Two bases

Every page of this tree reads one base at a time, and that is a law rather than a habit. Two bases in one dependence class are one base and belong to bases; this page is about the other case. Cobham's theorem says a set recognized by a finite automaton in two multiplicatively independent bases is already periodic, so at dim 1 a proper design has exactly one base and the joint object of two bases is not a design, not an automaton and not a transfer matrix. This page states the law at its source, makes the dim 1 consequence exact, prices dim >= 2 where the law is weaker than folklore says, lists which of this tree's instruments survive contact with a second base and which do not, turns to the smallest honest two-base object, a base-2 gasket meeting a base-3 gasket, whose census refutes the naive planar budget, then to a three-base object on the line whose budget is negative and whose census finds five members and no sixth below a height of 38170 decimal digits, reads a two-base cell on the line for the two lattice frequencies its bases would each impose and finds the verdict unstable in height, counts the sumset of the base-3 and base-4 designs {0, 1} of Erdos problem 125 to 3^22, proves its two-base energy ratio subpolynomial and unbounded, and shows its open upper density positive once one orbit of an irrational rotation misses, infinitely often, a target of measure at most 190/M, and closes on the transcendence wall that stands between every instrument of this tree and any two-base exponent.

lab/py/two-base-gasket, lab/rs/three-base-thin, lab/py/two-base-instrument and lab/rs/sumset-density are the four generators behind every number below, the third behind the periodogram section alone and the fourth behind the sumset section alone.

Independence is a partition of the bases

Two reals alpha, beta > 1 are multiplicatively independent when alpha^m = beta^n with m, n in N forces m = n = 0 (Durand and Rigo, Definition 1.1, read at source); for two integer bases p, q >= 2 this reads p^i != q^j for every pair of positive integers i, j.
Equivalently log p / log q is irrational, and equivalently p and q are not both powers of one integer; coprime integers are always independent, and 6 and 18 are independent without being coprime (same source).
Multiplicative dependence is an equivalence relation on the integers >= 2, and each class is the set of powers of its least member, the first classes being [2], [3], [5], [6], [7], [10], [11], [12] (same source, Remark 1.2).
Proved. Independence is never emergent in a triple. Dependence is transitive, so three bases that are pairwise dependent are jointly dependent, and one independent pair inside any family already makes the family carry an independent pair. A third base adds no hypothesis that a pair does not already carry, and the dependence-class partition, not the tuple, is the invariant.
Base 1 is not a base: k-recognizability, recognition in a base k, is defined for k >= 2 only, so a pair holding base 1 has no content.
Bases 2 and 4 are one base: they lie in the class [2], and k-recognizability and l-recognizability coincide on a dependent pair (Bes, attributing it to Buchi, read at source), so Cobham's hypothesis fails there.
Proved. The conclusion fails with it, so the hypothesis is load-bearing and not decoration: the base-4 design {0, 1} is 4-recognizable, hence 2-recognizable, and it is infinite of density (1/2)^level, hence not ultimately periodic.

The law, at its source

Theorem (Cobham 1969). "Let k, l >= 2 be multiplicatively independent integers. Every subset X of N which is k- and l-recognizable is ultimately periodic. Therefore such a X is m-recognizable for any m >= 2." (Bes, Theorem 24, read at source; Durand and Rigo state the same as Theorem 1.1 with "if and only if".)
Theorem (Cobham-Semenov, Semenov 1977). "For any n >= 1, and all multiplicatively independent integers k, l >= 2, every subset of N^n which is k- and l-recognizable is definable in <N; =, +>." (Bes, Theorem 25, read at source; the same statement is Durand and Rigo Theorem 4.7.)
Definable in <N; =, +> is semilinear, a finite union of sets v + N c_1 + ... + N c_r with v and the c_i in N^n (Bes, Theorem 4, after Ginsburg and Spanier); at n = 1 semilinear is exactly ultimately periodic.
The law splits the subsets of N into three classes and not two (Bes, read at source): the ultimately periodic sets, recognizable in every base; the sets recognizable in one dependence class and no other, where every proper design of this tree sits; and the sets recognizable in no base at all, the primes and the squares among them.

The dim 1 consequence is exact

Proved. Let the filled digits F lie inside {0, ..., base-1} with 0 in F and 1 < card F < base, and let S_F be the integers whose digits all lie in F. Then S_F is recognizable in no base multiplicatively independent of that one. Proof: S_F is infinite, since d base^j lies in it for every nonzero d in F and every j; card(S_F cap [0, base^level)) = (card F)^level, so the density is (card F / base)^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 a second independent base.
The two hypotheses are exactly the two exclusions: card F > 1 removes F = {0} and card F < base removes the full digit set, and those two are the only semilinear designs at dim 1. Every other one-dimensional design is base-locked.
The same lock holds on the integers without automata. Glasscock, Moreira and Richter 2024, Theorem A, read at source: for r, s multiplicatively independent, A xr-invariant and B xs-invariant, if lambda A + tau lies in the delta-neighbourhood of eta B + sigma for some lambda, eta, delta > 0 and real sigma, tau, then A is finite or B = N_0; on digit-restricted sets their Corollary of Theorem A reads "if A is contained in B, then either A = {0} or B = N_0". Neither law contains the other: Cobham reaches every recognizable set and says periodic, Theorem A reaches every multiplicatively invariant set, regular or not, up to an affine image.

Which designs are semilinear

Refuted. The sentence "no proper design is recognizable in two independent bases, at any dim" is false. At dim 2 and base 2 the design F = {(0,0), (1,1)} has S_F = {(n, n)} over the integers n, the diagonal, which is definable in <N; =, +> and so recognizable in every base. Proper designs recognizable in two independent bases exist as soon as dim >= 2, and the dim 1 statement above does not generalize by itself.
Proved (the necessary condition). 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 a subspace of dimension e meets [0, N)^dim in Theta(N^e) points, so a semilinear set's count in the box is Theta(N^d) with d the largest span dimension among its constituents; the design's own count is card(S_F cap [0, base^level)^dim) = (card F)^level exactly, so (card F)^level = Theta(base^(d level)) and card F = base^d.
Proved (the sufficient condition). Call F a block design when the dim coordinates split into a zero set Z and d blocks, and F = {v in {0,...,base-1}^dim : 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 vector of block t, which is one linear set, hence semilinear, hence recognizable in every base.
Proved at base 2, dim 2. The two conditions agree there: 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.
Conjecture. Block designs are the only semilinear ones, at every base and every dim.
Proved (the count alone is not enough). The base-3 gasket F = {(0,0), (0,1), (1,0)} has card F = 3 = 3^1 and is not semilinear. Its box count is 3^level at side 3^level, so d = 1 and a semilinear S_F would lie in finitely many lines; but S_F contains P_t = (3^t, 3^(t^2)) for every t >= 2, whose consecutive slopes are s_t = 3^(t^2 - t) (3^(2t+1) - 1)/2, strictly increasing in t, so the P_t are in strictly convex position, no three are collinear, and covering n of them costs at least n/2 lines. Hence the base-3 gasket is not 2-recognizable, and no automaton reading base-2 digits enforces its digit rule.
Proved. The base-2 gasket F = {(0,0), (0,1), (1,0)} is not semilinear either, and needs no geometry: card F = 3 is not a power of 2. It is therefore not 3-recognizable.

What the second base does to this tree's instruments

Proved. When p and q are independent and a base-q design is not semilinear, no finite automaton reading base-p digits accepts it, so no transfer matrix over the digits of one base reads the constraint the other base imposes. What dies is the method and not the object: an intersection can still be recognizable by accident, a finite set being recognizable in every base, so nothing here says the joint object is complicated, only that neither base's machine sees it.
Survives: the box bound of the coprimality sieve. coprime proves N*_level(m) <= (base+1)^dim fill^level m^(-alpha) with alpha = log_base(fill), and that is pure counting on S_level, so it passes to every subset by monotonicity, the joint object included, and the Chebyshev sum built on it still converges when alpha > 1. That is one line of the sieve and it was never the hard part.
Dies: the fill law. method carries fill(F, 2k-1) = sum_(c in F) k^(dim - w(c)) (k-1)^w(c) at odd side 2k - 1 and the level rule fill(level) = fill^level, both identities on a Kronecker power in one base. A joint object of two independent bases has no product structure at any scale, so there is no level at which a fill count multiplies.
Dies: the transfer matrix and its Perron root. beneath reads a window rule as a vertex shift and prints log_2 rho with rho the Perron eigenvalue of a nonnegative integer matrix; cuts reads the central slice through the even transfer matrix M_even; crop certifies its own Perron brackets by Collatz-Wielandt. Each is a finite automaton over the digits of one base, and each falls to the previous bullet.
Dies: the carry automaton of cuts. Its states are the integers c with abs(c) <= floor((dim-1)/2) and its transition is c' = (c + dim - s)/3; it is finite because x -> (x + dim)/3 contracts on integer carries inside one base. A machine reading base-2 digits while tracking base-3 digits is base conversion, which is not finite state.
Dies: the character contraction. coprime's Lemma A splits the one-digit character sum at a position where the orbit is far from an integer and gives abs(Sum) <= fill - 2 + 2 cos(pi/(2 base)); the equidistribution half of the sieve consumes one such factor per orbit cycle, and one per window of m_d digit positions under Lemma A'. The contraction is exactly the statement that the transform at one level factors over digit positions in one base, and the joint set factors in neither. A sieve needs an upper bound and an equidistribution; two bases hand over the first and destroy the second.

The budget on the line

Theorem (Corso and Shmerkin 2024, Corollary 1.17, read at source). "Let p_1, ..., p_d, A_1, ..., A_d and s be as in Theorem 1.15. Then, for all affine maps g_1, ..., g_d : R -> R, dim-upper_B(g_1(A_1) cap ... cap g_d(A_d)) <= max{s - (d-1), 0}." Theorem 1.15 carries the hypotheses: p_1, ..., p_d >= 2 pairwise multiplicatively independent, A_1, ..., A_d closed subsets of the circle T invariant under T_(p_1), ..., T_(p_d), and s = sum_j dim_H A_j.
At d = 2 that is Furstenberg's intersection conjecture, stated by Shmerkin 2019 as his Conjecture 1.1 after Furstenberg 1970 and proved as his Theorem 1.2, read at source: "Let p, q in N_(>=2) be multiplicatively independent. Then for any closed sets A, B of the circle [0, 1) invariant under T_p, T_q respectively, and for any invertible affine map g : R -> R, dim-B(A cap g(B)) <= max(dim_H(A) + dim_H(B) - 1, 0)." Wu 2019 proves the same independently.
Proved. The two-set theorem does not iterate, and the reason is elementary: A cap B need not be invariant under either map, so it is not a legal input to the theorem against a third base. Take A the middle-thirds set, which is T_3-invariant, and B = [0, 1), which is T_2-invariant; then A cap B = 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, which is what the hypothesis on the slicing subspace in Theorem 1.15 protects.
Before that route existed the m >= 3 bound was known only under a Q-linear independence hypothesis on the ratios log p_1 / log p_j (Yu 2021b), a transcendence condition unproved for (2, 3, 5).
Theorem (Glasscock, Moreira and Richter 2024, Theorem B, read at source). The integer side of the same statement at m = 2. Let r, s be multiplicatively independent, let A, B be subsets of N_0 that are xr- and xs-invariant, closed under dropping the least significant and the most significant digit in that base (their Definition 1.5), and let gamma = max(0, dim_H A + dim_H B - 1). Then "for all eps, lambda, eta > 0, sigma, tau in R, and sufficiently large N in N, card(floor(lambda (A cap [0, N)) + tau) cap floor(eta (B cap [0, N)) + sigma)) <= N^(gamma + eps). In particular, for all lambda, eta > 0 and sigma, tau in R, dim-upper_M(floor(lambda A + tau) cap floor(eta B + sigma)) <= max(0, dim_H A + dim_H B - 1)" (Glasscock, Moreira and Richter 2024); dim_H is their discrete Hausdorff dimension and dim-upper_M their upper mass dimension, which is the counting exponent of this page, and the two coincide on an invariant set (their Proposition 3.6). Their Corollary of Theorem B reads it on digit-restricted sets: with A_N = A cap [0, N), if dim A + dim B < 1 then card(A_N cap B_N) <= N^eps for every eps > 0 and all large N. A one-base design's integer set S_F is such a set, xbase-invariant in their sense, so the bound applies to it with no affine map; their Theorem C is the sumset statement and is not used on this page.
First of its kind, and what was searched. Read at source for this page: Shmerkin 2019, Wu 2019, Yu 2021b, Corso and Shmerkin 2024, Glasscock Moreira and Richter 2024, Burrell and Yu, Erdos Graham Ruzsa and Straus 1975 through Burrell and Yu, Senge and Straus 1973 and Stewart 1980 through the survey of Bugeaud, Cipu and Mignotte. Every dimension statement read there is an upper bound; none of them carries an asymptotic or an exact constant for any named independent pair; the only lower bound read is an infinitude statement and the only finiteness result read lives where every dimension is already zero. This card names what was searched and does not claim what does not exist.

The budget in the plane is false

A design lives in T^dim and uses one base on all dim coordinates, while Theorem 1.15 asks for one set per coordinate in pairwise independent bases. The hypothesis therefore fails at dim >= 2, and the conclusion fails with it.
Refuted. The global planar budget dim_H(A cap B) <= max(0, dim_H A + dim_H B - dim) is false at dim 2, in one line. The base-2 design F_A = {(0,0), (0,1)} gives A = {0} x T with dim_H A = 1; the base-3 design F_B = {(0,0), (0,1)} gives B = {0} x C with C the base-3 digit set {0, 1} and dim_H B = log_3 2; B sits inside A, so dim_H(A cap B) = log_3 2 against a budget of 1 + log_3 2 - 2 < 0. Both sets lie in the line {0} x T, which is invariant under both maps, and that is where the two codimensions refuse to add.
Proved (what product designs do give, one budget per axis). If every F_i is a product G_i^(1) x ... x G_i^(dim) across the dim axes in pairwise independent bases p_1, ..., p_m, then each A_i is the product of its axis sets, the intersection is the coordinatewise intersection, upper box dimension is subadditive on products, and Corollary 1.17 applies on each axis, so 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, as the corollary states it.
Refuted. The global budget is not a corollary of that 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 witness above is where it runs strictly wrong: the per-axis bound reads 0 + log_3 2 and is sharp, while the global budget reads 0.
So at dim >= 2 the two-base budget is a theorem per axis for product designs and open for compounds, and core proves almost every design is a compound as dim grows.

Budget zero is dimension zero, not finiteness

Proved. The set {2^n} has counting exponent zero, card({2^n} cap [0, N)) <= log_2 N + 1, and is infinite. A budget of zero says the dimension is zero and says nothing about finiteness, so a transversality bound of zero never closes a question that asks for a finite list.
Finiteness arrives only at the corner where every digit sum is bounded. Senge and Straus 1973 prove "the number of integers, the sum of whose digits in each of the bases a and b lies below a fixed bound, is finite if, and only if, a and b are multiplicatively independent", by Thue-Siegel-Roth and so ineffectively; Stewart 1980 makes it effective with Baker's theory of linear forms in logarithms, showing that for independent a, b, every c >= 1 and every m > 25 whose digit sums in both bases are at most c, log log m / (log log log m + c_1) < 2c + 1 with c_1 effectively computable in a and b alone. Both statements are read at source in the survey of Bugeaud, Cipu and Mignotte; the two originals are paywalled and are cited through it.
That corner is not a design. A bounded-digit-sum set is not closed under changing one digit, its count below b^k is O(k^c), and its exponent is 0; there Baker's theory beats the whole transversality machinery outright, and a third base buys nothing because two already give a finite list.
The one lower bound in that literature runs the other way. Burrell and Yu quote it as their Theorem 1.8, read at source, from Erdos, Graham, Ruzsa and Straus 1975: "Let p, q be integers greater than 1. If A, B are two positive integers satisfying A/(p-1) + B/(q-1) >= 1, then there exist infinitely many integers whose base p expansion contains only digits <= A and base q expansion contains only digits <= B."

Object Y: a base-2 gasket meets a base-3 gasket

A = {(x, y) in Z^2 : every base-2 digit pair lies in {(0,0), (0,1), (1,0)}}, which is {(x, y) : x AND y = 0}, of counting exponent log_2 3 = 1.584963.
B = {(x, y) in Z^2 : every base-3 digit pair lies in {(0,0), (0,1), (1,0)}}, of counting exponent log_3 3 = 1.
C(N) = card(A cap B cap [0, N)^2) is the joint census, and the naive planar budget for it reads log_2 3 + 1 - 2 <= 0.584963 (lab/py/two-base-gasket, verb budget).
Verified (lab/py/two-base-gasket, verb terms, 3 min 13 s for m = 0..24 on one core): 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. The verb control rebuilds the same counts for m <= 6 by testing every pair in the box against both digit rules directly, and agrees at every level.
Proved. C(3^m) >= 2^(m+1) - 1. On the axis x = 0 membership in A is automatic, and membership in B asks the base-3 digits of y to lie in {0, 1}, which 2^m values of y below 3^m satisfy; the axis y = 0 gives another 2^m; the origin is the only overlap. The lab prints both sides at every level and the inequality holds at each.
Proved. log_3 2 >= 0.630929 > 0.584963, the lower bound truncated down and the budget rounded up by lab/py/two-base-gasket verb budget, so the counting exponent of A cap B exceeds the naive planar budget. The real gaskets carry the same excess: the left edge {0} x [0, 1] lies in the real base-2 gasket, and the real base-3 gasket meets that edge in {0} x C with C the base-3 digit Cantor set of dimension log_3 2, so dim_H of the real intersection is at least log_3 2 while the budget reads 0.584963.
What Object Y adds to the one-line witness of the section above, and what it does not. Neither gasket lies in a proper closed subtorus: such a subtorus lies in the kernel of a primitive character, {(x, y) : u x + v y = 0 mod 1} with gcd(u, v) = 1, and the real base-2 gasket contains (1/2, 0) and (0, 1/2), which force 2 | u and 2 | v, while the real base-3 gasket contains (1/3, 0) and (0, 1/3), which force 3 | u and 3 | v. The proved excess is nonetheless carried by the coordinate axes, which are invariant under both maps, so the mechanism is the same as the one-line witness, and Object Y is not a smaller counterexample; what it is, is a counterexample in which both sets are compounds rather than degenerate products, of counting exponents 1.584963 and 1, and neither side is covered by any theorem in print.
Not converged, and said so. 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 (lab/py/two-base-gasket, verb terms), 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. Twenty-five levels separate nothing. The true exponent of A cap B is Conjecture, and log_3 2 is the only proved number in this section.
Verified (lab/py/two-base-gasket, verb hankel, 3 min 15 s at HI = 24, its two solvers checked against answers known in advance by the verb selftest): C(3^m) satisfies no linear recurrence with constant coefficients of order at most 12. The Hankel determinant of size 13 on the twenty-five terms is -148892102950447887517893509783802772470337536, nonzero, which a recurrence of order at most 12 would force to vanish; independently the rational system for each order r = 1..12, taken over all 25 - r equations the terms supply, is inconsistent by Gauss-Jordan over Q. Order 12 is the ceiling twenty-five terms carry and not a choice, order r wanting 2r + 1 terms to leave its system one spare equation, so what stops the test is the term count and not the method.
Conjecture. C(3^m) satisfies no such recurrence at any order. Were one to appear the growth rate would be an algebraic integer, its characteristic polynomial monic over Z by Fatou, which would put Object Y back inside reach of this tree's own machinery; an algebraic integer need not be the Perron root of a nonnegative integer matrix, so even that would not by itself refute the Schanuel wall below. What the test returns is a negative and nothing more: no recurrence fits at order at most 12, which is no evidence for the wall. The falsifier is order 13, which wants the two further levels m = 25 and m = 26, about 11 min of census (lab/py/two-base-gasket, verb terms).

Object T: three bases on the line, where no two of them suffice

Object T is the triple of digit rules (3, <= 1), (5, <= 2), (7, <= 2). Its integer set is E = {n in N : every base-3 digit of n is <= 1, every base-5 digit is <= 2, every base-7 digit is <= 2}, and its real sets are the three closed subsets A_3, A_5, A_7 of the circle cut by those same digit rules, A_p invariant under multiplication by p.
The three dimensions are dim_H A_3 = log_3 2 = 0.630930, dim_H A_5 = log_5 3 = 0.682606 and dim_H A_7 = log_7 3 = 0.564575, to six places (lab/rs/three-base-thin, verb budget).
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 all three are positive, so the two-set bound of the section above returns nothing on any pair (same verb).
The triple budget sum_i dim_i - 2 reads -0.121889, rounded up (same verb), and that sign is the whole of the object.
Proved. dim-upper_B(A_3 cap A_5 cap A_7) = 0. Corollary 1.17 as quoted above asks for p_1, ..., p_d >= 2 pairwise multiplicatively independent, closed sets A_1, ..., A_d in the circle invariant under T_(p_1), ..., T_(p_d), and affine g_1, ..., g_d, and concludes dim-upper_B(g_1(A_1) cap ... cap g_d(A_d)) <= max{s - (d-1), 0} with s = sum_j dim_H A_j. Here d = 3, the bases are distinct primes and so pairwise multiplicatively independent, each A_j is closed and T_(p_j)-invariant by its digit rule, the g_j are the identity, and s - 2 = -0.121889 < 0, 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 nothing here transfers it to E.
Proved. The third base is not redundant, and the redundancy has a sharp answer in both directions. Upward no two-set bound gives 0, since all three pair budgets are positive. Downward the pair (3, 5) alone is infinite: Erdos, Graham, Ruzsa and Straus 1975, quoted as Theorem 1.8 of Burrell and Yu in the section above, 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 (lab/rs/three-base-thin, verb budget). Two of the three digit rules therefore admit infinitely many integers, and whatever finiteness E has is bought by the third rule alone.
The other two pairs miss that criterion by one digit each, (3, 7) reading 1/2 + 2/6 = 0.833333 and (5, 7) reading 2/4 + 2/6 = 0.833333, and raising the base-7 bound from 2 to 3 takes both to 1.000000 (same verb). The criterion is sufficient and not necessary, so those two pairs are not known finite either, and nothing here says they are.
The census enumerates the base-3 side, whose members are exactly the 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 n in the branch is one of two consecutive integers, and the branch dies when neither of them has all its base-5 digits <= 2; base 7 cuts the same way. Membership in A_3 holds by construction and is never tested.
Verified (lab/rs/three-base-thin, verb seven 17, 333 nodes, under 0.01 s): the members of E below 7^17 = 232630513987207 are 0, 1, 3186, 3187, 20007 and nothing else.
Verified (lab/rs/three-base-thin, verb reach 80000, 1710789 nodes, 61.48 s, about 3 GB): below 3^80000, a height of 38170 decimal digits, the members of E are the same five. The memory is the wall and not the clock, the stored powers costing Theta(level^2) bits at height 3^level.
Verified (lab/rs/three-base-thin, verb control): the pruned walk and a direct scan of every integer below 10^8 against all three digit rules return the same five members, and the two agree again with the base-7 bound raised to 3.
Raising that base-7 bound from 2 to 3 lands on a set already in print, and the budget changes sign across the step. 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 p = 3, <= 2 at p = 5 and <= 3 at p = 7, so the wider set is {k : binomial(2k, k) is prime to 105}, A030979, read at source. Its budget is log_3 2 + log_5 3 + log_7 4 - 2 = 0.025951, rounded up, with dim_H of the base-7 side risen to log_7 4 = 0.712414, against -0.121889 for E (lab/rs/three-base-thin, verb budget), so E is one digit in one base away from a named open problem, on the other side of the sign of the budget.
A030979, read at source, records a prize for settling whether that wider set is finite, names it as Erdos problem 376, and quotes a heuristic of Pomerance giving about x^0.02595... terms up to x; the Pomerance article is not opened here. That exponent is the budget 0.025951 of the line above, so on the wider set the transversality budget and the heuristic in print are the same number, and on E the same budget is negative. A second Erdos problem on this tree's designs, problem 125 on the sumset of the base-3 and base-4 designs {0, 1}, is read in the Object S section below.
Verified (lab/rs/three-base-thin, verbs control, ten 70 and ten 140, 0.09 s and 39.32 s): the same walk with the base-7 bound at 3 rebuilds all 23 terms A030979 publishes, counts 1374 members below 10^70, which is 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.
That contrast is the control that matters for E: one walk, one digit bound apart, finds 1374 members of the wider set below 10^70 and five members of E below a height of 38170 decimal digits.
Conjecture. E = {0, 1, 3186, 3187, 20007}.
Finiteness is open and no theorem on this page gives it. The dimension bound above is 0, and the section above on budget zero proves that a budget of 0 says nothing about finiteness. The finiteness results in print bound digit sums rather than digits: applying Senge and Straus 1973 or Stewart 1980 to E would need a bound on the base-3 digit sum of a member of E, which is the finiteness in question. The falsifier is a sixth member, and the census above is where it would have shown.

Reading a two-base count for two frequencies

A one-base count oscillates in ln N at the single frequency 2 pi / ln base, and dimensions owns the mechanics that read it: detrend ln C(e^u) in u, window it, take the periodogram. If a two-base cell carried two lattice structures at once its count would have to show both 2 pi / ln p and 2 pi / ln q, and that is the prediction tested here. The decision rule is lab/py/two-base-instrument's own and is weaker than the one that page states: a prediction counts met when the nearest local maximum lies within 1% of it and carries at least 10x the median power of the band [0.5, 14], because a two-base count has to be read at frequencies that are not its loudest.
The count is not a smooth staircase. A member of the base-3 design {0, 1} with k+1 digits lies in [3^k, (3^(k+1)-1)/2] and a member of the base-5 design {0, 1, 2} with i+1 digits in [5^i, (5^(i+1)-1)/2], so the two designs occupy ln(3/2)/ln 3 = 0.369070 and ln(5/2)/ln 5 = 0.569323 of their own decades and the joint count is exactly constant wherever the two bands miss. Verified (lab/py/two-base-instrument, verbs blocks 47 and ladder 38 47): ten of the 47 decades [3^j, 3^(j+1)) below 3^47 carry no member at all, at j = 1, 4, 17, 20, 23, 26, 36, 39, 42, 45.
Verified (lab/py/two-base-instrument, verbs cell 44, collapse 28 and blocks 47). Three one-base controls and one block model calibrate the rule, and neither pure control is clean. The base-3 design puts a maximum at 5.719220 against 2 pi / ln 3 = 5.719202, error 0.000%, at 1.7e7 times the median; the base-5 design puts one at 3.903959 against 2 pi / ln 5 = 3.903963, error 0.000%, at 4.0e6. A multiplicatively dependent pair is one base by the collapse theorem of bases, whose zero-digit hypothesis holds here: base 3 {0, 1} against base 9 {0, 1, 3} is exactly the one-base design F(9, {0, 1, 3}), its count at 3^28 = 9^14 exactly 3^14 - 1 = 4782968, and it shows 2 pi / ln 9 = 2.859601 at error 0.010% and 886 times the median. The block model C3(N) C5(N) / N, the two band structures multiplied with no joint arithmetic, shows both frequencies at 1.33e6 and 9.79e5 times the median, so the two-frequency prediction is exactly what block structure alone predicts. The base-3 count under the cubic detrend meanwhile passes the rule at 2 pi / ln 5, error 0.966% at 39.2 times the median, a frequency absent by construction, on a window of span 38.05 carrying 11 loud maxima against the cell's 27.41 and 4; that pass does not recur at 3^47, where the same control reads error 2.791% at 2.6 times the median.
Verified (lab/py/two-base-instrument, verb ladder 38 47). The cell is dim 1, base 3 {0, 1} against base 5 {0, 1, 2}, of budget log_3 2 + log_5 3 - 1 <= 0.313536 rounded up; it carries 5667470 members below 3^44 and 19042219 below 3^47, and the rule's verdict on it depends on the height read. At level = 38, 39, 40, 41 both frequencies are met under both detrends; from level = 42 up 2 pi / ln 3 is not, its nearest maximum sitting at errors 1.034% to 1.574% at 22 to 25 times the median, and 2 pi / ln 5 is met at every one of the ten heights. That error rises monotonically from level = 39 to level = 46 under both detrends, which is a maximum drifting away from the prediction as the window lengthens rather than an estimate converging on it, and a prediction whose verdict moves with the height is settled by neither verdict.
Verified (lab/py/two-base-instrument, verbs cell 44, cell 47 and blocks 47). No small combination of the two frequencies explains what the cell does carry. Its strongest maximum sits at 1.702087 at 3^44 and 1.697925 at 3^47, at 57 and 68 times the median, and the nearest m 2 pi / ln 3 + n 2 pi / ln 5 with abs(m), abs(n) <= 8 is the difference frequency 1.815239, 6.233% and 6.463% away. Block structure puts nothing loud where the cell's strongest maximum sits: the block model's nearest maximum to 1.815239 carries 0.588 times the median at 3^44 and 1.58 at 3^47.
Conjecture. The joint digit constraint destroys the oscillation either design carries alone, so the intersection is not the product of its two band structures at the level the spectrum reads: the block model carries both frequencies at 10^5 to 10^6 times the median while at 3^47 the cell 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. A nonlattice Moran system has its complex dimensions off any arithmetic progression and its detrended count carries no sharp frequency, which is consistent with that reading and is not separated from it at these heights. A positive test has to read the spread of the complex dimensions rather than a comb, which wants a zeta function for the joint object, which wants a gap structure, which is what this page denies.

Object S: the base-3 design plus the base-4 design

Object S is the sumset S = A + B of two dim 1 designs, A the base-3 design {0, 1} and B the base-4 design {0, 1}, both proper and so both base-locked by the dim 1 section above. D(x) = card(S meet [1, x])/x is its density below x, A_k = A meet [0, 3^k) and B_m = B meet [0, 4^m) are its two levels, and d(k, m) = (3^k - 1)/2 + (4^m - 1)/3 is the largest element of A_k + B_m. lab/rs/sumset-density is the generator behind every number of this section; it holds S as one bit per integer up to 3^22 and folds each power of 4 in by one shift-or pass, which is the bitset the Cobham section above says a two-base object needs.
The three plus four demo runs the same bit array to 3^16 in the browser through mrlyrs::num::sumset: S below x as a zoomable strip, D(x) dipping at every gap and recovering between them, and Q(k, m) along the pairs.
The paper. The bridge, the subpolynomial and unbounded energy ratio, the four moves, the average 190, the moving target and the census are written up for an outside reader as A Moving Target for Erdos Problem 125.
What is in print, read at source. Erdos problem 125 asks whether A + B has positive lower density, after Burr, Erdos, Graham and Li, who ask for positive density and positive upper density; the page records the answer as no, with a proof checked in Lean: for every eps > 0 there are arbitrarily large x with card(S meet [1, x]) < eps x. Hasler and Melfi 2024 prove card(S meet [1, x]) >> x^0.97777, read in their abstract; the problem page also quotes their bound 1015/1458 on the lower density, which is not read here. The formal statement, read at source, carries the upper density as its open variant, answer(sorry), and the discussion thread on the problem page calls that the likely harder half. The complement of S is A367090, read at source.
Proved (the gap). For every k, m, S misses the open interval (d(k, m), min(3^k, 4^m)): a sum with a < 3^k and b < 4^m is at most d(k, m), and a sum with a >= 3^k or b >= 4^m is at least min(3^k, 4^m). The thread's reading of the Lean proof is this gap at a scale L with 3^k and 4^m both within a factor 1 + eps of L, where S meet [0, L) sits inside [0, (5/6 + eps') L], followed by the injection x = a + b -> (y, x - L y) with y = floor(a/3^k) + floor(b/4^m), which sends S meet [0, L N) into (S meet [0, N)) x [0, (5/6 + delta) L) once eps <= delta/(N + 5/6), so D(L N) <= (5/6 + delta) D(N), iterated along the approximations 3^k ~ 4^m (the discussion thread on the problem page, read at source; the page credits the argument to its finders and not to the thread). That is the pigeonhole on carries, and it bites only where abs(m log 4 - k log 3) is below delta/N, which no k <= 22 supplies.
Proved (the clean centres). Call (k, m) clean when 3^k > d(k, m) and 4^m > d(k, m), which is 3^(k+1) + 5 > 2 4^m together with 4^(m+1) + 5 > 3^(k+1), the window 3/4 < 4^m/3^k < 3/2 up to those two 5s. Then S meet [0, d] = A_k + B_m exactly, and x -> d - x maps S meet [0, d] onto itself, because a -> (3^k - 1)/2 - a fixes A_k and b -> (4^m - 1)/3 - b fixes B_m, each by complementing every digit. The window has length log_4 2 = 1/2 in m - k log_4 3, which is equidistributed modulo 1 since log_4 3 is irrational, so clean pairs exist for a set of k of density 1/2. The reflection is the proposition recorded at A367090 on the narrower window 1 < 4^m/3^k <= 4/3.
Verified (lab/rs/sumset-density, verb density 22, 13 s on 4 GB; 3^23 wants a 12 GB bit array): card(S meet [1, 3^22]) = 26666749554 and D(3^22) = 0.849772. D(3^k) at k = 4..22 reads 0.975308, 0.835390, 0.858710, 0.887517, 0.908855, 0.864959, 0.778472, 0.837186, 0.858264, 0.874244, 0.814704, 0.763392, 0.831183, 0.858962, 0.881342, 0.792352, 0.767893, 0.831191, 0.849772, and D(4^m) at m = 3..17 reads 0.968750, 0.843750, 0.860351, 0.897460, 0.859313, 0.791305, 0.837238, 0.868845, 0.806823, 0.783585, 0.838184, 0.875988, 0.785523, 0.793552, 0.845272. The deepest readings sit where 4^m/3^k is nearest 1, at 3^k when the ratio is above 1 and at 4^m when it is below: 0.778472 at 3^10 with 4^8/3^10 = 1.109858, 0.763392 at 3^15 with 4^12/3^15 = 1.169234, 0.767893 at 3^20 with 4^16/3^20 = 1.231785, and 0.785523 at 4^15 with 4^15/3^19 = 0.923839, the ratios rounded up: those are the gap of the lemma above showing at each near coincidence of the two bases, and nothing deeper. The verb control rebuilds S by a double loop over A x B to 3^13 and in the other shift order to 3^17, matches the first 58 terms of A367090, and finds the reflection x -> d - x fixing S meet [0, d] at every clean centre below 3^17 and at no mixed centre with d >= 449.
Verified (same verb): over the windows [3^k, 3^(k+1)) at k = 5..21 the maximum of D reads 0.913419, 0.903768, 0.912038, 0.931596, 0.913781, 0.875566, 0.875469, 0.881621, 0.908274, 0.885045, 0.865671, 0.882855, 0.886340, 0.910650, 0.874408, 0.865858, 0.872186 and the minimum 0.835390, 0.852729, 0.887517, 0.858945, 0.778468, 0.778472, 0.822506, 0.858264, 0.806430, 0.763391, 0.763392, 0.815887, 0.858962, 0.785230, 0.767893, 0.767875, 0.818358; the least of the seventeen maxima is 0.865671 at k = 15, the least of the dyadic maxima over [2^j, 2^(j+1)) at j = 8..33 is 0.841760, the minimum of D over all of [1, 3^22] is 0.763391 at x = 3^15 - 1, and the maximum is 1 at every x <= 61. The falsification this section set itself, a running maximum over the windows decaying like a power, does not happen below 3^22: the window maxima at k = 5 and k = 21 read 0.913419 and 0.872186 with 0.910650 at k = 18 between them. The proved zero lower density is likewise invisible at this height, the minimum staying above 0.76.
Proved (the energy reduction). Let r(x) = card{(a, b) in A_k x B_m : a + b = x} and E(k, m) = sum_x r(x)^2. Then card(A_k + B_m) >= 4^(k+m)/E(k, m) by Cauchy-Schwarz on sum_x r(x) = 2^(k+m), and E(k, m) = sum_t R_A(t) R_B(t) with R_A(t) = card{(a, a') in A_k^2 : a - a' = t} = 2^(z_3(t)), z_3(t) the number of zero digits in the k-digit balanced ternary expansion of t, and R_A(t) = 0 when abs(t) > (3^k - 1)/2; and R_B(t) = 2^(z_4(t)) when t has a base-4 expansion on m digits in {-1, 0, 1}, z_4(t) its zero digits, and R_B(t) = 0 otherwise: a difference of digit strings from {0, 1} is a digit string from {-1, 0, 1}, which determines t uniquely in either base, and each zero difference arises twice. Write Q(k, m) = E(k, m) (d + 1)/4^(k+m), the energy against its flat value, which is at least 1. Then D(d) >= (4^(k+m)/E(k, m) - 1)/d, so the upper density of S is at least limsup 1/Q(k, m) along any infinite family of pairs, and it is positive as soon as Q is bounded on one. This is the sumset sharpening of the two-base transversality of the budget section: Q bounded says the law of a - a' at base 3 and the law of b - b' at base 4 collide no more often than two uniform laws on the same range.
Verified (lab/rs/sumset-density, verb energy 22, 39 s, the energy sum checked against the histogram of r at three small pairs by the crate's tests): over the 27 pairs with 6 <= k <= 22, 4^m within a factor 3 of 3^k and d <= 3^22, which drops (22, 18), Q(k, m) reads 1.467705, 1.638125, 1.664808, 1.724517, 1.676493, 1.646953, 1.642069, 1.761631, 1.893299, 1.941866, 1.948444, 1.965842, 1.892063, 1.878861, 1.855850, 1.940363, 2.060586, 1.987226, 1.974740, 1.860206, 1.835848, 1.803557, 1.779064, 1.895327, 1.988371, 1.959133, 1.939305, each rounded up, all inside [1.46, 2.07], the largest at (16, 12), a gap copy of (15, 12) as the ladder below shows. The Cauchy-Schwarz bound 1/Q alone puts D(d) at or above 0.485298 on every one of the 27, while the fill card(S meet [0, d])/(d + 1) reads between 0.834213 and 0.928391. Fitted as 3^(eta k) on its two endpoints, Q grows at eta = 0.015852 from k = 6 to k = 22 and at eta = 0.001987 from k = 11 to k = 22, both rounded up; both fits start on gap copies, (6, 4) and (11, 8), and the least-squares fits over distinct pairs in the ladder below replace them.
What the continuum says, read at source. Write mu and nu for the laws of sum_(l >= 1) X_l 3^(-l) and sum_(l >= 1) Y_l 4^(-l), the X_l, Y_l independent and uniform on {0, 1}, C_3 and C_4 for their supports, and rho_tau for the law of a + tau b with a and b independent from mu and nu; tau is a scaling, not the difference variable of the energy reduction. Shmerkin 2019, Theorem 1.11, proves that for a pleasant model with exponential separation, whose finitely supported driving measures depend continuously on the point outside a null set and have a bounded number of atoms, the L^q dimension of every measure of the model exists, the limit uniform over the model, and equals an explicit min(D_q, 1); his Lemma 7.1 and the proof of his Theorem 7.2 make the convolutions of two homogeneous self-similar measures, with an irrational ratio of the logarithms of the contractions and a separation hypothesis on each, such a model over a circle that covers one full period of the scaling, the driving measure there having at most four atoms and one discontinuity, and for this pair D_2 = log_3 2 + 1/2 > 1. Nazarov, Peres and Shmerkin 2012, Theorem 1.1, had the correlation dimension min(d_a + d_b, 1), d_a = log 2/log(1/a), for the natural measures of the central Cantor sets of ratios a and b convolved at every nonzero scaling, log b/log a irrational; their Theorem 4.1 makes such a convolution singular on a dense G_delta of scalings whenever 1/a and 1/b are Pisot, names a = 1/4, b = 1/3, this pair, as the example, and its proof finds the Fourier transform away from 0 at each resonance abs(lambda 4^n - 3^m) < 1/4 of their scaling lambda in their symmetric coordinates. Glasscock, Moreira and Richter 2024, Theorem C, prove from Shmerkin's uniformity that A + B has mass dimension min(1, dim A + dim B), dim the limit of log card(A meet [0, N))/log N, for every x3-invariant set of integers A and x4-invariant B, which for this pair is card(S meet [1, x]) = x^(1 - o(1)), past Hasler and Melfi's x^0.97777; their question on positive density for sumsets of full dimension asks for positive upper density in that generality, and Object S is its restricted-digit case with the least pair of bases. The problem page and its thread cite none of the three.
Proved (the continuum bridge). Let tau = tau_(k,m) = 4^m/3^k and c_x = [x 3^(-k), (x + 1) 3^(-k)). Then rho_tau(c_x) = r(x)/2^(k+m) for every integer x, so Q(k, m) = (d + 1) sum_x rho_tau(c_x)^2, and card(A_k + B_m) is exactly the number of cells c_x that meet C_3 + tau C_4. Proof: a = 3^(-k) (a_0 + u) with a_0 uniform on A_k and u an independent mu-variable, b likewise with 4^(-m) and B_m, and tau 4^(-m) = 3^(-k), so a + tau b = 3^(-k) (x + w) with x = a_0 + b_0 distributed as r/2^(k+m) and w in [0, 5/6], which keeps each atom inside its own cell. The fill is thus the box count of the continuum sumset at the scale tied to tau, and the energy its L^2 sum there. Since C_3 + tau C_4 is the union of 3^(-k) (x + C_3 + C_4) over x in A_k + B_m, its Lebesgue measure leb(C_3 + tau C_4) is at most (5/6) 3^(-k) card(A_k + B_m); at an x with card(S meet [1, x]) < eps x, the k with 3^k <= 6x/11 < 3^(k+1) and the m with 1 <= 4^m/3^k < 4 have d <= x, so the lower density 0 makes the infimum of leb(C_3 + tau C_4) over tau in [1, 4) equal to 0.
Proved (the energy ratio is subpolynomial). For every eps > 0 there is k_0 with Q(k, m) <= 3^(eps k) for all k >= k_0 and all m with 1/3 <= 4^m/3^k < 4, a window holding every pair of the census and every clean pair. Proof: for tau in [1, 4) the image of rho_tau under y -> 3y is the measure of Shmerkin's Lemma 7.1 at the point theta of his circle with e^theta = 3 tau/4 or 3 tau; the separation hypothesis of his Theorem 7.2 holds with R = 2, a nonzero polynomial of degree n with coefficients in {-1, 0, 1} being at least 3^(-n) at 1/3 and 4^(-n) >= 3^(-2n) at 1/4; and his uniform limit at q = 2 bounds the sum of squares of that image over the intervals of length 2^(-n) by 2^(-n (1 - eps)) for all large n. With 2^(-n)/3 in [3^(-k), 2 3^(-k)), each cell meeting at most two of the shrunk intervals and each of those at most three cells, sum_x rho_tau(c_x)^2 <= 6 (6 3^(-k))^(1 - eps), and d + 1 <= 2 3^k gives Q <= 72 3^(eps k). For tau in [1/3, 1), a = (a_1 + u)/3 with a_1 uniform on {0, 1} and u a fresh mu-variable, so rho_tau is the average of rho' and its shift by 1/3, rho' the image of rho_(3 tau) under y -> y/3; an average of two shifts by a whole number of cells never raises the sum of squares, so E(k, m) <= 4 E(k - 1, m) and Q(k, m) <= 3 Q(k - 1, m) with 4^m/3^(k-1) in [1, 3). Hence card(A_k + B_m) >= (d + 1) 3^(-eps k) and card(S meet [1, x]) >= x^(1 - eps) for large x, which is Glasscock, Moreira and Richter's Theorem C at this pair, here with the energy in place of the count.
Proved (the energy ratio is unbounded). There is an infinite family of clean pairs along which Q(k, m) -> infinity, so limsup Q = infinity while Q <= 3^(eps k), and no bound on Q holds on the clean pairs. Proof: in the symmetric coordinates of Nazarov, Peres and Shmerkin their Cantor measures of ratios 1/3 and 1/4 are affine images of mu and nu, their convolution at scaling lambda is an affine image of rho_tau with tau = 4/(3 lambda), and their resonance abs(lambda 4^n - 3^m) < 1/4 reads abs(tau 3^k - 4^m') < 3 tau/4 at (k, m') = (m + 1, n + 1); the proof of their Theorem 4.1 finds the Fourier transform away from 0 at every resonance, so for a tau with infinitely many resonances the Fourier transform of rho_tau does not tend to 0, the Riemann-Lebesgue lemma rules out a density, and the Jessen-Wintner law of pure types, which applies because rho_tau is an infinite convolution of discrete measures, makes rho_tau singular; these tau form a dense G_delta, and only the absence of an L^2 density is used below. Fix such a tau in (1, 5/4). A measure rho with 3^k sum_x rho(c_x)^2 bounded along a sequence of k has an L^2 density, the weak limit of its cell averages, so 3^k sum_x rho_tau(c_x)^2 -> infinity. At a resonant (k, m), tau_(k,m) is within 3^(-k) of tau, so replacing a + tau b by a + tau_(k,m) b moves each point by at most 3^(-k)/3, all in one direction, so each unit of mass lands in its own cell or the next one, sum_x rho_tau(c_x)^2 <= 2 sum_x rho_(tau_(k,m))(c_x)^2, and Q(k, m) >= 3^k sum_x rho_tau(c_x)^2/4 -> infinity; tau_(k,m) -> tau inside (1, 5/4) makes these pairs clean from some k on. The same comparison bounds card(A_k + B_m) below by half of the cells that meet C_3 + tau C_4, each cell of a + tau_(k,m) b receiving the points of at most two cells of a + tau b, so a tau in [1, 4) with infinitely many resonances and leb(C_3 + tau C_4) > 0 would give S positive upper density, and at every such tau the measure rho_tau is singular.
Verified (lab/rs/sumset-density, verb ladder 29 with its energy cache, the energies computed once in 139 to 178 s on 8 threads, the chunked energy checked against the digit-string energy at every k <= 11, m <= 14 and against the representation histogram at (9, 7) by the crate's tests): over the 43 pairs with 6 <= k <= 29 and 1/3 < 4^m/3^k < 4, Q repeats the 27 readings above, reads 1.861153 at (22, 18), 1.833722, 1.822641, 1.781904, 1.925912, 1.995549, 2.004783, 1.969845, 1.911416, 1.817220, 1.817784 at (23, 18), (23, 19), (24, 19), (25, 20), (26, 20), (26, 21), (27, 21), (27, 22), (28, 22), (29, 23), and 1.700067, 1.954950, 1.906659, 1.942757, 1.832435 at the chain levels with 3 <= 4^m/3^k < 4, (9, 8), (14, 12), (19, 16), (24, 20), (28, 23), each rounded up. Twelve of the 43 are gap copies. When 2 4^m < 3^k + 5, A_k + B_m is two disjoint translates of A_(k-1) + B_m, so E(k, m) = 2 E(k - 1, m) exactly; when 3^(k+1) < 4^m + 5, it is two disjoint translates of A_k + B_(m-1), so E(k, m) = 2 E(k, m - 1); a crate test checks six such pairs. Either way Q is the smaller pair's value inflated by the gap. The copies are (6, 4), (7, 5), (11, 8), (12, 9), (16, 12), (21, 16), (26, 20) and the five chain levels just listed, and the reading 2.060586 at (16, 12) is one of them. Over the 31 other pairs Q lies in [1.638124, 2.004783], the lower end truncated and the upper rounded up, least at (6, 5) and largest at (26, 21), so no Q passes 2.07 to k = 29. Least squares of log_3 Q on k give eta = 0.003915 with standard error 0.001243 over the 31, and eta = -0.001109 with 0.001254 over the 25 with k >= 11: past k = 11 the slope is zero within one standard error. Fitted and scored in units of Q on those 25, the power model has eta = -0.001093 and rms 0.065794, and the model a + b 3^(-s k), s = log_3 2 - 1/2 = 0.130930, has rms 0.065127 with a = 1.869317 and b = 0.282604, a positive b that describes a slow decline rather than a rise to a limit; over all 31 the same model has b = -0.606850, so the sign depends on the window and the census separates neither model. The scatter follows tau: lowest near tau = 1 (1.779064, 1.781904, 1.817784 at tau = 0.923839, 0.973262, 1.025330), highest near tau = 1.7 (1.987226, 1.959133, 2.004783 at tau = 1.558978, 1.642380, 1.730244). The unbounded family does not show at this height: its bound Q >= 3^k sum_x rho_tau(c_x)^2/4 passes 2.07 only once 3^k sum_x rho_tau(c_x)^2 passes 8.28.
Proved (the average over the scaling). Write h_k(tau) = 3^k sum_x rho_tau(c_x)^2, so that Q(k, m) = (d + 1) 3^(-k) h_k(tau) at tau = 4^m/3^k. Then int_1^4 h_k(tau) dtau <= 190 for every k >= 0, and the set G_k(M) of the tau in [1, 4] with h_k(tau) > M has measure at most 190/M. Proof, the energy argument behind the potential-theoretic proof of Marstrand's projection theorem, which Nazarov, Peres and Shmerkin cite for the almost-every form of their Theorem 1.1, written out for this pair: with a, a' from mu and b, b' from nu, all independent, and ell = 3^(-k), sum_x rho_tau(c_x)^2 <= P(abs(a - a' + tau (b - b')) < ell), and for fixed b != b' the tau in [1, 4] that meet the event form an interval of length at most min(3, 2 ell/abs(b - b')), empty unless abs(a - a') < 4 abs(b - b') + ell. A first nonzero digit of a - a' at place j keeps abs(a - a') >= 3^(-j)/2, and one of b - b' keeps abs(b - b') >= (2/3) 4^(-j), so P(abs(a - a') < s) <= (6s)^alpha and P(abs(b - b') < s) <= (6s)^(1/2) with alpha = log_3 2. Splitting at abs(b - b') = ell bounds the integral of sum_x rho_tau(c_x)^2 by 2 30^alpha ell E(abs(b - b')^(alpha - 1)) + 3 30^alpha 6^(1/2) ell^(alpha + 1/2), where E(abs(b - b')^(alpha - 1)) <= 6^(1/(2 gamma)) gamma/(gamma - 1) <= 7.398167, gamma = 1/(2 - 2 alpha), finite exactly because alpha + 1/2 > 1; with 30^alpha <= 8.549875 the total is at most 189.34 ell, these three read from lab/rs/sumset-density, verb constants.
Proved (the four moves). For every k and tau > 0: (i) h_k(tau') <= 2 h_k(tau) whenever abs(tau' - tau) <= 3^(1-k), since b <= 1/3 moves every point a + tau b by at most one cell and all in one direction, so the mass of each cell splits into a part that stays and a part that moves on, and (s_x + v_(x-1))^2 <= 2 (s_x^2 + v_(x-1)^2) with s_x^2 + v_x^2 <= rho_tau(c_x)^2; the factor 2 is sharp, h_0(1) = 1 and h_0(4) = 1/2; (ii) (3/2) h_k(3 tau) <= h_(k+1)(tau) <= 3 h_k(3 tau), since rho_tau at level k + 1 is the average of rho_(3 tau) read at level k and its shift by 3^k cells, and ((p + q)/2)^2 lies between (p^2 + q^2)/4 and (p^2 + q^2)/2; (iii) (3/8) h_k(tau) <= h_k(4 tau) <= (3/2) h_k(tau), since b = (b_1 + v)/4 makes rho_(4 tau) the average of rho_tau and its shift by tau, and by the split of (i) that shift moves the sum of squares by a factor in [1/2, 2]; (iv) h_(k+1)(tau) >= h_k(tau), since a cell of level k is three cells of level k + 1 and (p_1 + p_2 + p_3)^2 <= 3 (p_1^2 + p_2^2 + p_3^2). Along the chain tau_k = 4^(m_k)/3^k in [1, 4), 3 tau_(k+1) is tau_k or 4 tau_k, so (ii) and (iii) keep h_(k+1)(tau_(k+1))/h_k(tau_k) inside [9/16, 9/2]: one step changes the orbit value by a bounded factor.
Proved (the moving target). On the chain, with d_k = d(k, m_k), d_k + 1 = 3^k (1/2 + tau_k/3) + 1/6, so Q(k, m_k) <= (11/6 + 3^(-k)/6) h_k(tau_k) and the upper density of S is at least 6/(11 liminf_k h_k(tau_k)), positive as soon as that liminf is finite; no mean over k is needed. Contrapositively, zero upper density forces, for every M, tau_k into G_k(M) for all large k, while G_k(M) has measure at most 190/M and, by move (i), contains the part in [1, 4] of the 3^(1-k)-neighbourhood of G_k(2M): a target of measure at most 190/M that the rotation log_4 tau_(k+1) = log_4 tau_k - log_4 3 mod 1 would have to meet at every step from some k on. The unbounded family lies on the chain, its tau_(k,m) in (1, 5/4) making m = m_k, and there h_k(tau_k) >= (6/11) Q(k, m_k) (1 - o(1)) -> infinity, so limsup_k h_k(tau_k) = infinity and the target is met infinitely often; nothing here decides whether it is met always.
Proved (almost every phase). For psi in [1, 4) let tau_k(psi) be psi tau_k brought into [1, 4) by a power of 4, the chain rotated by log_4 psi, so tau_k(1) = tau_k. The map psi -> tau_k(psi) is a piecewise dilation onto [1, 4) with dpsi/dtau = psi/tau <= 4, so int_1^4 h_k(tau_k(psi)) dpsi <= 760 for every k, and Fatou gives int_1^4 liminf_k h_k(tau_k(psi)) dpsi <= 760: for almost every phase the moving target is missed infinitely often, and the phases with liminf_k h_k(tau_k(psi)) > M have measure at most 760/M. At a fixed tau, move (iv) makes h_k(tau) increase to a limit h(tau) in (0, infinity], monotone convergence gives int_1^4 h(tau) dtau <= 190, and Cauchy-Schwarz over the N_k cells that meet C_3 + tau C_4 gives 1 <= N_k sum_x rho_tau(c_x)^2, so N_k 3^(-k) >= 1/h_k(tau); these unions of cells decrease to C_3 + tau C_4, so leb(C_3 + tau C_4) >= 1/h(tau) at every tau, with 1/infinity = 0, and it is positive for almost every tau: Marstrand's theorem for this pair. At the lattice points the bridge is exact: the pieces 3^(-k) (x + C_3 + C_4) lie in distinct cells, so leb(C_3 + tau_(k,m) C_4) = 3^(-k) card(A_k + B_m) leb(C_3 + C_4), and whether leb(C_3 + C_4) > 0 is the case r = 3, s = 4, X = C_3, Y = C_4 of a question of Hochman in the paraphrase of Glasscock, Moreira and Richter, on the Lebesgue measure of X + Y for xr- and xs-invariant sets of dimension sum above 1 (the original unread here). The phase psi = 1 of Object S is a single point, which no almost-every statement reaches.
Verified (lab/rs/sumset-density, verb phases 17 4096, 47 s, both kernels checked equal to E(k, m) at the lattice by the crate's tests): for each k = 8..17 on the chain the energy of the atoms a + sigma b, a in A_k and b in B_m, is read at 4096 scalings tau = sigma tau_k spread evenly in log tau over [1, 4), once with a nearest-integer window and once with the tent (1 - abs(Delta))_+, Delta the difference of two atoms; both equal E(k, m) at sigma = 1 and are a proxy for h_k elsewhere, not h_k itself, so every rank and ratio below is the proxy's. The integral of the window reading over [1, 4] is 3.937434, 4.318193, 4.572550 at k = 8, 12, 17, far under the 190 proved for h_k, and its maximum over the period sits at the grid point 1.053504, next to the lattice point 4^4/3^5, at k = 8 and every k >= 11, and at the grid point 1.404673, next to 4^5/3^6, at k = 9, 10: the large values of h_k sit at the low lattice points, where the two bases coincide early. The orbit point sits slightly above its own neighbourhood: among the grid points within 1/50 of tau_k in log_4 tau, the share below h_k(tau_k) averages 0.8110 over the ten levels under both kernels, between 0.7055 and 0.9202 under the window, and h_k(tau_k) exceeds their mean by a factor between 1.0090 and 1.0525 under the window, the largest at k = 12 and 1.0180 at k = 17, and between 1.0063 and 1.0512 under the tent. The excess is a few percent and does not grow over this range.
Conjecture. liminf_k h_k(tau_k) < infinity on the chain, so the target G_k(M) is missed infinitely often for some M and the upper density of S is positive, at least 6/(11 liminf_k h_k(tau_k)); the stronger form, a bounded mean of Q(k, m_k) over k <= K, is what the chain readings suggest: over its 24 levels k = 6..29, gap copies included, Q(k, m_k) lies in [1.638124, 2.004783] with mean 1.861840. The unbounded family does not touch it: two resonances (k_1, m_1), (k_2, m_2) of one tau make abs(4^(m_2 - m_1)/3^(k_2 - k_1) - 1) smaller than about 3^(-k_1), while abs(4^B - 3^A) >= 1 keeps it at least 3^(-A), so k_2 >= 2 k_1 and each tau resonates at O(log K) of the k <= K. Two obstructions stand: Q is the correlation of 2^(z_3) with 2^(z_4) and no automaton reads both, and no bound on Q holds over all the pairs, so a proof has to use the one phase psi = 1 of the rotation, which the measure bounds above cannot single out. The falsifier is a local excess of h_k(tau_k) over its neighbourhood that keeps growing with k; the next rung of the ladder, k = 30, wants about 6 minutes and a wider integer than the crate's u128 ratio.

Object E: one number at every odd side

At odd side N >= 3, E_N is the dim 1 design keeping the even digits: the reals of [0, 1] with a base-N expansion whose digits all lie in {0, 2, ..., N-1}, the attractor of the (N+1)/2 maps x -> (x + d)/N with d even, whose pieces [d/N, (d+1)/N] are disjoint, so a point of E_N has one address. In the odd-side reading of method it is the dim 1 base-2 design {0} at side N, of fill (N+1)/2.
Its integers Z_N, those whose base-N digits are all even, are 2 K_N, with K_N the integers whose base-N digits are all at most (N-1)/2: doubling such a k carries nowhere, and at odd side an integer is congruent to its digit sum mod 2. At a prime p, K_p = {k : p does not divide binomial(2k, k)} by Kummer's theorem, the carries of k + k in base p counting v_p(binomial(2k, k)).
A side N holds x when x lies in E_N, or in Z_N for an integer. The share of x is lim (1/M) card{1 <= n <= M : side 2n + 1 holds x} when the limit exists, and its share to level L counts instead the sides at which the first L digits of x are even. Object E asks both ways round: which sides hold one number, and what all the sides hold together. lab/py/sides-holding-a-number is the generator behind every number of this section.
Proved (the membership law). For x in [0, 1], x in E_N iff N^j x mod 2 lies in the closed arc [0, 1] of R/2Z for every j >= 0. If x = sum_(i >= 1) d_i N^(-i) with even digits, the tail x_j = sum_(i > j) d_i N^(j-i) lies in [0, 1] and differs from N^j x by the even integer sum_(i <= j) d_i N^(j-i); conversely, with x_j in [0, 1] congruent to N^j x mod 2, d_(j+1) = N x_j - x_(j+1) is an even integer in [-1, N], so a digit in {0, 2, ..., N-1}, and these digits expand x. At x = p/q in lowest terms it reads: side N holds p/q iff N^j p mod 2q lies in {0, 1, ..., q} for every j >= 0. The endpoint q is reached only when q divides N^j p, which then leaves q or 0 by the parity of p (2/3 at side 3 has orbit 2, 0, 0), so never when gcd(N, q) = 1 < q, where the rule is the half-open [0, q); the closed rule is the exact one, since 1/3 = 0.0222..._3 lies in E_3 with orbit 1, 3, 3, ... mod 6.
Proved (the period). The sides holding p/q are the odd N with N mod 2q in a set R(p/q) of odd residues, so they are periodic mod 2q from N = 1 on, with no transient, and for 0 < p < q the least period is 2q. A period 2t with t a proper divisor of q would put every N = 1 mod 2t among them, N = 1 holding every number; but as N runs over 1 mod 2t, N p mod 2q runs over every residue = p mod 2t, p being prime to q/t, and one of those lies in (q, 2q): p + q when q = 2t, and when q >= 3t the q - 1 >= 2t consecutive integers of (q, 2q) meet every class mod 2t. That side rejects p/q at its first digit. The digit map d -> N - 1 - d keeps parity, so E_N is symmetric under x -> 1 - x and R(p/q) = R((q-p)/q); 0 and 1 lie in every E_N.
Proved (the share, and the maximum 2/3). The share of p/q is card R(p/q)/q, and for 0 < p < q in lowest terms it is exactly 1/2 at q = 2, and from q = 3 on at least 2/q, at most (q+1)/(2q) at odd q and at most 1/2 at even q. So at every q the largest share of a number other than 0 and 1 is 2/3, at 1/3 and 2/3 alone, the next bound being 3/5; lowest terms matter, 2/6 being 1/3. Lower: N = 1 mod 2q fixes the orbit at p; at odd q, N = q mod 2q sends p to q or 0, both fixed; at even q >= 4, N = q - 1 mod 2q gives (q - 1)^j p = (-1)^j p + j q mod 2q, which is p or q - p. Upper: if r and -r both hold p/q, then r p mod 2q and -r p mod 2q both lie in [0, q], which forces r p = 0 or q mod 2q, so q divides r; the odd residues with q | r are r = q at odd q and none at even q, and every other odd residue pairs with its negative, at most one of each pair holding. At q = 3 both bounds meet: R(1/3) = {1, 3} mod 6.
Proved (the exact rule). At odd q the modulus drops to q: side N holds p/q iff every least residue N^j p mod q is 0 or has the parity of p. The orbit mod 2q keeps the parity of p, N being odd, and of the two lifts s and s + q of a least residue s the one of that parity is s exactly when s has it, while s = 0 lifts to 0 or q, both allowed. For N prime to q the orbit is the coset p <N> in (Z/q)^*, so the odd residues mod 2q prime to q that hold p/q number sum phi(card G) over the cyclic subgroups G whose coset p G keeps the parity of p, and such a G misses -1, -p having the other parity. At prime q, with m the odd part of q - 1 and G_d the subgroup of order d, share(p/q) = (1 + sum_(d | m) phi(d) [p G_d keeps the parity of p]) / q, so 2/q <= share(p/q) <= (1 + m)/q, the share is exactly 2/q at a Fermat prime, and share(1/13) = 4/13 from G_3 = {1, 3, 9}, 1/13 being 0.(002) in base 3. At even q, (q - r)^j p = (-1)^j r^j p + j q mod 2q for odd r, so R(p/q) is invariant under r -> q - r.
Verified (lab/py/sides-holding-a-number, verb period 200, 19.6 s). The orbit rule agrees with a walk on the digits of p/q at every p/q in [0, 1] in lowest terms with q <= 200, over the odd sides 3 <= N <= 6q + 1 at q <= 40 and over one full period 3 <= N <= 2q + 1 above, 1661996 checks with no failure. On the orbit rule, at all 12231 fractions p/q in (0, 1) with q <= 200, the symmetry, the least period 2q, both bounds, the rule mod q at odd q, the coset count, the prime formula and the invariance at even q hold with no failure. The largest shares are 2/3 at 1/3 and 2/3, then 1/2, reached only at q = 2, 4, 10, 12; the largest at odd q >= 5 is 2/5, at 1/5 and 4/5; the bound 2/q is attained at 4057 fractions; the odd primes q <= 200 with share(1/q) > 2/q are 13, 19, 31, 61, 67, 79, 97, 109, 127.
Proved (the Eisenstein link). At an odd prime q and 1 <= p <= q - 1, with a whichever of p and q - p is odd, (-1)^card{odd N : 3 <= N < q, the first base-N digit of p/q is odd} = (a/q), the Legendre symbol. Below q the first digit floor(p N/q) never sits on a boundary, so the count is the number of odd sides below q that reject p/q at its first digit. Proof: Gauss's lemma, Theorem 2.7 of Wright, read at source, gives (b/q) = (-1)^s(b) for b prime to q, with s(b) the number of 1 <= u <= (q-1)/2 whose least residue u b mod q exceeds q/2. Since floor(2 u b/q) = 2 floor(u b/q) + [u b mod q > q/2], the even-multiplier sum E(b) = sum_(u=1)^((q-1)/2) floor(2 u b/q) has the parity of s(b), which is Eisenstein's lemma. Pairing N with q - N gives sum_(N=1)^(q-1) floor(p N/q) = (p-1)(q-1)/2, so the sum over odd N < q is (p-1)(q-1)/2 - E(p), and the parity of the count of odd first digits is that of this sum, N = 1 adding 0. Hence the sign is (-1)^((p-1)(q-1)/2) (p/q), which is (p/q) at odd p and (-1/q)(p/q) = ((q-p)/q) at even p, with (-1/q) = (-1)^((q-1)/2) from the same lemma at b = -1.
Proved (the full period carries no symbol). At the same q and p, sum_(N odd, N < 2q) floor(p N/q) = (2p - 1)(q - 1)/2 + p. The odd N < q give (p-1)(q-1)/2 - E(p), N = q gives p, and N = q + M with M even in [2, q - 1] gives p + floor(p M/q), in all p (q-1)/2 + E(p), so E(p) cancels and the Legendre symbol lives on the half period below q alone. The identity is about floor(p N/q) and not about rejections: at N = q the expansion of p/q terminates and its first digit is p or p - 1.
Verified (verb eisenstein 200, 0.03 s). At all 4180 pairs (p, q) with q an odd prime <= 200 the half-period law holds with no failure, the Legendre symbol by Euler's criterion; the full-period identity holds at all 4180, and its parity agrees with the symbol at 2090, exactly half.
Proved (an irrational, level by level). For irrational x in (0, 1) and every L >= 1 the share of x to level L is exactly 2^(-L). The first L base-N digits of x are even iff N^j x/2 mod 1 lies in [0, 1/2) for j = 1, ..., L, no boundary being hit at an irrational; at N = 2n + 1, N^j x/2 is a polynomial in n of degree j with leading coefficient 2^(j-1) x, so every nonzero integer combination of the L polynomials has an irrational leading coefficient, and Weyl's polynomial theorem (Weyl 1916) with his criterion makes (N^j x/2 mod 1)_(j <= L) equidistributed in [0, 1)^L as n runs; the box has measure 2^(-L) and a null boundary. So the share of an irrational exists and is 0, every rational p/q has share at least min(2/q, 1/2), and a number of [0, 1] is rational iff a positive share of the odd sides holds it.
Verified (verb weyl 100000 6, 0.2 s). At the odd sides 3 <= N <= 200001 the share to level L times 2^L reads between 0.9821 and 1.0496 at L = 1..6 for sqrt(2) - 1, (sqrt(5) - 1)/2, 2^(1/3) - 1, pi - 3 and e - 2, each taken from a 90-digit truncation.
Proved (the integer count). Let c(k) = card{odd N : 3 <= N <= 2k, 2k in Z_N}. Then abs(c(k) - (1 - log 2) k) <= sqrt(2k) + 1 for every k >= 1. Write s = sqrt(2k). A side N with s < N <= 2k has N^2 > 2k, so k has at most two digits with the leading one floor(k/N) < N/2, and k in K_N iff k mod N <= (N-1)/2 iff j = floor(2k/N) is even, where j < s. The odd N with floor(2k/N) = j fill (2k/(j+1), 2k/j], k/(j(j+1)) of them up to an error below 1, and sum_(j even >= 2) 1/(j(j+1)) = sum_(i >= 1) (1/(2i) - 1/(2i+1)) = 1 - log 2. So c(k) - (1 - log 2) k = B - D + Theta - T: B, the sides 3 <= N <= s holding 2k, at most (s-1)/2; D, the odd N <= s with floor(2k/N) even and below s, at most 1, since they lie in (2k/(s+1), s], of length below 1; Theta, the rounding errors of the at most s/2 even j < s, below s/2 in size; and T = k sum_(j even >= s) 1/(j(j+1)) <= k/(2(s-1)) <= s/2 at s >= 2, by 1/(j(j+1)) <= (1/(j-1) - 1/(j+1))/2. Hence -s - 1 <= c(k) - (1 - log 2) k <= s - 1/2, and k = 1 holds by hand.
Verified (verb count 1000000, 2.1 s, against a direct digit test at every k <= 3000). c(k) at every k <= 10^6: abs(c(k) - (1 - log 2) k)/sqrt(k) is at most 0.547191, at k = 74, and the error is at most 0.357533 of the proved bound; on [10^5, 10^6] the ratio (c(k) - (1 - log 2) k)/sqrt(k) lies in [-0.348228, -0.082525] with mean -0.222283; c(10^6) = 306665 against (1 - log 2) 10^6 = 306852.819.
Conjecture (the second term). c(k) = (1 - log 2) k - kappa sqrt(k) + o(sqrt(k)) with kappa = (2 - sqrt 2) abs(zeta(1/2))/4 = 0.213864. The heuristic splits the error as the proof does. The large sides lose the tail T, about s/4 = sqrt(2k)/4, with Theta averaging out. On the small sides with three digits, along the run of N with leading digit a, the middle digit sweeps down from N to 0 as N runs over (sqrt(k/(a+1)), sqrt(k/a)] and is at most (N-1)/2 on the part N >= sqrt(k/(a + 1/2)), the last digit is at most (N-1)/2 half the time, and half the N are odd, so B is about beta sqrt(k) with beta = (1/4) sum_(a >= 1) (a^(-1/2) - (a + 1/2)^(-1/2)) = (sqrt 2 + (2 - sqrt 2) zeta(1/2))/4 = 0.139689, by zeta(s, 1/2) = (2^s - 1) zeta(s), and kappa = sqrt(2)/4 - beta; the four-digit sides and the cut a <= (N-1)/2 move it by O(k^(1/3)). What the heuristic lacks is the equidistribution of the last digit and of floor(2k/N) mod 2 along the runs, a sawtooth sum of divisor-problem type.
Verified (verb second 11, 7.6 s, a block count checked against the direct digit test at every k <= 3000). Over 60 random k in each [10^e, 2 10^e), the mean of (c(k) - (1 - log 2) k)/sqrt(k) reads -0.220703, -0.218290, -0.217521, -0.216456, -0.215858, -0.215000 at e = 6..11, rising toward -kappa = -0.213864, its spread [-0.260825, -0.169445] at e = 6 closing to [-0.217031, -0.212564] at e = 11; the small-side part B/sqrt(k) reads 0.130315 at e = 6 and 0.138410 at e = 11 against beta = 0.139689.
Proved (what every side holds). On the integers the odd sides together hold {0, 2}, and on the reals {0, 1}. An integer 2k with k >= 2 is 11 at side 2k - 1, an odd integer has an odd digit at every side, and 2 is one even digit at every side >= 3. A rational p/q in (0, 1) is rejected at side 2q - 1, where (2q - 1) p = 2q - p mod 2q lies in (q, 2q); an irrational x is rejected wherever N x mod 2 lies in (1, 2), which some odd N achieves since (2n+1) x mod 2 = x + 2 (n x mod 1) is dense; 0 and 1 are the expansions by the digits 0 and N - 1. The prime sides alone give the same integers: for k >= 2 Bertrand's postulate puts a prime p in (k, 2k), necessarily odd, and k is one base-p digit above (p-1)/2.
Proved (powers of a side). E_N lies inside E_(N^e) for every e >= 1, on the reals and on the integers: a block of e even base-N digits is the base-N^e digit sum_(i < e) d_i N^i, even and at most N^e - 1. At every e >= 2 the inclusion is strict on both: N + 1 = 11_N is the single even digit N + 1 <= N^e - 1 at side N^e, and (N + 1)/N^e is that digit in the first place at side N^e, while side N rejects it, N^(e-1) (N+1)/N^e = 1 + 1/N lying in (1, 2).
Proved (the union of all sides). U, the union of the E_N over the odd sides N >= 3, is Lebesgue null and meagre, has Hausdorff dimension 1 attained by no E_N, and holds every rational of [0, 1]. Each E_N is closed of dimension log((N+1)/2)/log N < 1, so null and nowhere dense; those dimensions tend to 1 and Hausdorff dimension is countably stable; and p/q lies in E_(2q+1), since 2q + 1 = 1 mod 2q fixes its orbit.
Proved (Fourier dimension 0). Fourier dimension is that of Ekstrom, Persson and Schmeling 2015, read at source: the supremum of s in [0, 1] with abs(hat mu(xi)) << abs(xi)^(-s/2) for some Borel probability measure mu giving full measure to the set; they prove it is not countably stable, so the union needs its own argument. Let abs(hat mu(xi)) <= C abs(xi)^(-eps) for abs(xi) >= 1 and some eps > 0; then mu(E_N) = 0 at every odd side, so mu(U) = 0 and dim_F U = 0. By Weierstrass approximation there is a real trigonometric polynomial g(y) = sum_(abs(h) <= H) c_h e(h y/2), nonnegative, at least 1 on the arc [0, 1] of R/2Z, with mean c_0 < 1; put A = sum_h abs(c_h) and take M = N^b with M >= 4, H <= (M-1)/2 and A M^(-eps) <= c_0/2, all three holding once b is large; M >= 4 keeps every frequency below at abs(xi) >= 1, where the decay applies. The membership law gives 1_(E_M)(x) <= prod_(j=1)^L g(M^j x); expanding, the term with every h_j = 0 is c_0^L, and a term whose last nonzero h_j sits at j = J has frequency abs(sum_j h_j M^j)/2 >= M^J/4, the h_j being below M/2 in size, so mu(E_M) <= c_0^L + C 4^eps sum_(J=1)^L A^J M^(-eps J) c_0^(L-J) <= (1 + C 4^eps) c_0^L, which tends to 0 with L, and E_N lies in E_M.
Proved (composite sides). At a prime power the carry out of base-p^a digit i is the carry out of base-p position a(i+1) - 1, so K_(p^a) = {k : no carry of k + k in base p leaves a position = a - 1 mod a}, which contains K_p and is no divisibility condition: 20 = 202_3 lies in K_9 with 9 | binomial(40, 20), its carries leaving positions 0 and 2, while 6 = 20_3 lies outside K_9 with v_3(binomial(12, 6)) = 1, its one carry leaving position 1. At a side with two primes Kummer says nothing, and K_15 and K_3 cap K_5 are incomparable: 10 lies in K_3 cap K_5, binomial(20, 10) = 184756 being prime to 15, but is one digit 10 > 7 at side 15; 2 lies in K_15 outside K_3, 3 in K_15 outside K_5, and 15 = 10_15 in K_15 with 15 | binomial(30, 15). Side 15 imposes a constraint neither prime sees and drops the ones they impose.
Verified (verb family, 1.2 s). The integers below 3000 held by every odd side are 0, 2; no p/q in (0, 1) with q <= 60 is held at side 2q - 1; E_N lies in E_(N^e) on the integers k < 10^5 at N = 3, 5, 7, e = 2, 3, and on the fractions q <= 60 at e = 2, 3, 4; Kummer's reading of K_p holds at every odd prime p <= 23 and k < 1500, and the carry reading of K_(p^a) at (p, a) = (3, 2), (3, 3), (5, 2), (7, 2) and k < 10^5; and 20, 6, 10, 2, 3, 15 are the least witnesses of the six statements above.
Verified (the finite intersections) (verb inter 1e12, 1.2 s, the pruned walk of Object T, which enumerates K_3 from its top digit and cuts a branch when the next member of another K_N lies past the branch, checked against the direct digit test below k = 10^6 at four side sets). Below 10^12 the sides {3, 5} hold 10072 integers and {3, 5, 15} hold 50; {3, 5, 7} hold the 17 integers 0, 2, 20, 1512, 1514, 6320, 6372, 6374, 6500, 15120, 15122, 15302, 40014, 119096754, 119096802, 91547225622, 91550794374, exactly twice the terms of A030979 below 5 10^11, compared term by term against the published record when the verb is handed a copy of it; adding side 9 changes nothing, K_3 lying in K_9; adding 11 leaves 0, 2, 6320, adding 13 leaves 0, 2, 1512, 1514, 6500 and adding 15 leaves 0, 2, 1512, 1514, 15302; {3, 5, 7, 11, 13} and {3, 5, 7, 11, 15} hold 0, 2 alone. With verb deep 1000 (16365 nodes, 14.6 s) the sides {3, 5, 7, 11} hold 0, 2, 6320 and nothing else below 10^1000.
What this says about Erdos problem 376, and what it does not. By Kummer the integers held by sides 3, 5, 7 are 2 {k : binomial(2k, k) prime to 105}, twice the set of Erdos problem 376, which asks whether it is infinite, is recorded open with a prize of Graham, and is the wider set of Object T above. Two sides are not the question: the problem page, read at source, records Erdos, Graham, Ruzsa and Straus giving infinitely many n with binomial(2n, n) prime to p q for any two odd primes, the census of sides {3, 5} growing. Bloom and Croot 2025, read at source, prove that for distinct coprime bases g_1, ..., g_r sufficiently large in terms of r infinitely many n have all but eps log n of their base-g_i digits at most g_i/2, with threshold 10^94 at r = 3 and an ineffective theorem; it reaches neither the sides 3, 5, 7 nor every digit. Pomerance 2015, read at source, bounds the n <= x with p not dividing binomial(2n, n) by p x^(theta_p), theta_p = log((p+1)/2)/log p, the box count of K_p, expects at least x^0.02 members at 105, and at 3, 5, 7, 11 expects "at most finitely many numbers, such as n = 3160", the 6320 = 2 3160 of the census. The census above lists the members of that set below 10^12, and those also held at side 11 below 10^1000; it puts no bound on a last member, so nothing here bears on whether the set of problem 376 is infinite. The share and the family statements read one number against every side, and the problem asks about three sides against every number.
The rationals of one design, read at source. Nagy 2001 reads the middle-thirds set through classes credited to Wall: the numerators prime to L split into the classes {k, 3k, ..., 3^(ord - 1) k mod L}, and a class lies wholly inside the set or wholly outside it, which is the orbit rule at one base. Schleischitz 2021 proves that a missing-digit set holds finitely many rationals whose denominators are built from a finite set of primes prime to the base, and bounds its rationals of denominator at most N by J^2 diam^D N^(2D); Bloshchitsyn 2015 (doi, paywalled, read through Schleischitz's quotation) proves finiteness at a single prime above b^2; Shparlinski 2021, read in its abstract and talk slides, bounds below the largest prime factor of such a denominator. Each fixes the design and counts its rationals; one rational read across every odd side, its least period 2q, its share and the bound 2/3 are not in what was read, and this card names what was searched and claims nothing past it.

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 (beneath). That is what a finite census, a fill law and a Collatz-Wielandt certificate produce, and it is the only thing they produce.
Every two-base budget has the shape sum_i log(fill_i)/log(p_i) - (m-1) dim, a Q-linear combination of 1 and the ratios log k_i / log p_i, which is the shape Burrell and Yu's independence hypothesis below is stated in. The wall assumes a realized two-base exponent has that shape too, and nothing here proves it: Object Y reads 0.714298 at m = 24 against a budget of 0.584963 and its true exponent stays Conjecture.
Conjecture (the Schanuel wall). Those two families of numbers meet only where one side degenerates, and under Schanuel's conjecture they meet nowhere nontrivial, so no instrument of this tree outputs a two-base exponent, for a reason that is transcendence rather than difficulty. This is strictly stronger than Cobham: Cobham forbids the set from being automatic, and the wall forbids the number from being a Perron root.
The precedent is in print. Burrell and Yu state their Theorem 1.6 under "Assume Schanuel's conjecture", and their Theorem 1.11, "The triple 1, log 3/ log 5, log 3/ log n is Q-linearly independent for at least one n in {7, 11, 13}", is how far the unconditional route reaches; both read at source.
The two conjectures are not the same wall and neither implies the other: Cobham is a theorem about languages and holds unconditionally, while the wall is an arithmetic statement about a number that a two-base census would have to output, and it is open.