zeros-of-the-design-zeta.md
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The zeros of the design zeta
- 2026-09-07 [Verified] This specific infinite design zeta has zeros in its own half-plane of absolute convergence, which the integers forbid, and the claim is the object and not the principle, since the positive-term Dirichlet series
1 + 2^(-s)has abscissa of absolute convergence-infinityand zeros at(2m+1) pi i/log 2. The census counts zeros of the Lyndon cofactorZ(s) = zeta_F(s)(1 - fill base^(-s)), analytic onRe s > alpha - 1, so it needs no pole-free strip and leaves no sliver against the pole line, on the single stripalpha - 0.92 < Re s < alpha + 3.02,0.02 < Im s < 60, split atRe s = alphaexactly. Base 3F = {0,1}atalpha = log_3 2carries 3 zeros right of the abscissa and 20 left of it inside that strip; base 10 missing the digit 9 atalpha = log_10 9carries 13 right and 25 left; base 3F = {0,2}carries 3 right, in the same three boxes as{0,1}. The largest surviving phase step on any census contour is0.9896and the largest propagated bound met at any census evaluation is9.99e-11, both printed beside every count. The base 2 full digit set is the control on both sides and each side names its object: zeros ofzetainalpha + 0.02 < Re s < alpha + 3.02count0, which is what the Euler product forbids, computed and not quoted; zeros ofzetainalpha - 0.98 < Re s < alpha - 0.02count13, the first thirteen belowIm s = 60; and the teeth of the cofactor1 - 2 base^(-s), which sit exactly ONRe s = alphaand are not zeros ofzeta, bring the one-strip count to19 = 13 + 6with6 = floor(60 log 2/2 pi). Right of the abscissa no continuation is used, since the positive series converges absolutely there and the ladder only rearranges it. The count is resolved and not certified: the largest surviving phase step is printed and nothing boundszeta_F'/zeta_Fon the contour, so a zero pair closer than the surviving spacing would stay invisible. Witness: lab/py/design-zeta. - 2026-09-07 [Proved] Scaled digit columns share a zero set exactly: for a positive integer
awitha max F <= base - 1, so thataFstays inside{0..base-1}, the carry-free bijectionm -> a mgiveszeta_(aF)(s) = a^(-s) zeta_F(s), an exponential factor with no zeros and no poles, sozeta_(aF)andzeta_Fhave the same zeros and residues in the ratioa^(-s_(m,j)), and the proof uses0 in Fnowhere. Base 3{0,2}against{0,1}agrees to5.6e-43at three points, and on what was censused, the stripalpha < Re s < alpha + 3.02,0.02 < Im s < 60, the two censuses coincide box for box: winding one inIm [22.01, 24.01], inIm [28.01, 30.01]and inIm [56.00, 58.00]for both, and zero in every other box. Left of the abscissa{0,2}is not censused and is inferred from the theorem. On the Mobius side the same bijection twists the meter by a sign, so the transfer is exact on both faces and trivial on one of them. Witness: lab/py/design-zeta, mobius.md. - 2026-09-07 [Proved] The Euler-product bridge between the two faces of RH is absent on a design:
zeta M = 1on the full set,S_Fis not multiplicatively closed for any properF, andzeta_F M_Fis not1, so no known route runs from a zero ofzeta_Ftotheta(F)and the zero census carries no bound on the square-root conjecture. What survives is not the zeros but the position product:G_levelpairs againsta = 1anda = mualike and the arithmetic sits entirely in the kernel. Coons 2010 Theorem 2.3 rules out the automatic-continuation route toM_Fand nothing wider. Witness: mobius.md, lab/py/design-zeta, lab/py/mrly-euler, REFS.md. - 2026-09-07 [Proved] The zeros of the design zeta are read off one analytic function and their positions near the pole lattice are forced by the residues. The Lyndon cofactor
Z(s) = zeta_F(s)(1 - fill base^(-s))is analytic onRe s > alpha - 1, since1 - fill base^(-s)cancels exactly them = 0line of the digit recursion's poles and no other; the poles ofZare thoses_(m,j) = alpha - m + 2 pi i j/log basewithm >= 1at whichzeta_Fhas a nonvanishing residue, the nearest line to that half-plane beingRe s = alpha - 1with residue-s_(1,j) gamma_1 r_j/fill, and on a full digit setZhas no pole at all, beingzeta(s)(1 - base^(1-s)), entire. One peel level givesZ(s) = E_1(s) + sum_(l >= 1) binom(-s,l) base^(-s-l) gamma_l zeta_F(s+l)withE_1(s) = sum_(a in F, a != 0) a^(-s)andgamma_l = sum_(a in F) a^l, checked against brute-force digit summation to1.6e-14with and without0inF, andZ(s) a_min^s -> 1to the right. A zero ofzeta_Fright ofalpha - 1is always a zero ofZ; conversely a zero ofZis a zero ofzeta_Fexcept at a poles_(0,j)withr_j = 0, whereZvanishes andzeta_Fis regular. Ats_(0,j)one hasfill base^(-s_(0,j)) = 1exactly for everyj, so withu = s - s_(0,j)the periodic factor is1 - base^(-u)with nojdependence andzeta_F(s) = Z(s)(1/(L u) + 1/2 + L u/12 - L^3 u^3/720 + ...),L = log base, giving residueZ_0/L, regular partZ_1/L + Z_0/2and its derivativeZ_2/L + Z_1/2 + Z_0 L/12from the Taylor coefficients ofZalone, on a disc of radius at least1and exactly1whenr_jdoes not vanish. Witness: lab/py/zeta-locus, lab/py/design-zeta, lab/py/burnol-residue. - 2026-09-07 [Verified] The zeros of the design zeta near the abscissa are a residue comb whose tooth position the residue and the regular part predict. A zero near the pole
s_(0,j)solvesu(R_j + R'_j u + ...) = -r_j, first orderu_1 = -r_j/R_jand second order the near root ofR'_j u^2 + R_j u + r_j = 0, both built from the Laurent data with nothing fitted. Over 20 designs toIm s = 40(every scaling class atbase = 3andbase = 4, two atbase = 5, base 9{0,1,2}, base 16{0,1,2,3}, base 10 missing 9, base 2 full set) one assignment radius0.45, fixed by the discs not overlapping and not by the tooth law so that every count is conditional on it, serves both the count and the tooth, discs never overlapping since the smallest period in the sweep is2.2662: the argument principle on that circle gives 164 poles carrying one zero ofZ, 40 none and 8 two, of which 21 are the residue-null pole centres of the three full-set columns and are zeros ofZthat are not zeros ofzeta_F, leaving 143 poles with one zero ofzeta_F, 61 with none and 8 with two. All 151 poles carrying a zero have their zeros located by a polar grid inside that same disc and not by the prediction, so no tooth is selected by the law it tests and no pole carrying a zero is left without one. Comparing prediction to tooth afterwards,miss2/miss1has median0.1637withmiss2 < miss1at 147 of the 151, and the accuracy is conditional on the tooth being close: the 43 teeth atabs(u) < 0.1have largest first-order miss0.01446and largest second-order miss0.00164, the 84 atabs(u) < 0.2have0.10815and0.01526, while the 31 atabs(u) >= 0.3reach1.64614and the prediction says nothing. The densest column is the sharpest: base 10 missing 9 atfill/base = 0.9locates 15 teeth to a largest first-order miss of0.013602and a median of0.000841. Witness: lab/py/zeta-locus. - 2026-09-07 [Verified] The critical line is the second family of the full digit set. On a full digit set
zeta_Fiszeta, whose only pole iss = 1 = alpha, so it is regular at everys_(0,j)withj != 0and the residue there vanishes as a one-line consequence rather than a measurement; the machinery reads those residues as1e-26to1e-33, which is a control of the engine, and the comb is empty. The winding of the cofactor overalpha - 0.92 < Re s < alpha + 3.02,0.02 < Im s < 40then splits exactly as six zeros ofzetaplusfloor(40 log base/2 pi)cofactor-only teeth, those teeth being the zeros of1 - base^(1-s)onRe s = 1by exact arithmetic:10 = 6 + 4atbase = 2,12 = 6 + 6atbase = 3and14 = 6 + 8atbase = 4. The six survivors readRe s = 0.5atIm s = 14.1347251417, 21.0220396388, 25.0108575801, 30.4248761259, 32.9350615877, 37.5861781588at all three bases, and the three columns share that zero set to1e-26because they are one arithmetic object. On a design the same split leaves a second family that is not a line atalpha/2: its real parts run-0.273079611to0.391038600over the 7 zeros belowIm 40at base 3{0,1}againstalpha/2 = 0.3154648768,-0.30495894to0.28101268over 6 zeros at base 4{0,1}against0.25, and0.060261843to0.97363028over 5 zeros at base 16{0,1,2,3}against0.25, the spread being the witness and no per-design mean claimed. Witness: lab/py/zeta-locus. - 2026-09-07 [Proved] The second family of the design zeta does not depend on which comb is stripped, and the next pole line's comb is computed from the first one's residues. For
m >= 1the cofactorZ_m(s) = zeta_F(s) prod_(i <= m)(1 - fill base^(-(s+i)))has exactly the zeros ofZ(s) = zeta_F(s)(1 - fill base^(-s))insidealpha - 1 < Re s < alpha + 3.02, since each extra factor vanishes only onRe s = alpha - ifori >= 1, so the survivors of the assignment are one set under every comb. WhatZ_madds is the level-icomb, and its Laurent data is forced by the level-zero data:Z(s) = E_1(s) + sum_(l >= 1) binom(-s,l) base^(-s-l) gamma_l zeta_F(s+l)is singular ats_(1,j) = alpha - 1 + 2 pi i j/log basethrough itsl = 1term alone, and withbase^(-s_(1,j)-1) = base^(-s_(0,j)) = 1/filland1 - fill base^(-s_(1,j)) = 1 - basethe residue ofzeta_Fthere isr_(1,j) = s_(1,j) gamma_1 r_(0,j)/(fill(base-1)), so the level-one comb is empty wherever the level-zero comb is, and at the full digit sets_(1,0) = alpha - 1 = 0makes it vanish, which iszetahaving no pole ats = 0; the lab printsabs r_(1,0) = 0.0with the null flag set andabs r_(1,1) = 8.89623e-29at the base 2 full set. Witness: lab/py/zeta-family, lab/py/zeta-locus, lab/py/design-zeta. - 2026-09-07 [Verified] The second family of the design zeta, split out and censused over twenty-two designs at a stated assignment radius, with no gap at that radius on any design. Stripping the level-zero and level-one combs at
rho = 0.45, a constant fixed only by the pole discs not overlapping and not by the tooth law, which is accurate only insideabs(u) < 0.2, the twenty designs of the locus sweep plus base 5{0,1,2,3}and base 10 missing two digits give 377 zeros wound by the argument principle, 351 located, 171 teeth of which 9 are level-one teeth, 19 cofactor-only zeros at null-residue poles and 161 second-family zeros, each design censused to its own printed height,40except the four base 3 designs at42.894, base 9{0,1,2}at41.464and base 10 missing two at25.923. There is no gap atrhoon a design: the distance from a second-family zero to the nearest live pole has minimum0.45510938at base 4{2,3},0.45909168at base 3{0,1},0.48696667at base 4{0,1,2}and0.50481072at base 4{1,3}, with base 4{2,3}putting five of its eight inside0.45 < abs(u) < 0.6, so every count is conditional onrhoand falls asrhorises,N_2reading8, 7, 13, 9, 14atrho = 0.45against7, 6, 12, 8, 7atrho = 0.6on base 3{0,1}and the four base 4 two-digit designs. The full digit set is where the gap exists: at base 2 the nearest live pole to a second-family zero is14.143566away and no radius below0.9moves any count. Where the located count falls short of the winding, base 4{2,3}at 12 of 18 being the worst,N_2is a lower bound. Witness: lab/py/zeta-family verb tests. - 2026-09-07 [Verified] Seventeen designs carry a lower bound on the Mertens exponent of their own Mobius, and the strongest bound is radius-robust. Nineteen of the twenty-two designs have a censused zero of
zeta_Fstrictly right ofalpha, twelve of them in the second family, and at the seventeen of them whose digit set contains1, so thatnu_Fexists, the transport theorem gives thatsum_(n <= x) nu_F(n)is notO(x^(Re rho - eps)); base 4{2,3}and{0,2,3}omit the digit1and carry a zero but nonu_F. Base 10 missing two digits has a zero at1.00151438765 + 2.77402670058 iagainstalpha = 0.903089987, a second base-10 column where the design's own Mobius has a Mertens exponent above 1 and so abovexitself; that zero is a level-zero tooth atabs(u) = 0.1083of thej = 1pole, deep inside every assignment radius tested, so the bound does not depend on where the comb is cut. Base 4{1,2}has a second-family zero at0.940012431696 + 13.0678968771 iagainstalpha = 1/2, an exponent of0.94against a design mass exponent of0.5, and base 3{0,1}reads0.720787601477atIm 28.6056765649againstalpha = 0.630929754. Witness: lab/py/zeta-family verb tests, lab/py/mrly-pairing verb inverse. - 2026-09-07 [Verified] What converges as a design fills is the ordinate set and not the real part. Against the derived null of a quarter of the mean gap between consecutive
zetaordinates in the range, the exact expectation for an equally spaced ordinate set of the same density and conservative for one with gap variance, the mean distance from a second-family ordinate to the nearestzetaordinate divided by that null falls monotonically inalpha:2.0495374at base 5{0,1}withalpha = 0.430676558,1.8953371at base 4{0,1}with0.5,0.75419266at base 3{0,1}with0.630929754,0.51648744at base 4{0,1,2}with0.792481250,0.32356636at base 5{0,1,2,3}with0.861353116,0.090501352at base 10 missing two with0.903089987and1.0429899e-23at the base 2 full set. The base andfillconfounds are dead: the fall is monotone at fixed base,2.0495374to0.32356636inside base 5 and1.8953371to0.51648744inside base 4, and at fixedfill = 2across bases,2.0495374, 1.8953371, 0.75419266, 1.0429899e-23atalpha = 0.430676558, 0.5, 0.630929754, 1; the nulls move only1.0425839to1.3595166across the ladder while the raw mean distance falls2.4032315to0.12303809, so the denominator does not drive it. Over the same designsmean abs(Re s - 1/2)reads0.36482392, 0.39426128, 0.3901396, 0.25540269, 0.31452367, 0.20473972and2.4065966e-23and does not fall monotonically, so atalpha = 0.903the heights are pinned to2.3percent of the mean gap while the real parts are still0.20off1/2.alphais a trend and not a function: the four base 4 two-digit designs at onealpha = 1/2spread0.79050661to2.8404536. The matching is nearest-ordinate and not injective, 3 distinct ordinates for 4 design zeros at base 10 missing two. Witness: lab/py/zeta-family verb limit. - 2026-09-07 [Proved] The ordinate shadow is a first-order perturbation and its constant-free form is a Newton step from the zeta zero. The discrete position identity
1_(D_level)(n) = base^(-level) sum_(a mod base^level) G_level(a/base^level) e(-n a/base^level)on0 <= n < base^levelgiveszeta_(F,level)(s) = base^(-level) sum_(a mod base^level) G_level(a/base^level) S_level(s, a/base^level)withS_level(s,x) = sum_(1 <= n < base^level) e(-nx) n^(-s), reproduced from the transform to8.326e-40atlevel = 2on ten designs, and sinceG_level(0) = fill^levelthea = 0fibre carries the weight(fill/base)^levelexactly against the partial sum ofzetatobase^level, with no arc and no limit. That identity splits the polynomial at levellevelagainst a TRUNCATED zeta while the object is the continuedzeta_Fagainst the fullzeta, and(fill/base)^levelfalls to0withlevelwhile both series tend to1on the right, so no level is forced andc = fill/baseis thelevel = 1reading and a definition. For any constantcthe splitzeta_F = c zeta + E_FgivesE_F(rho_0) = zeta_F(rho_0)at a zerorho_0ofzeta, an identity carrying no information aboutc, and a first-order zero ofzeta_Fatrho_0 - zeta_F(rho_0)/(c zeta'(rho_0)); readingc zeta'(rho_0)aszeta_F'(rho_0)removes the constant and givesrho_0 - zeta_F(rho_0)/zeta_F'(rho_0), Taylor at a simple zero ofzeta_F. The offset is one complex number, so at the zeros this law pairs the ordinate offset and the real-part offset are one quantity. The continuous form, the mass ofG_levelonabs(t) < 1/(2 base^level), is the exact sinc sum1/base^level + sum_(n in D_level, n > 0) sin(pi n/base^level)/(pi n)and equalskappa_level(F) (fill/base)^levelwithkappa_levelrunning0.6015221to0.96774464over the ladder atlevel = 1, 2, 3, so it adds no constant the fibre does not give. Witness: lab/py/zeta-shadow verb mass, lab/py/mrly-euler verb position. - 2026-09-07 [Verified] The constant-free first-order step predicts the design zero attached to each zeta zero, and it sharpens as the offset shrinks. Over nine designs at twelve zeta zeros to
Im s = 56.4462476971, six toIm s = 37.5861781588at the two densest so the rungs do not share one height, both predictions are computed fromzeta_F(rho_0),zeta_F'(rho_0),zeta'(rho_0)and the digit density alone and the zero is located afterwards by Newton fromrho_0, accepted only atabs(zeta_F) < 1e-16, within1.5ofrho_0and0.02clear of the pole lattice, largest ladder bound9.001e-23. The step's median ratio reads1.3843088, 1.284225, 1.2481449, 1.2060106, 1.2042502, 0.89075541, 1.0195598, 1.005076, 0.99741809atalpha = 0.430676558, 0.5, 0.630929754, 0.792481250, 0.861353116, 0.903089987, 0.954242509, 0.982877878, 0.994835739, with largestabs(ratio - 1)0.14041at base 20 missing one digit and0.01734at base 50 missing one digit, bands[0.94875, 1.14041]and[0.98266, 1.01144]; pooled over the ladder that largest deviation runs0.01734, 0.0508884, 0.193158, 0.83912, 3.32327over the bucketsabs off < 0.05,< 0.1,< 0.2,< 0.4and above, on7, 4, 14, 18, 44zeros. Thelevel = 1readingc = fill/baseis the looser column, median ratio1.4129353, 1.2842149, 1.0955991, 1.1806066, 1.1372453, 1.276577, 1.1347487, 1.0320127, 1.0507079with largestabs(ratio - 1)0.24964and0.0821168at the two dense rungs, five times looser than the step at base 50, and the couplingzeta_F'(rho_0)/zeta'(rho_0)does not select it either,median abs(coupling - fill/base)reading0.24057225and0.08291158there againstmedian abs(coupling - 1)0.27619434and0.079335871, a flip between the two rungs while the candidates differ only by0.05and0.02. Nine zeros at the three sparsest designs have no located zero inside the trust region, predicted offsets0.95618855to3.0967393, so those rungs' medians are conditioned on Newton succeeding. The base 2 full set is the exact control,abs(zeta_F(rho_0))between1.85e-34and1.329e-25at all twelve zeros, soE_F = 0and both offsets are0. Witness: lab/py/zeta-shadow verb predict. - 2026-09-07 [Verified] The paired shadow offset carries its exponent in the missing-digit density rather than in
1 - alpha, and two new rungs sample the interval betweenalpha = 0.954and1. The median paired offset divided bym/base = 1 - fill/basereads1.6463532, 1.2495026, 1.8345578, 1.5731321, 2.2102406, 1.6634381, 2.2424916, 1.8779239, 1.051349across the nine rungs and divided by1 - alphareads1.7350628, 1.2495026, 1.6569184, 1.8951686, 3.1883019, 3.432954, 4.9008186, 5.4839111, 4.0716338; a least squares in the logs, a fit and not a theorem, gives(m/base)^1.04544atR2 0.957842against(1-alpha)^0.71691atR2 0.944011, the first column spanning2.13297and the second4.38888, som/basecarries the exponent by a factor of2.05764inside the4.28797that(1-alpha)/(m/base)itself spans over this ladder, which is the whole discrimination the two normalisations admit here. The new rungs are base 20 missing its top digit atalpha = 0.9828778777and base 50 missing its top digit atalpha = 0.9948357391, all six zeros located at each, medianabs(E_F(rho_0))0.11830158and0.028066806and median offset0.093896196and0.021026979, so the PAIRED offset falls fast across that interval; this bounds no maximum over the whole second family and touches no jump clause, since the pairing selects zeros for closeness to azetazero and censuses nothing. Read in the form of the family row, the mean distance from a located design ordinate to the nearestzetaordinate over a quarter of the mean gap between consecutivezetaordinates in the range gives0.81218635, 0.57141859, 0.50488757, 0.37447728, 0.20954319, 0.29197634, 0.14794812, 0.052888241, 0.011098646and0at the full set; the pairing is zeta-zero-first where the family row's is design-zero-first, so this is a parallel ladder and not that row recomputed. Witness: lab/py/zeta-shadow verb rungs. - 2026-09-07 [Proved] A positive Rouche margin proves exactly one zero of the design zeta in a disc about a pole, with every input bounded from the digit recursion itself. Write
Z(s_0+u) = P(u) + T(u)at a poles_0 = s_(0,j)with nonvanishing residue, whereP(u) = (1 - base^(-u)) D_(P-1)(s_0+u) + E_P(s_0+u)is entire with Taylor coefficients the exact finite sumssum_n n^(-s_0)(-log n)^m/m!convolved against those of1 - e^(-L u), andTis thel >= 1part of the ladder numerator, bounded onabs(u) <= R_2byB_T = sum_(l >= 1) binom(abs(s_0)+R_2+l-1, l) base^(-sigma-l) gamma_l G(sigma+l)atsigma = Re s_0 - R_2withGthe peeled majorant. Thatlsum is closed by a majorant ratio and not by an observed one, the term ratio itself not being monotone: sincegamma_(l+1)/gamma_l <= a_maxandG(sigma+l+1)/G(sigma+l) <= base^(-(P-1))because every string in the pools is at leastbase^(P-1), the term ratio is at mostR_l = ((abs(s_0)+R_2+l)/(l+1)) a_max base^(-P), which decreases inlonceabs(s_0)+R_2 >= 1and is belowa_max base^(-P)otherwise, so stopping at the firstlwithR_l < 1and addingterm_l R_l/(1-R_l)is a proof. Thenabs(Z_n) <= B_T/R_2^nforn >= 2beyond the explicit part, so onabs(u) = rhoone hasabs(Z - (Z_0 + Z_1 u)) <= sum_(m >= 2) abs(P_m) rho^m + B_T tau^2/(1-tau)withtau = rho/R_2, whileabs(Z_0 + Z_1 u) >= abs(Z_1) rho - abs(Z_0); when the first is strictly less than the second the linear model andZhave the same zero count inabs(u) < rhoby Rouche, and that count is one becauseabs(Z_0/Z_1) < rhofollows from the same inequality. Since the residue does not vanish,Z(s_0) != 0and the zero is a zero ofzeta_F. No step uses a differenced quantity:Z_0is the ladder value with its propagated bound andZ_1is the first Fourier mode ofTon a circle of radiusR < R_2withNsamples, whose aliasing is at most(B_T/R_2)(R/R_2)^N/(1-(R/R_2)^N), plus an exactp_1. The peel depthPand the radiirhoandR_2are free parameters of the proof. Witness: lab/py/zeta-locus, lab/py/design-zeta. - 2026-09-07 [Verified] The residue comb carries exactly one zero of the design zeta at eleven certified poles, the peel depth is the lever that decides which, and the certificate fails at every pole carrying none or two. Running the Rouche margin with the peel depth raised at each pole until the certificate fires or the string pool caps, over 106 poles at base 3, base 5, base 9, base 16 and base 10 missing 9 to
Im s = 40inside a fifteen minute budget, gives 11 certified, 60 failed, 7 residue-null and excluded because there the model's zero is the pole centre, a zero of the cofactor that is not a zero ofzeta_F, and 28 skipped when a design spent its budget. The certified eleven, with depth, margin and the radius the proof used: base 3{0,1}j = 2atP = 7,0.13418242,rho = 0.205;j = 5atP = 7,0.028140545,rho = 0.16;j = 7atP = 9,0.00082974181,rho = 0.175; base 5{0,1}j = 4atP = 7,0.15035818,rho = 0.2775;j = 5atP = 7,0.12269904,rho = 0.295; base 9{0,1,2}j = 5atP = 5,0.038456894,rho = 0.26; and base 10 missing 9 atj = 1, 2, 3, 4, 7, all atP = 3, margins0.047105507,0.030062806,0.045802462,0.043508323and0.046292701at radii0.1275, 0.105, 0.1025, 0.09, 0.0725, each on 24 contour samples. Every certified disc agrees with the argument principle count of one and none disagrees; of the 19 poles carrying zero or two zeros inabs(u) < 0.45that the budget evaluated none is certified, the two double poles reached, base 3{1,2}j = 3andj = 5, both failing, while base 5{1,2}j = 7and base 10j = 15were skipped for budget. Base 10 is not closed by any sharper majorant but by peeling: at the automatic depthP = 2itsB_Truns1.08atj = 1to38.1atj = 15, and atP = 3it runs0.2096to1.2010over the eight poles reached, five of which certify. Proximity of the tooth is no threshold, the certifiedabs(Z_0/Z_1)running0.0282669to0.149708and base 3{0,1}j = 7at0.104443failing atP = 7and certifying atP = 9. The margins are evaluated in high precision and not in ball arithmetic, which is the one step between this row and Proved. Witness: lab/py/zeta-locus. - 2026-09-07 [Refuted] The locus of the zeros of the design zeta is no curve
Re s = f(Im s)shared by designs of equalalpha, no comb in the pole-period residue, and no law inalphaandfill/base. The witness against a shared curve is a pair of zeros of nearly equal imaginary part and very different real part on two designs of equalalpha, which a single curve cannot carry: base 4{1,2}and base 16{0,1,2,3}, bothalpha = 1/2, hold zeros0.015058apart inIm snearIm s = 4.72and0.817047apart inRe s; base 4{0,1}against{2,3}, equal inalphaand infill/base, gives0.0136014against0.719693nearIm s = 17.64; base 4{0,1}against{1,2}gives0.0063091against0.280397nearIm s = 22.87and base 4{1,2}against{2,3}gives0.00638631against0.198012nearIm s = 31.79, each pair drawn from censuses of the same box and the same height. Equality of bothalphaandfill/basetherefore fixes nothing. Within one design the worst real-part gap between two zeros of equalfrac(Im s log base/2 pi)runs0.077591803at base 10 missing 9 to0.65632474at base 3{0,1}, so the fractional part fixes nothing either, and the zeros per period atalpha = 1/2reads1.1897445at base 16,1.2868204at base 9 and1.586326, 2.0395621, 2.0395621, 2.2661801at base 4, so no counting law inalphaalone survives. The single exception isalpha = 1, where the full digit sets atbase = 2, 3, 4are one arithmetic object and do share every zero. Witness: lab/py/zeta-locus. - 2026-09-07 [Refuted] The second family of the design zeta is not symmetric about any vertical line
Re s = c_F. Readingc_Fas the midpoint of the real parts of the two second-family zeros of leastIm sand testing the rest, no second-family zero in any design has a reflection partner: the reflection branch needs two second-family zeros within the0.05test tolerance inIm s, and the smallest ordinate gap inside a design is far above that on every design tested, so the branch cannot fire at all. Every pair the sweep records is a self-pair, a real part landing within0.05ofc_F, and self-pairs occur below the chance rate: over the ten designs recensused the tally is 8 self-pairs and 0 reflection partners of 47 zeros tested, a rate of0.170213against the0.229904that drawing each real part uniformly from that design's own observed band predicts, and over the full sweep 22 of 117. The three full-set controls pair 13 of 13 atc_F = 1/2to1e-22, where the functional equation makes every zero its own partner.c_Fis not a quantity either:c_F - alpha/2runs-0.28413232to+0.47788515andc_F - 1/2runs-0.78413232to+0.28664994, so it is notalpha/2, nottheta(F)and not1/2. Witness: lab/py/zeta-family verb symmetry. - 2026-09-07 [Refuted] There is no counting law for the second family in
alphaor infill, at either assignment radius. The four base 4 two-digit designs sharealpha = 1/2andfill/base = 1/2exactly and giveN_2(40) = 7, 13, 9, 14atrho = 0.45and6, 12, 8, 7atrho = 0.6, withN_2(80) = 20, 30, 22, 29and17, 26, 21, 20: a factor of two at onealphaand onefill/baseat both radii, so the refutation is radius-robust even though the integers are not. The subject of the spread is comb occupancy and not the second family, base 4{0,1}and{2,3}differing by 29 percent in total winding, 14 against 18, and by a factor of two inN_2because 7 of 8 poles are occupied against 4 of 8. Read asN_2(T) = c_F T log T + d_F Tfrom the two heights,c_Fatalpha = 1/2is0.10820213, 0.072134752, 0.072134752, 0.018033688, spread0.09016844, against the base 3 and base 4 full-set controls0.15486803and0.16230319, the classical1/(2 pi) = 0.15915494and a control spread of0.0074351582. Every winding is the nearest integer to a numerically integrated phase whose largest surviving step runs0.9205to0.9998against a cap of1, so the counts are Verified and not Proved. Witness: lab/py/zeta-family verbs tests and count. - 2026-09-07 [Refuted] The real parts of the second family do not contract to
alpha/2as a design fills, so the critical line is not thealpha -> 1limit of MrlyMath. The refuted law is thatmax abs(Re s - alpha/2) -> 0asalpha -> 1. Undivided, that statistic stays flat along the ladder carryingalphatoward 1, reading0.2275679549at base 5{0,1}withalpha = 0.430676558,0.5549589411at base 4{0,1}with0.5,0.5885444877at base 3{0,1}with0.630929754,0.4233198337at base 4{0,1,2}with0.792481250,0.5365616661at base 5{0,1,2,3}with0.861353116and0.3151426744at base 10 missing two with0.903089987, then collapsing to1.43e-22,1.10e-21and1.76e-22at the base 2, 3 and 4 full sets. At base 10 missing 9,alpha = 0.954242509, the second family reads0.216084781875to0.70401657869aboutalpha/2 = 0.477121255, a band of width0.488against1 - alpha = 0.0458. Divided by1 - alphathe statistic runs0.39971647to5.7047812with no monotone inalpha, falling from3.8699872to3.2519104on the last two rungs, so the refutation rests on the undivided spread and not on the ratio. Witness: lab/py/zeta-family verb limit. - 2026-09-07 [Refuted] The ordinate shadow does not explain why a design's ordinates converge before its real parts, because at the zeros it pairs it separates neither. The first-order offset is one complex number, so for a paired zero the ordinate offset and the real-part offset are one quantity with no preferred phase: per zero
abs(Im off)/abs(Re off)spans0.137681to6.11895at base 20 missing its top digit and0.14167to18.7749at base 50 missing its top digit, and rung by rung the mediansmedian abs(Im off)againstmedian abs(Re s - 1/2)read0.55734029/0.43095421, 0.49670656/0.28603903, 0.48081533/0.22535435, 0.30633741/0.16748977, 0.12823995/0.36138728, 0.25093621/0.12181102, 0.11574693/0.19332005, 0.058239278/0.049215607, 0.011954894/0.010995712atalpha = 0.430676558to0.994835739, the ordinate offset larger on six rungs and smaller on three with both falling along the ladder. The law binds only the zeros Newton reaches from azetazero inside1.5of it and this lab enumerates no design zero, so it neither explains nor forbids what a design-zero-first census reports; the census contrast and this row are both consistent with a mixture in which the partnered zeros approach in both coordinates while the rest of the second family does not approach at all, and that mixture has no witness until the unpartnered count is measured. Witness: lab/py/zeta-shadow verb rungs, lab/py/zeta-family verb limit. - 2026-09-14 [Verified] The peeled continuation of a digit-design Dirichlet series runs in double precision inside the public crate.
mrlynum::laddercarriesDesign,zeta,cofactorandresidueonmrlynum::design::elementsandmrlynum::zeta::Complex, and returns every value beside a bound. Against the arbitrary-precision lab at the base 2 full set the gaps are7.4e-11ats = 2,4.6e-14ats = 0.3 + 40iand7.5e-12ats = -1 + 2iagainst reported bounds3.8e-10,1.3e-11and6.8e-10; the base 3 residues meet the certified enclosures to3.8e-15against bounds near8e-14. Five adversary breaks are repaired and no pinned number moved. Witness: mrlynum::ladder, lab/py/design-zeta. - 2026-09-14 [Proved] The carried scale of a double-precision ladder majorises every intermediate magnitude of the value recursion, term by term. Beside the propagated truncation bound the module carries
scale_j = (sum_(n in E_P) n^(-Re w) + cut + sum_l abs(binom(-w,l) base^(-w-l) gamma_l) scale_(j+l)) / abs(1 - fill base^(-w)). The induction is immediate: the base entries start attail >= 0,poly_scale(E_P, Re w) >= abs(poly(E_P, w))term by term, and every step applies the same nonnegative weights and the same divisor modulus, soscale_j >= abs(value_j)at every level. Witness: mrlynum::ladder. - 2026-09-14 [Verified] The rounding charge of the double-precision ladder is measured and not counted, and it holds with a factor of fifteen to spare. The reported bound is
truncation + ROUNDING * scalewithROUNDING = 1e-13; a count of about 75 roundings over up to 115 levels gives9.5e-13, an order above the constant, so the charge is not a standalone bound. Measured over four designs at 28 points each plus the real axis the worst ratio of true error to returned bound is0.0667, atbase = 3,F = {0,1},s = 0.5 + 40i, and the error never exceeded the bound anywhere probed; on a rejected rung the truncation half carries the bound and that ratio reaches451. Witness: mrlynum::ladder. - 2026-09-14 [Verified] The wall of the double-precision port is cancellation and not truncation, and it is visible at
Re s = -1. Ats = -1 + 2ion the base 2 full set the truncation bound falls to1.1e-20at shift12and to7.8e-206at shift114while the carried bound sticks at5.6e-10, because the peeled tailG_P(-1+2i)has modulus5.6e3againstzeta_F(-1+2i)of modulus0.183, a loss of four and a half digits; the module raises rather than returns at any tolerance under that. The lab reaches2.8e-32there only by lifting the working precision with the height. Witness: mrlynum::ladder, lab/py/design-zeta. - 2026-09-14 [Proved] The residue column below a pole of a digit-design zeta is a finite recursion in the peeled variables, Burnol's Proposition 7.1 in peeled form. With
R_mthe residue ats_(m,j) = alpha - m + 2 pi i j / log base, taking residues in the peel identity atw = s_(m,j)killsE_Pand leaves only the terms whose shifted argument is a pole:(1 - base^m) R_m = sum_(l = 1)^m binom(-s_(m,j), l) base^(-s_(m,j)-l) gamma_l R_(m-l)withR_0the numerator overlog base, sincefill base^(-s_(m,j)) = base^m. At base 3 onF = {0,1}them = 1,j = 1value0.6950303416383606 + 0.37779086109705695imeets the eight-node contour average on a circle of radius0.05. Witness: mrlynum::ladder, REFS.md Burnol 2026. - 2026-09-19 [Refuted] The pole lattice does not force the zeros of a design zeta: at base 3
{0,1}the two polished zeros right of the abscissa are0.665639628004 + 23.0347504431 iand0.720787601477 + 28.6056765649 i, an ordinate gap of5.5709261against the pole period2 pi/log 3 = 5.7192017, short by0.148, because the shifted termssum_(l >= 1) binom(-s,l) base^(-s-l) gamma_l zeta_F(s+l)of the digit recursion are not2 pi i/log baseperiodic although1 - fill base^(-s)is. Witness: lab/py/zeta-locus verb census.