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Erdős #265 · irrationality

Let 1≤a1<a2<⋯1\leq a_1<a_2<\cdots be an increasing sequence of integers. How fast can an→∞a_n\to \infty grow if

∑1anand∑1an−1\sum\frac{1}{a_n}\quad\textrm{and}\quad\sum\frac{1}{a_n-1}

are both rational?

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

Problem statement

Let 1≤a1<a2<⋯1\leq a_1<a_2<\cdots be an increasing sequence of integers. How fast can an→∞a_n\to \infty grow if

∑1anand∑1an−1\sum\frac{1}{a_n}\quad\textrm{and}\quad\sum\frac{1}{a_n-1}

are both rational?

Lean 4

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Background

Cantor observed that an=(n2)a_n=\binom{n}{2} is such a sequence. If we replace −1-1 by a different constant then higher degree polynomials can be used - for example if we consider ∑n≥21an\sum_{n\geq 2}\frac{1}{a_n} and ∑n≥21an−12\sum_{n\geq 2}\frac{1}{a_n-12} then an=n3+6n2+5na_n=n^3+6n^2+5n is an example of both series being rational.

Erdős believed that an1/n→∞a_n^{1/n}\to \infty is possible, but an1/2n→1a_n^{1/2^n}\to 1 is necessary.

This has been almost completely solved by Kovač and Tao [KoTa24], who prove that such a sequence can grow doubly exponentially. More precisely, there exists such a sequence such that an1/βn→∞a_n^{1/\beta^n}\to \infty for some β>1\beta >1.

It remains open whether one can achieve

lim sup⁡an1/2n>1.\limsup a_n^{1/2^n}>1.

A folklore result states that ∑1an\sum \frac{1}{a_n} is irrational whenever lim⁡an1/2n=∞\lim a_n^{1/2^n}=\infty, and hence such a sequence cannot grow faster than doubly exponentially - the remaining question is the precise exponent possible.

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 #265 link

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