Kovac 2026 erdos problem 251

../

theorem_1: Constructs integers g_n at least 2 with g_n = o(p_n) such that the sum of p_n over g_1...g_n equals 1, a claimed counterexample to the auxiliary statement quoted with problem 251; the proof sketch here is author-recorded and not independently reviewed.

---

ChatGPT 5.4 Pro (orchestrated by Vjeko Kovač), On the Erdős problem #251, two-page note, undated in the text; hosted on the author's web page at the University of Zagreb. The byline is the note's own; the PDF metadata date is 15 April 2026 and the server reports Last-Modified 2026-04-15 11:01:47 GMT. The note carries no license statement.

The copy read for this card is the hosted file: 260,418 bytes, as served at https://web.math.pmf.unizg.hr/~vjekovac/files/Erdos_problem_251.pdf on 2026-09-17 (UTC). The note prints no license statement on either page, and the hosting page (https://web.math.pmf.unizg.hr/~vjekovac/, read 2026-10-02) states no copyright, license or terms of use; the term is unstated.

Claim. The note's abstract (p. 1) says: "We disprove the assertion that, if $p_n$ is the $n$-th prime and $(g_n)_{n\ge1}$ is a sequence of integers with $g_n\ge2$ and $g_n=o(p_n)$, then $\sum_{n=1}^{\infty}p_n/(g_1\cdots g_n)$ must be irrational. In fact, we construct such a sequence for which the sum is exactly 1." This is the auxiliary statement that the catalog page of Problem 251 attributes to Erdős's 1988 survey, whose p. 103 reads: "It seems reasonable to expect that if $g_n\ge2$, $g_n/p_n\to0$ then $\sum_{n=1}^{\infty}p_n/g_1\ldots g_n$ (2) is irrational"; the survey is filed as erdos_1988_irrationality_certain_series_problems_results. The problem itself, the constant sequence $g_n=2$, is not touched: the construction produces one particular sequence with sum $1$. Theorem 1 carries the statement and a proof sketch.

Provenance and standing. The note was announced in the author's comment on the catalog's discussion thread at 11:13 on 15 April 2026 (site clock), which links a ChatGPT conversation and says: "I got ChatGPT 5.4 Pro figure out the proof and write up the details with only minimal orchestration from my side." A reply by another participant the same day (17:06) reports "I ran standard check which found no issues" with a link to another ChatGPT conversation, and remarks that the argument gives more than $g_n=o(p_n)$; that is a forum remark about an AI-run check, not a review. The community database's "AI contributions" wiki lists the note as "[251] | GPT-5.4 Pro | 15 Apr, 2026 | Solution to variant problem", a listing, not acceptance. The catalog page's text was unchanged on 2026-09-17. Standing here: claimed, elementary, non-refereed; no independent review is filed. The problem page therefore describes the auxiliary statement as having a claimed counterexample, not as disproved. The discussion record is filed as bloom_2026_erdos_problem_251_discussion.

Monotonicity. The author's comment adds: "It is possible that Erdős also wanted $(g_n)$ to be increasing, but then it would be weird to emphasize $g_n\geq2$ and switch the notation from $a_n$ to $g_n$ for this particular problem in [Er88c]." The construction neither assumes nor produces a monotone sequence, so the theorems on monotone denominators, Section 3 of Erdős's 1958 paper (card; $1<q_1\le q_2\le\cdots$ with a growth hypothesis, rational only when $q_n=qp_n+1$ eventually) and Theorem 5.1 of Hančl–Tijdeman 2004, as numbered in the authors' preprint (card; monotonic $a_n$ with $p_n=o(a_n^2)$, rational only when $p_n/(a_n-1)$ is eventually constant), are untouched; whether Erdős meant nondecreasing $g_n$ in (2) is not settled by the sources.

Construction. With $m_n=1+\lfloor\log\log(n+3)\rfloor$, choose $N$ so that $1+\sum_{k=N}^{n-1}m_k<p_n$ for $n\ge N$, put $c_1=\cdots=c_N=1$, choose $c_{n+1}$ as the unique element of $\{m_n,\dots,m_n+c_n-1\}$ with $c_{n+1}\equiv-p_n\pmod{c_n}$, and set $g_n=(p_n+c_{n+1})/c_n$. Then $g_n$ is an integer, $g_n\ge2$, $g_n=o(p_n)$, and $p_n/(g_1\cdots g_n)=c_n/G_{n-1}-c_{n+1}/G_n$ with $G_n=g_1\cdots g_n$, so the partial sums telescope to $1-c_{M+1}/G_M\to1$. The only input about primes is the prime number theorem, used through $p_n\sim n\log n$; $p_n/(n\log\log n)\to\infty$ would suffice.

Source: https://web.math.pmf.unizg.hr/~vjekovac/files/Erdos_problem_251.pdf.

Bears on. #251, as a claimed counterexample to the auxiliary variable-denominator statement quoted on the problem page; it is not a result on the problem itself.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.