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OP-9406OpenNumber theory

Erdős #1204 · number theory

We call a sequence of integers 0≤a1<⋯<ak0\leq a_1<\cdots <a_k admissible if it is missing at least one congruence class modulo every prime pp. Let A(k)=min⁡akA(k)=\min a_k. Estimate A(k)A(k) - in particular, is it true that

A(k)∼…A(k)\sim …

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Source reports OPENAs of 2026-09-08Source & evidence ↓

Problem statement

We call a sequence of integers 0≤a1<⋯<ak0\leq a_1<\cdots <a_k admissible if it is missing at least one congruence class modulo every prime pp. Let A(k)=min⁡akA(k)=\min a_k. Estimate A(k)A(k) - in particular, is it true that

A(k)∼klog⁡k?A(k)\sim k\log k?

Estimate

B(k)=min⁡a1+⋯+akk.B(k)=\min \frac{a_1+\cdots+a_k}{k}.

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Background

Erdős [Er80] attributes this problem to Elliott. The bounds are misstated in [Er80], but it is known that

(1/2+o(1))klog⁡k≤A(k)≤(1+o(1))klog⁡k.(1/2+o(1))k\log k \leq A(k)\leq (1+o(1))k\log k.

The upper bound Erdős attributes to Davenport, and follows simply by taking the kk smallest primes >k>k. The lower order terms have been improved by Hensley and Richard [HeRi73] (see Section 10 of [Po14c] for more details).

The lower bound is originally due to Elliott [El65], but was rediscovered by the Polymath project on bounded gaps between primes (see [Po14c]).

An upper bound of π(x+y)≤π(x)+(1+o(1))π(y)\pi(x+y)\leq \pi(x)+(1+o(1))\pi(y) (see [855]) together with the prime tuples conjecture would imply A(k)≥(1+o(1))klog⁡kA(k)\geq (1+o(1))k\log k.

It is trivial that B(k)<A(k)B(k)<A(k). Taking the first kk primes which are >k>k implies

B(k)≤(1/2+o(1))klog⁡k,B(k)\leq (1/2+o(1))k\log k,

and since aj≥A(j)a_j\geq A(j) we have B(k)≥1k∑1≤j≤kA(j)B(k) \geq \frac{1}{k}\sum_{1\leq j\leq k}A(j), whence if A(k)≥(c−o(1))klog⁡kA(k)\geq (c-o(1))k\log k we have B(k)≥(c/2−o(1))klog⁡kB(k) \geq (c/2-o(1))k\log k, and so it is likely that

B(k)∼(1/2+o(1))klog⁡k.B(k)\sim(1/2+o(1))k\log k.

In [Er80] Erdős also asks about the behaviour of aka_k where a1<a2<⋯a_1<a_2<\cdots is the greedy admissible sequence (i.e. a1=0a_1=0 and ai+1a_{i+1} is the smallest integer >ai−1>a_{i-1} such that a1,…,aia_1,\ldots,a_i is admissible).

Progress & source assessment

The source labels this entry OPEN in the 2026-09-08 snapshot. Editorial progress notes are reproduced as source commentary, not independently verified results. Prize amounts and Lean badges at the source are not local awards or verification.

References
  1. [1]Thomas Bloom (ed.), Erdős Problems #1204 link

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