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Source. Theorem 14, p. 417, of R. J. Simpson, On a conjecture of Crittenden and Vanden Eynden concerning coverings by arithmetic progressions, Journal of the Australian Mathematical Society (Series A) 63 (1997), 396-420, as identified on the source card.

Statement

Notation as on the Theorem 1 page: S(m,a)S(m,a) is the progression of integers congruent to aa modulo mm.

The conjecture (p. 397, quoted). The paper states the conjecture that Crittenden and Vanden Eynden posed in 1972 as: "If A\mathcal A is a collection of nn arithmetic progressions, each with modulus ≥k\geq k, such that ⋃A⊇[1,k2n−k+1]\bigcup\mathcal A\supseteq[1,k2^{n-k+1}], then ⋃A⊇Z\bigcup\mathcal A\supseteq\mathbb Z."

Minimal counterexample (p. 401). For an integer k≥3k\ge3, a minimal counterexample for kk is a collection A\mathcal A of nn arithmetic progressions, each of modulus at least kk, with ⋃A⊇[1,k2n−k+1]\bigcup\mathcal A\supseteq[1,k2^{n-k+1}] and ⋃A≠Z\bigcup\mathcal A\ne\mathbb Z, such that (a) nn is the least integer for which such a collection exists, and (b) every other collection of nn progressions with these properties has sum of moduli at least the sum of the moduli of A\mathcal A.

Notation of Theorem 13 (p. 416). π(x)\pi(x) is the number of primes less than xx, and Θ(x)=∑plog⁡p\Theta(x)=\sum_p\log p over all primes pp less than xx. Both use strict inequality.

Theorem 14 (p. 417). If A\mathcal A is a minimal counterexample for some k≥3k\ge3, then nn is less than

3(Θ(k)/log⁡2+k)−2π(k)+36log⁡2k+⌈log⁡2k⌉+(3log⁡23−4)⌈log⁡3k⌉−4.3\bigl(\Theta(k)/\log2+k\bigr)-2\pi(k)+36\log_2k+\lceil\log_2k\rceil +(3\log_23-4)\lceil\log_3k\rceil-4.

Consequence stated by the paper (pp. 397, 416, 418). If the conjecture fails for some k≥3k\ge3, then a minimal counterexample exists for that kk, and its nn is bounded as above; so for each fixed k≥3k\ge3 the conjecture can be settled by checking finitely many cases. The paper does not carry out this check. It reports that the check has been done for k=3k=3 in the author's doctoral thesis (its reference [10]; pp. 397 and 419), and it notes that the number of cases grows exponentially with kk (p. 397).

The theorem is restricted to k≥3k\ge3. The paper notes (p. 397) that the cases k=1k=1 and k=2k=2 of the conjecture coincide and are the theorem Crittenden and Vanden Eynden proved in 1970, and that the interval [1,k2n−k+1][1,k2^{n-k+1}] cannot be replaced by a shorter one: the collection of S(k,i)S(k,i) for 1≤i≤k−11\le i\le k-1 and S(2ik,2i−1k)S(2^ik,2^{i-1}k) for 1≤i≤n−k+11\le i\le n-k+1 covers [1,k2n−k+1−1][1,k2^{n-k+1}-1] but not Z\mathbb Z.

Proof pointer

Pages 416-418. Assume A\mathcal A misses 00. By Corollary 3 (§2), every modulus of a minimal counterexample is either a product of primes less than kk or a prime at least kk; write AL\mathcal A_L and AG\mathcal A_G for the two parts, of sizes nLn_L and nGn_G, and let PP be the least modulus of a progression S(P,A)S(P,A) disjoint from ⋃AL\bigcup\mathcal A_L. Theorem 1 gives nL≥g(P)n_L\ge g(P), inequality (41), and Theorem 8 gives nL≤log⁡2Pn_L\le\log_2P, inequality (42). Theorem 13 (p. 416) bounds log⁡2P−g(P)\log_2P-g(P) and 3log⁡2P−2g(P)3\log_2P-2g(P) by functions of kk alone. Since AG\mathcal A_G must cover S(P,A)∩[1,k2n−k+1]S(P,A)\cap[1,k2^{n-k+1}], reducing it via S(P,A)S(P,A) and applying the counting bound of Corollary 9 (p. 415) gives nG<3(k−1−nL+log⁡2P+12log⁡2k)n_G<3(k-1-n_L+\log_2P+12\log_2k), inequality (45). With n=nG+nLn=n_G+n_L and (41) this yields n<3k−3+36log⁡2k+3log⁡2P−2g(P)n<3k-3+36\log_2k+3\log_2P-2g(P), and part (c) of Theorem 13 turns this into the stated bound.

Dependencies

Theorem 1 (p. 397); Corollary 3 and Theorem 8 of §2; Corollary 9 (p. 415) of §3, which rests on the paper's Theorem 12 and its Lemmas 2 and 3; Theorem 13 (p. 416).

Read depth. Claims checked: the conjecture, the definition of a minimal counterexample, the notation of Theorem 13 and the statement of Theorem 14 were read clause by clause on pp. 397, 401, 416 and 417, and the discussion on pp. 418-419. The proof (pp. 416-418) was read for its structure only; §§2-3, on which it rests, were not checked.

Bears on

  • Problem 275: background only. The problem is the conjecture's case k=1k=1 (equivalently k=2k=2), with the interval [1,2n][1,2^n] replaced by any 2n2^n consecutive integers, which the paper notes changes nothing (p. 397); Crittenden and Vanden Eynden proved that case. Theorem 14 concerns only k≥3k\ge3 and gives no proof or bound for the problem.