Spectral Measures from Tiling Levels

MrlyProd

First published 2026-10-05

Take an integer base b and a set F of digits. The numbers written in base b with n digits from F form a finite set A_n, the level n; scaled by b^(-n) the levels converge to a Cantor-type set, and the uniform measures on them converge to the self-similar measure mu_{b,F}. The measure is spectral when some set of frequencies Lambda makes the exponentials e(lambda x) = exp(2 pi i lambda x) an orthonormal basis of L^2(mu_{b,F}), as the integers do for Lebesgue measure on [0, 1]. This paper compares that property of the limit with a property of the levels: whether each A_n tiles the integers by translation. Coven and Meyerowitz read tiling off two conditions on the cyclotomic factors of a mask polynomial, and an index lemma reduces those conditions at every level to finitely many checks on F. When every level tiles and |F| is a prime power or a product pq of two distinct primes, mu_{b,F} is spectral: the proofs build Hadamard triples, in the window of Łaba and Wang where F is a scaled complementing set mod b, and outside it through the product forms of An and Lai, cut from the p-adic tree of F and from a fibre decomposition of its residues. At base 12 the digit set {0, 1, 8, 9} tiles at every level, is no multiple of a complementing set mod 12, has no spectrum in 12^(-n) Z at any level, and its measure has the explicit spectrum {0, 3/4} + 3 Lambda(12, {0, 3, 6, 9}). In the other direction, a spectral measure whose levels are sets yields, for every L, |F|^L integers with orthogonal differences, which bounds |F| by a product over a covering set of primes; with 0 in F this proves that at five digits mu_{b,F} is spectral if and only if every level tiles, and that mu_{b,{0,1,3,5,6}}, the five-digit example An, He and Lai leave open, is spectral at no integer base.

1. Introduction

The figure shows what a spectrum does. Each band is one exponential e(lambda x) measured against a moving frequency xi; the bands never overlap in the sense of orthogonality, and stacked they fill a flat strip of height one at every xi for the whole spectrum (Theorem 4.3), the figure stacking a finite part, which is Parseval's identity for every exponential and so completeness. The measure is that of {0, 1, 8, 9} at base 12, and the spectrum is not of the form a complementing set gives: half of it is shifted by 3/4, and Section 4 explains why the shift is forced.

The question has a history. Jorgensen and Pedersen found that the Cantor measure of {0, 2} at base 4 is spectral while at odd base at most two exponentials are orthogonal. Łaba and Wang proved that a compatible pair, F scaled by b^(-1) together with a set L making a Hadamard matrix, gives a spectral measure (their Theorem 1.2), that a complementing set mod b whose size has at most two prime factors gives one (Theorem 1.4), and conjectured that a spectral mu_{b,F} has F a rational multiple of a complementing set mod b (Conjecture 3.1). Dai, He and Lau settled consecutive digits: mu_{b,{0..N-1}} is spectral if and only if N | b. An, He and Lai classified the spectral measures with at most four digits and left five or more open, naming {0, 1, 3, 5, 6}, whose transform vanishes only at irrational points. An and Lai built spectral measures from product-form Hadamard triples, among them digit sets that are no multiple of a complementing set. An and Wang, Theorem 1.7 as quoted by Liu, Wang and Zheng (their Section 1.1), classify the spectral measures of strict product-form digit sets with respect to pq, the sets E_0 ⊕ (pq)^l E_1 with E_0 ⊕ E_1 = {0..pq-1}, which overlaps Theorem 6.4 on that family to an extent this paper cannot determine without the original; their paper was not read for this one, and nothing below rests on it.

The guess tested here replaces residues by tiling: mu_{b,F} is spectral exactly when every level of F tiles the integers. The residue form fails for tiling already at base 12: {0, 1, 8, 9} tiles at every level and no integer multiple of it tiles Z/12 (Section 3). For the measure the paper proves the following, every hypothesis written into the statements of the sections named.

• If every level tiles and |F| is a prime power, mu_{b,F} is spectral (Theorem 5.2).
• If every level tiles and |F| = pq, mu_{b,F} is spectral (Theorem 6.4).
• In the Hadamard window, b >= 3, 0 in F and F/gcd(F) satisfying T1 and T2 with lcm(S_{F/gcd(F)}) dividing b, the spectrum is gcd(F)^(-1) Lambda(b, L) (Theorem 3.1); outside it, a level-shift lemma gives explicit spectra at base 12 and base 30 (Theorem 4.3), and at base 12 no spectrum of any level lies in 12^(-n) Z (Proposition 3.3).
• At four digits or fewer the classification of An, He and Lai is the every-level verdict (Proposition 7.1); the classification is theirs, the translation is the content here.
• If 0 is in F, |F| = 5 and every level is a set, mu_{b,F} is spectral if and only if every level tiles (Theorem 8.8), and mu_{b,{0,1,3,5,6}} is spectral at no integer base (Corollary 8.4).

A search of the later work of An, He and Lai, of the arXiv listings on spectral self-similar measures with more than four digits, and of the papers on product-form Hadamard triples found neither the integer case of {0, 1, 3, 5, 6} nor a converse at five digits in print; that is a search, not a proof of absence, and no priority is claimed. Section 2 sets up the objects and quotes the inputs at source, Sections 3 to 7 prove the forward direction, Section 8 the converse, Section 9 states what is open.

2. Definitions and the inputs at source

Throughout, b >= 2 is an integer and F a finite set of k = |F| >= 2 non-negative integers. The level n is A_n = {sum_{i<n} d_i b^i : d_i in F}, counted with multiplicity, and its mask is A_n(x) = prod_{i<n} F(x^(b^i)) with F(x) = sum_{d in F} x^d. When F lies in {0..b-1} every level is a set of k^n integers. mu_{b,F} is the infinite convolution of the uniform measures on b^(-j) F, j >= 1, the weak limit of the uniform measures on b^(-n) A_n, and hat mu(xi) = prod_{j>=1} hat F(b^(-j) xi)/k with hat F(t) = sum_{f in F} e(f t).

Definition 2.1 (zero orders). Z(F) = {m >= 2 : Phi_m divides F(x)}, so that hat F(a/m) = 0 with gcd(a, m) = 1 exactly when m is in Z(F). For a prime p, Z_p(F) = {c >= 1 : p^c in Z(F)} and e_p = v_p(b). For a finite set A of integers, S_A is the set of prime powers s with Phi_s | A(x).

Definition 2.2 (tiling, T1, T2). A finite set A tiles the integers when A + C = Z for some C, every integer written once. Coven and Meyerowitz define T1: A(1) = prod_{s in S_A} Phi_s(1), and T2: Phi_{s_1 ... s_m} | A(x) whenever s_1, ..., s_m in S_A are powers of distinct primes. Their theorems, read at source: T1 and T2 imply that A tiles (Theorem A); a tile satisfies T1 (Theorem B1); a tile whose size has at most two prime factors satisfies T2 (Theorem B2). Since prod_{s in S_A} Phi_s(x) divides A(x) in Z[x], prod_{s in S_A} Phi_s(1) always divides A(1). "Every level tiles" means that every A_n is a set and tiles the integers.

Lemma 2.3 (the index lemma). Phi_m | A_n(x) if and only if m/gcd(m, b^i) is in Z(F) for some i < n.

Proof. A primitive m-th root zeta makes zeta^(b^i) a primitive root of order m/gcd(m, b^i), and A_n(zeta) = prod_{i<n} F(zeta^(b^i)) vanishes exactly when one factor does. □

For a prime power the lemma reads: p^c is in S_{A_n} exactly when c is in Z_p(F), for p not dividing b, and exactly when c - e_p i is in Z_p(F) for some i < n, for p | b.

Lemma 2.4 (T1 at every level). Let every A_n be a set. T1 holds at every level if and only if (a) F satisfies T1, that is k = prod_p p^(|Z_p(F)|); (b) Z_p(F) is empty for every prime p not dividing b; (c) for every p | b the elements of Z_p(F) are pairwise incongruent mod e_p. Call (a) to (c) the T1 condition. If every level tiles, the T1 condition holds and k divides b; if moreover k is a prime power, every level tiles if and only if the T1 condition holds; if k has two prime factors and every level tiles, every level satisfies T1 and T2.

Proof. By Lemma 2.3 the exponents of p in S_{A_n} are Z_p(F) for p not dividing b, and the union of the translates Z_p(F) + e_p i, i < n, for p | b. T1 at level n compares v_p(k^n) = n v_p(k) with the number of those exponents, for every p. Off the base the number is constant in n, so T1 at every level forces v_p(k) = 0 = |Z_p(F)|. On the base the union has at most n |Z_p(F)| elements, with equality for every n exactly when no two elements of Z_p(F) differ by a positive multiple of e_p, and level 1 gives v_p(k) = |Z_p(F)|. Conversely (a) to (c) give equality at every p and n. If every level tiles, Theorem B1 gives T1 at every level, and (c) gives |Z_p(F)| <= e_p, so k | b. At prime-power size T2 is vacuous at every level, and Theorems A and B1 give the equivalence; at two prime factors |A_n| = k^n has two prime factors, and Theorems B1 and B2 apply to each tiling level. □

Lemma 2.5 (normalisation). Replacing F by F - t or by F/g with g = gcd(F), when 0 is in F, keeps mu_{b,F} spectral or not, every level a set or not, and every level tiling or not.

Proof. mu_{b,F-t} is a translate of mu_{b,F} and mu_{b,F/g} its image under x -> x/g, which carry spectra to spectra (scaled by g). The levels of F - t are translates of those of F, and those of F/g are those of F divided by g. A translate of a tile is a tile; if A + C = Z then gA + (gC + {0..g-1}) = Z, and if gA + C = Z then for each r the elements of C in r + gZ give A + (C_r - r)/g = Z. □

Definition 2.6 (spectra and the completeness test). A compactly supported probability measure mu is spectral with spectrum Lambda when {e(lambda x) : lambda in Lambda} is an orthonormal basis of L^2(mu). Orthogonality of e(lambda x) and e(lambda' x) is hat mu(lambda - lambda') = 0. For an orthogonal family put Q(xi) = sum_{lambda in Lambda} |hat mu(xi + lambda)|^2, the Bessel sum of e(-xi x) against it; Q <= 1 always, and Q = 1 everywhere if and only if Lambda is a spectrum, since the exponentials span a dense subspace of L^2(mu) (the Jorgensen-Pedersen lemma, Jorgensen and Pedersen 1998; Łaba-Wang Lemma 2.3(b) for Lambda(N, S)). For a finite set S the uniform measure on S has a spectrum exactly when some |S| frequencies are pairwise orthogonal.

Definition 2.7 (Hadamard triples). For an integer N >= 2 and finite sets D, L of integers with |D| = |L|, (N, D, L) is a Hadamard triple when hat D((l - l')/N) = 0 for distinct l, l' in L, that is |D|^(-1/2) [e(d l/N)] is unitary; Łaba and Wang call (N^(-1) D, L) a compatible pair. The elements of L are then distinct mod N, and the triple depends on L mod N only. Write Lambda(N, L) = {sum_{j<n} N^j l_j : n >= 1, l_j in L}.

Theorem 2.8 (Łaba and Wang, Theorem 1.2 with Lemma 2.2(b), (d), read at source). If (b^(-1) F, L) is a compatible pair, mu_{b,F} is spectral; if moreover gcd(F - F) = 1, 0 is in L and L lies in [2 - b, b - 2], then Lambda(b, L) is a spectrum of mu_{b,F}.

Theorem 2.9 (An and Lai, Theorems 2.2 and 2.4, read at source). A digit set in product form of their Definition 2.1, or in K-stage product form of their Definition 2.3, gives a spectral measure. Their Definition 2.1: D = ⋃_{s<n} (a_s + N^r B_s) with r >= 0, A = {a_s}, where (N, A, L_1) and every (N, B_s, L_2) are Hadamard triples with one L_2 for all s, and every (N, A ⊕ B_s, L_1 ⊕ L_2) is a Hadamard triple. Their Definition 2.3: for positive integers l_1, ..., l_k, D^(0) = E_0 and D^(j) = ⋃_{d in D^(j-1)} (d + N^(l_1 + ... + l_j) E_j(d)), where (N, E_0, L_0) and every (N, E_j(d), L_j) are Hadamard triples, and for every 1 <= m <= k and every choice of the d_j the prefix (N, E_0 ⊕ E_1(d_0) ⊕ ... ⊕ E_m(d_(m-1)), L_0 ⊕ ... ⊕ L_m) and the suffix (N, E_m(d_(m-1)) ⊕ ... ⊕ E_k(d_(k-1)), L_m ⊕ ... ⊕ L_k) are Hadamard triples, ⊕ meaning that every sum is written once. Below, N = b and every l_j = 1; the proofs supply the Hadamard condition for every range of consecutive windows, which contains the single windows, the prefixes and the suffixes, and the generator checks (i) and (ii) in exact arithmetic on every case it prints.

3. The Hadamard window and its edge

Theorem 3.1. Let b >= 3, 0 in F, g = gcd(F) and F' = F/g. If F' satisfies T1 and T2 and lcm(S_{F'}) divides b, then mu_{b,F} is spectral with spectrum g^(-1) Lambda(b, L), where L = {sum_{s in S_{F'}} k_s b/s : 0 <= k_s < p_s}, p_s the prime of s, with each element replaced by a representative mod b in [2 - b, b - 2].

Proof. Since 0 is in F', gcd(F' - F') = gcd(F') = 1. Each b/s is an integer since s | b, and the interval [2 - b, b - 2] holds 2b - 3 >= b integers, so the representatives exist, with 0 kept. Take distinct l, l' in L with coefficients k_s, k'_s. Then (l - l')/b = sum_s (k_s - k'_s)/s. For one prime p with exponents c_1 < ... < c_r in S_{F'}, let c* be the largest where the coefficients differ; the sum over p is u/p^(c*) with u = sum_i (k_i - k'_i) p^(c* - c_i), whose term at c* is prime to p and whose other terms are multiples of p. So (l - l')/b = r/m in lowest terms with m the product, over the primes where some coefficient differs, of the largest such p^(c*). That m is an element of S_{F'} or a product of elements at distinct primes, so T2 puts Phi_m in the mask and hat F'((l - l')/b) = 0. T1 gives |L| = prod_s p_s = |F'|. So (b^(-1) F', L) is a compatible pair, Theorem 2.8 makes Lambda(b, L) a spectrum of mu_{b,F'}, and mu_{b,F} is its image under x -> g x. □

This is the proof of Łaba-Wang Theorem 1.4 with T2 assumed in place of their two-prime count. Its hypotheses make every level tile: each c in Z_p(F') is at most e_p, so the T1 condition holds, and a T2 instance at level n reduces through Lemma 2.3 at its largest index to a T2 instance of F'.

Fact 3.2. Of the 609 digit sets with 0 at bases 2 to 16 whose every level tiles, 607 carry a Hadamard triple (b, F', L) with L in Z, found by a clique search on Z/b; the other two are {0, 1, 8, 9} and {0, 3, 8, 11} at base 12. Domain: every digit set with 0 and at least two digits in {0..b-1}, b = 2..16. Generator: lab/py/measure-spectra, verb census.

The window does not reach {0, 1, 8, 9} at base 12. Its mask is (1 + x)(1 + x^8) = Phi_2 Phi_16, so Z_2 = {1, 4}, e_2 = 2, and 1, 4 are incongruent mod 2: the T1 condition holds, the size 4 is a prime power, and every level tiles by Lemma 2.4. But lcm(S_F) = 16 does not divide 12, and no integer multiple vF tiles Z/12: scaling sends the exponents to {1 + v_2(v), 4 + v_2(v)} (Coven-Meyerowitz Lemma 1.4), while a tile of Z/12 has every element of S dividing 12 (their Lemma 2.1), and 2^(4 + v_2(v)) never does. So {0, 1, 8, 9} is no rational multiple of a complementing set mod 12, the form Łaba-Wang Conjecture 3.1(c) predicts for a spectral measure, while its measure is spectral (Theorem 4.3). The window also cannot be reached by passing to a power of the base:

Proposition 3.3. No level A_n of {0, 1, 8, 9} at base 12 has a spectrum inside 12^(-n) Z, so no Hadamard triple (12^n, A_n, L) with L in Z exists at any level.

Proof. A spectrum of the set A_n may be read mod 1, its points distinct there. For gamma - gamma' = d/12^n the order of d/12^n mod 1 is m = 12^n/gcd(d, 12^n) = 2^a 3^c, and orthogonality asks m in Z(A_n). By Lemma 2.3 that holds exactly when m/gcd(m, 12^i) = 2^((a-2i)_+) 3^((c-i)_+) is 2 or 16 for some i < n, that is a = 2i + 1 or a = 2i + 4 with c <= i; with a <= 2n the possible a form the set {1, 3, ..., 2n - 1} ∪ {4, 6, ..., 2n} of 2n - 1 values, all positive. So a spectrum Gamma, as a subset of Z/12^n, maps injectively to Z/2^(2n), two points with one image differing by an element of odd order, and the 2-adic valuations 2n - a of its differences take at most 2n - 1 values. A subset of Z/2^(2n) whose differences have their valuations in a set V has at most 2^|V| points, its binary tree branching only at the levels of V; and 2^(2n - 1) < 4^n = |A_n|. □

4. The level shift

Lemma 4.1. Let mu = nu * rho, where nu is the uniform measure on a finite set S with a spectrum Lambda_0 of |S| points, rho has a spectrum Gamma containing 0, and gamma (s - s') is an integer for every gamma in Gamma - Gamma and s, s' in S. Then Lambda_0 + Gamma is a spectrum of mu.

Proof. For gamma in Gamma - Gamma, e(s gamma) = e(s' gamma) for all s, s' in S, so hat nu(xi + gamma) = c_gamma hat nu(xi) with |c_gamma| = 1. Take lambda_0 + gamma != lambda'_0 + gamma'. If lambda_0 != lambda'_0, then hat nu(lambda_0 - lambda'_0 + gamma - gamma') = c hat nu(lambda_0 - lambda'_0) = 0; otherwise hat rho(gamma - gamma') = 0; either way hat mu = hat nu hat rho vanishes at the difference. The same computation with hat mu(0) = 1 shows that distinct pairs (lambda_0, gamma) give distinct sums. Since 0 is in Gamma, |hat nu(xi + lambda_0 + gamma)| = |hat nu(xi + lambda_0)|, and Q(xi) = sum_{lambda_0} |hat nu(xi + lambda_0)|^2 sum_gamma |hat rho(xi + lambda_0 + gamma)|^2 = sum_{lambda_0} |hat nu(xi + lambda_0)|^2 = 1, the inner sum being 1 because Gamma is a spectrum of rho, the outer because Lambda_0 is a spectrum of nu on the |S|-dimensional space L^2(nu). □

Lemma 4.2. If F = G ⊕ H, every element written once as g + h, then for every q >= 1, mu_{b,F} = nu * mu_{b, G + b^(-q) H} with nu uniform on S = sum_{j=1..q} b^(-j) H, counted with multiplicity.

Proof. The uniform measure on b^(-j) F is the convolution of those on b^(-j) G and b^(-j) H. Keep the factors b^(-j) H with j <= q in nu, and pair each remaining b^(-j-q) H with b^(-j) G: the pair is the uniform measure on b^(-j)(G + b^(-q) H). □

This is the decomposition in the proof of Liu-Wang-Zheng Theorem 3.5, an unrefereed preprint, here for any direct sum.

Theorem 4.3. mu_{12,{0,1,8,9}} is spectral with spectrum {0, 3/4} + 3 Lambda(12, {0, 3, 6, 9}), and mu_{30,{0,1,12,13,24,25}} is spectral with spectrum {0, 5/6, 5/3} + 5 Lambda(30, {0, 5, 10, 15, 20, 25}).

Proof. Base 12: take G = {0, 1}, H = {0, 8}, q = 1. Then S = {0, 2/3}, with spectrum {0, 3/4} since 1 + e(1/2) = 0, and G + H/12 = (1/3) D with D = {0, 2, 3, 5} = {0, 3} ⊕ {0, 2}, mask Phi_2 Phi_6 Phi_4. With L = {0, 3, 6, 9} the differences (l - l')/12 have orders 4 and 2, both in Z(D), so (12^(-1) D, L) is compatible; gcd(D - D) = 1 and L lies in [-10, 10], so Theorem 2.8 makes Lambda(12, L) a spectrum of mu_{12,D}, and 3 Lambda(12, L) one of its image mu_{12,D/3} under x -> x/3. That spectrum lies in 9Z, so gamma (s - s') = ± 2 gamma/3 is an integer, and Lemmas 4.1 and 4.2 conclude. Base 30: take G = {0, 1}, H = {0, 12, 24}, q = 1. Then S = {0, 2/5, 4/5} has spectrum {0, 5/6, 5/3}, since the differences 5/6, 5/3 send S to the three cube roots of unity; G + H/30 = (1/5) D with D = {0, 5} ⊕ {0, 2, 4}, mask Phi_2 Phi_10 Phi_3 Phi_6; with L = {0, 5, ..., 25} the differences (l - l')/30 have orders 6, 3, 2, all in Z(D); gcd(D - D) = 1 and L lies in [-28, 28]; the spectrum 5 Lambda(30, L) of mu_{30,D/5} lies in 25Z, where gamma (s - s') is an integer. □

The second design has pq = 6 digits and lies outside the window: its exponent of 3 is 2 > e_3 = 1. The control Lambda(12, {0, 3, 6, 9}) alone, a spectrum candidate read from the window, is not orthogonal for {0, 1, 8, 9}.

Fact 4.4. Both spectra of Theorem 4.3 are orthogonal in exact arithmetic, and 1 - min Q over 32 points of [-1, 1], namely 0, 1/3, 1/2, 1 and 28 drawn uniformly with seed 1, summed over the spectrum points with at most n digits in Lambda, is 9.64e-03, 4.08e-04, 1.73e-05 at n = 2, 4, 6 for base 12 and 9.68e-04, 9.55e-05, 9.30e-07 at n = 2, 3, 5 for base 30; the controls Lambda(12, {0, 3, 6, 9}) for {0, 1, 8, 9}, not orthogonal, and Lambda(4, {0, 3}) for {0, 2} at base 4, orthogonal with 1 - min Q = 1.00 at n = 4, 8, 12 through the cycle of Łaba-Wang Theorem 1.3, fail. Generator: lab/py/measure-spectra, verb spectra.

5. Prime-power digit counts

Lemma 5.1 (the tree lemma). For a finite set A of integers and j >= 1, Phi_(p^j) | A(x) if and only if, in each class mod p^(j-1), the p classes mod p^j above it hold equally many elements of A. If |A| = p^r and A satisfies T1, then S_A = {p^(c_1), ..., p^(c_r)} with c_1 < ... < c_r, each class of A mod p^(j-1) meets all p classes above it, equally, when j is some c_i, and exactly one of them otherwise.

Proof. A polynomial of degree below p^j is a multiple of Phi_(p^j)(x) = sum_{s<p} x^(s p^(j-1)) exactly when it is sum_{t < p^(j-1)} q_t x^t Phi_(p^j)(x), that is when its coefficients are constant on each coset of p^(j-1) Z/p^j Z; apply this to A(x) mod x^(p^j) - 1. T1 with |A| = p^r and Phi_(q^c)(1) = q forces S_A to be r powers of p. Follow one element: the count of its class mod p^j starts at p^r at j = 0, is divided by exactly p at each c_i, never grows, and is 1 for large j. The r divisions already bring p^r to 1, so at every other level the count is unchanged, and the class goes whole into one class above it. □

Theorem 5.2. Let F be a finite set of non-negative integers with |F| = p^r, p prime, whose every level at base b tiles. Then mu_{b,F} is spectral.

Proof. By Lemma 2.4, e = e_p >= 1 and the exponents c_1 < ... < c_r of Z_p(F) are pairwise incongruent mod e; let rho_i in 1..e be the residue of c_i and b = p^e b'. Cut each digit into base-p^e windows, f = sum_{j<=K} r_j(f) p^(ej) with 0 <= r_j(f) < p^e, K the last window over F, so that b'^K f = sum_{j<=K} b^j b'^(K-j) r_j(f). For a lower part d, a class mod p^(ej) met by F, let R_j(d) be the set of the distinct values r_j(f) over the f in that class, which are the classes mod p^(e(j+1)) that F meets inside it, and E_j(d) = b'^(K-j) R_j(d). By Lemma 5.1 the classes split p ways at the levels c_i and not elsewhere, so the number of distinct classes mod p^(e(j+1)) below a class at a level l in [ej, e(j+1)] depends only on l; hence the set R_j(d) splits equally, p ways, exactly at the levels c_i - je that lie in 1..e, which are the rho_i of the window's i, and Lemma 5.1 puts Phi_(p^(rho_i)) in its mask. The factor b'^(K-j) is prime to p, so Phi_(p^(rho_i)) divides E_j(d)(x) as well. Put L_j = {sum_i k_i b/p^(rho_i) : 0 <= k_i < p} over the window's i. For a range of consecutive windows m..m' and any choice of window sets, the mask of E_m(d_m) + ... + E_(m')(d_(m')) is the product of their masks and holds Phi_(p^(rho_i)) for every i of the range; those rho_i are distinct, so the argument of Theorem 3.1 at the single prime p makes the difference of two distinct elements of L_m ⊕ ... ⊕ L_(m') over b have order p^(rho) for some rho_i = rho of the range, a zero of the sum's mask. So the sum with L_m ⊕ ... ⊕ L_(m') is a Hadamard triple; the sizes match, p to the number of the range's i, and since a Hadamard matrix has rank equal to its size, the sum of window sets is direct. This is the K-stage product form of Theorem 2.9 for b'^K F = {sum_j b^j e_j}, so mu_{b, b'^K F} is spectral, and mu_{b,F} is its image under x -> x/b'^K. □

At {0, 1, 8, 9} and base 12 the windows are base 4: R_0 = {0, 1} with rho = 1, and R_1(d) = {0, 2} with rho = 2, for both lower parts.

Fact 5.3. Every gcd-1 tree set inside {0..b-1} of size 2^r with an exponent above e_2 at bases 12, 20, 24, 28, 36, 40, 44, 48, and of size 9 with an exponent above e_3 at bases 36, 45, 72, tiles at every level and passes conditions (i) and (ii) of An-Lai Definition 2.3 in exact arithmetic, 314 of 314. Generator: lab/py/measure-spectra, verb hunt.

6. Two-prime digit counts

Lemma 6.1 (the fibre lemma). Let N = p^alpha q^beta with p != q primes and alpha, beta >= 1. If Phi_N divides a polynomial P with non-negative integer coefficients, then the exponents of P mod N, with multiplicity, form a union of p-fibres {x + kN/p : k < p} and q-fibres {x + kN/q : k < q}.

Proof. Put zeta = e(1/N) and omega = zeta^(N/pq), of order pq. The degree of Q(zeta) over Q(omega) is phi(N)/phi(pq) = N/pq, so x^(N/pq) - omega is the minimal polynomial of zeta over Q(omega), and P(zeta) = sum_{r < N/pq} zeta^r G_r(omega) vanishes exactly when every G_r(omega) does, G_r(y) = sum_u g_r(u) y^u holding the counts g_r(u) of the exponents r + (N/pq) u, u in Z/pq. Its Galois conjugates vanish too, so the transform of g = g_r on Z/pq = Z/p x Z/q vanishes at every frequency with both coordinates nonzero, and g(i, j) = g(0, j) + (g(i, 0) - g(0, 0)). A function of j alone counts p-fibres, adding N/p to an exponent moving only i, and a function of i alone counts q-fibres. Let the row term g(i, 0) - g(0, 0) be least at i*; subtracting that least value from every row term and adding it to every column term leaves row terms at least 0 and the column term g(i*, j) >= 0, which is where non-negativity enters. □

The statement is Lam-Leung Theorem 3.3(2) with the reduction of their Theorem 3.1 to the subgroup of order pq, which they trace to de Bruijn; the proof is given to keep the paper self-contained.

Lemma 6.2 (the shape). Let |F| = pq with p != q primes, 0 in F, gcd(F) = 1, and every level of F at base b tiling. Then, naming the primes suitably, Z_q(F) = {1}, Z_p(F) = {c}, and F = ⋃_{j<q} (x_j + p^(c-1) q K_j) with the x_j distinct mod q and each K_j a set of p integers forming a complete residue system mod p.

Proof. By Lemma 2.4, Z_p(F) = {c}, Z_q(F) = {d}, and F satisfies T1 and T2, so Phi_(p^c q^d) divides F. By Lemma 6.1 at N = p^c q^d, F mod N is A p-fibres and B q-fibres with pA + qB = pq, so p | B, q | A and F is q p-fibres or p q-fibres. A q-fibre is one point mod p^c, so in the second case F mod p^c is at most p distinct residues; by Lemma 5.1 every class mod p^(c-1) met by F meets all p classes above it, so F lies in one class mod p^(c-1), that of 0, and p^(c-1) divides gcd(F) = 1: c = 1. Symmetrically the first case gives d = 1. Name the primes so that the exponent-1 prime is q and F mod p^c q is q p-fibres x_j + p^(c-1) q {0..p-1}. Assigning each element to its fibre writes it as x_j + p^(c-1) q k + p^c q t = x_j + p^(c-1) q (k + p t), which defines K_j. Each group lies in one class mod q, and Phi_q | F puts p elements in every class mod q, so the x_j are distinct mod q. □

Lemma 6.3 (three evaluations). For F of the shape of Lemma 6.2:

• for 1 <= alpha <= c - 1, Phi_(p^alpha q) | F if and only if the x_j lie in one class mod p^alpha;
• for beta >= 1, Phi_(p^c q^(1+beta)) | F if and only if every K_j lies in one class mod q^beta;
• for 1 <= alpha <= c - 1 and beta >= 1, Phi_(p^alpha q^(1+beta)) does not divide F.

Proof. Let zeta have the order in question and eta = zeta^(p^(c-1) q), so that F(zeta) = sum_j zeta^(x_j) K_j(eta). First case: eta = 1 and F(zeta) = p sum_j zeta^(x_j); write zeta = zeta_1 zeta_2 with orders p^alpha and q. The zeta_2^(x_j) are the q distinct q-th roots of unity, whose only linear relation over Q(zeta_1) has equal coefficients, Phi_q staying irreducible there; so the sum vanishes exactly when the zeta_1^(x_j) agree. Second case: eta has order p q^beta, the degree of Q(zeta) over Q(eta) is p^(c-1) q, so x^(p^(c-1) q) - eta is minimal and 1, zeta, ..., zeta^(p^(c-1) q - 1) are independent over Q(eta); the x_j, distinct mod q, are distinct mod p^(c-1) q, so F(zeta) = 0 exactly when every K_j(eta) = 0, which by the first case with p and q exchanged says that K_j lies in one class mod q^beta. Third case: eta = zeta_2^(p^(c-1) q) has order q^beta, with zeta = zeta_1 zeta_2 of orders p^alpha and q^(1+beta); over E = Q(zeta_1, zeta_2^q), which contains eta, zeta_2 has degree q, so its powers zeta_2^0, ..., zeta_2^(q-1) are independent over E, and zeta^(x_j) is zeta_2^(x_j mod q) times an element of E, with the x_j mod q distinct. So F(zeta) = 0 would need every K_j(eta) = 0, that is Phi_(q^beta) dividing the mask of a set of p integers, whose value p at 1 would then be a multiple of q. □

Theorem 6.4. Let F be a finite set of non-negative integers with |F| = pq, p != q primes, whose every level at base b tiles. Then mu_{b,F} is spectral. With F' the translate of F through 0 divided by its gcd and the primes named as in Lemma 6.2: either c <= e_p and Theorem 3.1 applies to F', or a e_p < c <= (a + 1) e_p for some a >= 1 and (b, (b/b_P)^a F', L_1 ⊕ L_2) is a product-form triple of An-Lai Definition 2.1 with r = a, b_P = p^(e_p) q^(e_q).

Proof. By Lemma 2.5 take F = F', of the shape of Lemma 6.2, and write e = e_p, f = e_q, both at least 1 by Lemma 2.4; pq | b gives b >= 6. If c <= e, then lcm(S_F) = p^c q divides b, F satisfies T1 and T2, and Theorem 3.1 applies. Let c > e and a >= 1 with a e < c <= (a + 1) e. The exponents of p and q in S_{A_(a+1)} include c (at i = 0) and 1 + f a (at i = a), and |A_(a+1)| has two prime factors, so Lemma 2.4 gives T2 there: Phi_(p^c q^(1 + f a)) divides A_(a+1). By Lemma 2.3 some l <= a has p^(c - e l) q^(1 + f(a - l)) in Z(F), the exponent c - e l being positive. At l = a the exponent c - e a lies in 1..c-1, and the first evaluation puts the x_j, and with them all of F, in the class of 0 mod p^(c - e a), against gcd(F) = 1. For 0 < l < a the exponent c - e l lies in 1..c-1 and f(a - l) >= 1, which the third evaluation excludes. So l = 0, and the second evaluation puts every K_j in one class mod q^(f a): K_j = kappa_j + q^(f a) K'_j with K'_j again complete mod p. With y_j = x_j + p^(c-1) q kappa_j, s = (b/b_P)^a and b^a = s p^(e a) q^(f a),

s F = ⋃_j (s y_j + b^a B_j), B_j = p^(c-1-a e) q K'_j.

The integer s is prime to p q, so A = {s y_j} is complete mod q and (b, A, L_1) is a Hadamard triple with L_1 = (b/q){0..q-1}. At a root of order p^(c - a e) or p^(c - a e) q, the power zeta^(p^(c-1-a e) q) has order p, so B_j(zeta) = K'_j(e(t/p)) with p not dividing t, which is 0 since K'_j is complete mod p. So (b, B_j, L_2) is a Hadamard triple with L_2 = (b/p^(c - a e)){0..p-1}, integers since c - a e <= e, and (b, A ⊕ B_j, L_1 ⊕ L_2) is one: a difference of L_1 ⊕ L_2 over b has order q, where A vanishes, or p^(c - a e) or p^(c - a e) q, where B_j vanishes. A ⊕ B_j is direct since A is distinct mod q and B_j divisible by q. Theorem 2.9 makes mu_{b, sF} spectral, and mu_{b,F} is its image under x -> x/s. □

The base-30 design of Theorem 4.3 is the case a = 1: {0, 1} ⊕ 12{0, 1, 2} has p = 3, c = 2, groups x_j = 0, 1 and K_j = {0, 2, 4}, in one class mod 2.

Fact 6.5. Over every digit set of size 6 with 0 in {0..b-1} at bases 30, 36, 42 and 48, the every-level designs number 3221, 8867, 17360 and 37559, of which 14, 0, 55 and 0 have no Hadamard triple after the gcd, and each of these 69 has the shape, the class condition and the product-form triple of the proof in exact arithmetic; at base 66, over the size-6 sets with 0 and every digit below 36, 8087 designs tile at every level, 31 have no Hadamard triple after the gcd, and all 31 pass the same checks. Generator: lab/py/measure-spectra, verb pq.

7. Four digits or fewer

An, He and Lai classify the spectral self-similar measures with at most four integer digits: for four digits their Theorem 4.2 with the sufficiency of their Section 4.1 says mu_{b,F} is spectral exactly when b = 2^beta m with beta >= 1 and m odd, and F is, up to scaling, {0, a, 2^t l, a + 2^t l'} with a, l, l' odd and beta not dividing t; for two and three digits their Theorem 5.1 gives the necessity, b even for {0, a}, and 3 | b with F a rational multiple of a complete residue system mod 3, and a compatible pair the sufficiency. The classification is theirs; what follows only translates it.

Proposition 7.1. Let 0 be in F, |F| <= 4 and F in {0..b-1}. The condition of An-He-Lai Theorems 4.2 and 5.1 holds exactly when every level of F at base b tiles, so by their classification mu_{b,F} is spectral if and only if every level tiles.

Proof. All three sizes are prime powers, so by Lemma 2.4 every level tiles exactly when the T1 condition holds. Size 2, F = {0, a}: Phi_m | 1 + x^a exactly when m | 2a and m does not divide a, so Z_2(F) = {v_2(a) + 1} and Z_q(F) is empty for odd q; the T1 condition reads b even. Size 3: T1 forces S_F = {3^c}, and by Lemma 5.1 F lies in one class mod 3^(c-1) and meets all three classes mod 3^c, so F = 3^(c-1) C with C a complete residue system mod 3; conversely such an F has Phi_(3^c) in its mask and T1, since the product of Phi_s(1) over S_F divides 3. A rational multiple t C' of a complete residue system containing 0 is of this form, its nonzero elements sharing the 3-adic valuation v_3(t) and staying distinct mod 3 after division by 3^(v_3(t)). So the T1 condition reads that F is such a multiple and 3 | b. Size 4: T1 forces S_F = {2^(c_1), 2^(c_2)}, and by Lemma 5.1, with v = v_2(gcd F) = c_1 - 1, F/2^v has two even and two odd elements, the evens differing by 2^t times an odd number and the odds likewise, t = c_2 - c_1 >= 1; this is {0, a, 2^t l, a + 2^t l'} with a, l, l' odd, and conversely. The T1 condition then reads that b is even and t is not a multiple of beta = v_2(b). □

Fact 7.2. Over every four-digit set {0, a, b, c} in {0..b-1} at bases 4 to 36, 58905 sets of which 2642 tile at every level, the every-level verdict equals the condition of An-He-Lai Theorem 1.7 on all 58905. Generator: lab/py/measure-spectra, verb four.

8. The converse

Write Theta for the finite set of zeros of hat F in [0, 1), so that the zero set of hat mu is ⋃_{j>=1} b^j (Theta + Z). An integer d is a zero of hat mu exactly when b^j/gcd(d, b^j) is in Z(F) for some j >= 1; say then that d reaches that order at level j. Every order an integer reaches has all its primes dividing b; call the orders of Z(F) with this property the b-orders of F, and j(m) the least j with m | b^j.

Lemma 8.1 (the class count). If A_n is a set of k^n integers, every spectrum Lambda of mu = mu_{b,F} meets at least k^n classes of R/b^n Z.

Proof. hat mu(xi) = hat mu_n(xi) hat mu(b^(-n) xi), with mu_n uniform on b^(-n) A_n; hat mu_n has period b^n and is 1 on b^n Z. Two points of one class Lambda_r differ by b^n z, so 0 = hat mu(b^n z) = hat mu(z): the set b^(-n) Lambda_r is orthogonal in L^2(mu), and Bessel's inequality gives sum_{lambda in Lambda_r} |hat mu(b^(-n)(x - lambda))|^2 <= 1. Splitting 1 = sum_lambda |hat mu(x - lambda)|^2 by classes gives 1 <= sum_r |hat mu_n(x - r)|^2 for every x. If the classes are finitely many, integrate over a period: the mean of |hat mu_n|^2 is k^(-n), the elements of A_n being distinct, so their number is at least k^n. □

Lemma 8.2 (rational cosets). A spectrum of mu_{b,F} through 0 meets finitely many cosets of Q.

Proof. In the Q-vector space R/Q, every point of the spectrum and every difference of two points maps into {0} or a set b^j x, j >= 1, with x the image of an irrational element of Theta. If infinitely many points had distinct images, infinitely many would map to b^j x for one x, with j in an infinite set K. For j > j' in K the difference maps to (b^j - b^(j')) x, which is b^i x' with x' the image of an element of Theta, necessarily x' = c x with c rational, and so b^(j - j') - 1 = c b^(i - j'). Finitely many c occur, and an integer prime to b equals c b^s for at most one s, so j - j' takes finitely many values on K, which is impossible. □

Theorem 8.3 (integer orthogonal sets). If every A_n is a set and mu_{b,F} is spectral, then for every L >= 1 some k^L integers, pairwise incongruent mod b^L, have every difference a zero of hat mu.

Proof. Take a spectrum Lambda through 0, C its number of cosets of Q (Lemma 8.2), Lambda_v its points in one coset, and M the lcm of Z(F) (M = 1 if Z(F) is empty). Differences in Lambda_v are rational zeros b^j(theta + i) with theta a rational zero of hat F, so they lie in M^(-1) Z, and Gamma = M(Lambda_v - lambda_v) is a set of integers whose classes mod M b^n are those of Lambda_v mod b^n Z. Two points c + M b^(tL) x and c + M b^(tL) x' of Gamma differ by M b^j (theta + i), theta + i not an integer, so j > tL and x - x' = b^(j - tL)(theta + i) is a zero of hat mu. So if g_L <= b^L is the largest number of integers, pairwise incongruent mod b^L, with all differences zeros of hat mu, a class of Gamma mod M b^(tL) holds at most g_L classes mod M b^((t+1)L), and Lambda meets at most C M g_L^(ceil(n/L)) classes mod b^n. Lemma 8.1 gives k^n <= C M g_L^(n/L + 1) for every n, so g_L >= k^L. □

Corollary 8.4. If every A_n is a set and mu_{b,F} is spectral, then k <= b and F has a b-order. If Z(F) is empty, mu_{b,F} is not spectral, with no hypothesis on the levels. In particular mu_{b,{0,1,3,5,6}} is spectral at no integer base b >= 2.

Proof. Theorem 8.3 at L = 1 gives k integers distinct mod b, so k <= b; and k >= 2 of them have a difference that is an integer zero of hat mu, which reaches a b-order. If Z(F) is empty, hat mu has no rational zero, so a spectrum through 0 holds at most one point in each coset of Q, and by Lemma 8.2 it is finite, while L^2(mu_{b,F}) is infinite-dimensional, the support being infinite. The mask 1 + x + x^3 + x^5 + x^6 has no cyclotomic factor (generator: lab/py/measure-spectra, verb converse), which covers the bases 5 and 6, where levels repeat points. □

An-He-Lai Example 5.2 takes a real ratio 0 < rho < 1 and expects no mu_{rho,{0,1,3,5,6}} to be spectral; Corollary 8.4 settles the integer ratios rho = 1/b, and a non-integer rho^(-1) stays open. The corollary also recovers the non-spectrality of {0, 2} at odd base, where Z = {4} has no b-order; Jorgensen and Pedersen prove more there, at most two orthogonal exponentials.

Theorem 8.5 (the covering bound). Let every A_n be a set and Q a set of primes meeting every b-order of F; for q in Q let rho_q count the classes mod e_q met by v_q(m) over the b-orders m divisible by q. If mu_{b,F} is spectral, then k <= prod_{q in Q} q^(rho_q).

Proof. Take the k^L integers of Theorem 8.3, and J the largest j(m) over the b-orders. A difference d reaching m at level j has m | b^j, so j >= j(m), and some prime q' of b has v_(q')(d) < L e_(q'). If q' does not divide m then v_(q')(d) >= j e_(q'), so j < L; if it does, v_(q')(m) = j e_(q') - v_(q')(d) > (j - L) e_(q') while v_(q')(m) <= j(m) e_(q'); either way j < J + L. For q in Q dividing m, v_q(d) = j e_q - v_q(m) lies in [0, (J + L) e_q) and in one of rho_q classes mod e_q, so it takes at most rho_q (J + L) values, a set V_q. Every difference reaches a b-order divisible by some q in Q, and two integers whose difference has q-adic valuation delta differ in their q-adic digit at position delta; so the digits at the positions V_q, over q in Q, tell all the integers apart, and k^L <= prod_q q^(rho_q (J + L)) for every L. Let L grow. □

Proposition 8.6. If F passes the T1 condition, in particular if every level tiles, its covering bound, the least prod_{q in Q} q^(rho_q) over sets Q meeting every b-order, is at least k.

Proof. Each p with Z_p(F) nonempty divides b, so the orders p^c, c in Z_p(F), are b-orders; a covering set contains p, and rho_p >= |Z_p(F)| since those c are pairwise incongruent mod e_p. So the bound is at least prod_p p^(|Z_p(F)|) = k. □

Lemma 8.7. Let |F| = p be prime, 0 in F and gcd(F) = 1. The only prime power that can lie in Z(F) is p, and it does exactly when F is a complete residue system mod p; then every level tiles exactly when p | b.

Proof. If Phi_(q^c) | F, Lemma 5.1 makes the count of every class mod q^(c-1) met by F a multiple of q; the total p forces q = p and one class, that of 0, so p^(c-1) | gcd(F) = 1 and c = 1, and Phi_p | F says that F meets every class mod p once. Then S_F = {p}, T1 holds, and the T1 condition reads p | b; at prime size Lemma 2.4 makes it the every-level verdict. □

Theorem 8.8 (five digits). Let F be a set of five non-negative integers with 0 in F, whose every level A_n at base b is a set. Then mu_{b,F} is spectral if and only if every level of F tiles.

Proof. If every level tiles, Theorem 5.2 applies. Conversely, by Lemma 2.5 take gcd(F) = 1 and suppose not every level tiles. By Lemma 8.7 no b-order is a prime power: the only candidate is 5, and if 5 is in Z(F) and divides b, every level tiles. Let m be a b-order. Lam-Leung Theorem 4.8, read at source, writes the residues of F mod m, a vanishing sum of m-th roots of unity with non-negative coefficients and weight 5, as minimal vanishing sums, each a rotated fibre of a prime of m or of weight above the third prime of m, at least 6. A single 5-fibre puts F in one class mod m/5, so m/5 | gcd(F) = 1 and m = 5, a prime power. So F = P ∪ T with a pair P whose difference d_P is m/2 mod m and a triple T whose differences are ± m/3 mod m; then a = v_2(m) = v_2(d_P) + 1, and T lies in one class mod 2^a. Let (P', T') be the splitting at a second b-order, with exponent a'. If P' = P then a' = a. If P' lies in T, then P lies in T', and a' - 1 = v_2(d_(P')) >= a and a - 1 >= a' cannot both hold. Otherwise P = {f_1, f_2} and P' = {f_1, f_3} with f_3 in T, f_2 in T', and some f_4 in T ∩ T'; in f_1 - f_2 = (f_1 - f_3) + (f_3 - f_4) + (f_4 - f_2) the terms have 2-adic valuations a' - 1, at least a, at least a', so a' < a would make the right side of valuation a' - 1 against a - 1 on the left; and in f_1 - f_3 = (f_1 - f_2) + (f_2 - f_4) + (f_4 - f_3), a < a' fails the same way. So every b-order is even with one exponent a of 2, Q = {2} covers them with rho_2 = 1, and Theorem 8.5 would give 5 <= 2. At odd b an even order is never a b-order, so F has none, and Corollary 8.4 applies. Either way mu_{b,F} is not spectral. □

Example 8.9. Theorem 8.3 at L = 1 alone does not decide the converse. F = {0, 1, 251, 375, 501} at base 510 = 2 * 3 * 5 * 17 has residues {0, 0, 1, 1, 1} mod 5, so not every level tiles; mod 750 it is the 2-fibre {0, 375} and the 3-fibre {1, 251, 501}, and 6, 30, 150, 750 are its b-orders. The integers 0, 17, 1734, 1751, 176868 are pairwise incongruent mod 510, and every difference reaches 30 at level 1, 150 at level 2 or 750 at level 3, so five integers with orthogonal differences exist at L = 1; yet all four orders have v_2 = 1, the covering bound with Q = {2} is 2, and mu_{510,F} is not spectral by Theorem 8.5. Likewise {0, 1, 9, 10} at base 12 fails T1, 4 against Phi_2(1) = 2, while 0, 2, 8, 10 are pairwise incongruent mod 12 with orthogonal differences; its orders 2, 6, 18 all have v_2 = 1, and its covering bound is 2 < 4. Generator for the orders and the integers: lab/py/measure-spectra, verb converse.

Fact 8.10. The covering bound is below p on every non-tiling digit set with 0 and gcd 1 in {0..b-1} of size 5 at bases 5 to 42, 7 at 7 to 26, 11 at 11 to 20 and 13 at 13 to 20, 795499, 653504, 167959 and 77519 sets; over the four-digit sets with 0 and gcd 1 at bases 4 to 36 it is below 4 on all 48897 that do not tile at every level, so on that box the bound alone gives the necessity half of the An-He-Lai classification. Generator: lab/py/measure-spectra, verb converse.

9. Open problems

The forward direction is open at a size p^alpha q^beta with alpha + beta >= 3 or with three or more primes, when an exponent exceeds e_p after the gcd: Lemma 6.1 then allows p-fibres and q-fibres together, and the groups need not be distinct mod q. The evidence is a census: of the 6898 designs that tile at every level among the digit sets with 0 at bases 2 to 16 of a size dividing the base and at bases 17 to 30 of size 4, 8, 9 or with two primes dividing the base, leaving out the cells size 8 and 12 at base 24, size 9 at base 27, size 14 at base 28 and sizes 10 and 15 at base 30, 6844 carry an integer Hadamard triple (b, F/gcd(F), L) and 54 are in An-Lai stage form, none neither (lab/py/measure-spectra, verb census). Conjecture 9.1. If every level of F at base b tiles, then F/gcd(F) carries an integer Hadamard triple at base b or F or F/gcd(F) is in An-Lai stage form with windows of base b_P, and mu_{b,F} is spectral. At three or more primes even the tiling side leans on the open necessity of T2; an unrefereed preprint (arXiv 2609.14026) claims that every tile satisfies T2, and nothing here rests on it. The converse is open beyond five digits: for p = 7 every non-prime-power b-order is even, but F mod m may split as 2 + 5 or 2 + 2 + 3, and the splittings are not shown to share one exponent of 2. Conjecture 9.2. The covering bound of F at base b is at least |F| exactly when F passes the T1 condition; it holds on every gcd-1 set with 0 in {0..b-1} of size 4 at bases 4 to 36, 5 at 5 to 42, 7 at 7 to 26, 11 and 13 at bases up to 20, size 6 at bases 30 and 42 and size 8 at base 24 (verb converse), and with Theorem 8.5 it would give the converse at every prime-power size when the levels are sets, leaving only T2 at the other sizes. Five digits with a level that repeats a point, and {0, 1, 3, 5, 6} at a non-integer ratio, are not decided. Joined, these are Conjecture 9.3: for an integer base b >= 2 and a finite set F of non-negative integers containing 0, mu_{b,F} is spectral if and only if every level is a set that tiles the integers; the direction from spectral to tiling levels is a question of An and coauthors quoted by Wu and Xiao, who settle it for equidifferent digit sets of prime size.

Reproducibility

One study prints every number of this paper: lab/py/measure-spectra, run as uv run python research/lab/py/measure-spectra/measure_spectra.py <verb> from the repository root, one core, exact integer and rational arithmetic for every verdict and floats only in Q. Its verbs: census (about 80 s), every digit set with 0 at bases 2 to 16 of a size dividing the base and at bases 17 to 30 of size 4, 8, 9 or with two primes dividing the base, cells of at most 130000 sets, Facts 3.2 and Section 9; spectra (about 2 s), the spectra of Theorem 4.3 and the controls, Fact 4.4; hunt (under 1 s), Fact 5.3; pq (about 172 s), Fact 6.5; four (about 16 s), Fact 7.2; converse (about 46 s, its full census longer), Corollary 8.4, Example 8.9, Fact 8.10 and the evidence of Conjecture 9.2. The study's README names the verb of every printed line; the every-level verdict and the zero sets come from lab/py/level-tiles, imported by path. The figure is figures/paper-spectral-tiling-levels.ts, pressed by bun figures/press.ts paper-spectral-tiling-levels: it sums |hat mu(xi - lambda)|^2 from the closed form |hat mu(xi)| = prod_j |cos(pi xi/12^j) cos(8 pi xi/12^j)| over the 2048 points of {0, 3/4} + 3 Lambda(12, {0, 3, 6, 9}) with at most five digits in Lambda, and stops unless the total lies within 1e-2 of 1 and never above it at every column.

References

• Coven and Meyerowitz 1999, Tiling the integers with translates of one finite set, J. Algebra 212, 161-174; read at source: T1, T2, Theorems A, B1, B2, Lemmas 1.4 and 2.1. arxiv.org/abs/math/9802122
• Łaba and Wang 2002, On spectral Cantor measures, J. Funct. Anal. 193, 409-420; read at source: Theorems 1.2 to 1.4, Lemmas 2.2 and 2.3, Conjecture 3.1. arxiv.org/abs/math/0109126
• Jorgensen and Pedersen 1998, Dense analytic subspaces in fractal L2-spaces, J. Analyse Math. 75, 185-228; read at source: Section 6 and Theorem 6.1. arxiv.org/abs/funct-an/9709007
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