthogonality, andE_r(level)is counted by a carry DP on the carry pairs of the two sides, so each moment is C-finite inlevelof order at mostr(r+1)/2and its growth constantLambda(2r) = base rhois an algebraic number,rhothe Perron root of the transfer matrix, certified in exact rationals (lab/rs/rho-decoupling, therieszmodule). - The fourth-moment constants. Verified.
Lambda(4) = 18at{0,1},{0,2}and{1,2}in base 3 (rho = 6),2(23 + sqrt 353) = 83.5766at{0,1,2}in base 4 (x^2 - 23x + 44),(275 + 5 sqrt 2369)/2 = 259.1809at{0,1,2,3}in base 5 (x^2 - 55x + 164),95at{0,2,4}in base 5, and6566.412to6567.410over the reflection classes of one excluded digit in base 10; every value sits strictly inside[max(fill^4, base fill^2), base fill^3]and a hair abovefill^4at the dense families (log_base(Lambda(4)/fill^4)is0.107at base 3,0.0004at base 10), and the sixth, eighth and tenth moments at{0,1}base 3 are39 + 3 sqrt 79,3(99 + sqrt 5265)/2and a cubic (lab/rs/rho-decoupling). - What a moment buys the bilinear sum. Proved. Holder with the
2r-th moment on the digit side and Parseval on the bilinear side bounds the Type II sum overmup toMandlup toN,4MN <= x, byx^(theta_p/p + 1/2 - 1/p)withtheta_p = log Lambda(p)/log base, which is at leastx^(alpha + 1/4)for every evenp >= 4and every digit set, above the trivialx^alpha; so no moment of the digit transform alone beats the trivial bound, and the route needs the bilinear sum on the minor arcs below its own root mean square, which random-sign coefficients defeat on the census (Verified,lab/rs/rho-decouplingthearcslines). - The multiplicative energy of a digit column has no exponent of its own. Proved. With
E_x(level) = #{(n_1, n_2, n_3, n_4) in D_level^4 : n_1 n_2 = n_3 n_4}andK = fill^level, the two diagonals give2K^2 - K <= E_x(level), andE_x(level) = sum_m r(m)^2 <= K^2 max_m r(m)withr(m) <= d(m)givesE_x(level) = fill^(2 level) x^(o(1))for every base and digit set (the census reads58760487at{0,1}base 3,level 12, the exponent1.356938falling toward2 alpha = 1.261860); so a Type II sum estimated through the energy obeys|Sigma| <= (2MN)^(1/2) x^(alpha/2 + o(1))and misses the trivial bound by(1 - alpha)/2; the excess over the diagonal is structure, not arithmetic: the shift family(base^i u, base^j v, base^(i') u, base^(j') v)withi + j = i' + j', counted in closed form when0is a digit, is0.44of it at{0,1}base 3,level 12(lab/rs/rho-decoupling, themenergymodule). - Above the sup the
L^pnorms of the transform buy nothing. Proved. The sandwichmax(fill^p, base fill^(p/2)) <= Lambda(p) <= base fill^(p-1)forcesLambda(p)^(1/p)/filldown to1, and it reads1.224744, 1.029883, 1.004288, 1.000653, 1.000107atp = 2, 4, 6, 8, 10for{0,1}atbase 3, so every higher norm is the supremum up to a factor tending to1and no ladder of moments reaches past the bullet above (lab/rs/rho-decoupling,riesz higher moments). - The unbalanced kernel carries no Type II estimate uniform over bounded coefficients at any digit set containing
0. Proved. Ata_m = b_l = 1the Type II sum is the box representation count and some admissible box carriesR >= x^(alpha - o(1)), so the trivial bound is attained and the only target left is the balanced sum; there the route returns the box's own trivial bound, the ratio of bound to trivial rising through1(1.0134atlevel 12and1.0730atlevel 14at{0,1}base 3) while the margin ofalphaover the achieved exponent falls from0.035675to0.027009, and the digit column is worse for the method than a random column of the same density at every cell of the arc regime (lab/rs/rho-decoupling,menergy type II). - Two box witnesses floor every coefficient-free route at
x^alpha. Verified at the dense cells. Ata_m = b_l = 1the Type II sum reads0.19to0.41offill^levelover the nine dense cells, and ata_m = 1_(base | m),b_l = 1the balanced sum still reads0.0024to0.104offill^levelthere, carried by the frequenciesa'/base^jat boundedj, which are exactly the major arcs the pair route below removes and computes; the statement for every digit set rests onR >= fill^level (log x)^(-C)and stays Conjecture, and the witness is void atalpha = 0.15, where the box is empty (lab/rs/rho-decoupling). - The Mobius signs cancel the column's correlation no better than random signs. Verified. The digit column carries a real off-diagonal multiplicative correlation, zero in the mean for a random column of the same density, and the Mobius and Liouville signs cancel it no better than an unstructured sign vector on the same support does,
|Sigma_mu|sitting at0.0913to0.7178of the random-sign root mean square against0.0359to1.5048for the support-matched controls over sixteen boxes, with the split against those controls3, 9, 4at chi-square0.375against the uniform-rank null; a sign vector built by greedy flips against a known column drives the same Cauchy-Schwarz bound to0.0265of its diagonal floor, so the census measures the arithmetic of the coefficients and not a limit of the method (lab/rs/rho-decoupling,menergy signedandmenergy signed engineered). - The coefficient the method is given is not the coefficient it would need. Verified. At seven of the eight swept boxes with both sides above
x^(2/5), the boxes a Vaughan decomposition actually produces, the coefficient sequence it hands the bilinear sum takes values in{-1, 0, 1}and needs no normalisation, and its full quadratic form sits between0.69and1.21of its own diagonal, where a sign vector engineered against the column reads0.13to0.21on the same boxes, so the sequence the method is given and the sequence the method would need are different objects (lab/rs/rho-decoupling,menergy signed vaughan). - The large-values refinement is the moment route itself. Verified. The large-values refinement of the moment route is costed out and is the
l^2route itself, exponent(1 + alpha)/2at every threshold (lab/rs/rho-decoupling,riesz large values chain, 66 cells over six families); the large frequencies are adjacent grid points (407in331runs at{0,1}base 3,level 12,eta = eta_4), so the grid offers no spacing gain, and at the dense families the bilinear sum ata_m = b_l = 1equals the box representation count,0.38 fill^levelthere, so no bound uniform over bounded coefficients holds at those families.
The exact transfer between designs
The census columns are tied together by one carry-free mechanism. Proved:
- Scaling. If every digit of
Fisatimes a digit ofF', soF = aF'inside{0..base-1}, thenm -> ammapsS_F'bijectively ontoS_Fpreserving digit length:am = sum (a d_j) base^jand eacha d_j <= base - 1, so no carry occurs and the digit string scales digitwise. HenceA_F(base^level)equals the string count ofF'at the same depth, andM_F(base^level) = sum of mu(am)overm in S_F'with at mostleveldigits. - Vanishing. If
ahas a square factor thenmu(am) = 0for everym, soM_Fis identically zero: atbase 5,F = {0,4} = 4 x {0,1}, the meter reads 0 at all 21 levels. A census that reads cancellation without factoring out the digit gcd reads this as infinite cancellation; the digit gcd must be squarefree beforethetameans anything. - Prime twist. If
a = pis prime thenmu(pm)is-mu(m)onp-freemand0otherwise, soM_(pF')(base^level) = -sum of mu(m)over them in S_F'not divisible byp. Atbase 3,F' = {0,1}: an elementm = sum of 3^jis odd exactly when its count of 1-digits is odd, and reading the digit string as a binary index that parity is the Thue-Morse sign, so the{0,2}column is the Thue-Morse-twisted{0,1}column. - Base-4 anti-symmetry. At
base 4,M_{0,2}(4^level) = -M_{0,1}(4^level)exactly: since4 | base, an element ofS_{0,1}is0or1 mod 4by its unit digit, so every even element is divisible by 4 and carriesmu = 0, and the odd-part twist above is minus the whole meter. Stronger,M_{0,2}(x) = -M_{0,1}(x/2)at every realx, and sinceS_{0,1}has no element strictly between(4^level - 1)/3and4^levelthe running maxima agree level by level as well. The census confirms both at all 22 levels, e.g. meters-110/110atlevel 15,-342/342atlevel 17,34/-34atlevel 22, andMmax = 1553for both atlevel 22.
Verified: the generator recomputes all eight scaled census families ({0,2} at base 3; {0,2}, {0,3} at base 4; {0,2}, {0,3}, {0,4}, {2,4}, {0,2,4} at base 5) from their primitive families through mu(am) and asserts equality at every level. The mechanism needs a common digit factor, so it partitions the census into primitive columns and their twists and says nothing across primitive columns.
The census
Every M_F(base^level) below is an exact integer: restricted families enumerated in ascending order with mu from deterministic factorization (trial division, Miller-Rabin on the twelve witnesses 2..37, Pollard rho), controls by a linear Mobius sieve; one family (base 3, F = {1,2}, level 16) is computed by both methods and asserted equal at every level. The base-10 control reproduces A084237 (-1, 1, 2, -23, -48, 212, 1037, 1928 at 10^1..10^8). Every table below is extracted by script from the generator's printed rows, never assembled by hand. Verified.
The three base-3 columns, checkpoint meter and running maximum Mmax = max of |M_F(x)| over x <= 3^level per row:
level | M_{0,1} | max | M_{0,2} | max | M_{1,2} | max |
|---|---|---|---|---|---|---|
| 4 | -2 | 3 | 2 | 3 | -8 | 8 |
| 6 | 2 | 5 | 0 | 3 | -8 | 11 |
| 8 | 2 | 8 | 3 | 7 | -31 | 33 |
| 10 | 5 | 13 | 0 | 11 | -14 | 38 |
| 12 | 56 | 61 | -37 | 40 | -35 | 88 |
| 14 | 11 | 105 | -10 | 67 | -205 | 230 |
| 16 | 149 | 173 | -124 | 152 | 4 | 281 |
| 18 | -30 | 312 | 67 | 249 | -1461 | 1582 |
| 20 | 496 | 539 | -382 | 485 | -3175 | 3255 |
| 21 | 533 | 866 | -194 | 617 | -2005 | 3855 |
| 22 | 1009 | 1089 | -1205 | 1324 | -690 | 3855 |
| 23 | 1824 | 2848 | -2242 | 2942 | -3214 | 3855 |
| 24 | -1886 | 3296 | -133 | 3843 | -3248 | 4113 |
The final checkpoint of every family, with thetamax = log(Mmax)/log A and its drift (max minus min) over the last five levels:
base | F | level | A_F(base^level) | M_F(base^level) | Mmax | thetamax | drift |
|---|---|---|---|---|---|---|---|
| 3 | 01 | 24 | 16777216 | -1886 | 3296 | 0.4869 | 0.0452 |
| 3 | 02 | 24 | 16777215 | -133 | 3843 | 0.4962 | 0.0596 |
| 3 | 12 | 24 | 33554430 | -3248 | 4113 | 0.4802 | 0.0754 |
| 4 | 01 | 22 | 4194304 | 34 | 1553 | 0.4819 | 0.0391 |
| 4 | 02 | 22 | 4194303 | -34 | 1553 | 0.4819 | 0.0391 |
| 4 | 03 | 22 | 4194303 | -541 | 1180 | 0.4638 | 0.0488 |
| 4 | 12 | 22 | 8388606 | -855 | 3965 | 0.5197 | 0.1056 |
| 4 | 13 | 22 | 8388606 | -712 | 2631 | 0.4940 | 0.0727 |
| 4 | 23 | 22 | 8388606 | -3255 | 3258 | 0.5074 | 0.0157 |
| 4 | 012 | 14 | 4782969 | -503 | 1057 | 0.4527 | 0.0475 |
| 4 | 013 | 14 | 4782969 | 2313 | 2899 | 0.5183 | 0.0487 |
| 4 | 023 | 14 | 4782968 | -753 | 1166 | 0.4591 | 0.0862 |
| 4 | 123 | 14 | 7174452 | -592 | 1644 | 0.4691 | 0.0729 |
| 5 | 01 | 21 | 2097152 | 153 | 849 | 0.4633 | 0.0485 |
| 5 | 02 | 21 | 2097151 | 250 | 889 | 0.4665 | 0.0268 |
| 5 | 03 | 21 | 2097151 | -116 | 700 | 0.4501 | 0.0311 |
| 5 | 04 | 21 | 2097151 | 0 | 0 | - | - |
| 5 | 12 | 21 | 4194302 | -128 | 1643 | 0.4856 | 0.0732 |
| 5 | 13 | 21 | 4194302 | -2875 | 3533 | 0.5358 | 0.0456 |
| 5 | 14 | 21 | 4194302 | -1511 | 2750 | 0.5193 | 0.0640 |
| 5 | 23 | 21 | 4194302 | 405 | 914 | 0.4471 | 0.0581 |
| 5 | 24 | 21 | 4194302 | 1065 | 2287 | 0.5072 | 0.0540 |
| 5 | 34 | 21 | 4194302 | -2137 | 2538 | 0.5141 | 0.0401 |
| 5 | 012 | 13 | 1594323 | -1016 | 1416 | 0.5080 | 0.0846 |
| 5 | 013 | 13 | 1594323 | -137 | 768 | 0.4652 | 0.0528 |
| 5 | 014 | 13 | 1594323 | 213 | 1005 | 0.4840 | 0.0821 |
| 5 | 023 | 13 | 1594322 | 759 | 858 | 0.4729 | 0.0455 |
| 5 | 024 | 13 | 1594322 | 686 | 959 | 0.4807 | 0.0240 |
| 5 | 034 | 13 | 1594322 | 501 | 1000 | 0.4837 | 0.0847 |
| 5 | 123 | 13 | 2391483 | 88 | 816 | 0.4565 | 0.0489 |
| 5 | 124 | 13 | 2391483 | 1036 | 1613 | 0.5029 | 0.0604 |
| 5 | 134 | 13 | 2391483 | -725 | 981 | 0.4690 | 0.0519 |
| 5 | 234 | 13 | 2391483 | -1926 | 2021 | 0.5182 | 0.1056 |
| 5 | 0123 | 11 | 4194304 | -474 | 1725 | 0.4887 | 0.0732 |
| 5 | 0124 | 11 | 4194304 | 426 | 1494 | 0.4793 | 0.0673 |
| 5 | 0134 | 11 | 4194304 | -644 | 1633 | 0.4852 | 0.0222 |
| 5 | 0234 | 11 | 4194303 | -362 | 2179 | 0.5041 | 0.0794 |
| 5 | 1234 | 11 | 5592404 | 145 | 1101 | 0.4508 | 0.0996 |
Base 10 with one digit excluded, the Kempner designs (fill = 9, the sets behind the convergent harmonic series of Kempner 1914, revisited at s = 1 in Allouche, Hu and Morin 2024), at x = 10^8:
| excluded | A_F(10^8) | M_F(10^8) | Mmax | thetamax |
|---|---|---|---|---|
| 0 | 48427560 | 6410 | 8177 | 0.5091 |
| 1 | 43046720 | 4108 | 6069 | 0.4956 |
| 2 | 43046721 | -183 | 3357 | 0.4619 |
| 3 | 43046721 | 455 | 3512 | 0.4644 |
| 4 | 43046721 | 56 | 4957 | 0.4841 |
| 5 | 43046721 | -7614 | 10601 | 0.5273 |
| 6 | 43046721 | -693 | 2564 | 0.4465 |
| 7 | 43046721 | -1411 | 6494 | 0.4994 |
| 8 | 43046721 | 2131 | 4495 | 0.4785 |
| 9 | 43046721 | 2181 | 5234 | 0.4871 |
The full-set controls at comparable depth: M(3^17) = -1423 with Mmax = 4610 (thetamax 0.4517), M(4^13) = 329 with 2845 (0.4413), M(5^11) = 617 with 2573 (0.4436), M(10^8) = 1928 with 3448 (0.4422). The Mertens function itself - limiting exponent exactly 1/2 if and only if RH, and at least 1/2 unconditionally - reads 0.4413..0.4517 at these depths, which calibrates every reading above: at census mass even the classical meter sits a few hundredths under 1/2.
The distribution of the apparent exponent across designs at fixed base, sorted by the generator: at base 3 the three columns read 0.4802, 0.4869, 0.4962; at base 4 the ten run 0.4527 to 0.5197; at base 5 the twenty-four with nonzero meter run 0.4471 to 0.5358; at base 10 the ten Kempner columns run 0.4465 to 0.5273. All 47 readings sit within 0.054 of 1/2, against cut readings (theta at the checkpoint alone) that scatter over 0.22..0.53 for the same data - the single-cut estimator is noise, the running maximum is the meter.
Digit strings across divisors
The meter weighs mu along a design; this section weighs the design itself against a divisor, the arithmetic input any multiplicative estimate over S_F has to have. Write N_F(level; d, r) for the number of digit strings of length level over F whose value sum_j f_j base^j is r mod d, and N_F(level; d) = N_F(level; d, 0). The value map is injective on strings of one length, so with 0 in F this counts the multiples of d below base^level whose padded digits lie in F, and with 0 outside F it is the block of S_F at length level, the blocks l <= level partitioning S_F below base^level. Write fill = |F|, e(x) = exp(2 pi i x), g_F(t) = sum_{f in F} e(f t), Delta_F for the gcd of the digit differences, gamma_F(d) = max over a not 0 mod d of |g_F(a/d)|/fill, and normalized error for d |N_F(level; d) - fill^level/d| / fill^level. Residue distribution of digit-restricted sets is the subject of Erdos, Mauduit and Sarkozy 1998; what follows is derived here from the transform, each statement carrying its own hypotheses, and alpha = log(fill) / log(base) is the design's dimension throughout. Every census number is printed by lab/rs/rho-decoupling; the rest is exact arithmetic carried out in the sentence that prints it.
- Orthogonality. Proved.
N_F(level; d, r) = (1/d) sum_{a mod d} e(-a r/d) prod_{j < level} g_F(a base^j/d): expand the divisibility indicator in additive characters modd; the digits are independent, so the character sum factors over positions. Thea = 0term isfill^level/dand every bound below is a bound on the rest. - The uniform geometric bound. Proved. For
fill >= 2,d >= 2,(d, base) = 1andgcd(d, Delta_F) = 1, everyrand everylevel >= 1:|N_F(level; d, r) - fill^level/d| <= ((d-1)/d) fill^level (1 - 8/(fill^2 d^2))^level <= fill^level exp(-8 level/(fill^2 d^2)). Coprimality tobasekeepsa base^jnonzero moddat every position,|g_F(a/d)|^2 = fill^2 - 4 sum_{f < f'} sin^2(pi a (f' - f)/d), and ifddivideda (f' - f)for every pair thend/gcd(a, d)would divideDelta_Fand forced | a, so one pair sits at distance>= 1/dfrom an integer andgamma_F(d)^2 <= 1 - 16/(fill^2 d^2). The census asserts the weaker form as an exact integer inequality at every cell where the hypotheses hold; the largest observed-to-bound ratio is0.187, atbase 100,F = {0,1},level 16. The exponentd^(-2)is not slack: atd | base - 1withFan arithmetic progression of common differencem', takinga m' = 1 mod dgives|g_F(a/d)|/fill = sin(pi fill/d)/(fill sin(pi/d)) = 1 - Theta(fill^2/d^2). Summed over a range it is microscopic and never a route on its own:D exp(-8 level/(fill^2 D^2)) < 1fails pastD ~ sqrt(level)/fill, so this bound alone certifies a level of distribution of that size and nothing like a power ofx. - The dense-digit bound. Proved. For
F = {0..base-1}minusEwithm = |E|,fill = base - mand(d, base) = 1:gamma_F(d) <= (d/2 + m)/fill, sinceg_Fis the full Dirichlet kernel lessg_E,|D_base(a/d)| <= 1/(2||a/d||) <= d/2and|g_E| <= m; hence ford/2 + m < fillthe error is at mostfill^level ((d/2 + m)/fill)^level, uniform inr. - A power saving at level
base^(1-eps). Proved. Fixeps in (0,1)and takebase >= 4^(1/eps),m <= base^(1-eps)/2,level >= 4/eps. Every2 <= d <= base^(1-eps)coprime tobasethen has per-digit factor(d/2 + m)/fill <= base^(-eps/2), sosum over those d of |N_F(level; d) - fill^level/d| <= fill^level base^(1 - eps level/2) <= fill^level x^(-eps/4)atx = base^level: a power saving over the whole block, not one divisor at a time. The saving is carried by the digit count and not by the base. At the fixed divisord = 7the per-digit error rate reads0.4869, 0.3312, 0.2484, 0.1104, 0.0167alongbase 3, 4, 5, 10, 100at the designs{0,1},{0,1,2},{0,1,2,3},{0..9}less7and{0..99}less37, asfillruns2, 3, 4, 9, 99, against per-factor ceilingsgamma_F(7) = max_a |sum_{f in F} e(a f/7)|/fillreading0.9010, 0.7490, 0.5617, 0.2002, 0.0221, the ceiling of the design and not offill, so the dense row's0.0221is attained ata = 2while the other four are attained ata = 1; forF = {0,1}atbase 100it stays0.4992with ceiling0.9010, the same ceilingF = {0,1}has atbase 3. - The split across the base's own divisors. Proved. For
d = d1 d2withd1 | base^mfor somem <= leveland(d2, base) = 1, the lowmdigits fix the value modd1and reach the rest only through the invertible multiplierbase^m mod d2, soN_F(level; d) = sum over w in F^m with d1 | val(w) of N_F(level - m; d2, r_w),r_w = -val(w) (base^m)^(-1) mod d2. The density splits exactly,rho_F(d1 d2) = (N_F(m; d1)/fill^m) (1/d2), and the base part is a digit-string count rather than1/d1: a divisor sharing a factor withbaseis read off the digits, never off a density. Checked against direct enumeration atbase 6,d = 10. - The digit-gcd hypothesis is a wall, not a convenience. Proved. If
gcd(d, Delta_F) > 1there is no equidistribution at all: atbase 3,F = {0,2},d = 2every value is even,N_F(level; 2) = fill^level, and the normalized error is exactly1at everylevel. Overd <= 200the unrestricted worst error for that family reads1.0483atlevel 32, pinned atd = 164, against0.019166oncedis required coprime toDelta_F. Such families reduce to a primitive one throughS_(aF') = a S_(F'), the scaling map of the transfer above. - The slow column at fixed digit count. Verified. The bound decays in
levelonly, at a rate the digit count controls, and the census sees nothing better:F = {0,1}atbase 100has worst normalized error28.593, 14.590, 9.0340, 7.2034atlevel 16, 32, 64, 96overd <= 500, per-digit factor0.9929, with argmaxd = 481 | base^3 - 1at the first two depths andd = 303 | base^2 - 1at the last. Sparse digit sets are outside the reach of every per-divisor estimate here, exactly as they are outside the reach of the exponent census above. - The worst divisor is pinned. Verified. The obstruction is small multiplicative order: the orbit
a base^j mod dvisits onlyord_d(base)points, so no averaging happens across positions, and at every family's deepest level the sweep argmax hasord_d(base) <= 8, hence dividesbase^t - 1witht <= 8(d = 164atbase 3,d = 143atbase 10,d = 101, 303, 481atbase 100); shallow depths can stray,d = 199withord = 99atbase 10,level 6. - The signed pinned sum does not cancel. Verified. Weight each squarefree pinned modulus
e = (base^t - 1)/g,g | base - 1,e >= 2,t <= level, bymu(e), withT_level(e) = N_F(level; e) - fill^level/e, and set the signed sumSigma_level = sum mu(e) T_level(e)against the absolute sumAbs_level = sum |T_level(e)|over the same moduli. Printed at everylevel 3..40by the generator, the ratioSigma_level/Abs_levelswings across[-1, 1](-1.00atlevel 5, 6in the first family) with no decay:-0.211, -0.123, +0.069, -0.498atlevel 10, 20, 30, 40forF = {0,1}atbase 3and+0.812, -0.495, -0.127, -0.192for one excluded digit atbase 10, whileAbs_level/fill^levelreads2.1 * 10^-4and3.9 * 10^-12atlevel 40; the sum rests on6of29terms in the first family and4of60in the second, the four largest att = 7, 9and att = 5, 7, 8, 10. The signed weight is itself a Mertens-type sum and that is why signing buys nothing here: withT_level(d) = N_F(level; d) - fill^level/dandP_level(e) = sum_{(a,e) = 1, 0 < a < e} hat F_level(a/e), real becauseaande - aconjugate, grouping each frequency by its reduced denominator givesT_level(d) = (1/d) sum_{e | d, e >= 2} P_level(e)and hencesum_{d <= U} mu(d) T_level(d) = sum_{e >= 2} (mu(e)/e) M_e(U/e) P_level(e)withM_e(y) = sum_{f <= y, (f,e) = 1} mu(f)/f, Proved; a pinned primitive modulus enters weighted byM_e, somu(e)fixes the sign and a Mertens-type sum fixes the size, and the signed route restates the wall one layer down rather than escaping it. No bound onM_e(y)uniform ineis available to lean on: at the primorialeof all primes up toPandy = Pthe onlyf <= ycoprime toeisf = 1, soM_e(P) = 1exactly, Refuted for any unrestricted uniformity. EveryN_F(level; e)is an exact integer of the carry count, which sums the digits in each residue class of positions modtand counts the targetsj eby carries, polynomial inlevelat everyt;mu(e)is read off a complete factorisation of everybase^t - 1tot = 40, cyclotomic factors first, then Pollard-Brent, every prime certified by deterministic Miller-Rabin below3.317 * 10^24. The sign ofmuacross the family does not organise the errors: a signed Type I sum over these moduli buys only a bounded factor over the absolute one,|Sigma_level|/Abs_levelreading0.498and0.192atlevel 40, and that factor does not grow with depth at any depth computed. - The worst orbit at a pinned divisor, two-sided. Proved. The full kernel's orbit product telescopes: for
t >= 1,d | base^t - 1withd >= 2andanonzero modd,prod_{j < t} |D_base(a base^j/d)| = 1exactly, sincea base^t = a mod dandd | base^t - 1forces(d, base) = 1, so no factor degenerates and the full digit set sees no closed shift orbit at all: every damping comes from the excluded digits. With|g_F| <= |D_base| + mand|D_base(a base^j/d)| <= B = min(base, d/2), convexity oflog(e^y + m)puts the maximum ofsum_j log(D_j + m)on{sum_j log D_j = 0, log D_j <= log B}at a vertex and givesprod_{j < t} |g_F(a base^j/d)| <= (B + m)^(t-1) (m + B^(1-t)). At one excluded digit that is sharp both ways: ford = base^t - 1,t >= 2,base >= 10and any single excluded digit,fill^(-1/t) (1 - 9/base) <= max_{a not 0 mod d} (prod_{j < t} |g_F(a base^j/d)|/fill^t)^(1/t) <= fill^(-1/t) (1 + 3/(base-1))uniformly int, the lower bound witnessed bya = 1through|e(f y) - 1| <= 2 pi f yat the firstt - 1positions andsin(pi y) >= 2 yat the last. So the orbit carriest - 1undamped positions and one damped by~ 1/fill, and the1 + o(1)is a two-sidedO(1/base)that does not grow witht; the folded constants9/baseand3/(base-1)are stated atbase >= 10and are recomputed before any smaller base quotes them. The census's pinned probes read a different quantity, the finite-depth error rate(|N_F(level; d) - fill^level/d|/fill^level)^(1/level) d^(1/level)atlevel 12andbase 100:0.1059atd = base^2 - 1and0.2369atd = base^3 - 1againstfill^(-1/2)andfill^(-1/3), with the proper divisord = 3367 | base^3 - 1at0.0549andd = 101 | base + 1pinned but harmless at0.0261. Thed^(1/level)of that normalisation is why a depth-12 rate sits beside the band and not inside it, and the probes corroborate the size rather than test the bound. Thea-average at the same modulus is exact, at every digit set and everym:sum_{a mod d} prod_{j < t} |g_F(a base^j/d)|^2 = d (fill^t + 2w)ford = base^t - 1, withw = 1when both0andbase - 1lie inFandw = 0otherwise, since congruent pairs of length-tstrings are the diagonal plus the one wraparound pair{0...0, (base-1)...(base-1)}when both endpoints are strings overF. Under the band's own hypotheses, one excluded digit andbase >= 10, the worst orbit therefore exceeds the average over alla, which isfill^t + 2w, byfill^(t-2) e^(O(t/base)). Most of that average is its owna = 0termfill^(2t)/d,970299/101of9803atbase 100andt = 2, so the average a second moment actually sees, overanonzero, is(d (fill^t + 2w) - fill^(2t))/(d - 1), smaller again by~ t/baseand980298/4999at the same cell: the spread is wider than the exponent states, never narrower. Fromt = 3on only afill^(2-t)fraction of residues can sit near the worst orbit: at a pinned divisor the bad mass is spread, and an average overais the one handle the supremum gives up. - The second moment across residues. Proved.
sum_{r mod d} (N_F(level; d, r) - fill^level/d)^2 = (1/d) sum_{a not 0 mod d} prod_{j < level} |g_F(a base^j/d)|^2, by Parseval moddon the orthogonality identity: the mean is thea = 0term, the variance is the rest, no cross terms survive. It gives up the supremum overrand buys an average overa, which is the one place a saving can survive at a pinned divisor, where every per-factor bound is flat. - No moment past the second helps at a pinned divisor. Proved. At
d = base^t - 1the2r-th orbit momentsum_{a mod d} prod_{j < t} |g_F(a base^j/d)|^(2r)is again an additive energy,dtimes the count of2r-tuples of length-tstrings overFwhose two halves have equal value sum modd, by the same orthogonality as the even moments above; and the pair-count certificate of the bisection bullet below places that energy a factor4 (base/fill)^(r t)above its own meanfill^(2 r t)/d, a loss growing inr. So the second moment is the only average overaa certificate delivers near that mean, and a chain built on the supremum overaand that second moment is already optimal for the two inputs it has. - The bisection bound, per divisor. Proved. For every
Fwithfill >= 1, everyd >= 2coprime tobase, everylevel >= 1and uniformly inr:|N_F(level; d, r) - fill^level/d| <= fill^(level/2) (1 + 2 base^((level+1)/2)/d). Cut the string in the middle and apply the second moment above to each half at depthsceil(level/2)andfloor(level/2): at a half's depthbthe variance isfill^b (1 + 2 base^b/d), since off the diagonal a congruent pair of length-bstrings needsval(f) - val(f') = j dwith0 < |j| <= (base^b - 1)/dandvalis injective on strings of one length, so each(f', j)fixes at most onef; then(d + 2 base^(b_1))(d + 2 base^(b_2)) <= (d + 2 sqrt(base) base^(level/2))^2. Two readings follow. Atd >= sqrt(base x)it givesmax_r |N_F(level; d, r) - fill^level/d| <= 3 fill^(level/2) = 3 fill^level x^(-alpha/2), asking nothing ofFand nothing ofdpast coprimality tobase, so the whole top range[sqrt(base x), x], where the orbit machinery says nothing, is covered by one line. And the pair count overshoots its own meanfill^(2b)/dby exactly the factor2 (base/fill)^b, the wraparound factor every route below inherits. - Level
alpha/2for the whole block, up to one factor. Proved. Summing that bound over2 <= d <= Dfor a divisor cutoffD, againstsum_{d <= D} 1/d <= 1 + log D, gives, for everybase >= 3, everyFwithfill >= 1, everylevel >= 1and everyD >= 2,sum_{2 <= d <= D, (d,base) = 1} max_r |N_F(level; d, r) - fill^level/d| <= fill^level (D fill^(-level/2) + 2 sqrt(base) (1 + log D) (base/fill)^(level/2)), everydand not only the squarefree ones, supremum over the target residue and not only the residue0. Read atD = x^thetawiththeta <= alpha/2, the first term is at most1and the whole sum is at most3 sqrt(base) (1 + log x) fill^level x^(m/(2 fill log base))on a set ofm = base - fillexcluded digits, since(base/fill)^(level/2) = e^((level/2) log(1 + m/fill)) <= e^(m level/(2 fill)) = x^(m/(2 fill log base)). That is a level of distributionx^(alpha/2 - o(1))at everythetaup toalpha/2at once, carrying a defectx^(m/(2 fill log base))that is sub-power inbaseand a positive power inx: the exponent is1/(2(base-1) log base)at one excluded digit, under0.0011atbase 100. The defect does not vanish aslevelgrows at fixedbase, so this is a level statement and not an equidistribution statement. - The assembled theorem across the whole divisor range. Proved. Fix
eps in (0,1)and an integerT_0 >= 2, and putbase_0(eps, T_0) = max(4^(1/eps), base_1)withbase_1any base satisfying3 base_1^(-eps)/log base_1 <= eps/(16 T_0). For everybase >= base_0, everyF = {0..base-1}minusEwith1 <= m <= base^(1-eps)/2, everylevel >= max(6 T_0, 4/eps)andx = base^level: (i) at every levelD <= x, the sum ofmax_r |N_F(level; d, r) - fill^level/d|over2 <= d <= Dcoprime tobasewithd <= base^(1-eps)orord_d(base) <= T_0is at most(T_0 + 2)(1 + log x) fill^level x^(-eps/(8 T_0)); (ii) at levelx^(alpha/2)the full sum is at most3 sqrt(base) (1 + log x) fill^level x^(m/(2 fill log base)); (iii) everydcoprime tobasewithsqrt(base x) <= d <= xhasmax_r |N_F(level; d, r) - fill^level/d| <= 3 fill^level x^(-alpha/2); and (iv) at one excluded digit the full pinned moduli are damped together,sum_{2 <= t <= level} max_r |N_F(level; base^t - 1, r) - fill^level/(base^t - 1)| <= level fill^(2 - sqrt(2 level)) e^(4 level/base) fill^level. Clause (i) is the level-base^(1-eps)saving and the bounded-order classes summed, proper divisors included; clause (ii) is the block above; clause (iii) is the top range; clause (iv) is the orbit bound of the pinned bullet against the exacta-average there, summed overt. No clause asksdsquarefree and every clause is a supremum over the target residue. Clause (iv) is superpolynomial inleveland not a fixed power ofx: its saving is worst att ~ sqrt(2 level), where asx^(-c)the exponentcfalls to0withlevel. - No clause in this norm can be a fixed power of
x, and clause (iv) is not slack. Proved. At one excluded digit andbase >= 10the two-sided orbit law above supplies the matching lower bound at a single modulus. Taked = base^t - 1witht = ceil(sqrt(level)), so the orbit closesfloor(level/t)times insidelevelpositions and the worstacarries|hat F_level(a/d)| >= fill^level x^(-O(1/sqrt(level))); the second moment across residues then givesmax_r |N_F(level; d, r) - fill^level/d| >= |hat F_level(a/d)|/d, andd <= base^(sqrt(level) + 1) = x^(O(1/sqrt(level)))costs only the same shape again. Hence for every fixedtheta > 0and everylevel >= max(9, 4/theta^2)the sum over2 <= d <= x^thetacoprime tobaseofmax_r |N_F(level; d, r) - fill^level/d|is at leastfill^level x^(-O(1/sqrt(level))), carried by that one modulus. Every clause above is in the supremum norm, so no assembly of them reaches a fixed power ofxand a level of distribution at a fixed power has to pass through signed sums. - What those clauses leave, named. Proved. One family survives them at a fixed level: the generic-order middle moduli
base^(1-eps) < d <= x^thetacoprime tobasewithord_d(base) > T_0, where no orbit period closes insideleveland the pinned machinery is silent. There only clause (ii) applies, so the block is certified at levelalpha/2up to the single factorx^(m/(2 fill log base)), and that factor is the whole distance between what is proved and a fixed power ofx. It is the wraparound overshoot and not slack in a constant: the pair-count certificate, the orbit moment and the additive large sieve over the Farey points share one diagonal, certify pair counts only to one-per-pair precision, and exceed the heuristicfill^(2b)/dby2 (base/fill)^b, which at the balanced depthb = level/2the certificate forces is exactly that defect. No rearrangement of cuts, no Cauchy-Schwarz and no divisor bookkeeping tried here removes it, and the reason is circularity rather than looseness: with the pair-count certificate alone both halves of a cut needbase^b <= d, sob <= 2 log_base dand the certificate reads3 d^(1 - alpha) > 1, missing the dip by exactly the sparsity ofF, while winning asks the depthb ~ (2/alpha) log_base dat whichfill^b ~ d^2strings meetdclasses, and equidistribution there is the statement being proved.
A power saving under GRH at large base
The census above measures cancellation and proves none of it. This section proves some, at the opposite end of the digit scale: not the sparse columns of the census but the dense ones, the base taken large and a single digit removed. There the mass exponent alpha_base = log(fill) / log(base) sits just under 1, the indicator of S_F opens into additive frequencies by the same orthogonality the divisor section uses, and each frequency carries a Mobius exponential sum, which under the generalized Riemann hypothesis is x^(3/4 + eps) uniformly in the frequency. The entire cost of the expansion is one l^1 norm, and past a computable base that cost is smaller than the mass. What comes out is a bound of Mertens shape read against the set's own counting function, conditional and dense-only, with both of those limits proved rather than assumed. Every constant, table row, margin and rung below is printed or test-pinned by lab/rs/mertens-numerology, and the sharpened one-excluded-digit constants of step 3, with the walls they move, by lab/py/mrly-pairing, verb onestep.
Proved under GRH below is the Proved tag with the hypothesis written inside the statement: steps 1, 2, 3 and 5 are derived here from definitions, and step 4 is one published theorem, quoted at its source and used exactly as stated.
- The setting.
Eis the excluded digit set withm = |E| >= 1,F = {0..base-1}minusE,fill = base - mandalpha_base = log(fill) / log(base), soA_F(x) >>_base x^(alpha_base)at everyxby the counting identities of the first section. The digit symbol isg_F(t) = sum_{d in F} e(d t)of the divisor section andD_base(t) = sum_{d = 0}^{base-1} e(d t)is the full Dirichlet kernel,|D_base(t)| = |sin(pi base t)/sin(pi t)|. WriteD_levelfor the digit strings of lengthleveloverFandhat F_level(t) = sum_{n in D_level} e(n t)for the transform at that level, which factors asprod_{j < level} g_F(base^j t)because the digits are independent. The one-step constant of the shifted-grid recursion isB_base(F) = sup_t sum_{r mod base} |g_F((t+r)/base)|, bounded above bybase PB_base(m)in step 3, with the proved constantPB_base(m) = sqrt(m) + Phi_base/base, wherePhi_base = (4/pi) base + (2 base/pi) H(ceil((base-2)/2)) + (1 - 2/pi)(base - 2) + 0.727andH(n) = ln n + gamma + 1/(2n); the exponent cost isc_base = log PB_base(m)/log base, defined from that proved bound and never from the exact supremum. Atm = 1andbase >= 17step 3 sharpens per excluded digit: withe_0the single excluded digit andc = e_0 - (base-1)/2, the sharpened bound of the step 3 bullet isbase PB_base(1, e_0)withPB_base(1, e_0) = PB_base(1) - 1/2 - (sec(pi c/base)/2 + 0.727 - 2(1 - 2/pi))/base, and the cost it carries isc_base(e_0) = log PB_base(1, e_0)/log base, again read from a proved bound and never from the exact supremum. Frombase >= 36the chord bullet of step 3 replaces the kernel constantPhi_baseby(4/pi) base + Psi'_base, which isPhi_base - base/2 + 2/piup to the0.00023954of0.727at evenbase, and the second per-digit form isbase PB'_base(1, e_0) = (4/pi) base + Psi'_base + base/2 - sec(pi c/base)/2, carrying the costc'_base(e_0) = log PB'_base(1, e_0)/log base, proved like the first and read like the first. Everything below withm > 1in it, the corollary and them-budget table included, runs onPB_base(m)alone. - Theorem. Proved under GRH. Assume the generalized Riemann hypothesis in its Dirichlet form:
L(s, chi)has no zero in the half planesigma > 1/2, for every Dirichlet characterchiof every modulus. LetFomit exactly one digite_0, letbase >= 1499, orbase >= 1032whene_0is0orbase - 1, and leteps > 0. Then|M_F(x)| <<_{base,eps} x^(3/4 + c'_base(e_0) + eps)for allx >= 2; and3/4 + c'_base(e_0) < alpha_baseat every suchbase, so withdelta_base(e_0) = (alpha_base - 3/4 - c'_base(e_0))/alpha_base > 0the same bound reads|M_F(x)| <<_{base,eps} A_F(x)^(1 - delta_base(e_0) + eps), a power saving against the set's own mass. The implied constant depends onbaseand onepsand on nothing else; no uniformity inbaseis claimed anywhere.delta_base(e_0)is fixed beforeepsis chosen, so the statement delivers every fixeddelta' < delta_base(e_0)and never the endpointA_F(x)^(1 - delta_base(e_0)). The two walls arem = 1only: atm > 1the corollary below carries the unsharpenedPB_base(m)and its own condition. - Step 1, orthogonality. Proved. For
0 <= n < base^level,1_{D_level}(n) = base^(-level) sum_{0 <= a < base^level} hat F_level(a/base^level) e(-n a/base^level)by completeness of the additive characters modbase^level, andhat F_levelfactors over digit positions: the identity of the divisor section with the modulusbase^levelin place ofd, read as an expansion rather than as a count. - Step 2, the
l^1recursion. Proved. Putc_level = sum_{0 <= a < base^level} |hat F_level(a/base^level)|and splita = a' + s base^(level-1). The transform peels at the positionj = 0,hat F_level(t) = g_F(t) hat F_{level-1}(base t), sosmoves that factor alone andhat F_{level-1}is1-periodic; the inner sum overs mod baseis a shifted grid ofbasepoints, andc_level = sum_{a'} |hat F_{level-1}(a'/base^(level-1))| sum_{s mod base} |g_F((a'/base^(level-1) + s)/base)| <= B_base(F) c_{level-1}. Hencec_level <= B_base(F)^leveland the normalizedl^1mass isbase^(-level) c_level <= (B_base(F)/base)^level: one constant per digit, no interaction between positions. - Step 3, the kernel bound. Proved.
|g_F| <= |D_base| + |g_E|splitsB_base(F)into a kernel part and an excluded part. Thebasepoints(t + r)/baseare spaced1/base; writingd_rfor the distance of each toZ,|D_base((t+r)/base)| = |sin(pi base d_r)|/sin(pi d_r), the two points nearest the singularity contribute at most(4/pi) base + 0.727by the two elementary inequalitiessin(pi v) <= 4v(1-v)on[0, 1/2]and1/sin x <= 1/x + 1 - 2/pion(0, pi/2](the first because4x(1-x) - sin(pi x)splits into a concave and a convex piece with the right signs, the second because1/sin x - 1/xincreases), and the remainingbase - 2points pair off at distances>= j/baseand contribute at most(2 base/pi) H(ceil((base-2)/2)) + (1 - 2/pi)(base - 2). Sosup_t sum_{r mod base} |D_base((t+r)/base)| <= Phi_base. The excluded part is exact rather than estimated: the excluded digits are distinct modbase, so Parseval on the shifted grid givessum_{r mod base} |g_E((t+r)/base)|^2 = base mfor everyt, and Cauchy-Schwarz turns that intosum_{r mod base} |g_E((t+r)/base)| <= base sqrt(m). HenceB_base(F) <= base PB_base(m)and the normalizedl^1mass of step 2 is at mostbase^(level c_base). - Step 3 sharpened at one excluded digit. Proved. At
m = 1,base >= 17andc = e_0 - (base-1)/2the triangle split of step 3 is lossy by a fixed share ofbase, and the loss is taken back with no new input:B_base(F) <= (4/pi) base + Psi_base + base/2 - sec(pi c/base)/2 = base PB_base(1, e_0), wherePsi_base = (2 base/pi) H(ceil((base-2)/2)) + (1 - 2/pi) baseis the kernel constant less its two-point part andbase PB_base(1) = (4/pi) base + Psi_base + base + 0.727 - 2(1 - 2/pi), so them = 1mass of step 2 is at mostbase^(level c_base(e_0)). At one excluded digit the shifted grid is exact,|g_F((t+r)/base)| = |A_r - e(c (t+r)/base)|withA_r = (-1)^r sin(pi t)/sin(pi (t+r)/base), becauseD_base((t+r)/base)factors as a unimodular phase times(-1)^r sin(pi t)/sin(pi (t+r)/base). Then|a - e(psi)|^2 = (a+1)^2 - 2a(1 + cos psi)withsqrt(1 - X) <= 1 - X/2produces a correction term, the phase identitysum_{r mod base} (1 + sign(A_r) cos(2 pi c (t+r)/base)) = base + cos(2 pi c (t - 1/2)/base)/cos(pi c/base)sums it and is at leastbase + 1since|2 pi c (t - 1/2)/base| <= |pi c/base| < pi/2,|A_r| >= sin(pi t) = sands/(1 + s) >= s/2weight it, step 3's own two-point and pairing estimates give thet-dependent kernel boundsum_{r mod base} |D_base((t+r)/base)| <= (4/pi) base + s Psi_base, andh(tau) = cos(pi tau)(Psi_base - base/2) - cos(pi tau) cos(2 pi c tau/base)/(2 cos(pi c/base))hash' <= 0on[0, 1/2]oncePsi_base >= (1 + pi) base/2, first true atbase 17, bysin(pi tau) >= 2 tau,sin y <= yandsec(pi c/base) <= base. The bound falls as|c|rises, so the two extreme digitse_0 in {0, base-1}carry the smallest constant,sec(pi c/base) = 1/sin(pi/(2 base))there, and the middle digit the largest,sec(pi c/base) = 1at oddbase(lab/py/mrly-pairing, verbonestep). - Step 3's kernel constant, replaced by its chord. Proved.
csc x - 1/xhas an all-positive Taylor series on(0, pi/2], so it is convex there and lies under its own chord,1/sin x <= 1/x + (2/pi)(1 - 2/pi) x, half the flat1 - 2/piof step 3 on average and equal only at the endpoint. Pairingrwithbase - 1 - rsends the shifted grid to the argument pairspi (t+r)/baseandpi (1-t+r)/baseatr < floor(base/2), every one inside(0, pi/2]fort in (0, 1/2], andK(t) = sum_{r mod base} |D_base((t+r)/base)| = sin(pi t) sum_{r mod base} 1/sin(pi (t+r)/base)is symmetric aboutt = 1/2, which carries the rest of the circle. The1/xhalf of ther = 0pair issin(pi t) base/(pi t (1-t)) <= (4/pi) baseand the1/xhalf of eachr >= 1pair hast-free maximum1/r + 1/(r+1), while the chord halves collapse to(1 - 2/pi) base/2because the paired argument sum is exactlybase^2/4at evenbase; at oddbasethe unpaired middle term makes itP(P+1) + t, which is where the odd form's extra1/(2 base)comes from. SoK(t) <= (4/pi) base + sin(pi t) Psi'_basewithPsi'_base = (base/pi)(2 H(P-1) - 1 + 1/P) + (1 - 2/pi) base/2at evenbase,P = floor(base/2)andHthe same harmonic upper bound, andPsi'_base = (base/pi)(2 H(P-1) - 1 + 2/P) + (1 - 2/pi)(base/2 + 1/(2 base))at oddbase; at evenbasethat is exactlyPsi_base - base/2 + 2/pi, and at oddbasePsi_base - Psi'_base = (2 base/pi)(H(P) - H(P-1)) + (base/pi)(1 - 2/P) + (1 - 2/pi)(base/2 - 1/(2 base)), whose first term is positive atP >= 2becauseH(P) - H(P-1) = ln(P/(P-1)) - 1/(2P(P-1))withln(P/(P-1)) > 1/(P - 1/2)by the midpoint rule on the convex1/x, and1/(P - 1/2) > 1/(2P(P-1))is2P^2 - 3P + 1/2 > 0, whose second is nonnegative atP >= 2and whose third is positive, soPsi'_base < Psi_baseat everybase >= 5with no scan. Nothing downstream changes, so the sharpening above runs on it verbatim:B_base(F) <= (4/pi) base + Psi'_base + base/2 - sec(pi c/base)/2 = base PB'_base(1, e_0)wherever the monotone step's hypothesisPsi'_base >= (1 + pi) base/2holds, first atbase 36on this page's readingH(n) = ln n + gamma + 1/(2n)and atbase 37on the harmonic number itself, an over-estimate that can only keep the kernel lemma true and that moves no printed wall, the lowest being1032. The hypothesis is sufficient and not necessary: the maximum ofh(tau)it exists to place attau = 0sits there at everye_0frombase 8up on the exhaustive scan4..79. The chord cuts the gap to the exact kernel supK_baseby a factor5.98, the up-rounded gap columns readingPsi_base - (K_base - (4/pi) base) <= 0.600121 baseagainstPsi'_base - (K_base - (4/pi) base) <= 0.100293 baseatbase 3690, the lemma's own slack floored to0.100292 basethere, the chord column flat to1e-5acrossbase 100, 1000, 2234and the gap attained at the seatt = 1/2; and nothing measured reaches the new bound, the worst ratio of the exactB_base(F)to it being0.902124over everye_0atbase 36..60,0.941239at the larger seats and0.936333over4000seeded draws (Verified for the measured columns, lab/py/mrly-pairing, verbonestep). - Step 4, the Mobius input. Quoted. Under GRH,
max_{theta in [0,1)} |sum_{n <= x} mu(n) e(n theta)| <<_eps x^(3/4 + eps). This is the casea = 1/2of Baker and Harman 1991, whose hypothesis is exactly thatL(s, chi)is zero-free insigma > afor every Dirichlet character, whose implied constant depends only oneps, and whose maximum is over all realtheta; the frequencies this proof uses are thea/base^l, well inside that uniformity. The statement is restated at source in Porritt 2018 and in Zhang 2024. This is the one step not derived here. - Step 5, assembly and the wall. Proved. If
0 in F, thenS_FbelowxisD_levelless{0}intersected with[1, x]atlevel = ceil(log_base(x+1)), and steps 1 to 4 apply once:|M_F(x)| <= (B_base(F)/base)^level max_theta |sum_{n <= x} mu(n) e(n theta)| <<_{base,eps} x^(3/4 + c_base + eps). If0is not inF, thenS_Fbelowbase^levelis the disjoint union of the exact-length blocksl <= level, and summing the per-block bounds is a geometric sum of ratiobase^(3/4 + c_base + eps) > 1, so the top block sets the exponent and the answer is the same; the statement holds at allx >= 2because the implied constant absorbs the bounded range wherelevelis small. Converting to theA_Fyardstick needs3/4 + c_base < alpha_base, equivalently the constant-space certificategap_base(m) = (base - m) base^(-3/4) - PB_base(m) > 0. Atm = 1that certificate is negative at every3 <= base < 3690and positive atbase 3690(Verified, exhaustive in the generator), and steps up at everybase >= 723by the monotone floor below, sobase >= 3690is a half line and not a window. Atm = 1three proved constants read that certificate and each prints its own wall. The step 3 constantPB_base(1)closes at3690(lab/rs/mertens-numerology); the phase-sharpenedPB_base(1, e_0)closes at2446at every excluded digit and at1812ate_0 in {0, base-1}, an up-set over the whole scan17..4 * 10^6and not a first crossing, theheldcolumn printing3997555 = 4000000 - 2446 + 1and the five like counts; and the chord formPB'_base(1, e_0)closes at1499and at1032, an up-set over its own scan36..4 * 10^6, itsheldcolumn printing3998502 = 4000000 - 1499 + 1and five like counts (Verified, lab/py/mrly-pairing, verbonestep). Above those scans the chord bullet'sPsi'_base < Psi_basecarries them:base PB_base(1) - base PB'_base(1, e_0) = (Psi_base - Psi'_base) + base/2 + sec(pi c/base)/2 + (0.727 - 2(1 - 2/pi)), every term positive, soPB'_base(1, e_0) < PB_base(1)at everybase >= 36and the step 3 certificate itself, positive from3690on, makes every sharpened wall a half line too. - What the proof does not use. Proved. No zero-density input, no restriction of
xto a power ofbase, no multiplicative structure ofS_F(there is none: the first section's4 x 13witness), and nol^1bound quoted from the literature. Step 3 is self-contained and explicit at every base, which is what the theorem needs and what the sharper base-10l^1bound behind Maynard 2019 does not offer at generalbase. - Corollary,
mexcluded digits. Proved under GRH. With|E| = m, the step 3 constantPB_base(m)and no sharpening, the five steps run unchanged wheneverPB_base(m) < (base - m) base^(-3/4), and give|M_F(x)| <<_{base,eps} A_F(x)^(1 - delta_base + eps)withdelta_base = (alpha_base - 3/4 - c_base)/alpha_base. Under the proved constants that condition holds form <= 6atbase 10^4,m <= 78atbase 10^5andm <= 451atbase 10^6(Verified, each maximum asserted maximal in the generator), againstsqrt(base) = 100, 316, 1000, and it holds asymptotically form <= base^(1/2)(1 - o(1))sincePB_base(m)issqrt(m)plus a term of size(2/pi) ln base. The squarefree-digit-gcd hypothesis carried by the exponent conjecture below is automatic in this regime and is not dropped:m < floor(base/2)leaves two consecutive digits inF, sogcd(F) = 1and the vanishing family of the transfer section cannot occur; at smallfillthe hypothesis must be stated. - The shape at large base. Proved under GRH.
Phi_baseis(2/pi) base ln baseup to lower order, soc_base = (ln ln base + ln(2/pi) + o(1))/ln base -> 0whilealpha_base -> 1, hencedelta_base -> 1/4and|M_F(x)| <<_{base,eps} A_F(x)^(3/4 + o(1)): the full-line GRH exponent transplanted verbatim onto the digit-restricted column, measured against that column's own mass. The convergence is logarithmic and nothing better;c_basetracks(ln ln base + ln(2/pi))/ln baseto within0.01atbase 10^12(Verified, the generator). - The constants. Verified. The generator prints
alpha_basetruncated down at six digits,c_baserounded up at five anddelta_baserounded down at five, each from the unrounded value with a directional guard of10^-12, so every printed digit is a true bound in its own direction andalpha_basenever prints as1.000000; the scientific rows carry a relative guard of10^-10. Both forms of the test,3/4 + c_base < alpha_baseandgap_base(m) > 0, are computed and their agreement asserted at every row and across3 <= base < 20000. Every column of the table is the step 3 constantPB_base(m): them = 1sharpenings move the closing base and not these rows.
base | alpha_base | c_base (proved, up) | delta_base (down) | closes, step 3 |
|---|---|---|---|---|
| 1000 | 0.999855 | 0.28087 | -0.03102 | no |
| 2000 | 0.999934 | 0.26335 | -0.01342 | no |
| 3000 | 0.999958 | 0.25430 | -0.00434 | no |
| 3689 | 0.999966 | 0.24997 | -0.00001 | no |
| 3690 | 0.999967 | 0.24997 | 0.00000 | yes |
| 5000 | 0.999976 | 0.24393 | 0.00605 | yes |
| 10^4 | 0.999989 | 0.23141 | 0.01858 | yes |
| 10^5 | 0.999999 | 0.19906 | 0.05094 | yes |
| 10^6 | 0.999999 | 0.17589 | 0.07411 | yes |
| 10^9 | 0.999999 | 0.13305 | 0.11695 | yes |
- The margin at the step 3 wall. Verified. The saving at
base 3690is far below the fifth printed digit, so the rounded columns cannot display its sign and never certify it. The certificate is the pair of scientific bounds printed from the cancellation-reduced formdelta_base = ln(1 + gap_base(m)/PB_base(m))/(alpha_base ln base), which never differences two numbers of size1to reach one of size10^-6:delta_base <= -2.395807653 * 10^-6andgap_base(1) <= -1.533059397 * 10^-4atbase 3689, againstdelta_base >= 5.863425182 * 10^-6andgap_base(1) >= 3.752213034 * 10^-4atbase 3690. Beyond the wall the gap rises at every one of the96310steps of3690..10^5, the smallest step being>= 0.00003172at the top of that range, where thebase^(-3/4)growth of the mass term is nearest the4/(pi base)jump of the harmonic term. - The ladder. Proved under
Z(a). WriteZ(a), for1/2 <= a < 1, for the hypothesis thatL(s, chi)has no zero insigma > afor every Dirichlet character;Z(1/2)is GRH. AssumeZ(a), letFomit exactly one digit and letbase >= base_0(a), the least base withPB_base(1) < (base-1) base^(-b(a)); at the three rungs them = 1sharpening is scanned at,PB'_base(1, e_0) < (base-1) base^(-b(a))from the two chord columns of the table on, and the hypothesis reads the smaller wall. Then for everyeps > 0and allx >= 2,|M_F(x)| <<_{base,eps} x^(b(a) + c + eps)withc = c_baseunder the step 3 wall andc = c'_base(e_0)under the smaller chord wall, andb(a) + c < alpha_basein each case, so|M_F(x)| <<_{base,eps} A_F(x)^(1 - delta + eps)withdelta = (alpha_base - b(a) - c)/alpha_base > 0; the cost in the conclusion is the one whose certificate the hypothesis reads, andc_baseis not available under a chord wall,b(a) + c_base >= alpha_baseat everybase < base_0(a)by that wall's own minimality; the corollary runs atmexcluded digits wheneverPB_base(m) < (base-m) base^(-b(a)). The proof is the one above with a single substitution: step 4 quotes the exponentb(a)thatZ(a)buys, and steps 1, 2, 3 and 5 never name an exponent, the geometric sum of step 5 still having ratio above1. Sincealpha_base -> 1andc_base -> 0whileb(a) < 1is fixed, every common zero-free half plane for Dirichlet L-functions buys a power saving over the dense column, and GRH is only its first rung: the price of a weaker hypothesis is paid entirely in the base. - The input
b(a), and where it comes from. Proved.b(a)is the smaller of two quoted tables: Baker and Harman 1991 givesa + 1/4on1/2 <= a < 11/20,4/5on11/20 <= a < 3/5and(a+1)/2on3/5 <= a < 1, and Zhang 2024, Theorem 1.1, gives(8a - 7a^2)/(4 - 2a)on1/2 <= a <= 4/7. Where both apply Zhang is smaller and the two meet exactly at the ends of the overlap, by two factorisations:Zhang(a) - (a + 1/4) = -5(a - 1/2)(a - 2/5)/(4 - 2a)is negative on(1/2, 11/20)andZhang(a) - 4/5 = -7(a - 4/7)(a - 4/5)/(4 - 2a)is negative on[11/20, 4/7), with equality ata = 1/2(both3/4) and ata = 4/7(both4/5); andb(a) >= 3/4on the whole range, Baker-Harman by inspection and Zhang byZhang(a) - 3/4 = -7(a - 1/2)(a - 6/7)/(4 - 2a) > 0on(1/2, 4/7]. Verified in the generator over every rational of denominator<= 200inside the overlap, in exact integer arithmetic,b(a)carried as a rational and compared by cross multiplication throughout. - The rungs. Verified. Each row names its
aand the table the exponent comes from;bothmeans the two tables agree there, and a rung is meaningless quoted without them.base_0(a)is the leastbase >= 3withgap_base(a, 1) = (base-1) base^(-b(a)) - PB_base(1) > 0andQ(b)the proved monotone floor below. A wall prints as an exact integer only when it sits below2^53and both neighbouring gaps exceed1024ulps of the terms differenced; otherwise the row prints<=and a scientific upper bound, which is a bound on the leastbaseand not the leastbase. Every wall below4 * 10^6is reproduced by an exhaustive scan frombase 3against the bisection. The two chord columns arem = 1only and readPB'_base(1, e_0)in place ofPB_base(1); each is the leastbaseof an exhaustive scan frombase 36and an up-set over that whole scan, and a-is a rung the chord has not been scanned at (lab/py/mrly-pairing, verbonestep). The three scans run36..4 * 10^6,36..8 * 10^6and36..4 * 10^7, each past that rung's ownbase_0(a), above which the step 3 certificate is positive on its own, so each chord wall is a half line and not a window.Q(b)is the step 3 floor and no chord column touches it.
a | b(a) | source | base_0(a), step 3 | chord, any e_0 | chord, e_0 in {0, base-1} | Q(b) |
|---|---|---|---|---|---|---|
| 1/2 | 3/4 | both | 3690 | 1499 | 1032 | 723 |
| 13/25 | 1417/1850 | Zhang | 8578 | 3525 | 2459 | 1486 |
| 11/20 | 913/1160 | Zhang | 33547 | 14078 | 10013 | 4754 |
| 4/7 | 4/5 | both | 92317 | - | - | 11221 |
| 3/5 | 4/5 | BH | 92317 | - | - | 11221 |
| 2/3 | 5/6 | BH | 3107080 | - | - | 216023 |
| 3/4 | 7/8 | BH | 6939524168 | - | - | 129458304 |
| 4/5 | 9/10 | BH | <= 3.09358e13 | - | - | 128606353005 |
| 9/10 | 19/20 | BH | <= 3.23663e34 | - | - | <= 1.73431e28 |
| 19/20 | 39/40 | BH | <= 9.24614e83 | - | - | <= 3.30712e68 |
- The floor is proved, not scanned. Proved. Per step
PB_{base+1}(1) - PB_base(1) < 1.291/(base-2)forbase >= 40: the harmonic term jumps by at most(4/pi)/(base-2), the(1 - 2/pi)(base-2)/baseterm adds under0.017/(base-2)and the0.727/baseterm falls, and a step that does not jump the harmonic term is net negative. The mass term(base-1) base^(-b)gains at least(1-b)(base+1)^(-b)per step, sogap_base(a, 1)steps up wherever(1-b)(base-2)(base+1)^(-b) >= 1.291, a quantity strictly increasing inbase;Q(b)is the leastbase >= 40where it holds, and the gap steps up at everybase >= Q(b). Below it nothing closes:gap_base(a, 1) < 0on3 <= base < 3690at every rung (exhaustive), and on[3690, Q(b)]the smooth majorantU(base) = base^(1-b) - PB_base^-(1)dominates the gap and has exactly one interior minimum, sinceU'(base) = (1-b) base^(-b) - (2/pi)/(base-2) - 2(1 - 2/pi)/base^2is positive exactly when a quotient falling strictly from+infto0drops below1, so its maximum on any interval sits at an endpoint and both endpoints are negative.Q(b) < base_0(a)at every rung, so each printed wall is the leastbaseand the gap steps up from it on, with no sweep needed at any rung. - The floor and the wall hold at every
bin[3/4, 1), not only at the printed rungs. Proved. Below3690the one exhaustive scan covers everybat once:base^(-b)falls inb, sogap_base(b, 1) <= gap_base(3/4, 1) < 0on3 <= base < 3690, that range being cleared exhaustively atb = 3/4. The floor itself rises withb, since(1-b)(base-2)(base+1)^(-b)falls inbat fixedbase, soQ(b) >= Q(3/4) = 723. At the floor, minimality ofQ = Q(b)bounds(Q-1) Q^(-b) < 1.291/u + 0.015above, the slack2 Q^(-b) < 0.015coming fromQ >= 723, andln(Q+1) > ln(1.291/u)/ubelow, both in terms ofu = 1 - balone; feeding them into the lower boundPB_Q^-(1)throughln((Q-2)/2) >= ln(Q+1) - ln(1448/721)and(Q-2)/Q >= 721/723givesgap_Q(b, 1) < [1.291 - (2/pi) ln(1.291/u) - 2.544 u]/u, the coefficient2.544assembled from those three ingredients,0.015,ln(1448/721)and721/723. Its bracket increases on(0, 1/4]and so is at most its value-0.39014atu = 1/4, hencegap_Q(b, 1) < -1.56. On[3690, Q]the majorant differs from the gap by under0.004, soU(Q) < -1.556, whileU(3690, b)falls inbwithU(3690, 1417/1850) < -0.95, and anybbelow1417/1850hasQ(b) <= 1486 < 3690and an empty range. Sogap_base(b, 1) < 0on[3, Q(b)]and steps up fromQ(b)on at everyb:base_0(a)exists and exceedsQ(b)at everya, printed rung or not. Constants Verified in the generator on theb-grid0.75..0.975in steps of0.005. - What a weaker half plane spends first. Verified. The
m-budget atbase 10^7is the largestmwithPB_base(m) < (base-m) base^(-b(a)), printed by the generator for the rungs whose wall lies below10^7. Each row carries itsaand its source, a rung quoted bybalone being meaningless: two rungs shareb = 4/5from different tables and the budget, not the theorem, is what a wider zero-free half plane costs. The budget is a statement aboutPB_base(m)atmexcluded digits, so them = 1sharpening never enters it.
a | b(a) | source | max m |
|---|---|---|---|
| 1/2 | 3/4 | both | 1971 |
| 13/25 | 1417/1850 | Zhang | 1002 |
| 11/20 | 913/1160 | Zhang | 365 |
| 4/7 | 4/5 | both | 176 |
| 3/5 | 4/5 | BH | 176 |
| 2/3 | 5/6 | BH | 8 |
- The cost-out against the Type I defect. Verified. A conditional Type I argument over
S_Fwould run against the level-x^(alpha_base/2)distribution bound of the divisor section above, whose error carries a defectx^(m/(2(base-m) ln base)), and would have to pay that defect out of the saving proved here, so the generator sets the two exponents side by side. They sit on different yardsticks and no derivation joins them:delta_baseis normalised to the mass, so as a power ofxthe saving isx^(alpha_base delta_base)withalpha_base >= 0.99993on every row compared, while the defect multipliesfill^level. On those rows the saving is below the defect at the wall (5.86342 * 10^-6against1.65022 * 10^-5atbase 3690, a factor above2.8) and above it frombase 3692on, the least such base in a scan of3690..10^5where the difference rises at all96310steps, monotonicity beyond the scan not being proved; bybase 10^9the saving1.16951 * 10^-1clears the defect2.41275 * 10^-11by over nine decades, and the tightest corollary row,base 10^6atm = 451, clears its own defect1.63296 * 10^-5at3.14081 * 10^-5. The whole failure at the wall is the two steps3690, 3691, so a sharper constant that moves the wall moves the comparison too and is re-costed rather than inherited: the sharpened cost-out block of lab/rs/mertens-numerology prints each proved wall beside its own crossing, the leastbaseat whichdelta_baseexceeds the defect1/(2(base-1) ln base), reading3690and3692at the step 3 constant,2446and2450at the phase sharpening,1812and1815at its extreme-digit form,1499and1502at the chord and1032and1036at the chord's extreme-digit form, each row scanned from its own floor,base >= 3,17and36, and each crossing an up-set to10^5within five steps of its own wall, so a lower wall costs out at once and no comparison is inherited. The comparison runs at the GRH rungb = 3/4and at no rung above it: no rung of the ladder pasta = 1/2is set against the defect anywhere here. - Conjecture. That such a defect is absorbed at all. The comparison above is two exponents from two unrelated statements on two yardsticks; it is not a necessary condition, no theorem about
M_Ffollows from it, and the string-to-interval bookkeeping and the bilinear half of any such argument are untouched here. - The
l^1floor, and what it forecloses. Proved. For every digit set,sum_{r mod base} |g_F((t+r)/base)|^2 = base fillexactly, sosum_{r mod base} |g_F((t+r)/base)| >= base fill / max_r |g_F| >= basefor everyt: the recursion of step 2 never contracts,B_base(F) >= base, andc_base >= 0at every base and every digit set, so a negativec_baseis an arithmetic error and not a discovery. At one excluded digit the floor is higher thanbaseand exact: lettingt -> 0on the shifted grid gives|g_F(0)| = base - 1and|g_F(r/base)| = 1at everyr != 0, soB_base(F) >= 2(base - 1)andc_base >= log_base(2 - 2/base) > 0at everybase >= 3, and that endpoint is the seat atbase 3, by hand and not by a grid: there|g_F((t+r)/3)| = 2 |cos(pi (t+r)/3)|, and withu = pi t/3in[0, pi/3)the three absolute values collapse to4 cos uon[0, pi/6)and to4 cos(u - pi/3)on[pi/6, pi/3), both at most4 = 2(base-1), attained atu = 0, soB_3(F) = 4exactly (Proved; the sup read on thet-grid of lab/py/mrly-pairing, verbonestepagrees and is a reading, never the certificate). So atm = 1no exact constant pushes this decomposition below the base where(base - 1) base^(-3/4) > 2 - 2/base, which isbase^(1/4) > 2and sobase >= 17: atm = 1the method needsfill > (2 - 2/base) base^(3/4)and notfill > base^(3/4). The other end of the same lever is measured and not proved: the exactB_base(F)read on the grid would close the GRH certificate at927at every excluded digit, last failurebase 926ate_0 = 462, and at304ate_0 in {0, base-1}, last failure303, against the proved1499and1032, but both are readings ofSigma(1/2)with no upper certificate on the supremum and no monotonicity inbase, so they bound nothing and enter no statement (Verified, lab/py/mrly-pairing, verbonestep). That higher floor is still too weak to askfill > 2 base^(3/4), since atbase 17the basefill = 16lies between(2 - 2/base) base^(3/4) = 15.759and2 base^(3/4) = 16.744. The consequence is a hard limit on this decomposition, not on the problem: at every digit set,m = 1included, it needsalpha_base > 3/4, that isfill > base^(3/4), so every column at fixed digit count is out of its reach,F = {0,1}at base 3 included, under GRH or without it. The dense columns this section proves something about and the sparse columns the census measures do not overlap. - Nothing unconditional follows in this decomposition. Proved. Put Davenport's unconditional
max_theta |sum_{n <= x} mu(n) e(n theta)| <<_A x (log x)^(-A), carried at source in Porritt 2018, into step 4: by thel^1floor the result is at best of sizex (log x)^(-A), which exceedsA_F(x)by the powerx^(1 - alpha_base). An unconditional power saving here would need an unconditional uniform power-saving input, which is itself of zero-free-strip strength; an unconditional route has to split arcs and use the structure ofmuin progressions modbase^j, which this decomposition never touches. - The
l^2route is worse than trivial. Proved. Cauchy-Schwarz with Parseval on both factors,sum_{a mod base^level} |hat F_level(a/base^level)|^2 = base^level fill^levelandsum_{a mod base^level} |sum_{n <= base^level} mu(n) e(n a/base^level)|^2of size(6/pi^2) base^(2 level), gives exponent(1 + alpha_base)/2 > alpha_base. The supremum over frequencies paid against thel^1mass is the only arrangement of this decomposition that saves anything. - The ceiling, and the endpoint. Proved. Even with the conjectured
x^(1/2 + eps)in step 4, the exponent Porritt 2018 records as the expected one, the floorc_base >= 0still forcesalpha_base > 1/2, that isfill > base^(1/2): the exponent conjecture below, which is about fixedfill, is beyond every version of this method and not merely beyond its conditional form. And within the dense regime the endpoint stays out:delta_baseis fixed beforeeps, so what is proved isA_F(x)^(1 - delta')for every fixeddelta' < delta_baseand neverA_F(x)^(1 - delta_base), a distinction that is the whole claim atbase 3690, wheredelta_base >= 5.863425182 * 10^-6. - What this is, against the literature. Verified. As far as the sources in REFS.md are read, none of them carries a Mobius or Mertens sum over a digit-restricted set: the nearest multiplicative function computed over a missing-digit set is the divisor function (Kim 2024), whose own framing is that the set's lack of multiplicative structure blocks the standard approaches, and the nearest arithmetic-function theorem over such a set is the prime count of Maynard 2019, which enters through the set's level of distribution and not through a Mobius bound. The theorem above is of Mertens shape: a power of the set's own counting function,
A_F(x)^(1 - delta')for every fixeddelta' < delta_base, at everyx >= 2, for one excluded digit at everybase >= 1499, and at everybase >= 1032when the excluded digit is0orbase - 1, and formexcluded digits under the stated condition, the bound it beats being the trivial|M_F(x)| <= A_F(x)on those columns. The card carries both halves: it is conditional on GRH, it yields nothing unconditional inside this decomposition, and it says nothing whatever in the sparse regimefill <= base^(3/4)where the census and the exponent conjecture live. - The lane. This section is written up on the shelf as sparse-mertens-under-grh: the theorem, the corollary, the ladder and the
l^1floor with full proofs, and itsscripts/verify.pyrecomputes every step 3 constant, table row, margin and rung above from the formulas alone, independently oflab/rs/mertens-numerology, in under three seconds. Them = 1sharpenings of step 3, their walls and the higherl^1floor at one excluded digit are not carried there.
The pair route
The section above buys a power saving on the dense columns under GRH and states plainly that nothing unconditional follows from that decomposition: an unconditional route has to split arcs and use the structure of mu in progressions mod base^j. This section is that route, laid out as far as it goes. It follows the only existing proof that counts a thin arithmetic sequence on a missing-digit set, Maynard 2019, and asks what changes when the sequence counted is mu rather than the primes. Most of that chain never looks at the sequence at all; the two steps that do are proved here; what is left is a region in the three exponents the digit set owns. No theorem about M_F comes out of it. A criterion does: its region is exactly two inequalities, its gate is fill >= base^(3/4), it is refuted at base 10 at every excluded digit, and the one design of the census that clears it is base 21 missing the digit 0.
Notation as in the divisor section, with x = base^level, D_level the digit strings of length level over F, hat F_level(t) = sum_{n in D_level} e(n t) = prod_{j < level} g_F(base^j t), alpha = log(fill) / log(base) the mass exponent, S_mu(t) = sum_{n <= x} mu(n) e(n t), rad(base) the product of the primes dividing base and omega(n) the number of them. Write alpha_1 for the l^1 exponent of the transform in its sup-over-shift form, sup_s sum_{a < Y} |hat F_l(s + a/Y)| << fill^l Y^(alpha_1), the supremum over real shifts s, which is 27/77 at base 10 with one digit excluded. That is the strength the source's own l^1 lemma carries and the strength Farey spacing consumes; it dominates the bare grid exponent, so every lower bound on the grid exponent below transfers up to it, and the threshold it is asked to clear is correspondingly the stronger ask. Write m_t for the l^t exponent of the normalised transform F_x(t) = fill^(-level) |hat F_level(t)| on the grid and beta = inf_{1 <= t < 2} m_t/(2 - t) for the exceptional-set threshold, the third exponent the design owns. Call e base-smooth when rad(e) divides rad(base), and write (E1) for the hypothesis that every prime dividing the gcd of the digit differences of F divides base: the one-dimensional form of condition (E) of coprime, and the hypothesis Lemma A' there consumes. Every exponent, region boundary, threshold certificate and census verdict below is printed by lab/py/mobius-region.
- The bilinear half of the chain never sees the coefficients. Verified. The Type II estimate of that proof is stated for arbitrary
1-bounded sequences with one support constraint, that every counted integer carries a divisor in a prescribed dyadic range; its proof applies Cauchy-Schwarz in the long variable first and then drops all four coefficient factors by the triangle inequality, leaving a sum over pairs of frequencies with no coefficient in it at all, which a geometry-of-numbers argument places near a rank-2 lattice or on a line. Residue sums of the coefficient side occur exactly once in that proof, on the major arcs at moduli below a fixed power oflog x. So a Type II estimate on a digit set is not a hypothesis about cancellation of the coefficients in progressions, and the whole range above that cut transfers from primes tomuunread, every sentence of this bullet read at its source. - The
l^1floor is the shifted-grid floor iterated. Proved. Thel^1floor of the GRH section,sum_{r mod base} |g_F((t+r)/base)| >= base fill / max_r |g_F| >= basefor everyt, is one digit position of the same statement; iterating it overlevelpositions through the peeling recursion of that section's step 2, or reading it off the grid directly bysum_a |z_a| >= (sum_a |z_a|^2)/max_a |z_a|with Parsevalsum_{a mod base^l} |hat F_l(a/base^l)|^2 = base^l fill^land the maximumfill^lata = 0, givessum_{a mod base^l} |hat F_l(a/base^l)| >= base^land hencealpha_1 >= 1 - alphaat every base and every digit set. One mechanism, stated once there per position and once here per exponent. Two consequences: anl^1exponent below1/2forcesfill > sqrt(base), so the sparse columns of the census are outside this route exactly as they are outside the route of the GRH section; andalpha + alpha_1 >= 1always, which is what makes the scale sum in the level-of-distribution statement below geometric with ratio at least1. That reachfill > sqrt(base)is true and unsharp: the same Parseval identity puts the same floor on the exceptional-set threshold, and the route's own window condition lifts the gate tofill >= base^(3/4)below. - No exceptional character sits at a base-smooth modulus. Proved. Every real primitive Dirichlet character of base-smooth modulus has conductor dividing
8 rad(base), and the conductors in play number exactly2^omega(base_1)at oddbaseand3 * 2^omega(base_1)at evenbase,base_1the odd part ofbase, while the characters number2^omega(base_1)at oddbaseand4 * 2^omega(base_1)at evenbase. A real primitive character of conductorf > 1is the Kronecker symbol of a fundamental discriminant of absolute valuef, so writingf = 2^u f_1withf_1odd,f_1is squarefree anduis0,2or3; base-smoothness forcesf_1 | rad(base), hencef | 8 rad(base). Conversely every2^u f_1of that shape occurs, and the two counts differ: exactly one of+-f_1is1 mod 4, giving one character atu = 0; exactly one of+-f_1is3 mod 4, giving one atu = 2; and both of+-2 f_1are2 mod 4and squarefree, giving two atu = 3, so four characters sit over three conductors for each odd squarefreef_1dividingrad(base), while at oddbaseonlyu = 0is available and the counts coincide. That is the whole content of the remark in the source that its major-arc moduli are too composite for Siegel zeros to matter: an exceptional zero belongs to a real primitive character, a real primitive character has a fundamental discriminant for a conductor, a fundamental discriminant is squarefree away from a factor4or8, and a power of the base is as far from squarefree as an integer gets. The conductors in play run over a set of size bounded in terms ofbasealone rather than to infinity, so Siegel's theorem is never invoked and the constants below are effective. - The major arcs for
mu, with the exponent they deliver. Proved. Letbase >= 3, letFsatisfy (E1), letC > 0, and putT = (log x)^CandM(C) = {a mod x : |a/x - b/d| <= T/x for some d <= T and some b coprime to d}. Then there arec > 0andx_0, both depending only onbase,fillandCand both effectively computable, with|x^(-1) sum_{a in M(C)} hat F_level(a/x) S_mu(-a/x)| <= fill^level exp(-c sqrt(log x))forx >= x_0. The proof splitsM(C)at the base-smooth denominators. Off them the modulus carries a factord_2 > 1coprime tobase, and the perturbed Lemma A' of coprime gives|hat F_level(a/x)| <= fill^level exp(-c' log x / log log x)against the trivial|S_mu| <= x. On themx = base^levelmakes every suchb/dan exact grid point, so the arcs are intervals of consecutive integers and no Dirichlet approximation enters; there|hat F_level| <= fill^levelis trivial, partial summation strips the shift, and what is left issum_{n <= u, n = r mod e} mu(n)at a base-smoothe <= T, which the classical zero-free region forL(s, chi)bounds byu exp(-c'' sqrt(log u))(Davenport, chapters 14 and 20) with the only ineffective ingredient, the exceptional real zero, removed by the conductor bound above and the effective Landau-Page bound of the same chapter 14. There is no main term at any arc, the frequencya = 0included, where the contribution isfill^level M(x)/x. The saving isexp(-c sqrt(log x)). It is not compared with the(log x)^(-C)the source states for the prime analogue, which is an asymptotic with a main term where this is a bound with none; what is worth stating is that the main term is absent at every arc and that the prime number theorem is what puts thea = 0term inside the error. - The level of distribution on an initial segment. Proved. Assume (E1) and the large sieve the design supplies,
sum_{d <= Q} sum_{(b,d) = 1} |hat F_m(b/d)| << fill^m (Q^(2 alpha_1) + Q^2 base^(-m(1 - alpha_1)))at everym <= level, which follows from thel^1exponent by Farey spacing alone and readsQ^(54/77) + Q^2 Y^(-50/77)at base 10. Then for everyB > 0there isCwithsum_{d <= Q, (d,base) = 1} max_{y <= x} |#{n in D_level : n <= y, d | n, (n,base) = 1} - (1/d) #{n in D_level : n <= y, (n,base) = 1}| <= fill^level (log x)^(-B)at everyQ <= x^(1 - alpha_1) (log x)^(-C). The initial segment costs nothing in the level and one power oflog xin the saving, for two reasons.D_levelbelowyis a disjoint union of blocks{P base^m + t : t in D_m}, at mostfillof them per scale whateveryis; and the error the transform gives for#{t in D_m : t = r mod d}is uniform in the target residuer, so a shifted target is exactly as cheap as the residue0the source asks for. Above the cut the large sieve pays, below it Lemma A' pays, and the scale sum is dominated by its top scale becausealpha + alpha_1 >= 1. - What the base's own divisors cost. Proved. For
d = d_1 d_2withd_1base-smooth and(d_2, base) = 1, the split of the divisor section carries the level tod: the low digits fixn mod d_1and reach the rest only through an invertible multiplier, so the count reduces to the same transform estimate ind_2. What does not carry is the main term. It is a digit-string count times1/d_2and not1/d, readingfill^(-v)against a naivebase^(-v)atd_1 = base^v, so a Type I sum with coefficientsc_dproducessum_d c_d rho_F(d)where the coprime case producessum_d c_d / d, and nothing here shows the first small. Nor is the coprimality peeled off in general:sum_{n in S_F, n <= x} mu(n) = sum_{w | rad(base)} mu(w) sum_{n' : w n' in S_F, (n', base) = 1} mu(n')is an identity, and whether it reduces the problem depends on the inner sets. Sometimes it does - atbase 10andF = {0,1}the carry-free scaling of the transfer section gives{n : 2 n in S_F} = S_{0,5}and{n : 5 n in S_F} = S_{0,2}, both designs, and that column sits below this route's ownl^1floor in any case,fill = 2 < sqrt(10). Sometimes it does not: atbase 10andF = {0,1,2}the set{n : 2 n in S_F}begins1, 5, 6, 10, 11, 50, 51, 55, 56, 60, 61, 100, 101, 105, and it is a digit design at no base tested, the base-10 digit set it forces being{1,5,6}, which misses10, or{0,1,5,6}, which wrongly admits15because30leavesS_F. The hypothesis(n, base) = 1therefore stays inside the criterion below. - From strings to the design. Proved. The two statements above count
D_level, the padded strings, while the criterion countsS_F. With0 in Fthe two agree belowbase^levelbut for the element0, which carriesmu(0) = 0. With0outsideF,S_Fbelowbase^levelis the disjoint union of the exact-length blocks, each of them aD_l, so both statements sum overlwith the top block setting the exponent, the geometric sum having ratiofill > 1; that the level-of-distribution statement holds on an initial segment is what makes the sum legitimate at everyl. - The window and the criterion, as arithmetic. Proved. The exceptional set is
E = {a : |hat F_level(a/x)| >= fill^level x^(-beta)}. Both places the source spends it reduce tom_t < (2 - t) betafor somet in [1, 2), so the least admissible threshold isinf_{1 <= t < 2} m_t/(2 - t), which at the source's ownt = 235/154andm_t = 59/433is9086/31609 = 0.287449, rounding up to its23/80. The Type II window is then[(5/4) beta, 1 - 2 beta], and by the symmetry of the phase in its two variables also[2 beta, 1 - (5/4) beta]; atbeta = 23/80that is[9/25, 17/40]. Decomposingmuby a Heath-Brown identity of order above1/alpha_1, a piece with a free variable abovex^(alpha_1)is Type I at the level above, and otherwise greedy accumulation lands in the window under two conditions,beta <= 1/4, which merges the two windows into one interval, andalpha_1 + (5/2) beta <= 1, the greedy overshoot. The bilinear estimate itself asks five more, listed in the lattice bullet below; every one of them is free underalpha_1 < 1/3,beta <= 1/4and thel^1floor. Takingt = 1, sobeta = alpha_1 + eps, the binding condition isalpha_1 < 1/4. - The hybrid bound the lattice branch needs holds at every base, with the exponent the digit set owns. Proved. The published proof reaches its bilinear estimate through one bound whose whole purpose is to beat the plain
l^1exponent in the modulus aspect: forD, E, Y, Q_1powers ofbasewithD E << Y,e_1 ~ Q_1coprime tobaseandd ~ Dbase-smooth,sum_{e_2 ~ Q_2, (e_2,base) = 1} sum_{a < d e_1 e_2, (a, d e_1 e_2) = 1} sum_{|eta| <= E/Y, (eta + a/(d e_1 e_2)) Y in Z} F_Y(a/(d e_1 e_2) + eta) << (D E)^(alpha_1) (Q_1 Q_2^2)^(1 - alpha) + E^(alpha_1 + alpha/2) D^(1 + alpha/2) Q_1 Q_2^2 Y^(-alpha/2), whereF_Yis the transform normalised by its own mass. The proof is the source's, carried in general parameters: the product identityF_(Y_1 Y_2)(t) = F_(Y_1)(t) F_(Y_2)(Y_1 t)and the monotonicityF_Y <= F_UforU <= Ysplit the sum, the Chinese remainder theorem sends the residues through complete reduced systems exactly once, thel^1exponent and the shifted large sieve it supplies by Farey spacing pay three of the four factors, and the fourth is pure Parseval on a windowR = base^r,int_0^1 F_R^2 = R^(-alpha)andint_0^1 (F'_R)^2 << R^2 R^(-alpha), the first exact when0is inFand two-sided up to constants otherwise, in the direction used either way. So the two exponents are the digit set's own dimension and nothing else: the modulus exponent is1 - alphaand the saving exponent isalpha/2. At base 10 with one digit excluded the source prints them as1/21and10/21on the single check20/21 < log 9 / log 10, and1 - alpha = 0.045757sits under1/21 = 0.047619whilealpha/2 = 0.477121sits over10/21 = 0.476190: both roundings are safe and both are lossy. By thel^1floor above,1 - alpha <= alpha_1at every base and every digit set, so this bound never loses to the plain one in the modulus aspect, and the reading that a general base must fall back onalpha_1there mistakes the floor's equality case for the general value. - The lattice branch transfers to every base, and the conditions it asks are free below
1/3. Proved. With that bound the whole lattice half of the bilinear estimate runs in general parameters: forx = base^leveland the windowN K >= x^(1 - 2 beta),delta >= N/x,Q <= x^(1/2), the sum ofF_x(a_1/x) F_x(a_2/x)over pairs whose large contribution comes from a rank-2 lattice is<< (log x)^5 (Q + E)^(-eps/4) x/(N K), the source's own log power, the count of pairs of base-smooth moduli beingO(Q_0^(eps/2))at every fixed base. Five inequalities close it:2 alpha_1 < alpha;(2 - alpha) 2 beta < 1 - alpha_1; someuin(0, min(1, 2 alpha_1/alpha)]has2 beta (alpha_1 (3 - u) + u - 1) < u alpha/2;5 beta < 1 + alpha/2; and2 beta < (1 - alpha_1)(1 - alpha_1 + alpha/2). Their provenance is not uniform: the source writes a numerical check for the second and the fourth, while the first, third and fifth are read off steps it performs silently under the phrase that the exponents have been simplified for an upper bound. Its third written check renames an exponent and imposes nothing past the fourth condition, sobeta < alpha/2is not a hypothesis of the branch. All five are monotone, worse asbetaoralpha_1grows and better asalphagrows, so the corneralpha = 1 - alpha_1,beta = 1/4decides them all, and there they readalpha_1 < 1/3,1/3,1/3,1/2and1 - 1/sqrt(3) = 0.422649. Underalpha_1 < 1/3,beta <= 1/4and thel^1flooralpha + alpha_1 >= 1every one of the five holds, and1/3is sharp: three of them are equalities there. The floor and the threshold onbetaalone are not enough, asalpha_1 = 0.40,alpha = 0.60,beta = 1/4shows, where the first three read0.8 < 0.6,0.7 < 0.6and0.4 < 0.3. So the criterion's ownalpha_1 < 1/4clears the branch with room, and the pairalpha_1 < 1/3and(1 + alpha_1) 2 beta < 1 - alpha_1, which a reading of the first two through the plainl^1exponent produces, is not a pair of separate demands. - The threshold obeys the same Parseval floor, and the gate is
fill >= base^(3/4). Proved. The normalised transform is at most1pointwise, som_tis non-increasing int; andm_2 = 1 - alphaexactly, since two digit strings of lengthlevelcongruent modulobase^levelare equal, which is Parseval on the grid again. Hencem_t >= 1 - alphafor everyt <= 2, and2 - t <= 1fort >= 1givesbeta >= 1 - alphaat every base and every digit set, the flooralpha + alpha_1 >= 1read on the third exponent. The route's window conditionbeta <= 1/4then forcesalpha >= 3/4, that isfill >= base^(3/4), with equality pinning the gridl^1exponent to1 - alphaas well, the floor's own equality case. Dropping that window condition does not widen the route, it narrows it, and the step that shows it isbeta <= alpha_1: takingt = 1in the infimum givesbeta <= m_1, and the grid sum is one shift of the supremum, som_1 <= alpha_1. The single-window branch carries the greedy step on its own wheneveralpha_1 <= 1 - (13/4) beta, which againstbeta <= alpha_1asksbeta <= 4/17 = 0.235294..., strictly under the1/4it replaces, and with the two floors asksalpha >= 13/17 = 0.764705..706, so the gate rises tofill >= base^(13/17). That step is load-bearing and thealphacoordinate alone does not replace it:alpha = 0.9,alpha_1 = 0.154,beta = 0.26meets both floors, all five lattice conditions,2 alpha_1 < alphaamong them, the greedy capbeta <= (2/5)(1 - alpha_1)and the branchalpha_1 <= 1 - (13/4) betawith slack on each, atbeta > 1/4, and onlybeta <= alpha_1kills it. The floorbeta >= 1 - alphais an equality of Parseval and improves at no design, so the only movable numbers in the gate are the window floormax((5/4) beta, (5 beta - 1/2)/3), which the line branch imposes, and the window ceiling1 - 2 beta; sharpening the first, or finding a greedy step that crosses the gap between the two windows atbeta > 1/4without merging them, is what would break3/4, and no other parameter of the region can. The gate arithmetic is exact from the two floors; verbcheckoflab/py/mobius-regionprints those floors at four designs andbeta <= alpha_1att = 1at the base-21 recompute, and verbboundaryprints the branch capbeta <= 4/17and the gatealpha >= 13/17themselves, with that witness beside them. This3/4is also not thefill > base^(3/4)of the GRH section, which comes from that section'sl^1floorB_base(F) >= base: the two numbers meet for different reasons and share no proof. - The region the route asks for is exactly two inequalities. Proved. Of the eight inequalities the whole chain asks, one is a ceiling on
alpha_1alone,2 alpha_1 < alpha, and seven are caps onbetaat fixed(alpha, alpha_1); four of the seven fall inalpha_1and three are constant in it, so each takes its minimum over the region at the wallalpha_1 = alpha/2. At that wall the cap(2 - alpha) 2 beta < 1 - alpha_1reads exactly1/4at everyalpha, the cap2 beta < (1 - alpha_1)(1 - alpha_1 + alpha/2)reads(2 - alpha)/4and the cap5 beta < 1 + alpha/2reads(1 + alpha/2)/5, three identities inalphaand not roundings, the last two strictly above1/4foralphain(1/2, 1). Theu-condition2 beta (alpha_1 (3 - u) + u - 1) < u alpha/2splits atalpha = 2/3and is1/4only above it: at the wall its coefficient isc(u) = (3 alpha/2 - 1) + u (1 - alpha/2), andu/c(u)rises inuexactly when3 alpha/2 - 1 > 0, so foralpha >= 2/3the bestuisu = 1withc(1) = alphaand the cap is exactly1/4, while foralphain(1/2, 2/3)the coefficient is negative at small admissibleu, everybetapasses and the condition is vacuous rather than1/4. Vacuous or1/4, it never cuts, and none of the four ever cuts. Foralphain(1/2, 1)the region is thereforealpha_1 < alpha/2together withbeta <= min(1/4, (2/5)(1 - alpha_1)), the greedy cap cutting below1/4exactly fromalpha_1 = 3/8and from nowhere else, a threshold free ofalpha; the admissibility capbeta <= 2/5never binds beside the window cap. The three identities and theu-condition's split are proved and need no sweep; a sweep ofalphain[67/100, 999/1000]by1/1000andalpha_1in(0, alpha/2]byalpha/400, which lies entirely above2/3, corroborates them at0of66000cells and0of264000cap tests, with both wall equalities holding at each of330rationalalpha(Verified,lab/py/mobius-region, verbsregionandboundary). The headlinealpha_1 < 1/4is thet = 1proxy of the window cap and clears the whole region with room. - The line branch and the two bookkeeping steps, at every base. Proved. With
x = base^level, a thresholdbetaadmissible and at most2/5, which with the Parseval floorbeta >= 1 - alphaalready asksalpha >= 3/5of the design,delta >= N/x,N K >= x^(1 - 2 beta),Kabove the absolute constant of the pair dichotomy, andN >= x^(eps + max((5/4) beta, (5 beta - 1/2)/3)), the pair sum ofF_x(a_1/x) F_x(a_2/x)over the pairs whose large contribution lies on a line is<< (log x)^(O(1)) x^(-eps') x/(N K)forxpast a point depending onbase,fillandepsalone, witheps'a function ofepsand the implied constant depending on those three alone. The statement asks nothing of thel^1exponent and asks of the dimension only what the admissibility ofbetaalready encodes, so the wholel^1content of the route sits in the lattice half and the greedy step. Two bookkeeping steps complete it, and neither is a new idea. For coefficients bounded by thej-fold divisor function, orthogonality on the grid withtau_j^2 <= tau_(j^2)gives#{a mod x : |sum_n c_n e(n a/x)| >= x/C} <<_j C^2 (log x)^(j^2 - 1), so a Heath-Brown piece costs a power oflog xwhere a1-bounded sequence costs none. And Cauchy-Schwarz in the long variable gives|Sigma|^2 <= (x/N) sum_{(a_1, a_2) in E^2} F_x(a_1/x) F_x(a_2/x) T(a_1, a_2)withT(a_1, a_2) = sum_{l_1, l_2 <= N} min(x/N, ||(a_1 l_1 - a_2 l_2)/x||^(-1)), which is the exact step at which all four coefficient factors leave by the triangle inequality; the dyadic split into level sets and pair-mass classes costs two more log powers. Those two steps are derived here; the line estimate itself is the source's, carried in general parameters, and its close is read once here and owes a second reading. - Nothing in the bilinear half is base-10 mathematics. Refuted. No step of it needs
base 10numerically, and no step of it needs the coefficient side past1-boundedness. The lattice and line sections never see the polytope, the coefficients or the sequence weighting the frequencies; those are spent one section earlier and survive only as the two window numbers. Base 10 enters in exactly three places and all three are names rather than arithmetic: the set of integers all of whose primes divide the base, the coprimality to the base, and dyadic parameters that are powers of the base. Every printed exponent in those sections is a rounding ofalpha,alpha_1orbeta:1/21rounds1 - alphaup,10/21roundsalpha/2down,27/77isalpha_1,50/77is1 - alpha_1,9/8rounds1 - alpha_1 + alpha/2down,3/16roundsalpha/2 - betadown,17/40is1 - 2 beta,9/25rounds(5/4) betaup, and23/80isbeta. - The criterion. Conjecture. A digit set satisfying (E1) whose
l^1exponent obeysalpha_1 < 1/4hassum_{n in S_F, n <= x, (n,base) = 1} mu(n) = O_B(A_F(x) (log x)^(-B))for everyB. This is a program, not a theorem, and every part of it is named. The two steps that are aboutmurather than about primes are the two Proved statements above. The lattice and line estimates, the geometry of numbers and the exceptional-set bookkeeping are set-only or coefficient-free and transfer as read, and the lattice half, the line half and both bookkeeping steps are written out at general base in the bullets above, so the three write-outs the program lists as owed are written. What stands in their place is arithmetic rather than machinery, and it is one item: the level of distribution at base-divisible moduli, needed only to drop(n, base) = 1. The level itself carries, by the split of the divisor section, but the main term does not, being a digit-string count times1/d_2and not1/d, so a Type I sum with coefficientsc_dproducessum_d c_d rho_F(d)where the coprime case producessum_d c_d/d, and nothing here shows the first small. The hypothesis(n, base) = 1is therefore exactly what keeps every Type I modulus coprime to the base, since a Heath-Brown factorisation of an integer coprime tobasehas every factor coprime tobase. Of the source reading itself, the lattice branch is re-derived line by line and the line branch's close is read once; a second reading of that close is the only source work left. - Base 10 is refuted at every excluded digit. Refuted. A published or certified moment exponent is an upper bound, so it bounds
betafrom above and can never show the criterion fails: the publishedbeta <= 23/80 = 0.2875against1/4prices a gap of3/80and refutes nothing. The threshold is bounded from below by two monotonicities.F_x <= 1pointwise makesm_tnon-increasing int, so on a cell[t_0, t_1]everytin it hasm_t/(2 - t) >= m_(t_1)/(2 - t_0); and above a cut the Parseval valuem_2 = 1 - alphaalone forces the ratio past1/4. With the moment bounded below by the infimum window matrix, adaptive chains of25to53cells certifybeta > 1/4at all ten one-missing-digit sets of base 10, the certified lower bounds running0.2502716to0.2541480, so no admissible threshold clears the window condition there. The single-window branch is no escape and needs no second certificate: it asksalpha_1 <= 1 - (13/4) beta, which withbeta <= alpha_1readsbeta <= 4/17 < 1/4, the harder of the two asks, so one certificate ofbeta > 1/4kills both branches at once and the route is dead at base 10 at every digit rather than merely unreached. The refuting certificates do not order the columns: the brackets[0.2510933, 0.2625620]at the digit9and[0.2515026, 0.2875159]at the digit4overlap. Run at the target0.2626the same chain certifiesbeta >= 0.2632014at each of the eight non-extreme digits, up to0.2645208at the digit7, above both extreme upper bounds, while the digits0and9come back undecided as they must, and that settles the two extreme digits as strictly the cheapest columns of the base. The miss is at most0.0125620at the cheapest column and at least0.0139557at the digit4, and the factor2.99between the two printed excesses over1/4,0.0375159against0.0125620, is a ratio of upper bounds and not of misses (lab/py/mobius-region, verbsthresholdandthreshold 0.2626). A second miss is Refuted. Reading thel^1exponent against1/3substitutes the plainl^1exponent for the modulus exponent of the hybrid bound; with the true modulus exponent1 - alpha = 0.045757base 10 clears every one of the five lattice conditions as published, at0.701299 < 0.954242,0.601311 < 0.649350,0.304275 < 0.350649atu = 0.734926,1.4375 < 1.477121and0.575 < 0.731475, all five arithmetic from the three brackets thatlab/py/mobius-region, verbparamsprints, with theuadmissible againstalpharounded up. The greedy cap asks onlybeta <= 20/77 = 0.259740there, so the window cap1/4is what binds, and it is the cap base 10 provably fails: the obstruction there is the exceptional-set threshold alone. - The census of the criterion, and the least base that clears. Verified. Over the 38 proper digit sets of
base 3, 4, 5, the ten base-10 one-missing-digit columns and base 21 missing0, one design clears the criterion,47are refuted and one is open, the open cell beingbase 5withF = {0,1,3,4}, where the transform vanishes inside a window cell, the infimum matrix loses a row and the machine returns no positive lower bound rather than a false one. A pass is decided at the pessimistic corner,alphalow andalpha_1,betahigh, and a failure at the optimistic one, every cap being monotone in each parameter (Proved). The one design that clears is base 21 missing the digit0, atalpha_1 in [0.2499715, 0.2499822]from a second implementation of the window method against the five-digit[0.2499715, 0.2499821]already certified: the two agree on the lower bound to all seven printed digits and differ by one unit in the last on the upper, so the agreement witnesses transcription and the upper-bound gap is the only independent information (lab/py/mobius-region, verbscriterionandparams). The least base whose whole one-missing-digit family clears isbase 34, certified with its brackets and the bases it beats in coprime. - The shape the route can deliver, at best. Proved. The conclusion is a log saving and not a power, and the binding step is the level of distribution rather than the arcs: the major-arc lemma gives
exp(-c sqrt(log x))and the minor arcs give a power, while the Type I input gives(log x)^(-B). So this route decides whetherM_F(x) = o(A_F(x))on the columns it reaches and says nothing whatever about the exponenttheta(F)below. It reaches only the columns withfill >= base^(3/4), by the Parseval floor on the threshold above, and only under(n, base) = 1, by the split above: three limits stated rather than assumed. The weaker readingfill > sqrt(base), which follows fromalpha_1 < 1/2alone, stays true and is simply not sharp. - What would break this route against the census, and that it does not. Verified. The criterion's conclusion is a log saving, so it caps
theta(F)at1inA_Funits, far above every measured running-maximum exponent of the census below,0.4465to0.5358: neither statement can break the other. The chain that does print an exponent is the GRH one of coprime,theta(F) <= 1 - (1/4 - alpha_1)/alpha, and it reads above1at every base-10 column,1.1054746at the digit4, and0.9999819atbase 21missing0, a saving under2 * 10^(-5)in the exponent against the trivial bound. So no proved conditional bound sits below a measured exponent at any design of the census, in either chain, and a design whose measured exponent rose above its own proved ceiling would refute one of them (lab/py/mobius-region, verbcriterion).
The unconditional dissection
The GRH section buys a power saving on the dense columns and the pair route buys a log saving through the Type I level of distribution. A third route splits the frequency grid by Dirichlet approximation instead of by arithmetic, imports the minor-arc machinery built for the primes on a missing-digit set, and reaches a saving of classical zero-free-region shape with no hypothesis at all. Its whole price is paid in the base.
- Conjecture. Let
Fomit exactly one digit, letbase >= 92317, the ladder rungb = 4/5of the table above, letx = base^leveland letlevelbe past a point depending onbasealone. Then|M_F(x)| <= C(base) A_F(x) exp(-c(base) sqrt(log x))withC(base)andc(base) > 0both effective: an unconditionalo(A_F(x))on the dense column, the saving a classicalexp(-c sqrt(log x))rather than a power. The dissection below is written out and every region but one is derived here; the tag is Conjecture and not Proved because the one load-bearing analytic input is carried from memory rather than read at source, and because four bookkeeping items named below are unwritten. - The dissection, and which region binds. Approximate each grid frequency
a/base^levelby a reducedl/dwithd <= x^(3/5), writeh = |a d - l base^level|for the height and cut atZ = exp(C sqrt(log x)). Region A,d >= x^(2/5), pays the fulll^1mass of the transform against an unconditional minor-arc bound of strengthx^(4/5), so it asksc_base + 4/5 < alpha_base, which is exactly the ladder rungb = 4/5and its wall92317. Region B,d < x^(2/5)withmax(d, h) >= Z, asks the design'sl^1exponent to sit under1/4, the same bar the pair route's headline asks, and closes lower down the base scale than region A does, so region A is the binding one. Region C1,d < Zcarrying a prime factor outsidebase, runs the perturbed Lemma A' of coprime against the trivial|mu| <= 1and reads nothing aboutmuat all. Region C2,ddividing a power ofbase, is the only place the arithmetic ofmuenters, and no exceptional zero can live there by the conductor bullet of the pair route above. - What it has not got. The minor-arc bound in
d-form formuat an arbitrary arc denominator,|S_mu(theta)| << (x d^(-1/2) + x^(4/5) + (x d)^(1/2)) (log x)^(O(1))on|theta - l/d| <= 1/d^2with(l, d) = 1, is quoted from memory and not read at source, and it is the one load-bearing input; the twin of that shape for the von Mangoldt function is read at source in Maynard 2022, so what is owed is the substitution ofmuforLambdathrough Vaughan's identity with1/zetain place of-zeta'/zeta, one page and not a new idea. Unwritten besides: the contour bound and the truncated Perron constants,xnot a power ofbase, the region C1 transfer from an exact fractionl/dto the grid point nearest it, and everym > 1. - The shape is in print for the primes, the object is not. Maynard 2022 remarks that Siegel zeros play no role at these highly composite moduli, so the error terms there could be made effective of
exp(-c sqrt(log x))shape; that puts the shape over a missing-digit set in print for the von Mangoldt function. What is not in print over such a set is the object, a Mobius or Mertens sum, exactly as the GRH section's literature bullet records. The card would be first in its object and never in its shape. - Dead at fixed digit count. Proved. At
F = {0,1}in base 3 the design'sl^1exponent islog 2 / log 3 = 0.630929, above every bar the dissection sets, region B's1/4and region A's own ask included: the route is dead there at every strength of every input. This is thel^1floor of the GRH section again, read arc by arc instead of position by position, and it is why the wall sits in the base and not in the depth. - Its reach against the divisor section. The imported saving is
exp(-c(base) sqrt(log x))withc(base)falling in the base like a fixed negative power of it, while the level-x^(alpha_base/2)bound of the divisor section saves a power ofxoutright, so the imported bound does not overtake the divisor section untillevelis past a threshold that grows like a fixed power ofbase: at every depth any computation reaches, the divisor section is the better number. The import is also silent on everyFwhosel^1exponent is at least1/2, at every modulus sharing a factor withbase, and at any fixedlevel. - Vaughan's identity is circular here at power strength. Proved. In the
mu_{<=U} * mu_{<=U} * 1piece of Vaughan's identity the main term overS_Fisfill^level M_1(U)^2withM_1(y) = sum_{f <= y} mu(f)/f, so bounding the pieces one by one at power strength already forcesM_1(U) << U^(-delta), which continues1/zetaintosigma > 1 - delta. The method is blocked at power strength by the same circularity that leaves Vaughan's identity for the classical Mertens function at log strength. Nothing aboutM_Fitself is decided by this: what it asks for is an identity whose pieces do not isolateM_1(U), and this page carries none.
The exponent, tagged honestly
- Conjecture. For every digit set
Fwith2 <= fill <= base - 1whose digit gcd is squarefree,theta(F) = 1/2: square-root cancellation against the set's own mass, the RH shape transplanted to the sparse column. The census is consistent with this and proves none of it: the 47 running-maximum exponents sit in[0.4465, 0.5358]with per-family drifts of0.0157..0.1056over the last five levels, and the full-set controls - whose limiting exponent is1/2under RH - read0.4413..0.4517at the same depths. A slope is a fit; the exact integers above are the claim, the exponent is not. - The believable refutation targets are one family with a proved exponent below
1/2(excess cancellation) or a proved omega-result (a family whose meter provably tracks its mass). The scaling mechanism produces neither: the vanishing family{0,4}atbase 5is total cancellation for the trivial reason4 | n, and its reduced column{0,1}carries the open question unchanged. - The base-free carrier of the two-digit column is exact. Proved. For
F = {0,1}the meter is base-free: it is one functionnu*on the nonzero polynomials ofZ[x]up to units, supported on the monoidM*of products of0/1polynomials and summed in the order the base fixes, and the whole ladder rests on the classes of nonzero constant term, sincenu*vanishes at every polynomial divisible byx^2andnu*(x b) = -nu*(b). Its design Mertens at the level boundary,M(base^level) = sum_(deg P < level) nu*(P) + nu*(x^level), is pinned to-1at everylevel >= 2from the graded mass1 - 2t, and reads-1at every level2to23where it is computed. The running maximum reads 1, 1, 2, 3, 4, 7, 15, 23, 45, 86, 162, 331, 741, 1665, 3173, 7508, 17753, 36147, 79645, 182432, 427806, 858703, 2026147 atlevel 1..23, always far under the design mass2^level: the ratiomax/2^levelbottoms at 0.079102 atlevel 11, falls for the last time atlevel 15, and rises at every step from there to 0.241536 atlevel 23. What runs ahead of the mass is the rate, and its estimate is window-unstable: at depth 23 the geometric mean step reads 2.245836, 2.202419, 2.242075 over the last 4, 6, 8 levels but 1.996559 over the last 20, a hull straddling the mass rate 2 because the long windows open inside the levels where the ratio was still falling, while at the short windows every reading from depth 20 to 23 sits above 2.18 and above its own depth-18 value. That the rate exceeds the mass rate 2, on a census of twenty-three levels withlog_2(max)/levelclimbing to 0.910883 without settling, stays Conjecture. - The technology gap is real: distribution of digit-restricted sets in residue classes is Erdos, Mauduit and Sarkozy 1998, the ellipsephic almost-primes rest on it (Dartyge and Mauduit 2000), and primes in one-excluded-digit sets took the full circle method at large base in Maynard 2019, whose Type I input (Proposition 7.1) is the set's own level of distribution in base 10: moduli coprime to 10 up to
X^(50/77), the residue0, a saving of any power oflog X, with50/77 = 1 - 27/77for thel^1exponent27/77of the digit transform (Lemma 10.3), itself the Markov eigenvalue boundlambda_(1,4) < 2.24190 < 10^(27/77)of that paper's (10.5). Among Maynard 2019, Maynard 2022, Nath 2024 and the sources REFS.md lists, that proposition is the one Type I statement for such a set, and it is stated for base 10 with one excluded digit: the general-base multi-digit version is a substitution sketched in its Section 16, reachings <= base^(23/80)excluded digits ands <= base - base^(57/80)when they are consecutive, and Maynard 2022 reachess < base^(1/5 - eps)andbase - s >= base^(4/5 + eps)through the four Fourier norms of its Section 5 and the sketch of its Section 9. On the prime side Nath 2024 proves Bombieri-Vinogradov theorems forLambda(n) 1_A(n)at large base: unweighted with a maximum over residues only to levelX^(1/3 - delta), and nearX^(1/2)only against well-factorable weights, never unweighted; the set enters that proof through four norms of its transform (l^1, large sieve, hybrid,l^infinity) and never through a progression count, its only set-level fact being the count of the set in one last-digit class. Leng and Sawhney 2025 settle ternary Goldbach on the one-missing-digit set with thel^1boundg^(eps fill)of the digit transform. The nearest multiplicative function computed over a missing-digit set is the divisor function (Kim 2024), and a provedthetafor any restricted column sits at or beyond that frontier; a whole-text search of the three circle-method sources finds the wordMobiusonce, as an inversion step inside the proof of Proposition 7.1,Liouvillenowhere, andMertensonly as Mertens' theorem on a product over primes, so none of them carries a Mobius or Mertens sum over the set. What the divisor section above adds against that is a power saving inxwhere Proposition 7.1 saves a power oflog X, uniform over up tobase^(1 - eps)/2excluded digits, on the modulus ranged <= base^(1 - eps): sharper in saving type and in digit count, far shorter in range, and not a first level-of-distribution statement for such sets. Verified at source for Maynard 2019, Maynard 2022, Nath 2024 and Leng and Sawhney 2025. The census stands as the falsifiable record the eventual theorem must match.
Generators
- lab/rs/mobius-designs prints every row, identity check, slope, distribution and band above:
CARGO_BUILD_JOBS=4 cargo run --release -p mobius-designs. - lab/rs/rho-decoupling prints every divisor-section number:
CARGO_BUILD_JOBS=4 cargo run --release -p rho-decoupling. - lab/rs/mertens-numerology prints every constant, table row, margin, rung and cost-out number of the GRH section:
CARGO_BUILD_JOBS=4 cargo run --release -p mertens-numerology; its 34 tests pin every rendered row as a string, the sign change of the certificate at3689 -> 3690, the exhaustive sweep of3690..10^5with its smallest step, the kernel bound against the exact shifted-grid sum on a4001-point grid, each ladder wall below4 * 10^6against a scan frombase 3, and the constants of the general-bfloor bound. - lab/py/mrly-pairing prints the exact one-step constant at one excluded digit, the phase identity behind the sharpening, the two sharpened step 3 bounds and the walls they move:
uv run python research/lab/py/mrly-pairing/pairing.py onestep, its runtime printed in that study's README. Its sup is read on at-grid of cut1/4000over[0, 1/2], where the seat sits att = 1/2at every family printed frombase 100up and interior atbase 11andbase 13withe_0 = 0, so every measuredB_base(F)there is a reading and never a certificate; the statements above use the sharpened bound and not the reading. - lab/py/mobius-region prints the pair route's three exponents, its region and corner identities, its threshold certificates and its census:
uv run python research/lab/py/mobius-region/mobius_region.py checkin 27 seconds,regionandboundaryin under a second each,thresholdin 41 seconds andthreshold 0.2626in 80,criterionin 155 seconds over the 49 designs, andparams 21 123456789abcdefghijkin 12. The caps are exact rational functions evaluated inFraction,alphais bracketed by integer comparison offill^bagainstbase^aat wholeaandband never by a float logarithm, and a window cell whose infimum falls to zero prints no lower bound rather than a false one. The rest of the section's numbers are exact rational arithmetic carried out in the sentence that prints them, exponents quoted from the source named there, or certified base thresholds whose generator is named on coprime. - The Mertens control on the farey page is rendered by
lab/py/mertens-meter; the checkpoint controls here are the same function read at powers of the base. - lab/rs/carry-free-mobius prints the base-free ladder, its
M(base^level)by level, its running maxima, the census ofM*by degree and the rate above:uv run python research/lab/rs/carry-free-mobius/carry_free.py exponentis the pinned reference to level 18 andCARGO_BUILD_JOBS=4 cargo run --release -p carry-free-mobius -- ladder 22 2.5carries it to level 23, asserting the Python numbers at every level to 18 before it prints a deeper one.