memory-dial.md
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The memory dial
- 2026-09-13 [Proved] A width-
kruleWover the2^dimdigit vectors hasN_W(level) = 1^T A_W^(level-k+1) 1forlevel >= k - 1, where the states ofA_Ware the2^(dim(k-1))windows of widthk - 1andA_W[s][t] = 1iffsandtoverlap ink - 2digits and thek-window they form is allowed, with thek = 1case reading one state andA_W = [#W]; an accepted word of lengthlevelis exactly a path oflevel - k + 1steps (witness: lab/py/memory-census README THE TRANSFER MATRIX). - 2026-09-13 [Proved] The memory number
kappa(W) = log_2(#W)/k - log_2 rho(W)is nonnegative for every width-krule, since an accepted word of lengthmksplits intomdisjoint allowed windows and soN_W(mk) <= #W^m, givingrho^k <= #W; it is zero on every product ruleW = F^kwithFnon-empty, whereN_W(level) = #F^level, while the empty rule has#W = 0andrho = 0and carries nokappaat all (witness: lab/py/memory-census README THE MEMORY NUMBER). - 2026-09-13 [Proved] Every element of
G_(dim,k), the signed permutationsB_dimapplied diagonally to thekdigits of a window together with window reversal, preservesN_W(level)for everylevel: a diagonalB_dimelement conjugatesA_Wby a permutation matrix and reversal transposes it, andAandA^Tshare a characteristic polynomial (witness: lab/py/memory-census README THE GROUP). - 2026-09-13 [Proved] At
dim = 1the classes of width-krules under diagonalB_1alone number2^(2^k - 1) + 2^(2^(k-1) - 1), because the digit flip acts on the2^kwindows asw -> 2^k - 1 - win2^(k-1)two-cycles (witness: lab/py/memory-census README THE CENSUS, Burnside). - 2026-09-13 [Proved] The set of Perron roots occurring at width
kis contained in the set occurring at widthk + 1: the ruleW' = {(d_1..d_(k+1)) : (d_1..d_k) in W and (d_2..d_(k+1)) in W}accepts the same words of lengthk + 1and above, which is allrhoneeds, while atlevel = kexactly it has no window and accepts every word (witness: lab/py/memory-census README THE PERRON ROOTS). - 2026-09-13 [Proved]
G_(1,4) < G_(2,2) < B_4as permutation groups of the4-cube: flipping all four bits is the diagonalB_2element flipping both axes, and the width-4 window reversal is the(2,2)block swap composed with the diagonal axis swap (witness: lab/py/memory-census README THE GROUP). - 2026-09-13 [Verified] Width-
krules at base 2 in dimension 1 fall into3, 9, 88, 16960classes underG_(1,k)fork = 1..4, out of4, 16, 256, 65536rules, the orbit walk agreeing with an independent Burnside average on every row (witness: lab/py/memory-census census.csv, orbit walk and Burnside). - 2026-09-13 [Verified] Under diagonal
B_dimalone with no window reversal the counts are3, 10, 136, 32896atdim = 1andk = 1..4and6, 8548atdim = 2andk = 1, 2, against3, 9, 88, 16960and6, 4660with reversal (witness: lab/py/memory-census census.csv, orbit walk and Burnside). - 2026-09-13 [Verified] The memory dial at
k = 1is the plain design census:3classes atdim = 1and6atdim = 2, A000616 at1and2, with the two groups agreeing because reversal is trivial (witness: lab/py/memory-census census.csv, the(dim,1)rows). - 2026-09-13 [Verified] A width-
krule in dimensiondimis a subset of thek dim-cube counted with a smaller group, so the class counts meet or exceed A000616 atk dim, by factors1, 3/2, 4, 42.19atdim = 1andk = 1..4and1, 11.59atdim = 2, equal atk = 1where reversal is trivial and the two censuses coincide (witness: lab/py/memory-census census.csv column a000616). - 2026-09-13 [Verified] Code
7atdim = 1andk = 2, the rule forbidding the window11, has Perron root the golden ratio, minimal polynomialx^2 - x - 1andrho = 1.618033988749(witness: lab/py/memory-census classes.csv, PARI factor and polrootsreal). - 2026-09-13 [Verified] At
dim = 1andk = 3the four named roots land on the four expected codes, each the least code of its class: code127gives tribonaccix^3 - x^2 - x - 1at1.839286755214, code55golden, code23supergoldenx^3 - x^2 - 1at1.465571231876, code54plasticx^3 - x - 1at1.324717957244(witness: lab/py/memory-census classes.csv, PARI). - 2026-09-13 [Verified] The same four named polynomials occur in dimension 2 at width 2, on least codes
327tribonacci,19golden,323supergolden and326plastic, carrying121, 588, 54, 48classes, so the named roots are not a dimension-one accident (witness: lab/py/memory-census classes.csv). - 2026-09-13 [Verified] Across the whole census
kappa(W) = 0holds on exactly the non-empty product classes,2at every(1,k)and5at every(2,k), with zero counterexamples over19563live classes; the test is exact,kappa = 0iff the minimal polynomial ofrhodividesx^k - #W(witness: lab/py/memory-census census.csv columns classes_kappa0 and kappa0_nonproduct). - 2026-09-13 [Verified] For
k >= 2the largest memory number in the census is attained atrho = 1, by the largest rule of zero entropy, on a tie of1, 3, 4, 3classes whose least codes arelog_2(3)/2 = 0.792481on code11at(1,2),log_2(6)/3 = 0.861654on code175at(1,3),log_2(13)/4 = 0.925110on code49071at(1,4)andlog_2(10)/2 = 1.660964on code36079at(2,2); atk = 1every live rule is a product,kappais identically0and the maximum is attained on every live class, the full rule atrho = 2^dimincluded (witness: lab/py/memory-census census.csv columns kappa_max, kappa_max_code and kappa_max_ties). - 2026-09-13 [Verified] The window budget of zero entropy, the largest number of windows a live class with
rho = 1allows, is1, 3, 6, 13atdim = 1andk = 1..4and1, 10atdim = 2: a rule may allow that many windows and still accept subexponentially many words, code11at(1,2), which allows00, 01, 11, accepting every0^a 1^bwithN_W(level) = level + 1(witness: lab/py/memory-census census.csv column rho1_windows). - 2026-09-13 [Verified] The distinct characteristic polynomials number
3, 6, 23, 431atdim = 1andk = 1..4and5, 333atdim = 2andk = 1, 2, and the distinct minimal polynomials ofrhonumber3, 4, 10, 177and5, 185(witness: lab/py/memory-census census.csv, exact Faddeev-LeVerrier and PARI factor). - 2026-09-13 [Verified] The Perron roots that fail to dominate their conjugates strictly number
12of177at(1,4)and5of185at(2,2), and every one of them isp(x^m)for somem >= 2withpthe minimal polynomial of a strict root already in the census, for instancex^4 - x^2 - 1andx^6 - x^3 - 1for the golden ratio; strictness is decided numerically, PARI complex roots against a1e-20gap, and thestrictcolumn is left empty on the dead rowxso that the live count reads12and5(witness: lab/py/memory-census census.csv columns weak_perron_polys and weak_are_radicals). - 2026-09-13 [Verified] Burnside extends the class counts past the orbit walk at no cost: under
G_(1,k)fork = 1..8they are3, 9, 88, 16960, 1074036736, 4611686053860868096, 85070591730234617055658644612208132096, 28948022309329048855892746252171977006958709724020498949042189405102555529216(witness: lab/py/memory-census memory.py Burnside extension, 0.01s). - 2026-09-13 [Verified] None of
3, 9, 88, 16960,6, 4660,3, 4, 10, 177or3, 6, 23, 431appears in the local OEIS dump;3, 10, 136, 32896, 2147516416greps three hits, A055708, A056006 and A191363, each a list of integers with a sigma property agreeing only through the closed form2^(m-1)(2^m + 1)atm = 2^(k-1)(witness: grep of the local dump for each comma-delimited string, names read at source). - 2026-09-13 [Verified] The census, the crate and the demo read a corner the same way:
mrlymath::bang::universe::corners(dim)emits the corner vector row first andcorner_indexfolds it most significant first, so a crate design's corner integer atdim = 2isc = x + 2y, bit0the column and bit1the row, and no code label moves between the three (witness: crates/mrlydemo/tests/memory.rs::width_one_is_the_plane_design_cell_for_cell, which pins codes11and13, exchanged by the axis swap and drawn differently). - 2026-09-13 [Verified] The width-one memory rule is the plane design of the same code cell for cell, for codes
1, 7, 9, 11, 13, 14at every level one to six (witness:mrlydemo::memory::memory_sheetagainstmrlydemo::two::two_grid, testcrates/mrlydemo/tests/memory.rs::width_one_is_the_plane_design_cell_for_cell) - 2026-09-13 [Verified] The golden rule,
dim = 1width2code7, which forbids the window11, accepts2, 3, 5, 8, 13, 21, 34, 55words at levels one to eight, the Fibonacci numbers (witness:mrlynum::memory::counts, testthe_golden_rule_counts_the_fibonacci_numbers) - 2026-09-13 [Verified] The golden rule has Perron root
1.618034, growth exponent0.694242and memory numberkappa = log_2(3)/2 - log_2(phi) = 0.098239(witness:mrlynum::memory::perron,exponent,kappa, check rowmemory golden growth) - 2026-09-13 [Verified] The supergolden rule,
dim = 1width3code23, the sponge code read as a window rule, allows at most one1a window and accepts2, 4, 4, 6, 9, 13, 19, 28words at levels one to eight, the Narayana cow recurrencea(level) = a(level - 1) + a(level - 3)holding fromlevel = 2k = 6on and failing atlevel = 5, where the count is9againsta(4) + a(2) = 10(witness:mrlynum::memory::counts, testthe_supergolden_rule_counts_the_narayana_cows) - 2026-09-13 [Verified] The supergolden rule has Perron root
1.465571, the supergolden ratio, the real root ofx^3 = x^2 + 1(witness:mrlynum::memory::perron, check rowmemory cow root) - 2026-09-13 [Verified] The rule that forbids a digit twice in a row,
dim = 2width2code31710, accepts4, 12, 36, 108words at levels one to four, root exactly3since its transfer matrix isJ - Ion the four digits with minimal polynomialx - 3, and memory numberkappa = log_2(12)/2 - log_2(3) = 0.207519(witness:mrlydemo::memory::memory_read, check rowmemory no repeat, lab/py/memory-census classes.csv rowx - 3at(2,2)) - 2026-09-13 [Verified] The full rule accepts
2^(dim level)words at every dimension one to three and every width its span allows, and its growth exponent is the dimension (witness:mrlynum::memory::countsandexponent, testthe_full_rule_counts_every_word) - 2026-09-13 [Verified] The empty rule accepts every word shorter than its window and nothing at or past it, and its Perron root is exactly zero (witness:
mrlynum::memory::countsandperron, testthe_empty_rule_dies_past_its_window) - 2026-09-13 [Proved] The memory number
kappa(W) = log_2(card W) / k - log_2 rho, forcard Wthe allowed windows, is the bits per digit a rule spends on memory and is zero on every memoryless designW = F^kwithFnon-empty, so every width-one rule reads zero (witness:mrlynum::memory::kappaandallowed_windows, testwidth_one_is_the_memoryless_design) - 2026-09-13 [Verified] The plastic rule,
dim = 1width3code54, accepts2, 4, 4, 5, 7, 9, 12, 16words at levels one to eight, Perron root1.324717957the plastic number, the real root ofx^3 - x - 1, and memory number0.260981(witness:mrlydemo::memory::memory_read, testthe_width_three_presets_name_the_plastic_and_tribonacci_roots, check rowmemory plastic root) - 2026-09-13 [Verified] The tribonacci rule,
dim = 1width3code127, which forbids only the window111, accepts2, 4, 7, 13, 24, 44, 81, 149words at levels one to eight, Perron root1.839286755the tribonacci constant, the real root ofx^3 - x^2 - x - 1, and memory number0.056639(witness:mrlydemo::memory::memory_read, testthe_width_three_presets_name_the_plastic_and_tribonacci_roots, check rowmemory tribonacci root) - 2026-09-13 [Verified] The supergolden rule has memory number
kappa = 2/3 - log_2(1.465571232) = 0.115204(witness:mrlynum::memory::kappa) - 2026-09-13 [Proved] A width-
kdigit ruleWin dimensiondimat basebasecounts its accepted words by a path count: with states thebase^(dim(k-1))words ofk-1digit vectors andA[x,y]the number of allowed windows with prefixxand suffixy,N_W(level) = 1^T A^(level-k+1) 1for everylevel >= k-1, and atk = 1the matrix is[card W]so the count iscard W^level, today's fill law. (witness: beneath.md, The transfer matrix) - 2026-09-13 [Proved] A width-
krule in dimensiondimis a subset of the corners of thek dim-cube, so the raw census2^(base^(k dim))is the design count at dimensionk dimand transports unchanged, while the quotient does not: cube symmetry acts diagonally on thekwindows, so the group isB_dimof order2^dim dim!and notB_(k dim)of order2^(k dim) (k dim)!. (witness: beneath.md, Width k in dimension dim is a subset of the k dim-cube) - 2026-09-13 [Proved] The coupling
kappa(W) = log(card W)/(k log base) - log(rho(A))/log(base)of a width-krule is nonnegative, by cutting an accepted word of lengthmkinto itsmdisjoint windows so thatN_W(mk) <= card W^m, whilerho^(level-k+1) <= N_W(level)by the entry sum ofA^(level-k+1); andkappa = 0on every productW = G^kwithGnon-empty, whereN_W(level) = card G^levelandrho = card G, so every memoryless design with a non-empty rule sits at coupling zero. (witness: beneath.md, The coupling) - 2026-09-13 [Proved] Memory does not leave the lattice class: the counting series
sum_level N_W(level) x^levelof a width-krule is rational with denominatordet(I - x A), so atx = base^(-s)its poles sit on finitely many vertical linesRe s = log(abs(lambda))/log(base)over the nonzero eigenvalueslambdaofAand the pole set is invariant unders -> s + 2 pi i / log base, the same period as the memoryless case; one line can carry a finer progression, as atdim = 1,k = 2, code6, whose eigenvalues1and-1put poles at gappi / log baseonRe s = 0. (witness: beneath.md, What the dial does not buy; lab/py/memory-census, the class of code 6) - 2026-09-14 [Proved] The class count of width-
kbinary rules underG_(1,k)isa(2m) = 2^(2^(2m)-2) + 2^(2^(2m-1)-2) + 2^(2^(2m-1)+2^(m-1)-1)form >= 1anda(2m+1) = 2^(2^(2m+1)-2) + 2^(2^(2m)-1) + 2^(2^(2m)+2^m-2)form >= 0, by Burnside over the order-4 group: the digit flip fixes no window, reversal fixes the2^ceil(k/2)palindromes, and flip-reversal fixes the2^(k/2)antipalindromes at evenkand none at oddk; the form reproduces3, 9, 88, 16960and every Burnside extension term throughk = 8(witness: lab/py/memory-census verbburnside, the cycle index and the closed form agreeing atk = 1..11, and beneath.md, The memory dial). - 2026-09-14 [Proved] At
dim = 1andk >= 2every transfer matrix has determinant in{-1, 0, 1}, so the constant term of every characteristic polynomial is0,1or-1: rowssands + 2^(k-2)are both supported on the columns2sand2s+1taken modulo2^(k-1), those column pairs partition the columns assruns over0..2^(k-2)-1, and the matrix is therefore a row permutation of a block diagonal matrix with2^(k-2)blocks of size2 x 2over{0,1}; checked over all16, 256, 65536rules atk = 2, 3, 4(witness: lab/py/memory-census verblemmas, a Bareiss determinant and the signed block product agreeing in{-1,0,1}on every rule atk = 2, 3, 4, and beneath.md, The memory dial). - 2026-09-14 [Proved] Distinct minimal polynomials of the Perron root are distinct Perron roots: every conjugate of
rho(W)is a root of the characteristic polynomial ofA_Wand so an eigenvalue ofA_W, hence at mostrho(W)in modulus, so two conjugate Perron roots are equal in modulus and, both being nonnegative, equal; the3, 4, 10, 177minimal polynomials atdim = 1andk = 1..4are therefore3, 4, 10, 177distinct growth rates (witness: lab/py/memory-census verblemmas, no conjugate aboverhoon any of the 463 characteristic polynomials atk = 1..4, the 3, 4, 10, 177 minimal polynomials carrying 3, 4, 10, 177 distinctrho, and beneath.md, The memory dial). - 2026-09-14 [Verified] Second generators reproduce the memory census whole: a cycle-index Burnside counter gives
3, 9, 88, 16960, 1074036736, 4611686053860868096underG_(1,k)atk = 1..6and6, 4660, 1152921592116822016underG_(2,k)atk = 1..3, with the remaining extension terms atk = 7, 8andk = 4agreeing as well, and an exact integer Faddeev-LeVerrier enumeration over all rules, factored in PARI, gives3, 6, 23, 431characteristic polynomials and3, 4, 10, 177minimal polynomials atdim = 1andk = 1..4(witness: lab/py/memory-census verbburnsideatdim = 1, the census run's Burnside extension atdim = 2, verblemmasfor the polynomial counts, every cell reproduced). - 2026-09-14 [Verified] None of the memory-census counts is in the local OEIS dump:
3, 9, 88, 16960with every Burnside extension term throughk = 8,6, 4660with its extension throughk = 4,3, 4, 10, 177and3, 6, 23, 431each grep to zero hits as comma-delimited runs, while3, 10, 136, 32896hits A055708, A056006 and A191363 through the coincidence2^(2^k-1) + 2^(2^(k-1)-1) = A007582(2^(k-1)), and9and16960recur inside A367526, a grid tiling count with different neighbours (witness: grep of the local OEIS dump on the runs printed by lab/py/memory-census verbburnside, each hit read at source). - 2026-09-14 [Verified] The binary words of length
ncounted under the same group that the width-nrules are counted under, reversal together with bitwise complementation, are A005418 atn:1, 2, 3, 6, 10, 20, 36, 72atn = 1..8(witness: lab/py/memory-census verbburnside, the word orbits printed beside the class count, agreeing with the entry in the local OEIS dump). - 2026-09-14 [Verified] The Perron root of a transfer matrix is the largest root over the strongly connected components of its digraph, each component being irreducible with a simple root, so it is exact against that component's integer characteristic polynomial; at
dim = 1,k = 3, code5gives1, codes62,125and190give the plastic number1.324717957, code91gives1.380277569, code95gives the golden ratio1.618033989, and the coupling of codes125and190islog_2(6)/3 - log_2(1.324717957) = 0.455969; a stop comparing one scalar across two sweeps halts on a plateau of theI + Amass ratio and misses all six (witness: mrlynum::memory::perron and its test, lab/rs/memory-meter). - 2026-09-14 [Verified] The memory meter's control column is the Mertens function: the width-
1rule of code3at base2accepts every integer, and its meter reads-1, 1, 2, -23, -48, 212, 1037, 1928at10^1..10^8, asserted inside the run, which is A084237 (witness: lab/rs/memory-meter, thecontrol mertensline, and beneath.md, The memory meter). - 2026-09-14 [Verified] The same linear sieve reproduces the memoryless base-
3design meters exactly, so the memoryless row of the dial is pinned against the existing census: digits{0,1}read(M, max abs M) = (11, 105)atlevel = 14,(149, 173)atlevel = 16and(-30, 312)atlevel = 18, digits{1,2}read(-1461, 1582)atlevel = 18, each asserted; digits{0,2}atlevel = 20wants3^20, past the2^30sweep, and is printed unpinned atlevel = 14, 16, 18as(-10, 67),(-124, 152)and(67, 249). The four pinned pairs are read at source in lab/py/design-meter, which computes them and cites lab/rs/mobius-designs as their census (witness: lab/rs/memory-meter, thecontrol designlines, and beneath.md, The memory meter). - 2026-09-14 [Verified] The window-profile recurrence,
profile(n)beingprofile(n >> 1)unioned with the windown mod 2^k, reads the accepted integer set of every width-1,2and3rule atdim = 1, base2: on all276rules its masses agree with a direct digit recount below2^20, profile containment agrees withmrlynum::memory::Rule::acceptsbelow2^12, and at every one of the89phases the subset-sum transform of the per-profilemusums equals the independently carried per-rule meter (witness: lab/rs/memory-meter, thecontrol recountline, and beneath.md, The memory meter). - 2026-09-14 [Verified] Code
7at(dim, k) = (1, 2), the golden rule forbidding the window11, opens exactly the fibbinary integers A003714 without its zero, and its mass below2^levelis a Fibonacci number,A = 2178309below2^30; code11, forbidding10, opens exactly the Mersenne numbers A000225 without its zero and holds30elements below2^30, one per level (witness: lab/rs/memory-meter, thecontrol code 7andcontrol code 11lines, and beneath.md, The memory meter). - 2026-09-14 [Verified] The Mobius meter of every width
1,2,3rule atdim = 1, base2, read to2^30at the89phasesx = floor(2^(level + j/4)),level = 8..30,j = 0..3, no exponent fitted, each ratio at a named phase: at30.00the full line readsA = 1073741824,M = -10374,max abs M = 11173, ratios-0.316589and0.340973; the golden rule code7atk = 2,kappa = 0.098239, readsA = 2178309,M = 551,max abs M = 716, ratios0.373329,0.485125; thek = 3least codes23,54,127read normalised peaks0.731125,0.677943,1.239625atkappa = 0.115204, 0.260981, 0.056639(witness: lab/rs/memory-meter, theruleandrowlines, and beneath.md, The memory meter). - 2026-09-14 [Verified] A rule's
rhoandkappaare read off the transfer matrix of its word language whileAandMare read off its integer set, and on a rule that is not zero-closed those are different objects: atk = 3code5the word language grows on the self-loop000and carriesrho = 1, while the integer set holds3elements below2^30(witness: lab/rs/memory-meter, therule k=3 code=5line, and beneath.md, The memory meter). - 2026-09-14 [Verified] Over the census of
53rules of all three widths holding at least10^4integers below2^30, the floor fixed before any reading, the normalised peakmax abs M_W/sqrt(A_W)spans[0.293624, 1.239625]at phase30.00, least onk = 3code125and largest onk = 3code127, and spans[0.500000, 2.169240]over all89phases, the top onk = 3code232at phase16.75; the last-phase leader and the sweep-wide leader are different rules, so no rule is the dial's widest excursion (witness: lab/rs/memory-meter, thespanlines, and beneath.md, The memory meter). - 2026-09-14 [Verified] Read against the full line at the same phase, which needs no band and no grid, the factor
max abs M_W/sqrt(A_W)overmax abs M/sqrt(x)runs[0.861136, 3.635552]at phase30.00over the53census rules and reaches6.375774onk = 3code190at phase12.75over all phases (witness: lab/rs/memory-meter, thefactorlines, and beneath.md, The memory meter). - 2026-09-14 [Verified] Grouped by the coupling over the same
53rules at phase30.00, the mean normalised peak reads0.340973onkappa = 0with3rules, then0.622171on6,0.581446on14,0.644840on12and0.645966on18for the bands[10^-9, 0.1),[0.1, 0.2),[0.2, 0.3)and[0.3, 0.5); every band of positive coupling sits above thekappa = 0band, whose three rules are the full line under three codes, and among the positive bands the means are not monotone inkappa. Restricted tok = 3the same bands read0.340973, 0.698386, 0.581446, 0.644840, 0.645966on populations1, 4, 14, 12, 18(witness: lab/rs/memory-meter, thekappabandlines, and beneath.md, The memory meter). - 2026-09-14 [Verified] The full line's own normalised peak runs
[0.272410, 0.500000]over the grid, the ceiling at phase8.00, and that ceiling is a property of where the grid starts and not of the full line: below the grid the same ratio reads1.000000atx = 1,0.894427at5,0.832050at13,0.718421at31and0.565685at200(witness: lab/rs/memory-meter, thegridstartline, and beneath.md, The memory meter). - 2026-09-14 [Verified] The falsification fires. Five of the
53census rules never enter the full line's band at any phase where they hold10^4elements, all of them above it:k = 3codes159,182,190,218and250, holding211116,13607,31535,59860and4126645integers. The band's ceiling being grid-dependent, the same-phase factor is the instrument that carries the reading, and it is read rule by rule on the generator'sfactorlines (witness: lab/rs/memory-meter, theband outsidelines, and beneath.md, The memory meter). - 2026-09-14 [Verified]
16of the53census rules attain their sweep-wide normalised peak in the last quarter of the phases, from24.75on, so most of the dial peaked earlier and is not growing at the end of the sweep. No rule at any width holding at least1000elements hasM_W/sqrt(A_W)ormax abs M_W/sqrt(A_W)rise at every one of the last eight phases, but that test asksmax abs M_Wto grow about9%per quarter-level across two whole levels, so its empty answer carries little and the late-peak count is the informative statistic (witness: lab/rs/memory-meter, thelatepeakandclimbinglines, and beneath.md, The memory meter). - 2026-09-14 [Verified] Leading zeros move most rules' integer sets. A rule is zero-closed when prepending one zero changes no membership below
2^20: the zero-closed codes number3of4atk = 1,8of16atk = 2and64of256atk = 3, and are the codes allowing0with the empty code, those allowing01, and those allowing both010and011, asserted code for code. Under any number of zeros the word language agrees with the integer set on2of4,4of16and16of256codes, tested on every word to length14, so atk = 3the two readings part company on240of256(witness: lab/rs/memory-meter, thezeroclosedandreadinglines, and beneath.md, The memory meter). - 2026-09-14 [Verified] Equal mass is not the same set: code
14atk = 2, forbidding00, and code126atk = 3, forbidding000and111, each hold28655integers below2^20and each carryrho = 1.618033989, yet they share only1077of them, the symmetric difference is55156, and4is the least integer the second holds and the first does not; below2^30both hold3524576integers while their meters read-466and435and their normalised peaks0.454355and0.594444at phase30.00(witness: lab/rs/memory-meter, thepairline, and beneath.md, The memory meter). - 2026-09-14 [Verified] The matrix ladder of a memory design runs in double precision in the public crate.
mrlynum::automatoncarriesAutomaton,zeta,cofactor,residueanddenominator, every value beside its bound. The full rules, code15atk = 2and255atk = 3, meetmrlynum::ladderon the base 2 full design to7.18e-11ats = 2and2.03e-14ats = 0.3 + 40i, inside bounds; the product rule code8meets a direct Mersenne sum to1.12e-16; the golden rule code7meets a direct fibbinary sum with its Fibonacci tail bound, and every pinned golden reading meets the arbitrary-precision control inside its own bound, the fourzeta_Wvalues to3.0e-15and the largest of the fourteen rows1.134e-11against7.348e-11(witness: mrlynum::automaton, lab/py/memory-zeta, beneath.md, The memory zeta). - 2026-09-14 [Proved] The polynomial
beneath.mdnames the string equation of a memory rule is also the denominator of the rule's Dirichlet series, strictly more than that page proves.beneath.mdprovesdet(I - x A)denominates the counting series and namesdet(I - base^(-s) A) = 0the Moran replacement; the peel carries it tozeta_W, each level of(I - base^(-w) T) G_P(w) = E_P(w) + sum_(l >= 1) binom(-w,l) base^(-w-l) Gamma_l G_P(w+l)dividing bydet(I - base^(-w) T). So the poles ofzeta_Wlie inbase^(-s) lambda_i = base^m,lambda_ia nonzero eigenvalue andm >= 0whole, and that determinant is them = 0level's denominator, not the whole one (witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Proved] Right of the abscissa the matrix ladder carries its bound as a nonnegative vector and needs no norm and no primitivity, which settles the Conjecture row the ladder unit left open. For nonnegative
y,abs((I - base^(-w) T)^(-1)) y <= sum_(i >= 0) (base^(-Re w) T)^i yentrywise, sinceabs(base^(-w)) = base^(-Re w)andTis nonnegative; the remainder closes on the guidev = (I + T)^60 1, which meetsT v <= mu vfor the bracket's upper endmu, assum_(i > N) A^i y <= (max_u y_u/v_u) theta^(N+1)/(1 - theta) v,theta = base^(-Re w) mu < 1. The seed is exact:sum_(j >= P) E_j(sigma) <= base^(-(P-1)sigma) (I - base^(-sigma) T)^(-1) c_P(witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Proved] The residue of a memory zeta at a simple pole needs no eigenvector: the adjugate is the spectral projector in polynomial form. Faddeev-LeVerrier on
Tgives integer matricesM_kand integer coefficientsc_kwithadj(I - x T) = sum_(k < n) x^k M_kanddet(I - x T) = sum_(k <= n) c_k x^k, so withx = base^(-w)andNthe ladder numerator,Res_(w0) zeta_W = 1^T adj(I - x_0 T) N(w0) / (-x_0 log base det'(x_0)). The form is stable at the pole, wheredet(I - x_0 T) (I - x_0 T)^(-1)is not, and it dies exactly where the ladder unit said it would, at a multiple root, wheredet'(x_0) = 0and the pole order exceeds one (witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Verified] The golden rule carries two genuine pole combs interleaving at half a tooth.
Tof code7atk = 2is[[1,1],[1,0]]anddet(I - x T) = 1 - x - x^2, so one comb sits atRe s = log_2 phi = 0.6942419136306174withIm sin2 pi Z / log 2and one atRe s = -log_2 phiwithIm sin(2Z + 1) pi / log 2, the argumentpiof the negative eigenvalue shifting it half a period. The six residues atm = 0are under THE POLE COMB and none is zero. Comb two comes from the left-of-abscissa branch, so its path is the arbitrary-precision control and not the contour average: the six agree to6e-16and4.3e-12(witness: mrlynum::automaton, lab/py/memory-zeta, beneath.md, The memory zeta). - 2026-09-14 [Verified] Burnol's Proposition 5.1 survives the memory dial verbatim: the residue at the abscissa is the limit of the level digit sums. For code
7the direct sum ofn^(-alpha)over the fibbinary integers of exactlylevelbits, overlog 2, reads0.946747043404283,0.946743630023052and0.946743410426742atlevel = 16, 20, 24against the ladder's0.946743395641970, the gap falling like2^(-level)(witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Verified] The matrix Lyndon cofactor is the determinant with its adjugate, and it reads on the whole
m = 0comb where the series is singular.Z_W(s) = det(I - base^(-s) T) zeta_W(s)is carried asdet(I - base^(-s) T) D_(P-1)(s) + 1^T adj(I - base^(-s) T) N(s), so it never divides by the vanishing determinant; for code7it reads0.991729890316722ats = 3,0.973380053858285ats = 2and0.913335748872126ats = 0.8, each to a bound near1e-13, whilezeta_W(0.8) = 9.536379694275015is already climbing the pole at0.6942419136306174(witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Proved] Left of the abscissa a matrix ladder has no free denominator bound, and the module buys one with the residual of its own inverse.
abs(1 - k base^(-w))has no matrix analogue and the Neumann majorant diverges oncebase^(-Re w) rho >= 1, so the level closes on the computed inverseCcertified againstR = I - (I - base^(-w) T) C: for nonnegativey,abs((I - base^(-w) T)^(-1)) y <= abs(C)(y + (max_u y_u) r/(1 - r) 1)withrthe max row sum ofabs(R), and the module raises whenr >= 1. The second comb of code7is read only through that branch, at bounds near4e-9against1.3e-13on the first (witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Verified] The matrix Lyndon cofactor
Z_W(s) = det(I - 2^(-s) T) zeta_W(s)of the golden rule, code7at width2, has exactly20zeros in the box-0.95 < Re s < 2,0.02 < Im s < 43.1: the determinant strips bothm = 0combs in one factor, leavingZ_Wmeromorphic there with exactly4simple poles, the level-one teeth onRe s = -0.305758086, so each cell count is its argument-principle winding plus the level-one teeth the cell holds; all20are located with largestabs(Z_W)9.694e-12, largest surviving phase step0.999894radians against a cap of one, largest propagated bound1.474e-10, nothing within0.02of an outer box edge, and identical cell rows and zeros at contour seeds0.1,0.05and0.025, at7298,11777and21599evaluations, and with0 < Im s < 0.02,Im s > 43.1andRe s < -0.95uncounted,20is exact on the box (witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Verified] The
4poles that census adds back are read and not assumed: a48-point circle mean ofZ_Wat each level-one tooth of code7gives residues-1.990368154340-0.795661945868i,-0.350975872907-0.436714265460i,-3.135030562964-2.032376530959iand-1.028888122837+2.036528502776i, radius0.05against radius0.02agreeing to7.3e-14, each simple to5.1e-05against(s - s_0) Z_Wat1e-5, while a blank point on the same line reads4.6e-16, and the same read at code23gives four residues of modulus2.231820,3.226224,2.438482and1.779516, the two radii agreeing to5.1e-14, against a blank point at6.3e-16(witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Proved]
Z_Whas no zero inRe s >= 2on code7: the least element ofS_Wis1and the coefficients are nonnegative, soabs(zeta_W(s) - 1) <= zeta_W(2) - 1 < 1there fromzeta_W(2) = 1.415825532885 < 2, anddet(I - 2^(-s) T)has no root right of the abscissalog_2 phi, which makes the census box's right edge a wall and not a choice (witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Verified] On code
7the first comb carries a zero comb and the second carries none: at the design family census radius0.45all4teeth of the comb onRe s = log_2 phibelowIm s = 43.1carry a zero, at distances0.317490225,0.045406362,0.143076282and0.070233751, while0of the5teeth onRe s = -log_2 phido, least distance0.666213518and largest0.758440773, and the emptiness holds over every point of every disc, the radius0.45disc reachingRe s = -1.144241913631, since the same census on-1.2 < Re s < 2, which admits no new pole line before-1.305758086369, returns the same20zeros, nineteen to twelve decimals and the twentieth to eleven, the same4of4and0of5and the same five distances (witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Verified] The first-order tooth law
u_1 = -r/Rholds on code7's first comb to0.000951131,0.003342909,0.020403749and0.062287774and misses on the second by0.214394685,0.645682670,0.408924841,0.489190529and0.519744494, and it is a reading beside the census and not a second falsification:Ris a circle mean of radius0.3, three of the five second-comb predictions ofabs(u_1),0.501975708,0.398920848and0.301764481, are read outside that disc, and on the first comb the one prediction past0.3carries the worst miss (witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Verified] The second comb's line carries zeros where its teeth do not: three of code
7's20zeros sit within0.05ofRe s = -log_2 phi, at0.000322593,0.014258173and0.043369824from that line, while their distances to the nearest tooth of the same comb are4.104099275,0.747618761and4.283371881, and stripping the4first-comb teeth leaves a second family of16with real parts in[-0.737611737911, 0.540957439322]; the excess is4.4times the0.68that20uniformly spread real parts would put in a window of width0.1on a box2.95wide, on a sample of20(witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Verified] Exactly
9of the88width-3rule classes underG_(1,k)carry two pole lines, none with a repeated eigenvalue, and the cofactor's zeros are a resolved census on one printed box for all nine,S_Wread off the minimal base-2string:-1.15 < Re s < 2,0.02 < Im s < 20, contour seed0.05, which holds every radius0.45occupancy disc of every second line, the deepest reachingRe s = -1.144241913631; the nine read9to14zeros in20to36cells, every zero located, largest residual3.236e-11, largest surviving phase step0.999909radians against a cap of one, largest propagated bound9.110e-09, and no pole ofZ_Wwithin0.02of any contour, the least clearance being exactly0.02, from the cutIm s > 0.02to the level-mpole on the real axis. Resolved and not certified: nothing boundsZ_W'/Z_Won the contour, and two zeros sit within0.02of the left contour,-1.134547677+3.580553251ion code127and-1.143621954+17.814806003ion code63(witness: lab/py/memory-zeta, verbteeth, and beneath.md, The memory zeta). - 2026-09-14 [Verified] The occupancy reading is invariant across two boxes whose contours fail a
0.02guard in disjoint ways. On-1.2 < Re s < 2the pole linesRe s = -1.202842615688of codes54and62andRe s = -1.188629537248of code223sit0.002842615688and0.011370462752from the left contour, the first pair outside the box and so never added back; on-1.15 < Re s < 2every pole clears0.02and two zeros do not. Both boxes read50teeth,34occupied at radius0.45,23teeth predictingabs(u_1) < 0.3occupied22,15predicting at or above0.45occupied4,12between occupied8, and the same occupancy column on all nine rules; only the totals off the discs move, code23from13zeros to11and code31from14to13(witness: lab/py/memory-zeta, verbteeth, at--left -1.2and--left -1.15, and beneath.md, The memory zeta). - 2026-09-14 [Proved] With
S_Wread off the minimal base-2string, so that a word shorter than the window holds no window and is accepted, the width-3rule55accepts exactly the set of the width-2rule7with the single integer3adjoined, hencezeta_55(s) = zeta_7(s) + 3^(-s). Code55forbids exactly the windows011,110and111, which is exactly the ban on an adjacent pair of ones inside a3-window, and for length at least3the window starting atmin(i, level-3)holds the pair at(i, i+1), while the word11carries no window and is accepted; checked over1 .. 262143with3the only difference either way. Padded to the window width instead the two sets are equal and the claim is empty (witness: lab/py/memory-zeta, verbbridge, and beneath.md, The memory zeta). - 2026-09-14 [Verified] The width-
3ladder meets the controlled width-2one across that gap, which is the control on a new rule:Z_55(s) - Z_7(s) - det(I - 2^(-s) T) 3^(-s)reads at most1.168e-13over seven points including three teeth and one located zero, every point inside the sum of its own two bounds, so the4-state ladder, adjugate and peel meet the2-state ones that carry the stored arbitrary-precision control, itself met to1.134e-11on14rows with none outside its bound (witness: lab/py/memory-zeta, verbsbridgeandcontrol, and beneath.md, The memory zeta). - 2026-09-14 [Verified] One tooth of the
50carries two zeros inside the occupancy radius, so34occupied teeth hold35zeros: code54, second line, toothRe s = -0.202842615688,Im s = 14.612532469, holds0.213711738933+14.629308261174iat0.416892021and-0.597460129789+14.668266829134iat0.398533940, andabs(u_1) = 0.631828588there puts it in the bin the first-order law reads as empty. The other49teeth hold at most one, and the count is printed by the census itself astooth zeros 6against5occupied teeth on that rule (witness: lab/py/memory-zeta, verbsteethandcensus --width 3 --code 54, and beneath.md, The memory zeta). - 2026-09-14 [Proved] For a finite set
Fof positive integers disjoint fromS_W, the setS_W + F = S_W u Fhaszeta_(W+F)(s) = zeta_W(s) + P_F(s)withP_F(s) = sum_(n in F) n^(-s)a Dirichlet polynomial and so entire, hence the two series carry the same poles, the same orders and the same residues at every point of the plane, andZ_(W+F)(s) = Z_W(s) + det(I - base^(-s) A) P_F(s); adding a finite set is a knob on the zero set alone (witness: lab/py/memory-zeta verbdial, the off-tooth identityZ_(W+F)(s) - Z_W(s) - det(I - 2^(-s) T) P_F(s)missing by at most2.384e-15at code7and4.003e-16at code23against an added part of up to1.912203and1.708983). - 2026-09-14 [Proved] The knob cannot move the cofactor at a tooth: at every
m = 0toothtthe determinant vanishes, soZ_(W+F)(t) = Z_W(t)exactly, and the principal part ofzeta_Wattis fixed while the constant term becomesR + P_F(t), so the first-order zero position isu_1(F) = -r/(R + P_F(t)); every higher coefficient of the regular part moves too, the linear one byP_F'(t) = -log(n) n^(-t)(witness: lab/py/memory-zeta verbcensus, code55at width3readingr = 0.210170579-0.581938843iatIm s = 9.064720digit for digit against code7's andR = 1.313430833+1.119663028iagainst1.714940435+0.882338583i, a difference of-0.401509602+0.237324445iwhich is3^(-t)to nine decimals). - 2026-09-14 [Verified] The residue and tooth probes of the dial cannot falsify the perturbation's entirety and the off-tooth identity can: a
48-point circle mean annihilates an entire addition and the determinant vanishes at a tooth, so the printed gaps are an aliasing floor and a determinant residual and not a measurement, while the identity read off the teeth agrees to fifteen decimals against an added part of order one (witness: lab/py/memory-zeta verbdial,largest residue gap over every probe 1.776e-15, largest tooth value gap 1.250e-13, largest off-tooth identity miss 2.384e-15 against an added part of up to 1.912203at code7). - 2026-09-14 [Verified] The dial's disc probe reproduces the zero census it is read against: on code
7at width2it reads4of4teeth occupied on the comb atRe s = log_2 phiand0of5on the comb at-log_2 phi, and on code23at width3it reads3of4on the first comb and7of10on the second line, each occupancy an argument-principle count on a circle of radius0.45about the tooth with the level-mpoles inside added back (witness: lab/py/memory-zeta verbdial,edge guard splits 0on both baselines, a guard that covers the baselines and not the perturbed grid). - 2026-09-14 [Verified] Occupancy at radius
0.45under the knob is undetermined wherever a zero sits within the0.02guard of the occupancy circle and the inner count is0, which is9of the110cells of code7's second comb,10of the108of code23's abscissa comb and9of the270of code23's second line, so a minimum taken over a tooth's candidate row has two readings and the reading must be named (witness: lab/py/memory-zeta verbdial, per-line rowsoccupancy undetermined 9,occupancy undetermined 10andoccupancy undetermined 9). - 2026-09-14 [Verified] Every empty tooth of code
7's second comb is occupied by a single added integer from the22integers of2 .. 40outsideS_W, so the smallestFthat occupies a tooth of the empty comb has one element and that element is at most11under either reading of the seam, while the least singleton itself is radius-dependent at two of the five teeth:{3},{6},{7},{11},{11}on the inner reading against{3},{6},{3},{11},{6}on the outer, atIm s = 4.532360,13.597080,22.661801,31.726521and40.791241(witness: lab/py/memory-zeta verbdial, per-tooth rowssmallest singletonandouter reading occupied ... smallest singleton). - 2026-09-14 [Verified] Occupancy moves both ways on code
23and the count is stable under either reading of the seam: four empty teeth are filled,Im s = 9.064720by{7}or{6},20.807773by{5},29.872493by{7}and38.937214by{15}or{11}, and two occupied second-line teeth are emptied by a singleton off the seam,2.678332by{6}and42.645269by{6}and by{7}, so6of the14teeth of the box change occupancy under a one-element perturbation (witness: lab/py/memory-zeta verbdial,occupied teeth emptied by a singleton 2, both emptied teeth printing an empty seam list). - 2026-09-14 [Verified] The dial meets the cross-width control exactly where one exists:
S_7 + {3}isS_55, and the dial's grid at code7with the added element3reads the second comb's tooth atIm s = 4.532360occupied and the tooth at13.597080empty, which is the1of2the width-3four-state ladder prints for code55(witness: lab/py/memory-zeta verbsdial,bridgeandcensus,bridgereadingZ_55 - Z_7 - det(I - 2^(-s) T) 3^(-s)at most1.168e-13over seven points and the code55census readingzeros in the disc 1at4.532360and0at13.597080). - 2026-09-14 [Verified] The exact minimum of
abs(R + P_F(t))over all4158861subsetsFof size at most16of the22integers of2 .. 40outsideS_Wis1.095277075,0.784350607,1.295268436and1.662786204at code7's four abscissa-comb teeth, the greedy chain attains every one of them, and the disc at each exact minimiser keeps its zero off the seam,1/1,2/2,1/1and1/1, the tooth atIm s = 18.129441gaining a second zero rather than losing its first (witness: lab/py/memory-zeta verbdial --deep 16, rowsexact minimiser over the 4158861 subsets of size at most 16). - 2026-09-14 [Conjecture] Every width-
krule atdim = 1, base2, withrho > 1hasM_W(x) = O(A_W(x)^(1/2 + eps))for everyeps > 0: over all89phases every one of the53census rules has its sweep-wide maximum ofmax abs M_W/sqrt(A_W)inside[0.500000, 2.169240]and its same-phase factor against the full line inside[0.861136, 3.635552]at phase30.00, the sweep-wide maximum being6.375774on code190at phase12.75, with16of53peaking in the last quarter of the grid, against Mullner 2017, which givesM_W(x) = o(x)for an automatic set and no rate at all. A band at finite depth is not a rate and nothing here bounds the constant (witness: lab/rs/memory-meter, Mullner 2017, and beneath.md, The memory meter). - 2026-09-14 [Conjecture] What selects an occupied tooth is the first-order quantity
u_1 = -r/R, the residue ofzeta_Wat the tooth against the regular part ofZ_W/det, and not the spectrum. Over the50teeth of the nine two-line classes at width3,34teeth are occupied at radius0.45, the23whose predictionabs(u_1)falls below0.3, inside the radius0.3disc that buildsRand so where the reading is self-consistent, are occupied22times, and the15withabs(u_1)at or above0.45are occupied4times; the one exception inside0.3is code55's second-line tooth atIm s = 13.597080,abs(u_1) = 0.267301885with the nearest zero at0.497761908. Occupancy is a per-tooth Boolean and the law is a law onabs(u_1): the largest modulus missabs(d - abs(u_1))is0.739013203and the largest vector missabs(z - t - u_1)is1.287895060, both at code63's second-line tooth atIm s = 13.597080,abs(u_1) = 0.520202113against a nearest zero at1.259215316(witness: lab/py/memory-zeta, verbteeth, and beneath.md, The memory zeta). - 2026-09-14 [Conjecture] The knob's strength at a tooth
tisabs(n^(-t)) = n^(-Re t)and its direction the phase-Im(t) log nmod2 pi, so on a line withRe t < 0the strength grows withnand the largest candidate still reading empty rises with the tooth height, while on the abscissa comb, whereRe t = log(rho)/log(base) > 0, the strength decays and the flippers are confined to a bounded range ofnthat the phase selects inside: code7's second comb reads11,25,28,35on the inner seam convention and11,24,28,35on the outer, increasing under both, over a candidate range stopping at40(witness: lab/py/memory-zeta verbdial, per-tooth rowslast candidate reading emptyon both readings). - 2026-09-14 [Refuted] The Euler wall of
zeta.mdstands over the memory dial and its construction does not. The conclusion transfers:S_Wfor code7, the fibbinary integers, holds the coprime pair5and9whose product45 = 101101carries adjacent ones and leaves the set, so the indicator ofS_Wis not multiplicative and no Euler product over primes exists;45is the least such product over all coprime pairs ofS_Wbelow2^16. The construction does not:zeta.mdbuilds its witness from the repunitsR_candR_(c+1)of the least missing digitc, and a memory rule has no missing digit to take the least of (witness: mrlynum::automaton, beneath.md, The memory zeta). - 2026-09-14 [Refuted] "A memory rule's second pole comb carries no zero comb": the supergolden rule, code
23at width3, hasdet(I - x T) = 1 - x - x^3, one comb onRe s = log_2 psi = 0.551463089746and two interleaved onRe s = -0.275731544873from the conjugate eigenvalue pair of moduluspsi^(-1/2), and its census on-0.75 < Re s < 2,0.02 < Im s < 43.1reads24zeros in70cells, largest phase step0.998514, of which at radius0.45the first comb holds3of4teeth and the second line holds7of10, least distance0.170257380(witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Refuted] "Symmetric pole combs give a symmetric zero set": code
7's two combs sit symmetrically aboutRe s = 0and code23's aboutRe s = 0.137865772436, yet under reflection in that line no zero of either census has a partner other than itself within0.05in both coordinates,0of20and0of24, there being no functional equation on either side; the exclusion bites once, code7's zero-0.023033432741+33.122746617086isitting0.046066865482from its own reflection and being the only self-match inside the tolerance on either census, code23's nearest missing at0.075316339787(witness: lab/py/memory-zeta, verbcensus, and beneath.md, The memory zeta). - 2026-09-14 [Refuted] No function of the spectrum selects an occupied comb, which is the falsification L7 named. Codes
55and63at width3and code7at width2all carrydet(I - x T) = 1 - x - x^2, so all three have the same two combs,Re s = log_2 phiat argument0andRe s = -log_2 phiat argumentpi, and the same teeth; on the second line at radius0.45they read1of2,0of2and0of2occupied, least tooth-to-zero distances0.264392586,1.259215316and0.702616482. Codes54and62share the whole spectrum,det(I - x T) = 1 - x^2 - x^3, and differ on both lines,2against1and3against4. Two rules with one spectrum reading two occupancies kills every function of it, monotone, threshold or otherwise; the ratioabs(lambda_2)/rhois neither, code223at0.430159709002reading0of4while code127at the smaller0.400890564601reads2of4and code62at0.655865618097reads4of4against code31at0.563624162161reading2of4(witness: lab/py/memory-zeta, verbteeth, and beneath.md, The memory zeta). - 2026-09-14 [Refuted] The pole data cannot select occupancy at all, the residue included, and one line proves it:
zeta_55 - zeta_7 = 3^(-s)is entire, so codes55and7carry the same poles, the same orders and the same residues at everym >= 0, while their zero sets differ, code55having a zero at-0.442302243578+4.612546440182iwhere code7reads-0.097731686660-0.868473160333iand reading1of2against0of2on the second line. Read atm = 0through the determinant the residues agree digit for digit,-0.259501222742937-0.592535006433179iat-log_2 phi + pi i/log 2and0.896350590641921+1.403072744223695iat-log_2 phi + 3 pi i/log 2from both rules (witness: lab/py/memory-zeta, verbbridge, and beneath.md, The memory zeta). - 2026-09-14 [Refuted] The first-order quantity
u_1(F) = -r/(R + P_F(t))selects occupancy under perturbation on the abscissa comb: at code7's toothIm s = 9.064720the exact minimiserF = [3, 6, 11, 12, 19, 22, 23, 35, 38, 39]drivesabs(R + P_F)to1.095277075, below the emptying thresholdabs(r)/rho = 1.374951382, so the law predictsabs(u_1) = 0.564905571and an empty disc, and the disc reads1/1with no seam (witness: lab/py/memory-zeta verbdial --deep 16, rowexact minimiser ... predicted abs(u1) 0.564905571 emptying threshold abs(r)/rho 1.374951382 zeros in the disc 1/1). - 2026-09-14 [Refuted] The first-order quantity is a selector across the perturbed family on a subdominant line: on code
7's second comb the grid holds110cells of which77read occupied, the law calls75right and the constantoccupiedpredictor77, and on code23's second line,270cells and222occupied, the law calls218against222; the deficit only widens on the outer reading of the seam,80against86and219against231(witness: lab/py/memory-zeta verbdial, per-line rowsfirst-order law agreesagainstthe constant occupied predictor agreeson both readings). - 2026-09-14 [Refuted] Only the smallest added integers flip a tooth of the abscissa comb: code
23's empty tooth atIm s = 9.064720is occupied by{7},{13}and{14}and by none of the smaller candidates5,6,10,11and12, so the flipping set is not an initial segment of the candidate list, and the two occupied teeth a singleton empties are emptied by{6}and{7}while the smaller candidate5occupies both (witness: lab/py/memory-zeta verbdial, code23tooth rowsmeasured 001000110000000000000000000 occupied 3 of 27,measured 101111111111111111111111111andmeasured 100111111111111111111111111). - 2026-09-19 [Proved]
G_(1,4) < G_(2,2) < B_4as permutation groups of the4-cube, flipping all four bits being the diagonalB_2element that flips both axes and the width-4window reversal being the(2,2)block swap composed with the diagonal axis swap, so one cube carries three nested groups and its class counts nest the other way,16960 > 4660 > 402, the last A000616 at4. Witness: lab/py/memory-census.