Erdoes #264 - working report - Erdős Problem a Day

Erdős problem #264 — wave 8n

Date: 2026-07-28 UTC

Claim labels used below:

(a) elementary-rigorous: proved here from elementary facts, or a direct observation from the cited live page/source.
(b) rigorous-modulo-named-theorem: a stated consequence of an identified theorem in a verified primary source.
(c) plausible/structural-unverified: heuristic diagnosis only.
(d) computational-only: certified by the exact standalone program, but not promoted to a uniform theorem.

0. Mandatory live-page gate

I fetched the rendered live page and its expanded discussion thread through the Bright Data browser path on 2026-07-28. I also fetched the page's LaTeX-source view so that the wording below is not reconstructed from memory.

Verbatim current statement

Let $a_n$ be a sequence of positive integers such that for every bounded sequence of integers $b_n$ (with $a_n+b_n\neq 0$ and $b_n\neq 0$ for all $n$) the sum\[\sum \frac{1}{a_n+b_n}\]is irrational. Are $a_n=2^n$ or $a_n=n!$ examples of such a sequence?

Source: live problem #264 and its LaTeX-source view.

(a; live-page facts) The page is OPEN, was last edited 20 January 2026, lists 0 claimed proofs, and has “Interested in collaborating: None” and “Currently working on this problem: None.” Thus the mandatory collision/skip gate passes.

(a; live-page facts) The known results listed on the page are:

$a_n=2^{2^n}$ is an example.
Kovač and Tao proved that $2^n$ is not an example. More generally, a

strictly increasing positive integer sequence with convergent $\sum 1/a_n$ is not an example when \[ \liminf_{n\to\infty} a_n^2\sum_{k>n}\frac1{a_k^2}>0. \] In particular, bounded successive ratios imply a negative answer.

For every $F$ with $F(n+1)/F(n)\to\infty$, Kovač and Tao construct an

example satisfying $a_n\sim F(n)$.

(a; direct reading of all 15 comments) The comments concern Lean formalizations of the already-known negative result for $2^n$ and the positive result for $2^{2^n}$. Vjekoslav Kovač says that an Aristotle run on the $n!$ case did not succeed and that he believes this part remains open; Boris Alexeev confirms that it did not resolve the factorial problem. The remaining comments discuss the correctness and packaging of the formalizations. No comment claims a factorial proof, a factorial counterexample, or current work on #264.

1. Primary-source literature check

(b) The paper actually exists as Vjekoslav Kovač and Terence Tao, “On several irrationality problems for Ahmes series,” arXiv:2406.17593 (2024). Its full HTML text gives the page's definition as “Type 3 irrationality sequence” in §2.1.3. Theorem 2.5 is the negative criterion quoted above, Corollary 2.6 handles bounded successive ratios, and Theorem 2.7 is the positive asymptotic-existence theorem.

(a; primary-source check) Erdős's paper “On the irrationality of certain series: problems and results,” pp. 102–109, primary PDF, says on p. 105 that this bounded-perturbation definition leaves both $2^n$ and $n!$ undecided. On p. 102 it records the stronger expectation that $\sum 1/(n!+t)$ is transcendental for every integer $t$.

(a; live-page check) The special case

E_-:=\sum_{n\ge2}\frac1{n!-1} \tag{1}

is itself Erdős problem #68. Its live page was still OPEN on 2026-07-28, with 0 claimed proofs. That related page does list a current worker, memeister27; I use #68 only as a hardness reduction below and did not attempt or claim its resolution.

(a; search report) Targeted searches of arXiv and the primary sources for “Type 3 irrationality sequence,” “Erdős problem 264,” bounded factorial perturbations, and $\sum1/(n!+t)$ found no later primary paper resolving the factorial question. The 2026 paper “Irrationality of rapidly converging series: a problem of Erdős and Graham,” arXiv:2601.21442, concerns a different series $\sum1/(a_na_{n+1})$ and does not supply the missing result here. This is a reported search miss, not a claim that no uncatalogued result exists.

2. An exact reduction to the still-open problem #68

(a) Set

b_1=1,\qquad b_n=-1\quad(n\ge2).

This is a bounded, everywhere nonzero integer sequence; also $1!+b_1=2$ and $n!-1>0$ for $n\ge2$. Its sum is exactly

\sum_{n\ge1}\frac1{n!+b_n} =\frac12+\sum_{n\ge2}\frac1{n!-1} =\frac12+E_-. \tag{2}

Consequently:

if $E_-\in\mathbb Q$, then (2) is an explicit counterexample to the

factorial claim in #264;

if $n!$ is a Type 3 irrationality sequence, then $E_-$ must be irrational.

Thus an affirmative answer for $n!$ necessarily contains a solution of open problem #68. Irrationality of (1) alone would not settle #264, since #264 quantifies over every bounded perturbation.

(a) More generally, for every fixed nonzero integer $t$, taking $b_n=t$ after repairing the finitely many forbidden early indices shows that a positive solution of #264 would imply irrationality of the corresponding constant-shift series $\sum1/(n!+t)$, up to a rational finite correction.

3. Why the Kovač–Tao negative criterion misses factorials exactly

Define the quantity in their Theorem 2.5 for $a_n=n!$:

T_n=(n!)^2\sum_{k>n}\frac1{(k!)^2}.

(a) The first tail term gives

T_n\ge \frac1{(n+1)^2}.

For $k=n+j$,

\frac{(n!)^2}{((n+j)!)^2} =\frac1{\bigl((n+1)\cdots(n+j)\bigr)^2} \le\frac1{(n+1)^{2j}},

and hence

\frac1{(n+1)^2}\le T_n \le\sum_{j\ge1}\frac1{(n+1)^{2j}} =\frac1{n(n+2)}. \tag{3}

Therefore $T_n\to0$, whereas Theorem 2.5 requires a positive liminf. This is not a borderline numerical failure: the theorem's hypothesis fails by a factor asymptotic to $n^2$.

(b) Conversely, Theorem 2.7 applies to $F(n)=n!$, since $F(n+1)/F(n)=n+1\to\infty$. It therefore constructs some Type 3 irrationality sequence $a_n\sim n!$. Thus asymptotic growth alone cannot distinguish the exact factorial centers; any resolution must exploit their arithmetic structure.

4. Fixed-bound separation theorem

The next elementary result formalizes the opposite geometry from the interval-filling argument used for $2^n$.

Lemma

(a) Fix $B\ge1$. Let two admissible perturbation tails $(b_k)_{k\ge n}$ and $(c_k)_{k\ge n}$ satisfy $1\le |b_k|,|c_k|\le B$, and suppose their first difference is at $n$. If

n!\ge2B,\qquad n(n+2)>6B, \tag{4}

then

\sum_{k\ge n}\frac1{k!+b_k} \ne \sum_{k\ge n}\frac1{k!+c_k}. \tag{5}

Proof

All denominators from index $n$ onward are positive. Assume $b_n<c_n$. The first difference has magnitude at least

\frac{c_n-b_n}{(n!+b_n)(n!+c_n)} \ge \frac1{(n!+B)^2} \ge\frac4{9(n!)^2}. \tag{6}

At a later index $k$, the largest possible difference is at most

\frac1{k!-B}-\frac1{k!+B} =\frac{2B}{(k!)^2-B^2} \le\frac{8B}{3(k!)^2}. \tag{7}

Also $(n+j)!\ge n!(n+1)^j$, so the sum of all later differences is at most

\frac{8B}{3(n!)^2} \sum_{j\ge1}\frac1{(n+1)^{2j}} =\frac{8B}{3n(n+2)(n!)^2} <\frac4{9(n!)^2} \tag{8}

by (4). The first difference cannot be cancelled by the entire future tail, proving (5).

(a) For $B=1$, both conditions (4) already hold at $n=2$. At $n=1$, $b_1=-1$ is forbidden, so $b_1=1$ is forced. Hence the map

(b_n)_{n\ge2}\in\{-1,+1\}^{\mathbb N} \longmapsto \frac12+\sum_{n\ge2}\frac1{n!+b_n} \tag{9}

is injective.

This does not imply irrationality: an injective Cantor coding can still hit countably many rational points.

5. A dimension-zero structural consequence

(a) For fixed $B$, after fixing a prefix through $N$, the diameter of all possible tails is at most

\begin{aligned} W_N &\le \sum_{k>N} \left(\frac1{k!-B}-\frac1{k!+B}\right)\\ &=\sum_{k>N}\frac{2B}{(k!)^2-B^2}\\ &\le \frac{8B}{3(N!)^2N(N+2)} \qquad(N!\ge2B). \end{aligned} \tag{10}

There are at most $(2B)^N$ length-$N$ prefixes. For every $s>0$, the $s$-dimensional content of this cover is bounded by

(2B)^N W_N^s\longrightarrow0,

because $\log(N!)\sim N\log N$ dominates the linear term $N\log(2B)$. Therefore the compact set of all sums with $1\le|b_n|\le B$ has Hausdorff dimension $0$ (and in particular Lebesgue measure $0$).

This strengthens the separation picture, but dimension zero does not rule out rational points because $\mathbb Q$ itself has dimension zero.

6. Exact finite computation for all $\{-1,+1\}$ perturbations

Standalone verifier:

runs/erdos264_wave8n_verify.py

It uses only Python's exact fractions.Fraction; no floating-point number is used in the certification.

For a fixed prefix $b_1,\ldots,b_N$, with $b_1=1$ and $b_2,\ldots,b_N\in\{-1,+1\}$, every continuation lies in

P_N+ \left[ \sum_{k=N+1}^{M}\frac1{k!+1},\ \sum_{k=N+1}^{M}\frac1{k!-1}+\frac2{M!\,M} \right], \tag{11}

where $P_N=\sum_{n\le N}1/(n!+b_n)$. (a) This is an outer interval because

\sum_{k>M}\frac1{k!-1} \le2\sum_{k>M}\frac1{k!} \le\frac2{M!\,M}. \tag{12}

(a) For each rational interval in (11), the least denominator of any rational it contains is found by the standard continued-fraction “simplest fraction in an interval” recursion. Its correctness follows from the Farey-neighbor fact: if $a/b<c/d$ and $bc-ad=1$, every rational strictly between them has denominator at least $b+d$, with the mediant attaining $b+d$. The recursion follows the common Stern–Brocot path of the two endpoints until the first rational enters the closed interval. The verifier also cross-checks this routine against brute force on an exhaustive small rational grid.

(d) With $M=38$, exhaustive exact enumeration gives:

prefix depth $N$	cylinders $2^{N-1}$	minimum reduced denominator in the outer union (11)
5	16	179
8	128	59,980
10	512	1,918,102
12	2,048	94,284,193
14	8,192	34,268,048,309
16	32,768	2,784,133,960,729
17	65,536	30,550,050,369,307

It follows rigorously from the finite cover that if any sum in (9) is a rational $p/q$ in lowest terms, then

q\ge 30\,550\,050\,369\,307. \tag{13}

The minimum for the outer cover at level 17 is witnessed by the prefix

\[ (1,1,1,1,-1,-1,-1,1,-1,-1,1,-1,1,1,-1,-1,1) \]

and by the rational

\frac{31\,350\,796\,816\,639}{30\,550\,050\,369\,307}

lying in that prefix's outer interval. This does not assert that the rational is attained by an infinite perturbation; it only shows that (13) is the sharp output of this particular certified cover.

Reproduction commands:

python runs/erdos264_wave8n_verify.py
python runs/erdos264_wave8n_verify.py --levels 17 --cutoff 36

Two full $M=38$ runs completed in 113.989 and 114.617 seconds on this VM. An independent $M=36$ run completed in 67.567 seconds and returned the identical level-17 minimum, witness prefix, and rational.

7. Precise wall

(a) The interval-filling machinery that disproves the $2^n$ case reverses direction for factorials. A local digit change is of order $1/(n!)^2$, while all later digit freedom is only of order

\frac{B}{((n+1)!)^2} \asymp \frac{B}{(n+1)^2}\frac1{(n!)^2}.

The ratio tends to zero, producing separated cylinders rather than an interval containing a rational.

(a) The usual proof of the irrationality of $e$ also does not transfer. Multiplication by $N!$ clears $1/n!$ for $n\le N$, but not $1/(n!+b_n)$. In the already-unresolved case $b_n=-1$, every prime divisor of $n!-1$ is larger than $n$: a prime $p\le n$ divides $n!$ and therefore cannot divide $n!-1$. Thus the current denominator is never cleared by the same factorial multiplier.

(c) A generic least-common-multiple attack has the wrong crude scale. The product bound for $\operatorname{lcm}\{n!+b_n:n\le N\}$ has logarithm on the order of $N^2\log N$, while the reciprocal tail has only logarithmic size on the order of $-N\log N$. Closing that approach would require strong, presently unavailable common-factor information for the shifted factorials; the crude LCM bound cannot make a rationally cleared tail lie between 0 and 1.

(a) The exact missing lemma is visible before any arbitrary perturbation is considered: prove or disprove the irrationality of (1), Erdős problem #68. For the full #264 problem, one additionally needs either:

a uniform Diophantine argument excluding rational points from every

fixed-$B$ dimension-zero coding set, with no bound assumed on a rational denominator; or

an explicit bounded digit sequence whose nested factorial cylinders can

be certified to converge to a specified rational.

The finite calculation only raises the possible denominator and cannot supply this uniformity step.

(d; cost estimate) Depth 17 has 65,536 cylinders. Depth 25 would have 16,777,216 and, at the observed exact-arithmetic rate plus larger integer sizes, would cost roughly 6–10 one-core hours. Depth 30 has 536,870,912 cylinders and would cost at least about 150 core-hours, plausibly over 200. Neither finite depth can prove irrationality, so those runs were not made.

PARTIAL: Proved fixed-bound factorial-tail separation and Hausdorff dimension zero, reduced a positive factorial answer to open problem #68, and exactly certified that every rational sum for all admissible $\{\!-1,+1\}$ perturbations has reduced denominator at least $30,550,050,369,307$; the required uniform irrationality step remains open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger