Erdős problem 293 — live-page audit and exact finite computation
Access date: 2026-07-26 (UTC).
Claim labels used below:
- (a) elementary-rigorous: a complete argument is given here.
- (b) rigorous-modulo-named-theorem: the named published result is used.
- (c) plausible/structural-unverified: not asserted as a theorem.
- (d) computational-only: certified by the accompanying exact program, but not a hand proof of every finite case.
0. Mandatory live-page gate
I fetched the live problem page, its LaTeX view, and the discussion thread through the Bright Data browser path rather than datacenter curl.
The live page says OPEN, 0 claimed proofs, and Currently working on this problem: None. It lists Woett and Quanyu_Tang under both “Likes this problem” and “Interested in collaborating”; an interest marker is not the page's current-worker marker. Thus none of the requested stop conditions is present.
The exact statement in the page's LaTeX view is:
Let $k\geq 1$ and let $v(k)$ be the minimal integer which does not appear as some $n_i$ in a solution to\[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}\]with $1\leq n_1<\cdots <n_k$. Estimate the growth of $v(k)$.
Definition caveat
(a) Taken literally, “minimal integer” is defective: over all integers there is no minimum missing integer, and over positive integers one has $v(k)=1$ for every $k\geq2$. The third live-page comment says that the intended quantity is “the smallest integer larger than 1,” and the 2026 van Doorn--Tang paper explicitly uses that definition. All results below use
This is the only interpretation consistent with all of the live page's stated bounds.
Results and comments recorded on the live page
The page states the following.
- (b) Results of Bleicher and Erdős are said to imply
$v(k)\gg k!$; the page speculates that the growth could be doubly exponential in $\sqrt{k}$ or even in $k$.
- (a) It records the elementary bound $n_k\leq k u_k$, where
$u_1=1$ and $u_{i+1}=u_i(u_i+1)$, hence $v(k)\leq k c_0^{2^k}$ with Vardi's constant $c_0=1.26408\ldots$.
- (b) It records van Doorn and Tang's
$v(k)\geq \exp(c k^2)$ and their connection with problem 304. It says that an estimate $N(b)\ll\log\log b$ in problem 304 would likely upgrade their method to $v(k)\geq\exp(\exp(c k))$.
There are three comments, which the site itself warns are not verified.
- Quanyu Tang, 2025-12-29: announces joint work with Wouter van Doorn,
arXiv:2512.22083, proving $v(k)\geq\exp(c k^2)$ and describing the two-way connection with problem 304.
- Quanyu Tang, 2025-12-08: uses the optimality of Sylvester's
underapproximation to show $n_k\leq s_k-1$, hence $v(k)\leq s_k=\lfloor c_0^{2^k}+\tfrac12\rfloor$.
- Woett, 2025-12-08: notes
$v(k)\leq kF(k)+2\leq c_0^{(1/5+o(1))2^k}$ using the bound for $F(k)$ in problem 148, and clarifies the intended “larger than 1” definition.
1. Literature check
- (b) The paper named in the live comment exists:
W. van Doorn and Q. Tang, The smallest denominator not contained in a unit fraction decomposition of 1 with fixed length, arXiv 2512.22083v2 (24 May 2026). It was published online by Cambridge University Press on 8 July 2026. Theorem 1.1 is exactly the $\exp(c k^2)$ lower bound. Lemma 2.1 proves $D_k\subseteq D_{k+1}$, and the proof uses Vose's $N(b)\ll\sqrt{\log b}$ theorem.
- The same paper says that extracting the older claimed $v(k)\gg k!$ bound
from the Bleicher--Erdős papers “does not seem straightforward” to its authors. I located the primary 1975 paper, The number of distinct subsums of $\sum_{i=1}^N1/i$, but did not reconstruct that implication from scratch. I therefore do not use it in any new claim here.
- I did not obtain a full searchable copy of the 128-page 1980
Erdős--Graham monograph in this run. The recent peer-reviewed paper quotes its page 35 definition as “the smallest integer $>1$,” which resolves the live statement's omission.
- As an external cross-check only, [OEIS
A006585](https://oeis.org/A006585) gives $1,0,1,6,72,2320,245765$ for the numbers of fixed-length decompositions through length 7, exactly matching the independent enumeration below. OEIS A097048 and Hugo van der Sanden's source and output concern the related but different question of the simplest rational requiring a given number of unit fractions. I do not use that database as a proof about $v(k)$.
The searches found no primary source tabulating the exact values of this specific $v(k)$ through $k=7$. This is a report of the search miss, not a claim that no such table exists.
2. Exact finite result
Let $F(k)$ be the number of increasing $k$-term decompositions of 1. The standalone standard-library verifier exhausts every such decomposition for $1\leq k\leq7$.
(d) Exact table:
| $k$ | $F(k)$ | $ | D_k | $ | $\max D_k$ | $v(k)$ | checked witness containing $v(k)-1$ |
|---|---|---|---|---|---|---|---|
| 1 | 1 | 1 | 1 | 2 | $(1)$ | ||
| 2 | 0 | 0 | -- | 2 | -- | ||
| 3 | 1 | 3 | 6 | 4 | $(2,3,6)$ | ||
| 4 | 6 | 15 | 42 | 11 | $(2,3,10,15)$ | ||
| 5 | 72 | 77 | 1806 | 17 | $(2,3,10,16,240)$ | ||
| 6 | 2320 | 1152 | 3263442 | 103 | $(2,3,7,72,102,8568)$ | ||
| 7 | 245765 | 83231 | 10650056950806 | 733 | $(2,3,7,45,732,4522,24825780)$ |
For orientation, the exhaustive list at $k=4$ is
The code stores a first checked witness for every $2\leq m<v(k)$, not merely the last witness displayed in the table. The SHA-256 digest of the complete ordered solution stream for $k=1,\ldots,7$ is
4dd616335e4ef7df9f44012a8caf8b30d75e0d9c393ae8a4276d1317d1696417.
A certified bracket at $k=8$
(a) Whenever no collision is created, replacing a term by
turns a verified seven-term solution into an eight-term solution.
(d) Applying every such collision-free split to all 245765 verified seven-term solutions, then adjoining the following three independently checked eight-term certificates,
covers every denominator $2\leq m\leq2306$. Thus $v(8)\geq2307$.
(a) For any $m>1$ there is an exact equivalence
Indeed, remove $1/m$ in one direction and adjoin it in the other.
(d) An exhaustive search under this equivalence finds no seven-term decomposition of $27538/27539$ avoiding denominator 27539. Therefore $27539\notin D_8$ and
The upper endpoint agrees with a value suggested by the related OEIS data, but the assertion here comes from the independent forbidden-denominator search.
3. Why the enumeration is exhaustive
This is the elementary proof behind the checker, not a heuristic pruning rule.
Suppose the already chosen denominators end at $p$, the remaining rational is $a/b>0$ in lowest terms, and $t>1$ terms remain. If the next denominator is $x$, then:
- (a) Positivity after taking $1/x$ gives $1/x<a/b$, so
\[ x\geq \left\lfloor\frac ba\right\rfloor+1. \]
- (a) All $t$ remaining denominators are at least $x$, so
\[ \frac ab\leq\frac t x, \qquad x\leq\left\lfloor\frac{tb}{a}\right\rfloor. \]
- (a) Increasing order also requires $x\geq p+1$.
The program tries every integer in exactly this finite interval, replaces the residual by
reduces it by a gcd, and recurses. When one term remains, the only possible completion is $1/b$, and it is accepted only if the reduced numerator is 1, $b>p$, and $b$ is not forbidden. Induction on $t$ proves that no solution is missed; strict ordering gives a unique recursion path, so none is counted twice.
For each emitted tuple the verifier separately recomputes its reciprocal sum with fractions.Fraction, checks positivity and strict order, records all denominators, checks every prefix witness, and compares counts and hashes with hard-coded expected values. It also reruns the missing-value test in the complementary-fraction formulation above.
Reproduction command:
python3 runs/erdos293_wave5r_verify.py
The complete final run took about 2.4 CPU-minutes and under 55 MB RSS on this VM; it ended with ALL CHECKS PASSED.
4. What remains and the precise wall
The finite table does not narrow the asymptotic gap. The current published lower bound is $\exp(c k^2)$, while the strongest upper estimate quoted on the live page is still doubly exponential in $k$.
(b) The van Doorn--Tang method currently uses Vose's bound for the minimum length $N(b)$ of unit-fraction representations. (c) The authors say that the conjectural improvement
would likely, through their method, yield a doubly exponential lower bound for $v(k)$. No argument for that estimate was found here. (b) Conversely, their paper proves that a stronger lower bound for $v(k)$ would improve the corresponding bound for $N(b-1,b)$. This is the exact analytic bottleneck, not a missing amount of small-case search.
(d) A full Python enumeration at $k=8$ would emit the database value $151182379$ solutions. (c) At the measured rate for $k=7$, merely processing that many solutions would cost about 9 core-hours, plausibly 10--15 core-hours including the larger-integer search overhead. I did not run it. The targeted construction and forbidden-denominator search above obtain the stated bracket within the allowed few-minute budget, but proving the exact value of $v(8)$ still requires certificates covering every integer from 2307 up to a missing candidate, or a uniform lemma that supplies those certificates.
PARTIAL: exact v(1..7) = 2,2,4,11,17,103,733 and the independently verified bracket 2307 <= v(8) <= 27539; the asymptotic problem remains open.