---

Source. Theorem 1, p. 1, with its proof on pp. 1--2, of the two-page note On the Erdős problem #251, bylined "ChatGPT 5.4 Pro (orchestrated by Vjeko Kovač)", hosted at https://web.math.pmf.unizg.hr/~vjekovac/files/Erdos_problem_251.pdf (PDF metadata dated 15 April 2026). The edition is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause, and every step of the proof was read and checked while writing the sketch below. No independent review is filed.

Statement

Here $p_n$ is the $n$-th prime. Theorem 1 (p. 1): "There exists a sequence of integers $(g_n)_{n\ge1}$ with $g_n\ge2$ and $g_n=o(p_n)$ such that

$$\sum_{n=1}^{\infty}\frac{p_n}{g_1\cdots g_n}=1.$$

"

The theorem's last sentence draws the consequence that the irrationality statement of the note's abstract is false. That statement asserts irrationality of $\sum_{n\ge1}p_n/(g_1\cdots g_n)$ for every integer sequence with $g_n\ge2$ and $g_n=o(p_n)$; it is the variable-denominator statement that the problem page of Problem 251 quotes from Erdős's 1988 survey (p. 103), where the growth condition is written $g_n/p_n\to0$.

Proof sketch

The idea is to choose slowly growing integers $c_n\ge1$, with $c_1=1$, and to set $g_n=(p_n+c_{n+1})/c_n$. Then $p_n=c_ng_n-c_{n+1}$, so each term $p_n/(g_1\cdots g_n)$ is the difference of consecutive values of $c_n/(g_1\cdots g_{n-1})$, and the $M$-th partial sum is $1-c_{M+1}/(g_1\cdots g_M)$.

The note takes $m_n=1+\lfloor\log\log(n+3)\rfloor$, so that $\sum_{k\le n}m_k=O(n\log\log n)$, which is $o(p_n)$ by the prime number theorem; it fixes $N$ with $1+\sum_{k=N}^{n-1}m_k<p_n$ for all $n\ge N$. It puts $c_n=1$ for $n\le N$ and, for $n\ge N$, picks $c_{n+1}$ in the block of $c_n$ consecutive integers starting at $m_n$ so that $c_n$ divides $p_n+c_{n+1}$. The divisibility makes $g_n$ an integer, and by induction $c_n\le1+\sum_{k=N}^{n-1}m_k<p_n$ for $n\ge N$, which makes $g_n>1$ (for $n<N$, $g_n=p_n+1$). Since $c_{n+1}\ge m_n\to\infty$ and $c_{n+1}\le c_n+m_n-1$, the ratio $g_n/p_n$ is at most $1/c_n+1/p_n+m_n/p_n$, which tends to $0$. Finally $g_1\cdots g_M\ge2^M$ while $c_{M+1}=O(M\log\log M)$, so the partial sums tend to $1$.

Dependencies

The prime number theorem, used only through $p_n\sim n\log n$; the argument needs no more than $p_n/(n\log\log n)\to\infty$, and the bound $p_n\ge n$ alone would not do, since $\sum_{k\le n}m_k$ has order $n\log\log n$.

Scope

The construction neither assumes nor yields a monotone $(g_n)$, and it produces one particular sequence, depending on the choice of $N$. It says nothing about the constant sequence $g_n=2$, that is about $\sum p_n/2^n$, which is the question of Problem 251. The theorems on monotone denominators named on the source card are not contradicted.

Bears on

• Problem 251: a claimed counterexample to the auxiliary variable-denominator statement quoted on the problem page; not a result on the problem itself. The note is unrefereed and AI-generated by its byline, and no independent review is filed.