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 >= 2 whose 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: base 1 is not a base, k-recognizability being defined for k >= 2 only, and bases 2 and 4 are one base, so Cobham's hypothesis fails there and so does its conclusion, the base-4 design {0, 1} being 4-recognizable, hence 2-recognizable, infinite and of density (1/2)^level at level level, 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 F inside {0, ..., base-1} with 0 in F and 1 < card F < base, the set S_F of integers whose digits all lie in F is infinite, since d base^k lies in it for every nonzero d in F, and has density (card F / base)^level at level level, which tends to 0; an infinite ultimately periodic set has positive density, so S_F is not ultimately periodic, and Cobham's theorem forbids any base multiplicatively independent of base. The two hypotheses are the two exclusions and nothing else: card F > 1 removes F = {0} and card F < base removes the full digit set, and those are the only two semilinear designs at dim = 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_F is semilinear then card F = base^d for an integer 0 <= d <= dim, and S_F lies in a finite union of d-dimensional affine subspaces: a linear set v + N c_1 + ... + N c_r whose generators span dimension e meets [0, N)^dim in Theta(N^e) points, so a semilinear set counts Theta(N^d) in the box with d the largest span dimension among its constituents, while the design counts (card F)^level exactly at side base^level, forcing card F = base^d. Consequence with no geometry: the base-2 gasket {(0,0), (0,1), (1,0)} has card F = 3, not a power of 2, so it is not semilinear and not 3-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. Call F a block design when the dim coordinates split into a zero set Z and d blocks with F = {v : v_i = 0 on Z, and v_i = v_j whenever i and j share a block}; then card F = base^d and S_F = N c_1 + ... + N c_d with c_t the 0/1 indicator of block t, one linear set, hence semilinear and recognizable in every base. At base = 2, dim = 2 the characterization is complete: of the eight designs containing 0, the five of cardinality 1, 2, 2, 2, 4 are exactly the block designs and are semilinear, and the three of cardinality 3 are 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 is 3^level at side 3^level, so d = 1 and a semilinear version would lie in finitely many lines; but it contains P_t = (3^t, 3^(t^2)) for every t >= 2, whose consecutive slopes are exactly s_t = 3^(t^2 - t) (3^(2t+1) - 1)/2, strictly increasing, so the P_t sit in strictly convex position, no three are collinear, and covering n of them costs at least n/2 lines. 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) with alpha = log(fill)/log(base), which is pure counting on S_level and passes to any subset by monotonicity, the Chebyshev sum built on it still converging when alpha > 1. Dies: the fill law of method.md, an identity on a Kronecker power in one base; the transfer matrix and its Perron root, the vertex shift of beneath.md, the even slice matrix M_even of cuts.md and the Collatz-Wielandt brackets of crop.md, each a finite automaton over the digits of one base; the carry automaton of cuts.md, finite only because x -> (x + dim)/3 contracts on integer carries inside one base, while a machine reading base-2 digits and tracking base-3 digits is base conversion; and the character contraction abs(Sum) <= fill - 2 + 2 cos(pi/(2 base)), which is the statement that the level-level transform 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 B need not be invariant under either map. With A the middle-thirds set, T_3-invariant, and B = [0, 1), T_2-invariant, the intersection is A, which is not T_2-invariant, since 1/4 = 0.020202..._3 lies in it and T_2(1/4) = 1/2 = 0.1111..._3 does not. The m-fold bound is proved instead by rewriting the intersection as one slice of the product A_1 x ... x A_d inside T^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_i is a product G_i^(1) x ... x G_i^(dim) across the dim axes in pairwise independent bases, each A_i is 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, giving dim-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), reads 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, 151133095 at m = 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 for m <= 6 by 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-gasket verbs terms and control.
  • 2026-09-14 [Proved] C(3^m) >= 2^(m+1) - 1, and that already breaks the naive planar budget. On the axis x = 0 membership in the base-2 gasket is automatic and membership in the base-3 gasket asks the base-3 digits of y to lie in {0, 1}, which 2^m values below 3^m satisfy; the axis y = 0 gives another 2^m; the origin is the only overlap. Hence the counting exponent is at least log_3 2 >= 0.630929, against a budget log_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 dimension log_3 2. Witness: cobham.md:78 and cobham.md:79, lab/py/two-base-gasket verbs terms and budget.
  • 2026-09-14 [Proved] Budget zero is dimension zero and not finiteness: {2^n} has counting exponent 0, its count below N being at most log_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 below b^k is O(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_7 the 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 at d = 3 asks for pairwise multiplicatively independent p_j >= 2, closed T_(p_j)-invariant A_j and affine g_j, and gives dim-upper_B(g_1(A_1) cap g_2(A_2) cap g_3(A_3)) <= max{s - 2, 0} with s = sum_j dim_H A_j; the bases are distinct primes, each set is closed and invariant by its digit rule, the g_j are the identity, and s - 2 = -0.121889 rounded up, so the bound is 0 and 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-thin verb budget, 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 - 1 read 0.313536 at (3, 5), 0.195505 at (3, 7) and 0.247182 at (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-p digits <= A and base-q digits <= B whenever A/(p-1) + B/(q-1) >= 1, and 1/(3-1) + 2/(5-1) = 1.000000 exactly. The other two pairs miss the criterion by one digit each, both reading 0.833333, and both reach 1.000000 when the base-7 bound rises from 2 to 3; 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-thin verb budget, 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 = 232630513987207 are 0, 1, 3186, 3187, 20007 and nothing else, found in 333 nodes. The walk enumerates the base-3 side as subset sums of distinct powers of 3 from the top power down and cuts a branch by a proved bound: once the powers 3^k and above are chosen the remaining addition is at most (3^k - 1)/2, so with j least such that 5^j > (3^k - 1)/2 the high part floor(n / 5^j) of every reachable n is one of two consecutive integers, and the branch dies when neither has all its base-5 digits <= 2; base 7 cuts the same way and membership in the base-3 set is never tested. Witness: cobham.md:94, lab/rs/three-base-thin verb seven 17.
  • 2026-09-14 [Verified] The same five members and no sixth below 3^80000, a height of 38170 decimal digits, in 1710789 nodes and 61.48 s on about 3 GB. The node count grows near 21 level at height 3^level and the stored powers cost Theta(level^2) bits, so memory and not the clock is what stops the census. Witness: cobham.md:95, lab/rs/three-base-thin verb reach 80000.
  • 2026-09-14 [Verified] The pruned walk is checked against a direct scan of every integer below 10^8 against all three digit rules, at base-7 bound 2 and again at base-7 bound 3, and the two agree in both. Witness: cobham.md:96, lab/rs/three-base-thin verb control.
  • 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 to p exactly when every base-p digit of k is below p/2, which reads <= 1 at 3, <= 2 at 5 and <= 3 at 7, so raising the base-7 bound from 2 to 3 gives {k : binomial(2k, k) is prime to 105}, OEIS A030979, whose triple budget log_3 2 + log_5 3 + log_7 4 - 2 reads 0.025951 rounded up against -0.121889 for 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 about x^0.02595... terms up to x, which is the same number as the budget. The same walk at base-7 bound 3 rebuilds all 23 terms A030979 publishes, counts 1374 members below 10^70, exactly the length of the table that entry calls complete to 10^70, and counts 216020 below 10^140; the effective exponents log(count)/log(height) read 0.044828 and 0.038103, truncated down, both above 0.025951 and falling. Witness: cobham.md:97 and cobham.md:98 and cobham.md:99, lab/rs/three-base-thin verbs budget, control, ten 70 and ten 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, 151133095 at m = 0..24 are printed from scratch in this run by the verb terms and again by the verb hankel, 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 verb control, which tests every pair of [0, 3^m)^2 against both digit rules with a separate routine and agrees at every m <= 6. Witness: lab/py/two-base-gasket verbs terms, hankel, control; cobham.md:77.
  • 2026-09-14 [Verified] No linear recurrence with constant coefficients of order r <= 12 fits all twenty-five terms of C(3^m). Two exact tests over the rationals agree. The Hankel determinants of the matrix with entries a[i+j], computed by fraction-free Bareiss over Z, are nonzero at every size k = 1..13, reading 1, -2, 4, 8, 5360, -1259267712, -5516990041856, 112485023878830080, 5697070341654551514880, -1283183611235610612560681984, 512735847124145678895522067154944, -30665906970422092677442692752788846592, -148892102950447887517893509783802772470337536; a recurrence of order r would force every Hankel determinant of size above r to vanish, so det H_13 != 0 alone kills every order at most 12. Independently, for each order r = 1..12 the linear system a(n) = c_1 a(n-1) + ... + c_r a(n-r) taken over every one of the 25 - r available equations is inconsistent by Gauss-Jordan over Q, order by order with no order left consistent and none underdetermined. Order 12 is the largest the data can test: it leaves 13 equations against 12 unknowns, one spare, while order 13 leaves 12 equations against 13 unknowns, one short of determined, so that system is consistent for trivial reasons and tests nothing; 12 is exactly the bound twenty-five terms carry and not a choice. The two solvers behind the negative have their own generator: the verb selftest fits Fibonacci first at order 2 with x^2 - x - 1, 2^n + 3^n + 1 first at order 3 with dominant root 3.0000000, and n^3 + 2^n at no order below 5 and at 5 with (x - 1)^4 (x - 2), and checks bareiss against five determinants computed by hand, eleven checks in 0.05 s. Witness: lab/py/two-base-gasket verbs hankel, 3 min 15 s at HI = 24, and selftest.
  • 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 dump research/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 at 5.719220 against 2 pi / ln 3 = 5.719202, error 0.000%, at 1.7e7 times the median power of the band [0.5, 14] under the linear detrend and 1.9e7 under the cubic; the base-5 design {0, 1, 2} puts one at 3.903959 against 2 pi / ln 5 = 3.903963, error 0.000%, at 4.0e6 times the median. At 3^44 the base-3 control's nearest maximum to 2 pi / ln 5 sits at error 1.115% and 35.9 times the median under the linear detrend and at 0.966% and 39.2 under the cubic, which passes the rule at a frequency absent by construction; the base-5 control's nearest maximum to 2 pi / ln 3 sits at 1.060% and 2.21 under the linear detrend and at 0.963% and 1.95 under 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: span 38.05 with 11 maxima above 10x the median covering 0.113 of the band, and span 38.49 with 4 covering 0.048, against the cell's span 27.41, 4 maxima and 0.020. Witness lab/py/two-base-instrument verb cell 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 of bases.md, 0 lying in both digit sets, so the joint set is exactly the one-base design F(9, {0, 1, 3}) of exponent log_9 3 = 1/2; at the aligned height 3^28 = 9^14 its count is exactly 3^14 - 1 = 4782968, the fitted exponent is 0.497809, and its spectrum shows 2 pi / ln 9 = 2.859601 at 2.859889, error 0.010%, at 886 times the median, with the first harmonic 2 pi / ln 3 at error 0.003% and 2.5e3 times the median, on a window of span 17.47 carrying 4 maxima above 10x the median covering 0.042 of the band. Witness lab/py/two-base-instrument verb collapse 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+1 digits lies in [3^k, (3^(k+1)-1)/2] and a member of the base-5 design with i+1 digits in [5^i, (5^(i+1)-1)/2], so the designs occupy ln(3/2)/ln 3 = 0.369070 and ln(5/2)/ln 5 = 0.569323 of their own decades and the cell's count is exactly constant wherever the two bands miss: of the 47 decades [3^j, 3^(j+1)) below 3^47, the 10 with j = 1, 4, 17, 20, 23, 26, 36, 39, 42, 45 carry no member at all, proved by the bands and witnessed by the ladder, whose new hits are 0 at level = 40, 43, 46. The block model C3(N) C5(N) / N, the same two band structures multiplied with no joint arithmetic, shows 2 pi / ln 3 at error 0.001% and 1.33e6 times the median and 2 pi / ln 5 at 0.004% and 9.79e5, on a window of span 31.55 with 4 loud maxima covering 0.043; the cell's own count on a window of span 30.70 with the same 4 loud maxima and 0.020 reaches 22.5 and 15.7. Witness lab/py/two-base-instrument verbs blocks 47 and ladder 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 3^38 to 3^47 returns both frequencies present at level = 38, 39, 40, 41, 2 pi / ln 3 at errors 0.473%, 0.073%, 0.342%, 0.704% under the linear detrend and 0.325%, 0.036%, 0.484%, 0.862% under the cubic at 19 to 24 times the median, and returns 2 pi / ln 3 absent from level = 42 up, at errors 1.034% to 1.574% and 22 to 25 times the median, while 2 pi / ln 5 is present at every one of the ten heights. The error at 2 pi / ln 3 rises monotonically from level = 39 to level = 46 under 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 at 3^44 the nearest maximum to 2 pi / ln 3 sits at error 0.825% under the linear detrend and 0.791% under the cubic, inside the tolerance and held out by the power gate alone at 5.55 and 6.1 times the median, and at 3^47 at 2.430% and 2.293%. No height rule is stated in advance, so no height is entitled to the verdict. Witness lab/py/two-base-instrument verbs ladder 38 47, cell 44 and cell 47.
  • 2026-09-14 [Verified] The cell's strongest maximum matches no small combination of the two lattice frequencies. It sits at 1.702087 at 3^44 and 1.697925 at 3^47, at 57 and 68 times the median on windows of span 27.41 and 30.70 carrying 4 maxima above 10x the median covering 0.020 of the band, and the nearest m 2 pi / ln 3 + n 2 pi / ln 5 with abs(m), abs(n) <= 8 is (1, -1) = 1.815239 at errors 6.233% and 6.463%; 2 pi / ln 15 misses by 3.750% and 3.457% and the sum frequency by 1.528% and 1.977% at 2.2 and 2.1 times the median, while 2 pi / ln 5 is met at 0.461% and 0.594%. Block structure does not account for it: the block model's nearest maximum to 1.815239 sits at 1.895843 and 0.588 times the median at 3^44 and at 1.697497 and 1.58 at 3^47. The same verbs read the local counting exponent falling through the budget 0.313536, 0.323301 over the full window and 0.295978 over its upper half at 3^44, 0.318728 and 0.292184 at 3^47, which is a fit at finite height and decides nothing about the limit. Witness lab/py/two-base-instrument verbs cell 44, cell 47 and blocks 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 gasket G_n is the base-3 digit pairs from {(0,0),(1,0),(0,1)}, the word binary inside 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 at coprime.md:150 and the base-4 and base-5 simplex probes at coprime.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 pair B(s,t), per band direction and per modulus 3^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 by 3^k (coprime.md:228) carry no uniform state bound, while B(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 like ab in 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 = 3^1 and 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) / m reads 0.754141, 0.749634, 0.746312, 0.743739, 0.738025, 0.732546, 0.729032, 0.724371, 0.721151, 0.717651, 0.714298 at m = 14..24, falling by about 0.004 a level on the mean of those ten steps and still 0.129 above the budget at the last level, consistent with any limit from 0.630929 upward. The cheap falsifier is a linear recurrence for 1, 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-gasket verb terms.
  • 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 shape sum_i log(k_i)/log(p_i) - (m-1) dim, a Q-linear combination of 1 and the ratios log 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 is 0 and 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-thin verb reach 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 not log_3 of a Perron root and the Schanuel wall holds in data for Object Y. What is proved is the order at most 12 case above; the step to every order is the conjecture, and nothing here rules out a recurrence of order 13 or more. Falsification: order r is testable once the terms leave its system a spare equation, which wants 2r+1 of them, so order 13 wants 27 terms, the two levels m = 25 and m = 26 beyond what is run, about 11 min by the measured per-level factor; the same verb tests any order the census reaches. Witness: lab/py/two-base-gasket verbs hankel and selftest, extrapolated from the r <= 12 negative.
  • 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 10^5 to 10^6 times the median while at 3^47 the cell's own count carries 2 pi / ln 5 at 15.7 times the median and puts nothing nearer to 2 pi / ln 3 than a maximum 1.488% 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. Witness lab/py/two-base-instrument verbs cell 44, cell 47, ladder 38 47 and blocks 47.
  • 2026-09-14 [Refuted] The sentence "no proper design is recognizable in two independent bases, in any dim" is false. At dim = 2 and base = 2 the design F = {(0,0), (1,1)} has S_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 as dim >= 2, so the dim = 1 lock 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 at dim = 2, in one line and with product designs. The base-2 design F_A = {(0,0), (0,1)} gives A = {0} x T of dimension 1, the base-3 design F_B = {(0,0), (0,1)} gives B = {0} x C of dimension log_3 2, B sits inside A, so the intersection has dimension log_3 2 while the budget reads max(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-gasket verb budget.
  • 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 reads 0 + log_3 2 and is sharp there, while the global budget reads 0. So in dim >= 2 the two-base budget is a theorem per axis for product designs and open for compounds, and core.md proves almost every design is a compound as dim grows. Witness: cobham.md:62.