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, and closes on the transcendence wall that stands between every instrument of this tree and any two-base exponent.

Every claim carries a tag. Proved means a proof is given or restated here; Verified means recomputed from scratch by a lab study; Conjecture means neither; Refuted means shown false. lab/py/two-base-gasket, lab/rs/three-base-thin and lab/py/two-base-instrument are the three generators behind every number below, the last of them behind the periodogram 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.

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).
The integer side of the same statement at m = 2 is Glasscock, Moreira and Richter 2024, whose main results include "integer analogues of two of Furstenberg's transversality conjectures pertaining to the dimensions of the intersection A cap B and the sumset A+B of xr- and xs-invariant sets A and B when r and s are multiplicatively independent" (abstract read at source). A one-base design's integer set is xq-invariant in that sense at dim 1.
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.
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.

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.