digit-designs-and-the-euler-product.md
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Digit designs and the Euler product
- 2026-09-07 [Proved] The indicator of
S_F, the integers whose base digits all lie in the digit setF, is multiplicative exactly at the full digit set.1 in Fis forced byf(1) = 1; if a digitc >= 2is missing take the least, andR_c R_(c+1)has no carry because itsbase^mcoefficient ismin(m+1, c, 2c-m) <= base-1, so its digit set is exactly{1..c}whilegcd(R_c, R_(c+1)) = R_1 = 1; if only0is missing then oddbasegives the coprime pair(2, (base^2+1)/2)with productbase^2 + 1 = 101, and evenbasegives(base^2-1, base^2+1), coprime and odd, whose productbase^4 - 1has every digitbase-1whilebase^2+1does not lie in the set. Over all 8177 sets with2 <= base <= 12the constructed witness is asserted at each of the 4083 sets that passf(1) = 1and are not full, and an independent search finds a minimal witness for every one, hardestbase = 12,F = {1}, pair(5, 377). No design outside the full set carries an Euler product over primes;0excluded and a single digit both fail. Witness: lab/py/mrly-euler verb wall. - 2026-09-07 [Proved] For every
Fstrictly inside{0..base-1}the design zeta and the design Mobius series obey a disjunction and not a universal: if1is outsideFthe constant coefficient ofzeta_F M_Fis0; if a primepofS_Fhasp^2outsideS_Fthe coefficient atp^2is-1, since(p,p)is the only admissible factorisation; and otherwisezeta_F M_F = 1forces the least elementg > 1ofS_Fto be prime with every powerg^jinS_F, a necessary condition on an escapee and not a contradiction. At the full digit set the two are inverse,zeta_F = zetaandM_F = 1/zeta. Over 257 sets the leastn > 1with a nonzero coefficient is at most50, first atn = 4for base 3{0,1}andn = 9for base 10 missing9, while the eight full sets have none below4000. Witness: lab/py/mrly-euler verb pair. - 2026-09-07 [Proved] The position product. With
G_level(t) = prod_(i<level) sum_(d in F) e(d base^i t) = fill^level hat F_level(t), uniqueness of the digit expansion givesint_0^1 G_level(t) e(-nt) dt = 1_(D_level)(n)for every integern, hencesum_(n in D_level, n >= 1) a(n) n^(-s) = int_0^1 G_level(t) A(s,t) dtfor every absolutely convergent Dirichlet series, withA(s,t) = sum_(n >= 1) a(n) e(-nt) n^(-s);a = 1is the periodic zeta of DLMF 25.13.1 anda = muthe Lerch-Mobius series, sozeta_FandM_Fare pairings of one set-only product against one arithmetic-only kernel. The set enters through the digit positions and never through the primes. Checked to1.95e-16and2.04e-16at base 10 missing9,level = 3andlevel = 4, and2.9e-16at base 3{0,1},level = 3, 4, 5. Witness: lab/py/mrly-euler verb position. - 2026-09-07 [Proved] The tree's pair route is Holder on the position identity. When
0is inF, atx = base^levelthe identity is finite on both sides,M_F(base^level) = int_0^1 G_level(t) S_level(t) dtwithS_level(t) = sum_(n < base^level) mu(n) e(-nt), soabs(M_F(base^level)) <= (int_0^1 abs(G_level)) max_t abs(S_level)is at mostfill^level base^(level(alpha_1 - 1)) x^b = x^(alpha + alpha_1 - 1 + b); when0is outsideFthe same upper bound holds after summing the levels, a geometric sum of ratiobase^(alpha + alpha_1 - 1 + b) > 1by the flooralpha + alpha_1 >= 1of mobius.md. It sits under the trivialx^alphaexactly whenalpha_1 < 1 - b, which is the bar of coprime.md and mobius.md derived rather than posited, withb = 3/4 + epsunder GRH from Baker and Harman 1991. Witness: lab/py/mrly-euler verb position. - 2026-09-07 [Proved] The fibres of the Lerch-Mobius series are inverse Dirichlet L-functions. Splitting
nbyg = gcd(n,Q)and expanding on the characters of(Z/(Q/g))^*givesM(s, a/Q) = sum_(g divides Q) mu(g) g^(-s) phi(Q/g)^(-1) sum_(chi mod Q/g) tau_a(chi) L(s,chi)^(-1) prod_(p divides Q not Q/g) (1 - chi(p) p^(-s))^(-1), soM(s, a/Q)continues toCwith singularities inRe s > 0only at zeros ofL(s,chi)of modulus dividingQ, andM(s,0) = 1/zeta(s). SinceG_level(a/base^j) = fill^(level-j) G_j(a/base^j)are the largest values the position product takes, the design's major arcs are thebase-power rationals, and on that family holomorphy inRe s > 1/2is exactly GRH forbase-power modulus. Coefficient identity checked to2.6e-12at eleven pairs(Q,a)includingQ = 3, 9, 27, 100, the Euler-factor step to7.4e-16. Witness: lab/py/mrly-euler verb fibre. - 2026-09-07 [Proved] The reflection moves the kernel and not the design. Solving Hurwitz's formula DLMF 25.13.3 at
x = tandx = 1-tgivesZ(s,t) = ((2 pi)^s Gamma(1-s)/(2 pi i))(e^(pi i s/2) zeta(1-s,t) - e^(-pi i s/2) zeta(1-s,1-t))forsnot a positive integer, the derivation dividing by2i sin(pi s); this is DLMF 25.13.2 recovered, the gain being the rangeRe s > 0in place ofRe s > 1. The position identity turns it into a dual integral of the sameG_levelagainst Hurwitz zetas at1-s, never a relation betweenzeta_F(s)andzeta_F(1-s); the design's own symmetry is thebase-adic scalingG_level(t) = g(t) G_(level-1)(qt), whose transfer eigenvaluefill base^(-s)is what makes the vertical pole lattice. Formula checked to2.1e-30ats = 3.3,2.7 + 1.9iand0.6 + 4.1i. Witness: lab/py/mrly-euler verb dual. - 2026-09-07 [Proved] The design's multiplicative shadow is a Lyndon Euler product with no RH content. On the free monoid over
Fwith normN(w) = base^(abs(w)),sum_w N(w)^(-s) = 1/(1 - fill base^(-s)) = prod_(level>=1) (1 - base^(-level s))^(-c_fill(level))withc_fill(level)the Lyndon count, by Chen-Fox-Lyndon: every word factors uniquely as a non-increasing product of Lyndon words, so the free monoid onFis equinumerous by norm with the free abelian monoid on Lyndon words and is not equal to it. The primes are the Lyndon words, the zeta is zero-free, its Mobius is supported on the empty word and the letters so its Mertens is1 - fillbeyond norm1, and its poles are exactlys = alpha + 2 pi i m / log base, the design pole lattice. All RH content ofzeta_Ftherefore sits in the cofactorzeta_F(s)(1 - fill base^(-s)). Expansion verified throughu^16atfill = 2, 3, 4, 9, 10,c_2(level)being A001037. Witness: lab/py/mrly-euler verb word, A001037. - 2026-09-07 [Proved] The Beurling system of a design with non-unit digit gcd is finitely generated, on two branches. If
gcd(F) = a > 1every element ofS_Fis a multiple ofa; whenais prime the primes of the design are{a},N_Fis the powers ofaandM_B(x) = 0forx >= a, and whenais compositeS_Fholds no prime at all,N_F = {1}andM_Bis identically1, witnessbase = 10,F = {0,4,8}. Either way the eight scaled census families of mobius.md are exactly the columns the Beurling route cannot see, while the scaling transfer reads them exactly. Witness: lab/py/mrly-euler verb beurling. - 2026-09-07 [Verified] The Beurling census on the primes of a design, to
x = 10^6. Base 3{0,2}has the single prime2andM_Bidentically zero past2; base 3{0,1}has525primes,N_F(920483) = 2198, runningmax abs(M_B) = 98and exponent0.3339againstalpha/2 = 0.3155; base 10 missing9has35139primes,N_F(10^6) = 488864againstx^alpha = 531441,M_B(10^6) = 1860, running max1866, and exponentlog(running max)/log xreading0.4203, 0.4882, 0.5452at10^4, 10^5, 10^6againstalpha/2 = 0.4771, where full base 10 as control reads0.4084, 0.4241, 0.4276at the same points against its ownalpha/2 = 0.5. What the census reads is the level and not a trend:+0.068overalpha/2for the design against-0.072for the control, a running maximum climbing in both. There is cancellation,0.545against the trivialalpha = 0.954, and it is abovealpha/2, so the census supports cancellation and does not support the square-root conjecture onN_F;N_Fis notS_F. Full base 10 reproduces-23, -48, 212at10^4, 10^5, 10^6, A084237. Witness: lab/py/mrly-euler verb beurling, A084237. - 2026-09-07 [Proved] The identity that replaces
zeta M = 1on a design. For every(base,F)with1 in Fthe indicator1_(S_F)has a Dirichlet inversenu_F, given bynu_F(1) = 1andnu_F(n) = -sum_(d divides n, d > 1, d in S_F) nu_F(n/d), sozeta_F(s) N_F(s) = 1withN_F(s) = sum nu_F(n) n^(-s); the support ofnu_Flies inside the multiplicative semigroup generated byS_Fand strictly inside it, since9,27and36lie in the semigroup withnu_F = 0while16,48and52lie in the semigroup and outsideS_F, so the semigroup is a third set besideS_Fand the Beurling integers on the primes of the design and the support is a fourth, andnu_Fismuexactly at the full digit set, where the classical identity is the special case. Ifrhois a zero ofzeta_FwithRe rho > alphathensigma_c(N_F) >= Re rho, by the identity theorem on the connected pole-free half planeRe s > max(sigma_c(N_F), alpha), sosum_(n <= x) nu_F(n)is notO(x^(Re rho - eps))for anyeps > 0; the converse boundsigma_c(N_F) <= sup Re rhois not claimed. Checked againstmuterm for term on the full digit set ton = 131072atbase = 2andn = 177147atbase = 3, and the partial sums ofN_F(sigma)meet1/zeta_F(sigma)to1.60e-3atsigma = Re rho + 0.08 = 0.8008and1.96e-4atsigma = Re rho + 0.20 = 0.9208at base 3{0,1}, and to1.72e-2atsigma = 1.0816and2.39e-3atsigma = 1.2016at base 10 missing9, both offsets sitting aboveRe rho. Witness: lab/py/mrly-pairing verb inverse, lab/py/design-zeta. - 2026-09-07 [Proved] The design's own Mobius has anti-cancellation, and that is what makes the decoupling a blessing. Winding boxes on
zeta_Fby the argument principle certify one zero each and pinRe rhoto the box edges: winding1onRe in [0.72074, 0.72084],Im in [28.60563, 28.60573]at base 3F = {0,1}with contour minimumabs(zeta_F) = 8.298e-4against the engine bound6.284e-30, and winding1onRe in [1.00150, 1.00168],Im in [2.73915, 2.73925]at base 10 missing9with contour minimum6.865e-4against2.798e-23, while the control rectangleRe in [0.99900, 1.00050],Im in [2.73810, 2.74030]there returns winding0. Both boxes lie strictly right ofalpha = 0.6309297536and0.9542425094, sosum_(n <= x) nu_F(n)is notO(x^(0.72074 - eps))and notO(x^(1.00150 - eps))respectively: the limsup of the design's own Mertens function exceeds the design's own massA_F(x), and at base 10 missing9exceedsxitself, the box lying right ofRe s = 1. The square-root conjecture in thealpha/2shape is therefore false fornu_Fand can only be carried bymurestricted toS_F; the sibling's decoupling theorem is what protects it. Pointwise the census is far below both limsups,max/A_F = 0.0738at base 3level = 16andmax/x = 0.0847at base 10level = 7: the running maximum ofsum nu_F(n)grows by9.4474, 11.5000, 10.2220, 10.0354per level at base 10 missing9,level = 4..7, againstbase^(Re rho) = 10.036661and the trivialfill = 9, only the last of the four landing on the predicted rate, withmax/A_F(base^level)rising0.1043, 0.1094, 0.1398, 0.1588, 0.1771; at base 3{0,1}the geometric mean of the four stepslevel = 12..16is2.059against2.207512and2while the arithmetic mean of the five printed level ratios is1.9972, below the trivial2, a census too short to separate them. Witness: lab/py/mrly-pairing verbs box and inverse, lab/py/design-zeta. - 2026-09-07 [Verified] The pair
zeta_F M_F = 1 + D_Fgains nothing:D_Fhas abscissa exactlyalpha. Absolute convergence ofzeta_F^2putssigma_a(D_F) <= alphaand that half is proved; for the other half, ifsigma_c(D_F)were belowalphathenM_F(sigma) = (1 + D_F(sigma))/zeta_F(sigma)would tend to0assigma -> alpha+, sincezeta_Fhas nonnegative coefficients and is singular at its abscissa by Landau, sozeta_F(sigma) -> +infinity, and there is no circularity in the argument becauseM_Fis dominated termwise byzeta_Fand so converges absolutely at everysigma > alphawith no hypothesis ontheta(F). That half rests on a measurement, unconditional in shape sincesigma_c(D_F) < alphawould forceP(x) = o(x^alpha):P(x) = sum_(n <= x) c_F(n)divided byx^alphais bounded away from0and from infinity, reading0.493767, 0.699235, 0.758519, 0.587055at four sampling phases at base 3{0,1}, the four phases being needed becauseP(x)/x^alphais log-periodic and sampling only atx = base^levelaliases every Fourier mode onto one number. TheM_F(sigma) -> 0limit test is not a witness here: the tail the generator prints beside it isbase^(-level alpha/2), which assumes the square-root conjecture, and against the unconditional tail(fill-1) base^(-level eps)/(1 - base^(-eps))fromA_F(base^l) = fill^lno printedM_Fvalue at base 10 missing9is distinguishable from0. SinceM_F = (1 + D_F) N_Fandsigma_c(N_F) > alpha, the glue is not neutral but lossy. Witness: lab/py/mrly-pairing verb glue. - 2026-09-07 [Proved] The position pairing is exact on the grid and its
l^1mass sits at the top level, which kills the per-denominator split. For0 in F,M_F(base^level) = base^(-level) sum_(a mod base^level) G_level(a/base^level) S_level(a/base^level)exactly, both factors being trigonometric polynomials of degree belowbase^level; writinga = base^v a'withbasenot dividinga'andj = level - vgivesG_level(a/base^level) = fill^(level-j) G_j(a'/base^j)and the exact level decompositionC_level = sum_(j=0)^level fill^(level-j) c_jof thel^1mass, withC_j = fill C_(j-1) + c_j. Thel^1floorC_j >= base C_(j-1)forces the top-level sharec_level/C_level >= 1 - fill/base = m/baseat every base and digit set, measured0.485846, 0.602606, 0.687994, 0.510055against floors0.333333, 0.500000, 0.600000, 0.100000, with levelsj >= level/2carrying0.995116, 0.996061, 0.997043, 0.942350. Since the Baker-Harman Proposition beats the uniformx^(3/4)only belowj = level/2, weighting the Mobius input per denominator saves exactlylog(C_level/c_level)/(level log base), a constant factor capped bybase/m: the numerator is0.657068at base 3{0,1}, identical at everylevel = 6..14. The split exponents are0.988106, 0.912502, 0.905006, 1.012881uniform and0.941173, 0.879287, 0.879188, 0.964150per denominator againstalpha = 0.630930, 0.500000, 0.430677, 0.954243, while the Cauchy-Schwarz split is(alpha+1)/2exactly sinceint abs(G_level)^2 = fill^level; the base 2 and base 3 full-set controls return0.500000, the classical RH exponent. Witness: lab/py/mrly-pairing verb split. - 2026-09-07 [Proved] The principal fibre of the grid pairing has exponent
alpha - 1/2under RH, below the conjecturedalpha/2, and that is an asymptotic statement only. Thea = 0term of the grid pairing isbase^(-level) fill^level M(base^level), of exponentalpha - 1/2under RH, andalpha - 1/2 < alpha/2for everyalpha < 1, so in the limit the classical Mertens function cannot carry the conjectured size of the design meter. At finite depth it carries a great deal: thea = 0term reads-0.31857of11,0.11133of6,-0.05924of9and112.66549of276at base 3{0,1}level = 14, base 4{0,1}level = 11, base 5{0,1}level = 9and base 10 missing9level = 6, shares-0.028961, 0.018555, -0.006583, 0.408208, and exactly all of the meter on the two full-set controls. So at base 10 missing9the principal fibre carries40.8percent of the meter at the only measuredlevel, which refutes any claim that the square-root conjecture lives entirely off the principal fibre at finite depth: the exponent gap there is0.454243against0.477121, and a factor of10between them needsx = 10^44. Witness: lab/py/mrly-pairing verb split. - 2026-09-07 [Proved] The one-step constant of the digit transform never exceeds the triangle-split bound, and is strictly below it at every family measured beyond
level = 1. WithH(t) = sum_(r mod base) abs(g_F((t+r)/base))andB_base(F) = sup_t H(t), the identityC_level = sum_(a mod base^(level-1)) abs(G_(level-1)(a/base^(level-1))) H(a/base^level)givesC_level <= B_base(F) C_(level-1), soC_level/C_(level-1) <= B_base(F)at everyleveland every family with no computation at all; the inequality is not strict in general and equality is attained,C_1/C_0 = 4 = B_base(F)exactly at base 3{0,1}, so strictness needslevel >= 2. Verified there:C_level/C_(level-1)reads3.889888518, 5.032783116, 6.410132461, 18.369402635at base 3{0,1}, base 4{0,1}, base 5{0,1}(alllevel = 9) and base 10 missing9(level = 6), againstB_base(F) = 4.000000000, 5.226251860, 6.472135955, 19.888543820, and the ratio agrees between the two consecutivelevelthe generator prints to8.5, 7.4, 10, 5.0digits by family, so the stability is family by family and two values oflevelare all that is measured. Witness: lab/py/mrly-pairing verb split. - 2026-09-07 [Proved] The design Mobius of the two-digit design is base-free. Let
S*be the nonzero0/1polynomials ofZ[x],M*the monoid they generate,nu*the Dirichlet inverse of1_(S*). ForF = {0,1}at everybase >= 2,nu_F(n) = sum over P in M* with P(base) = n of nu*(P). Evaluation is a bijectionS* -> S_F, a monoid homomorphism, and of finite fibres, since an element ofM*has nonnegative coefficients soP(base) = ncaps every coefficient bynanddeg Pbylog(n)/log(base); the pushforwardgtherefore exists,g(1) = 1because1is the only element ofM*of value1, and grouping the pairs(D, Q)inS* x M*withD(base) Q(base) = nbyP = DQturns1_(S*) * nu* = deltainto1_(S_F) * g = delta, where the Dirichlet inverse is unique. Hencesum_(n <= x) nu_F(n) = sum over P in M* with P(base) <= x of nu*(P)at everyx: the base enters only as the order in which one base-free function is summed, and atbase = 2the classicalmuis that pushforward. Checked term for term with 0 mismatches ton <= 2^15,3^10,4^8and5^7, where 108978 elements ofM*collapse onto 32768 integers at base 2. Witness: lab/rs/carry-free-mobius verb lemma. - 2026-09-07 [Proved] The degree-graded mass of the base-free design Mobius is
1 - 2texactly. Degree is a monoid homomorphismM* -> Nwith finite fibres because1is the only constant inM*, which holds forF = {0,1}and for no design carrying a digit at least 2, where a constantc >= 2makes the degree-zero fibre{c^k}infinite; pushing1_(S*) * nu* = deltaalong it with2^dpolynomials of degreedgivesA(t)/(1 - 2t) = 1, so the graded sums are1, -2, 0, 0, ...andsum over deg P < level of nu*(P) = -1for everylevel >= 2. The design Mertens function is therefore pinned to-1at every level boundarylevel >= 2inside the carry-free window, and reads0atlevel = 1. Witness: lab/rs/carry-free-mobius verb sequence. - 2026-09-07 [Proved] The design zeta has an explicit zero free half plane, and it closes the census right of the abscissa. Let
a_minbe the least nonzero digit ofF, hence the least element ofS_F, every element of two digits or more exceedingbase. If a realsigma > alphasatisfiesa_min^sigma zeta_F(sigma) < 2thenzeta_Fhas no zero inRe s >= sigma: the coefficients are nonnegative and the series converges forsigma > alpha, so forRe s = sigma' >= sigmaone hasabs(a_min^s zeta_F(s) - 1) = abs(sum_(n in S_F, n > a_min) (n/a_min)^(-s)) <= sum_(n > a_min) (n/a_min)^(-sigma) = a_min^sigma zeta_F(sigma) - 1 < 1. The hypothesissigma > alphais load bearing and the test is one real evaluation carrying the ladder's own error bound. On the gridalpha + 0.05 nthe edgesigma_1reads0.5at base 4{1}to1.75at the three full digit sets over twenty-four designs, witha_min^sigma zeta_F(sigma)in[1.8635, 1.9995]and largestsigma_1 - alphaequal to0.95, so a census of the zeros right ofalphaneeds no hand chosen right edge and thealpha + 3.02strip of the locus sweep is three times wider than the zeros need. Witness: lab/py/transport-census verb census. - 2026-09-07 [Verified] The transport census: every proper design censused carries zeros right of its abscissa, the full digit set alone carries none, and each rightmost is certified by a winding box. On the box
alpha + 1e-6 < Re s < sigma_1, where the cofactorZ = zeta_F(s)(1 - fill base^(-s))is analytic and its zeros right ofalphaare exactly those ofzeta_F, the transfer failing only at residue null poles which sit on the lineRe s = alpha, the argument principle counts157zeros right ofalphabelowIm s = 40over twenty-three designs, all157located, plus2at base 50 missing one digit belowIm s = 4. The count is exact on the box and a lower bound for the half plane, since the sliveralpha < Re s <= alpha + 1e-6, the band0 < Im s < 0.02, everything above the census height and the conjugate half plane are uncounted. Twenty-one of the twenty-four designs carry such a zero; the three that do not are the base 2, 3 and 4 full digit sets, whose windings read-1.97e-33,1.73e-33and1.53e-33. Every rightmost carries the height it is read below, because the teeth of the level zero comb drift right with the pole index: base 20 missing one digit reads1.000285484146atIm s = 2.0988,1.000549674321at4.1971and1.002685494779at14.6920. BelowIm s = 40the rightmost real parts run0.441505537191at base 5{0,1}to1.002685494780at base 20 missing one digit, each certified by a winding1box onzeta_Fof half width5e-5inRe sand inIm swhose sampled contour minimum,1.2e-4to6.1e-3, beats the engine's error bound by at least eight orders of magnitude and whose distance to the pole lattices_(i,j) = alpha - i + 2 pi i j/log baseis at least0.00517845, four hundred box half widths. The two published boxes of lab/py/mrly-pairing reproduce at their own edges, winding1and1with contour minima8.298e-4and6.865e-4, and its control rectangle returns winding0. Witness: lab/py/transport-census verb census. - 2026-09-07 [Proved] A certified zero right of the abscissa refutes every square-root-shaped bound for the design's own Mobius, and the digit
1is the hypothesis that bites. Let1 in F, letrhobe a zero ofzeta_Fcertified by a winding1box with left edgex_0 > alphacontaining no pole, and letnu_Fbe the Dirichlet inverse of1_(S_F). The transport theorem givessigma_c(N_F) >= Re rho >= x_0 > alpha, sosum_(n <= x) nu_F(n)is notO(x^(x_0 - eps))for anyeps > 0; sinceA_F(x)has exponentalphathe square-root exponent isalpha/2 <= alpha < x_0, so the design's own Mobius satisfies no square-root-shaped bound and misses even the trivialO(x^(alpha - eps)), the first inequality failing to be strict only at the two designs withalpha = 0, whereA_F(x)grows likelog x. Nineteen of the twenty-four designs censused meet all three hypotheses and get a bound, and seventeen of the twenty-two the locus and family sweeps censused, the bounds runningtheta(nu_F) >= 0.4414555at base 5{0,1}totheta(nu_F) >= 1.0026354at base 20 missing one digit. Four designs haveRe rho > 1, so their own Mobius outruns the count of all integers belowx: base 10 missing two digits, base 10 missing the digit9, base 20 missing one digit and base 50 missing one digit, atfill/base = 0.8, 0.9, 0.95, 0.98andalpha = 0.9030900, 0.9542425, 0.9828779, 0.9948357; only the SIGN ofRe rho - 1is read and never its size, three of the four being censused toIm s = 40and base 50 toIm s = 4. Thefill/basereading dies on its control, base 5{0,1,2,3}at the samefill/base = 0.8with rightmost0.989748105861. Two designs carry a zero right ofalphaand no bound: base 4{2,3}and base 4{0,2,3}omit the digit1, so1is outsideS_F, the indicator vanishes there andnu_Fdoes not exist. Witness: lab/py/transport-census verb law. - 2026-09-07 [Refuted] The gain of a design's rightmost zero over its abscissa is not a function of
alphaandfill/base. The refuted functional is new: the locus row already refutes a law for the POSITION of the zeros, this refutes one for the single statistic the transport theorem reads, the rightmost real part lessalpha. Four equal key families, one base and one digit count each soalphaandfill/baseagree exactly and not to a rounding, read unequal gains: atalpha = 1/2,fill/base = 1/2the four base 4 two digit designs give0.0853043873,0.4400124317,0.2706238545,0.3439264581, a spread of0.35470804; base 5 atalpha = 0.4306766,fill/base = 0.4spreads0.37474232; base 3 two digit atalpha = 0.6309298spreads0.17605693; base 4 three digit atalpha = 0.7924813spreads0.060972003, which is still six hundred box widths. The two columns disagree in direction: the gain is largest at the sparsest designs,0.5291214025and0.4485242462atalpha = 0, while the rightmost real part itself is smallest there. What rises withalphais the floor, the least rightmost real part at eachalphareading0.4485242462, 0.4415055372, 0.5853043873, 0.7207876015, 0.9126562295, 0.9897481059, 1.0015143877, 1.0015892753, 1.0026854948, 1.0000614750up ten rungsalpha = 0, 0.4307, 0.5, 0.6309, 0.7925, 0.8614, 0.9031, 0.9542, 0.9829, 0.9948, rising at every step but the first and the last, the last being where the census height drops from40to4; one design per rung abovealpha = 0.86against six atalpha = 0.5, and no fit is taken. Witness: lab/py/transport-census verb law. - 2026-09-11 [Proved]
nu*vanishes at every polynomial divisible byx^2, andnu*(x b) = -nu*(b)at everybof nonzero constant term. The convolution runs overZ[x]divisors andM*is not divisor-closed,1 + x^2 + x^4 = (1 + x + x^2)(1 - x + x^2), so the claim lives onA, the nonzero polynomials mod units, wherenu*vanishes offM*by induction.Asplits asN x R,Rthe classes of nonzero constant term and divisor-closed;x^k blies inS*exactly whenblies inS*_odd, so1_(S*)is the outer product of the all-ones function onNwith1_(S*_odd), inversion factors, and the inverse of the all-ones function onNis1 - t. Witness: lab/rs/carry-free-mobius verb ladder. - 2026-09-11 [Proved] Every element of
M*of degreedhas its coefficient ofx^iat mostbinomial(d, i), so the maximum coefficient at degreedis exactlybinomial(d, floor(d/2)), A001405, attained by(1+x)^d. A product of0/1polynomials of degrees summing todis dominated coefficientwise byprod_k (1 + x + ... + x^(d_k)), each factor by(1 + x)^(d_k), and domination survives products of nonnegative polynomials, so the product is under(1 + x)^d, itself inM*. This retires the measured clause and the crude cap2^(L-1)of the carry-bound row. Checked at every degree to 22, maximum 705432. Witness: lab/rs/carry-free-mobius verb ladder. - 2026-09-11 [Proved] On
R, the classes of nonzero constant term among the nonzero polynomials up to units,nu*is fixed by the reciprocalb -> x^(deg b) b(1/x). There the reciprocal is degree-preserving, multiplicative and involutive, hence a monoid automorphism, and it carriesS*_oddonto itself by reversing the bitmask, so it preserves1_(S*_odd)and its Dirichlet inverse. It is no invariance on all ofZ[x]: the reciprocal drops thexpower andnu*(x) = -1againstnu*(1) = 1. Checked with 0 mismatches over the 35121747 classes of degree 1 to 21. Witness: lab/rs/carry-free-mobius verb ladder. - 2026-09-11 [Proved] The carry-free window of a two-digit design is
(base+1)^(level-1) < base^level. A0/1polynomial of degreedhasP(base) <= (1+base)^dand degrees add over a product, so the maximum ofP(base)overM*at degree belowlevelis exactly(base+1)^(level-1), attained by(1+x)^(level-1), and the least base holding every such element underbase^levelis the leastbasewith(base+1)^(level-1) < base^level. Verified by enumeration atlevel = 3..14, reading 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7. The windows are 4, 7, 9, 12 and 15 at bases 3 to 7, andsum_(n <= base^level) nu_F(n)leaves-1at level 5, 8 and 10, one level past the window each time. Witness: lab/rs/carry-free-mobius verb lemma. - 2026-09-11 [Conjecture] The base-free Mertens maximum of the two-digit design gains on the design's mass without reaching it. The running maximum of
sum nu*over degree belowlevelreads 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, and 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. A turn-down at a deeper level kills the trend and none is seen to 23. Witness: lab/rs/carry-free-mobius verb ladder. - 2026-09-11 [Conjecture] The rate of that maximum exceeds the design's mass rate 2, and its estimate is window-unstable. At depth 23 the geometric mean step reads 2.245836, 2.202419, 2.242075, 2.206405 over the last 4, 6, 8, 10 levels but 2.194975, 2.149760, 2.092480, 2.074545, 1.996559 over the last 12 to 20, a hull of [1.996559, 2.245836] straddling 2, the long windows opening inside the levels where the ratio still fell. Every short-window reading at depths 20 to 23 sits above 2.18 and above its depth-18 value. No constant is claimed;
log_2(max)/levelreaches 0.910883 atlevel = 23unsettled. Witness: lab/rs/carry-free-mobius verb ladder. - 2026-09-19 [Proved] Weil's theorem reaches a design zeta along no evaluation bridge:
P -> P(base)on polynomials with coefficients in{0..base-1}is a bijection onto the nonnegative integers and additive only where no carry occurs, a coefficient of the polynomial productP Rreaching(base-1)^2 (min(deg P, deg R) + 1)against the digit capbase - 1, so an integer product's digit string is the carry reduction of the polynomial product andR_c R_(c+1)is the shortest coprime pair whose carry-free product shows the missing digitc. Witness: lab/py/mrly-euler verb wall. - 2026-09-19 [Proved] Under the hypothesis
alpha < 1the large sieve does not rescue the per-denominator split on the major arcs: writinga = base^v a'withbasenot dividinga'andj = level - v, thel^2mass of the grid pairing over the levelsj <= Jis exactlyfill^(2 level - J) base^J, and the spacing form of the large sieve on thosebase^(-J)spaced points gives(base^level + base^J) base^level, so the levels belowJ = u levelcostx^(alpha + u(1-alpha)/2), strictly abovealphaat everyu > 0and equal toalphaonly atu = 0. Witness: lab/py/mrly-pairing verb split, Montgomery and Vaughan 1973.