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
(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:
1. $a_n=2^{2^n}$ is an example.
2. 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.
3. 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:
1. a uniform Diophantine argument excluding rational points from every
fixed-$B$ dimension-zero coding set, with no bound assumed on a rational
denominator; or
2. 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.