cobham.md
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Cobham
- 2026-09-14 [Proved] Multiplicative independence is a property of a pair and never of a triple: dependence is an equivalence relation on the integers
>= 2whose classes are the powers of one least member, so three bases that are pairwise dependent are jointly dependent and a third base adds no hypothesis a pair does not carry; the invariant is the dependence-class partition and not the tuple. The two degenerate readings are settled the same way: base1is not a base,k-recognizability being defined fork >= 2only, and bases2and4are one base, so Cobham's hypothesis fails there and so does its conclusion, the base-4 design{0, 1}being4-recognizable, hence2-recognizable, infinite and of density(1/2)^levelat levellevel, hence not ultimately periodic. Witness: cobham.md:9 to cobham.md:15, Durand and Rigo Definition 1.1 and Remark 1.2 read at source, Bes on Buchi read at source. - 2026-09-14 [Proved] Every proper one-dimensional design is base-locked, exactly. For
Finside{0, ..., base-1}with0 in Fand1 < card F < base, the setS_Fof integers whose digits all lie inFis infinite, sinced base^klies in it for every nonzerodinF, and has density(card F / base)^levelat levellevel, which tends to0; an infinite ultimately periodic set has positive density, soS_Fis not ultimately periodic, and Cobham's theorem forbids any base multiplicatively independent ofbase. The two hypotheses are the two exclusions and nothing else:card F > 1removesF = {0}andcard F < baseremoves the full digit set, and those are the only two semilinear designs atdim = 1. Witness: cobham.md:26, Bes Theorem 24 and Durand and Rigo Theorem 1.1 read at source. - 2026-09-14 [Proved] A necessary condition on the digit set, sharp enough to settle every gasket. If
S_Fis semilinear thencard F = base^dfor an integer0 <= d <= dim, andS_Flies in a finite union ofd-dimensional affine subspaces: a linear setv + N c_1 + ... + N c_rwhose generators span dimensionemeets[0, N)^diminTheta(N^e)points, so a semilinear set countsTheta(N^d)in the box withdthe largest span dimension among its constituents, while the design counts(card F)^levelexactly at sidebase^level, forcingcard F = base^d. Consequence with no geometry: the base-2 gasket{(0,0), (0,1), (1,0)}hascard F = 3, not a power of2, so it is not semilinear and not3-recognizable. Witness: cobham.md:32 and cobham.md:38. - 2026-09-14 [Proved] A sufficient condition, and the two conditions agree at
base = 2, dim = 2. CallFa block design when thedimcoordinates split into a zero setZanddblocks withF = {v : v_i = 0 on Z, and v_i = v_j whenever i and j share a block}; thencard F = base^dandS_F = N c_1 + ... + N c_dwithc_tthe0/1indicator of blockt, one linear set, hence semilinear and recognizable in every base. Atbase = 2, dim = 2the characterization is complete: of the eight designs containing0, the five of cardinality1, 2, 2, 2, 4are exactly the block designs and are semilinear, and the three of cardinality3are excluded by the count. Witness: cobham.md:33 and cobham.md:34. - 2026-09-14 [Proved] The base-3 gasket is not semilinear and so not
2-recognizable. Its box count is3^levelat side3^level, sod = 1and a semilinear version would lie in finitely many lines; but it containsP_t = (3^t, 3^(t^2))for everyt >= 2, whose consecutive slopes are exactlys_t = 3^(t^2 - t) (3^(2t+1) - 1)/2, strictly increasing, so theP_tsit in strictly convex position, no three are collinear, and coveringnof them costs at leastn/2lines. Hence no automaton reading base-2 digits enforces the base-3 gasket's digit rule. Witness: cobham.md:37. - 2026-09-14 [Proved] Five of this tree's instruments, checked one by one on the page that carries each, split as follows under a second independent base; the audit covers those five and asserts nothing about the instruments it did not check. Survives: the box bound of the coprimality sieve,
N*_level(m) <= (base+1)^dim fill^level m^(-alpha)withalpha = log(fill)/log(base), which is pure counting onS_leveland passes to any subset by monotonicity, the Chebyshev sum built on it still converging whenalpha > 1. Dies: the fill law ofmethod.md, an identity on a Kronecker power in one base; the transfer matrix and its Perron root, the vertex shift ofbeneath.md, the even slice matrixM_evenofcuts.mdand the Collatz-Wielandt brackets ofcrop.md, each a finite automaton over the digits of one base; the carry automaton ofcuts.md, finite only becausex -> (x + dim)/3contracts on integer carries inside one base, while a machine reading base-2 digits and tracking base-3 digits is base conversion; and the character contractionabs(Sum) <= fill - 2 + 2 cos(pi/(2 base)), which is the statement that the level-leveltransform factors over digit positions in one base. A sieve needs an upper bound and an equidistribution: two bases hand over the first and destroy the second. Witness: cobham.md:40 to cobham.md:46. - 2026-09-14 [Proved] The two-set transversality theorem does not iterate, for an elementary reason:
A cap Bneed not be invariant under either map. WithAthe middle-thirds set,T_3-invariant, andB = [0, 1),T_2-invariant, the intersection isA, which is notT_2-invariant, since1/4 = 0.020202..._3lies in it andT_2(1/4) = 1/2 = 0.1111..._3does not. Them-fold bound is proved instead by rewriting the intersection as one slice of the productA_1 x ... x A_dinsideT^d. Witness: cobham.md:52, Corso and Shmerkin 2024 Theorem 1.15 and Corollary 1.17 read at source. - 2026-09-14 [Proved] What product designs do give is one budget per axis, not one budget in total. If every
F_iis a productG_i^(1) x ... x G_i^(dim)across thedimaxes in pairwise independent bases, eachA_iis the product of its axis sets, the intersection is the coordinatewise intersection, upper box dimension is subadditive on products, and Corso and Shmerkin Corollary 1.17 applies on each axis, givingdim-upper_B(cap_i A_i) <= sum_(j=1)^dim max(0, sum_i dim_H A_i^(j) - (m-1)), upper box on the left and Hausdorff on the right. Witness: cobham.md:61. - 2026-09-14 [Verified] The joint census of the base-2 gasket
{(x, y) : x AND y = 0}against the base-3 gasket,C(N) = card(A cap B cap [0, N)^2), readsC(3^m) = 1, 3, 7, 19, 45, 111, 241, 467, 1175, 2443, 5285, 11939, 25281, 53477, 109001, 231737, 498083, 1077727, 2179165, 4372741, 9051805, 18107943, 37126191, 75050077, 151133095atm = 0..24. Each level is computed by a pruned walk over the base-3 exponents whose cut is a proved bit-length bound on the remaining addition, and rebuilt form <= 6by scanning every pair in the box against both digit rules with one shared membership routine; the two agree at every level and the axis bound is asserted at each. Witness: cobham.md:77,lab/py/two-base-gasketverbstermsandcontrol. - 2026-09-14 [Proved]
C(3^m) >= 2^(m+1) - 1, and that already breaks the naive planar budget. On the axisx = 0membership in the base-2 gasket is automatic and membership in the base-3 gasket asks the base-3 digits ofyto lie in{0, 1}, which2^mvalues below3^msatisfy; the axisy = 0gives another2^m; the origin is the only overlap. Hence the counting exponent is at leastlog_3 2 >= 0.630929, against a budgetlog_2 3 + 1 - 2 <= 0.584963, the lower bound truncated down and the budget rounded up. The real gaskets carry the same excess: the left edge lies in the real base-2 gasket and the real base-3 gasket meets it in a Cantor set of dimensionlog_3 2. Witness: cobham.md:78 and cobham.md:79,lab/py/two-base-gasketverbstermsandbudget. - 2026-09-14 [Proved] Budget zero is dimension zero and not finiteness:
{2^n}has counting exponent0, its count belowNbeing at mostlog_2 N + 1, and is infinite, so a transversality bound of zero never closes a question that asks for a finite list. Finiteness in two bases is reached only at the bounded-digit-sum corner, by Senge and Straus 1973 ineffectively through Thue-Siegel-Roth and by Stewart 1980 effectively through Baker; that corner is not a design, its set is not closed under changing one digit and its count belowb^kisO(k^c). Witness: cobham.md:67 and cobham.md:68, both statements read at source in the survey of Bugeaud, Cipu and Mignotte. - 2026-09-14 [Proved] The three-base thin set has real upper box dimension zero, by the three-set theorem and not by any pair. With
A_3, A_5, A_7the closed subsets of the circle whose base-3, base-5 and base-7 digits lie in{0,1},{0,1,2}and{0,1,2}, Corso and Shmerkin 2024 Corollary 1.17 atd = 3asks for pairwise multiplicatively independentp_j >= 2, closedT_(p_j)-invariantA_jand affineg_j, and givesdim-upper_B(g_1(A_1) cap g_2(A_2) cap g_3(A_3)) <= max{s - 2, 0}withs = sum_j dim_H A_j; the bases are distinct primes, each set is closed and invariant by its digit rule, theg_jare the identity, ands - 2 = -0.121889rounded up, so the bound is0and upper box dimension is nonnegative. The conclusion is about the three real sets and about upper box dimension, and it does not transfer to the integer set. Witness: cobham.md:90,lab/rs/three-base-thinverbbudget, Corso and Shmerkin 2024 Corollary 1.17 read at source. - 2026-09-14 [Proved] The third base is not redundant, and the redundancy is sharp in both directions. Upward, the three pair budgets
dim_i + dim_j - 1read0.313536at(3, 5),0.195505at(3, 7)and0.247182at(5, 7), each rounded up and each positive, so the two-set bound returns nothing on any pair. Downward, the pair(3, 5)alone is infinite: Erdos, Graham, Ruzsa and Straus 1975, quoted as Theorem 1.8 of Burrell and Yu 2021, give infinitely many integers with base-pdigits<= Aand base-qdigits<= BwheneverA/(p-1) + B/(q-1) >= 1, and1/(3-1) + 2/(5-1) = 1.000000exactly. The other two pairs miss the criterion by one digit each, both reading0.833333, and both reach1.000000when the base-7 bound rises from2to3; the criterion is sufficient and not necessary, so neither pair is claimed finite. Witness: cobham.md:88 and cobham.md:91 and cobham.md:92,lab/rs/three-base-thinverbbudget, Burrell and Yu 2021 Theorem 1.8 read at source. - 2026-09-14 [Verified] The members of the three-base thin set below
7^17 = 232630513987207are0, 1, 3186, 3187, 20007and nothing else, found in333nodes. The walk enumerates the base-3 side as subset sums of distinct powers of3from the top power down and cuts a branch by a proved bound: once the powers3^kand above are chosen the remaining addition is at most(3^k - 1)/2, so withjleast such that5^j > (3^k - 1)/2the high partfloor(n / 5^j)of every reachablenis one of two consecutive integers, and the branch dies when neither has all its base-5 digits<= 2; base7cuts the same way and membership in the base-3 set is never tested. Witness: cobham.md:94,lab/rs/three-base-thinverbseven 17. - 2026-09-14 [Verified] The same five members and no sixth below
3^80000, a height of38170decimal digits, in1710789nodes and61.48s on about3GB. The node count grows near21 levelat height3^leveland the stored powers costTheta(level^2)bits, so memory and not the clock is what stops the census. Witness: cobham.md:95,lab/rs/three-base-thinverbreach 80000. - 2026-09-14 [Verified] The pruned walk is checked against a direct scan of every integer below
10^8against all three digit rules, at base-7 bound2and again at base-7 bound3, and the two agree in both. Witness: cobham.md:96,lab/rs/three-base-thinverbcontrol. - 2026-09-14 [Verified] The budget changes sign one digit away, and on the far side it is a heuristic already in print. By Lucas's theorem
binomial(2k, k)is prime topexactly when every base-pdigit ofkis belowp/2, which reads<= 1at3,<= 2at5and<= 3at7, so raising the base-7 bound from2to3gives{k : binomial(2k, k) is prime to 105}, OEIS A030979, whose triple budgetlog_3 2 + log_5 3 + log_7 4 - 2reads0.025951rounded up against-0.121889for the thin set; A030979 read at source records a prize for settling whether it is finite, names it as Erdos problem 376, and quotes a heuristic of Pomerance giving aboutx^0.02595...terms up tox, which is the same number as the budget. The same walk at base-7 bound3rebuilds all23terms A030979 publishes, counts1374members below10^70, exactly the length of the table that entry calls complete to10^70, and counts216020below10^140; the effective exponentslog(count)/log(height)read0.044828and0.038103, truncated down, both above0.025951and falling. Witness: cobham.md:97 and cobham.md:98 and cobham.md:99,lab/rs/three-base-thinverbsbudget,control,ten 70andten 140, OEIS A030979 read at source. - 2026-09-14 [Verified] The twenty-five terms
C(3^m) = 1, 3, 7, 19, 45, 111, 241, 467, 1175, 2443, 5285, 11939, 25281, 53477, 109001, 231737, 498083, 1077727, 2179165, 4372741, 9051805, 18107943, 37126191, 75050077, 151133095atm = 0..24are printed from scratch in this run by the verbtermsand again by the verbhankel, which recomputes the census before its algebra, and the two agree term for term with each other and with the page. Both call the same pruned walk, so that agreement is determinism and not corroboration: the one independent rebuild is the verbcontrol, which tests every pair of[0, 3^m)^2against both digit rules with a separate routine and agrees at everym <= 6. Witness: lab/py/two-base-gasket verbsterms,hankel,control;cobham.md:77. - 2026-09-14 [Verified] No linear recurrence with constant coefficients of order
r <= 12fits all twenty-five terms ofC(3^m). Two exact tests over the rationals agree. The Hankel determinants of the matrix with entriesa[i+j], computed by fraction-free Bareiss overZ, are nonzero at every sizek = 1..13, reading1, -2, 4, 8, 5360, -1259267712, -5516990041856, 112485023878830080, 5697070341654551514880, -1283183611235610612560681984, 512735847124145678895522067154944, -30665906970422092677442692752788846592, -148892102950447887517893509783802772470337536; a recurrence of orderrwould force every Hankel determinant of size aboverto vanish, sodet H_13 != 0alone kills every order at most12. Independently, for each orderr = 1..12the linear systema(n) = c_1 a(n-1) + ... + c_r a(n-r)taken over every one of the25 - ravailable equations is inconsistent by Gauss-Jordan overQ, order by order with no order left consistent and none underdetermined. Order12is the largest the data can test: it leaves13equations against12unknowns, one spare, while order13leaves12equations against13unknowns, one short of determined, so that system is consistent for trivial reasons and tests nothing;12is exactly the bound twenty-five terms carry and not a choice. The two solvers behind the negative have their own generator: the verbselftestfits Fibonacci first at order2withx^2 - x - 1,2^n + 3^n + 1first at order3with dominant root3.0000000, andn^3 + 2^nat no order below5and at5with(x - 1)^4 (x - 2), and checksbareissagainst five determinants computed by hand, eleven checks in0.05s. Witness: lab/py/two-base-gasket verbshankel,3min15s atHI = 24, andselftest. - 2026-09-14 [Verified] The sequence
1, 3, 7, 19, 45, 111, 241, 467, ...is absent from the local OEIS dump, and so are eight simple transforms of it: the sequence from its second term, the first differences, the partial sums,a(n) - 1,a(n) + 1,(a(n) - 1)/2,2 a(n), and the halved first differences. Each was searched as its seven-term window starting at the second term. Witness: grep of the local dumpresearch/data/oeis/stripped, nine patterns, zero hits. - 2026-09-14 [Verified] The decision rule is calibrated on one-base counts before any two-base object is read, and neither one-base control is clean. The base-3 design
{0, 1}puts a maximum at5.719220against2 pi / ln 3 = 5.719202, error0.000%, at1.7e7times the median power of the band[0.5, 14]under the linear detrend and1.9e7under the cubic; the base-5 design{0, 1, 2}puts one at3.903959against2 pi / ln 5 = 3.903963, error0.000%, at4.0e6times the median. At3^44the base-3 control's nearest maximum to2 pi / ln 5sits at error1.115%and35.9times the median under the linear detrend and at0.966%and39.2under the cubic, which passes the rule at a frequency absent by construction; the base-5 control's nearest maximum to2 pi / ln 3sits at1.060%and2.21under the linear detrend and at0.963%and1.95under the cubic, inside the tolerance and held out by the power gate alone. Both control windows are longer than the cell's and give a prediction more room to be met by position alone: span38.05with11maxima above10xthe median covering0.113of the band, and span38.49with4covering0.048, against the cell's span27.41,4maxima and0.020. Witnesslab/py/two-base-instrumentverbcell 44. - 2026-09-14 [Verified] The multiplicatively dependent pair is a one-base control and not a two-base count. Base 3 digits
{0, 1}against base 9 digits{0, 1, 3}satisfies the zero-digit hypothesis of the collapse theorem ofbases.md,0lying in both digit sets, so the joint set is exactly the one-base designF(9, {0, 1, 3})of exponentlog_9 3 = 1/2; at the aligned height3^28 = 9^14its count is exactly3^14 - 1 = 4782968, the fitted exponent is0.497809, and its spectrum shows2 pi / ln 9 = 2.859601at2.859889, error0.010%, at886times the median, with the first harmonic2 pi / ln 3at error0.003%and2.5e3times the median, on a window of span17.47carrying4maxima above10xthe median covering0.042of the band. Witnesslab/py/two-base-instrumentverbcollapse 28. - 2026-09-14 [Verified] The two-frequency prediction is what the two band structures alone predict, and the cell does not meet it at their scale. A member of the base-3 design with
k+1digits lies in[3^k, (3^(k+1)-1)/2]and a member of the base-5 design withi+1digits in[5^i, (5^(i+1)-1)/2], so the designs occupyln(3/2)/ln 3 = 0.369070andln(5/2)/ln 5 = 0.569323of their own decades and the cell's count is exactly constant wherever the two bands miss: of the47decades[3^j, 3^(j+1))below3^47, the10withj = 1, 4, 17, 20, 23, 26, 36, 39, 42, 45carry no member at all, proved by the bands and witnessed by the ladder, whose new hits are0atlevel = 40, 43, 46. The block modelC3(N) C5(N) / N, the same two band structures multiplied with no joint arithmetic, shows2 pi / ln 3at error0.001%and1.33e6times the median and2 pi / ln 5at0.004%and9.79e5, on a window of span31.55with4loud maxima covering0.043; the cell's own count on a window of span30.70with the same4loud maxima and0.020reaches22.5and15.7. Witnesslab/py/two-base-instrumentverbsblocks 47andladder 38 47. - 2026-09-14 [Verified] The rule's verdict on the cell is not stable in height, so this run settles the two-frequency prediction neither way. On the full window under both detrends the ladder from
3^38to3^47returns both frequencies present atlevel = 38, 39, 40, 41,2 pi / ln 3at errors0.473%, 0.073%, 0.342%, 0.704%under the linear detrend and0.325%, 0.036%, 0.484%, 0.862%under the cubic at19to24times the median, and returns2 pi / ln 3absent fromlevel = 42up, at errors1.034%to1.574%and22to25times the median, while2 pi / ln 5is present at every one of the ten heights. The error at2 pi / ln 3rises monotonically fromlevel = 39tolevel = 46under both detrends as the window lengthens, which is a nearest maximum drifting away from the prediction rather than an estimate converging on it. On the upper half of the window at3^44the nearest maximum to2 pi / ln 3sits at error0.825%under the linear detrend and0.791%under the cubic, inside the tolerance and held out by the power gate alone at5.55and6.1times the median, and at3^47at2.430%and2.293%. No height rule is stated in advance, so no height is entitled to the verdict. Witnesslab/py/two-base-instrumentverbsladder 38 47,cell 44andcell 47. - 2026-09-14 [Verified] The cell's strongest maximum matches no small combination of the two lattice frequencies. It sits at
1.702087at3^44and1.697925at3^47, at57and68times the median on windows of span27.41and30.70carrying4maxima above10xthe median covering0.020of the band, and the nearestm 2 pi / ln 3 + n 2 pi / ln 5withabs(m), abs(n) <= 8is(1, -1) = 1.815239at errors6.233%and6.463%;2 pi / ln 15misses by3.750%and3.457%and the sum frequency by1.528%and1.977%at2.2and2.1times the median, while2 pi / ln 5is met at0.461%and0.594%. Block structure does not account for it: the block model's nearest maximum to1.815239sits at1.895843and0.588times the median at3^44and at1.697497and1.58at3^47. The same verbs read the local counting exponent falling through the budget0.313536,0.323301over the full window and0.295978over its upper half at3^44,0.318728and0.292184at3^47, which is a fit at finite height and decides nothing about the limit. Witnesslab/py/two-base-instrumentverbscell 44,cell 47andblocks 47. - 2026-09-14 [Proved] The dimension-one lane of coprime, Conjectures O, W and Z with the shelf lane
lemma-b-pincer, is a one-base problem whose automata come in families indexed by the object read, and never meets Cobham's hypothesis. Every object of the lane is read in base 3 alone: the gasketG_nis the base-3 digit pairs from{(0,0),(1,0),(0,1)}, the wordbinaryinside the lane means a base-3 expansion with digits in{0, 1}and never base 2, and the only other bases in the section are the general simplex corollary atcoprime.md:150and the base-4 and base-5 simplex probes atcoprime.md:253, each a separate design in its own single base and not a second base on one object. The machines are one automaton per primitive ray, per multiplier pairB(s,t), per band direction and per modulus3^k; the band family, whose states are the integers in[-(z_2-1)/2, (z_1-1)/2](coprime.md:233), and the digit-congruence family indexed by3^k(coprime.md:228) carry no uniform state bound, whileB(s,t)carries only the upper bound(s+1)(t+1)(coprime.md:214) and a measured 167 live states at(25,52)and(31,40)(coprime.md:197,gasket-ray-machine), and the ray automata grow likeabin the shelf lane's own words. Cobham asks one set recognized by a finite automaton in two multiplicatively independent bases, which the lane never presents, so the wall of cobham, that no transfer matrix over the digits of one base reads the constraint a second independent base imposes, does not touch the lane. Witness:coprime.md:187; the shelf lane README reading "counting the multiples of a ray inside the gasket is a finite automaton on the base-3 digits of the multiplier, with growth rate rho(a,b). There are infinitely many rays and the automata grow like ab, so no computation settles them";cobham.md:41. - 2026-09-14 [Conjecture] Block designs are the only semilinear designs, at every base and every dimension. The count alone is not enough, which is what makes the conjecture nontrivial: the base-3 gasket has
card F = 3 = 3^1and is not semilinear. Witness: cobham.md:35. - 2026-09-14 [Conjecture] The true counting exponent of the two-gasket intersection is unknown and twenty-five levels decide nothing.
log_3 C(3^m) / mreads0.754141, 0.749634, 0.746312, 0.743739, 0.738025, 0.732546, 0.729032, 0.724371, 0.721151, 0.717651, 0.714298atm = 14..24, falling by about0.004a level on the mean of those ten steps and still0.129above the budget at the last level, consistent with any limit from0.630929upward. The cheap falsifier is a linear recurrence for1, 3, 7, 19, 45, 111, 241, 467, ..., which would make the growth rate algebraic and put the object back inside this tree's machinery; it is not run here. Witness: cobham.md:81 and cobham.md:82,lab/py/two-base-gasketverbterms. - 2026-09-14 [Conjecture] The Schanuel wall. Every growth exponent this tree prints has the shape
log(algebraic)/log(base), the Perron root of a nonnegative integer matrix read in its own base, and every two-base budget has the shapesum_i log(k_i)/log(p_i) - (m-1) dim, aQ-linear combination of1and the ratioslog k_i / log p_i, the wall assuming that a realized two-base exponent has that shape too, which nothing here proves; those two families meet only where one side degenerates, and under Schanuel's conjecture they meet nowhere nontrivial, so no instrument of this tree ever outputs a two-base exponent. This is strictly stronger than Cobham, which forbids the set from being automatic while the wall forbids the number from being a Perron root, and neither implies the other. Witness: cobham.md:108, Burrell and Yu Theorem 1.6 and Theorem 1.11 read at source. - 2026-09-14 [Conjecture] The three-base thin set is exactly
{0, 1, 3186, 3187, 20007}. Finiteness is open and no theorem on the page gives it: the dimension bound is0and the page already proves that a budget of zero says nothing about finiteness, while the finiteness results in print, Senge and Straus 1973 and Stewart 1980, bound digit sums rather than digits, so applying either to the thin set would need a bound on the base-3 digit sum of a member, which is the finiteness in question. The falsifier is a sixth member and the census is where it would have shown. Witness: cobham.md:101 and cobham.md:102,lab/rs/three-base-thinverbreach 80000. - 2026-09-14 [Conjecture]
C(3^m)satisfies no linear recurrence with constant coefficients of any order, so the counting exponent of Object Y is notlog_3of a Perron root and the Schanuel wall holds in data for Object Y. What is proved is the order at most12case above; the step to every order is the conjecture, and nothing here rules out a recurrence of order13or more. Falsification: orderris testable once the terms leave its system a spare equation, which wants2r+1of them, so order13wants27terms, the two levelsm = 25andm = 26beyond what is run, about11min by the measured per-level factor; the same verb tests any order the census reaches. Witness: lab/py/two-base-gasket verbshankelandselftest, extrapolated from ther <= 12negative. - 2026-09-14 [Conjecture] The joint digit constraint destroys the oscillation that either design carries alone. The block model of the same two designs carries both lattice frequencies at
10^5to10^6times the median while at3^47the cell's own count carries2 pi / ln 5at15.7times the median and puts nothing nearer to2 pi / ln 3than a maximum1.488%away, so the intersection is not the product of its two band structures at the level the spectrum reads. A nonlattice Moran system has its complex dimensions off any arithmetic progression and its detrended count carries no sharp frequency, which is consistent with the cell's flat reading, and nothing in this run separates that reading from the height instability the ladder prints. A positive test has to read the spread of the complex dimensions rather than a comb, and that needs a zeta function for the joint object, which needs a gap structure, which is what Cobham denies. Witnesslab/py/two-base-instrumentverbscell 44,cell 47,ladder 38 47andblocks 47. - 2026-09-14 [Refuted] The sentence "no proper design is recognizable in two independent bases, in any
dim" is false. Atdim = 2andbase = 2the designF = {(0,0), (1,1)}hasS_F = {(n, n)}, the diagonal, which is definable in<N; =, +>and so recognizable in every base by the easy half of Cobham-Semenov. Proper designs recognizable in two independent bases exist as soon asdim >= 2, so thedim = 1lock does not generalize by itself and the theorem must be stated as Cobham-Semenov states it, with semilinear in place of ultimately periodic. Witness: cobham.md:31, Bes Theorem 25 read at source. - 2026-09-14 [Refuted] The global planar budget
dim(A cap B) <= max(0, dim A + dim B - dim)is false atdim = 2, in one line and with product designs. The base-2 designF_A = {(0,0), (0,1)}givesA = {0} x Tof dimension1, the base-3 designF_B = {(0,0), (0,1)}givesB = {0} x Cof dimensionlog_3 2,Bsits insideA, so the intersection has dimensionlog_3 2while the budget readsmax(0, 1 + log_3 2 - 2) = 0. Both sets lie in the line{0} x T, invariant under both maps, and that is exactly where the two codimensions refuse to add. Witness: cobham.md:60,lab/py/two-base-gasketverbbudget. - 2026-09-14 [Refuted] The global budget is not a corollary of the per-axis bound, and the step that fails is
sum_j max(0, x_j) >= max(0, sum_j x_j), which runs the wrong way. The one-line witness above is where it runs strictly wrong: the per-axis bound reads0 + log_3 2and is sharp there, while the global budget reads0. So indim >= 2the two-base budget is a theorem per axis for product designs and open for compounds, andcore.mdproves almost every design is a compound asdimgrows. Witness: cobham.md:62.