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ERDőS #373 · PARTIAL

Erdős problem 373 — wave8y report

Date of live-page access and computation: 2026-07-28 UTC.

Claim labels

The requested labels are used as follows.

source; I did not re-prove the named theorem.

route, never used as a theorem.

proof of the algorithm is (a), but the fact that the program actually

exhausted its stated range is (d).

Step 0: mandatory live-page audit

I fetched both the problem page and its discussion thread through the Bright

Data browser, not datacenter curl:

The page was last edited 29 January 2026. Its status was OPEN. It showed

0 claimed proofs, Interested in collaborating: None, and **Currently

working on this problem: None**. Therefore the mandatory stop condition did

not apply.

Verbatim current statement

> Show that the equation

> \[n! = a_1!a_2!\cdots a_k!,\]

> with $n-1>a_1\geq a_2\geq \cdots \geq a_k\geq 2$, has only finitely many solutions.

The page attaches [Er76d,p.28], [ErGr80,p.70], and [Er97e,p.537] to the

statement.

Results and qualifications listed on the live page

The following is a faithful audit of the mathematical material on the page;

these are attributed page claims, not new claims of this report.

1. The result would follow from

\(P(n(n+1))/\log n\to\infty\), where \(P(m)\) is the largest prime factor.

The page also says Erdős [Er76d] showed it would follow from

\(P(n(n-1))>4\log n\).

2. The strict condition \(a_1

obtained when \(n=a_2!\cdots a_k!\).

3. Surányi conjectured that the only nontrivial two-factor identity is

\(6!7!=10!\). Hickerson conjectured the complete nontrivial list

\[ 9!=2!3!3!7!,\quad 10!=6!7!,\quad 10!=3!5!7!,\quad 16!=14!5!2!. \]

4. (b) Luca [Lu07b] proved finiteness conditional on the ordinary

\(abc\) conjecture. Unconditionally, the page gives the exceptional-set

estimate

\[ \#\{n\le x:n!\text{ has a nontrivial representation}\} \le \exp\!\left(f(x)\frac{\log x}{\log\log x}\right) \]

for every \(f(x)\to\infty\).

5. Guy discusses the problem as B23.

6. (b) For \(k=2\), the page attributes to Erdős the bound

\(a_1\ge n-5\log\log n\) for sufficiently large \(n\). It says Bhat and

Ramachandra replace 5 by

\((1+o(1))/\log 2\), and extend the bound to arbitrary \(k\ge2\).

7. (b) For \(k=2\), the page says Caldwell and Habsieger performed

numerical investigations and no solution other than \(10!=6!7!\) occurs

for \(n\le10^{3000}\).

The live LaTeX bibliography has an internal metadata inconsistency: its

[Er93] entry is a graph-theory paper, while the discussion comment and later

primary papers identify the factorial result as Erdős, item 6669, *A

consequence of a factorial equation, American Mathematical Monthly* 100

(1993), 407–408. I did not silently merge those two records.

All three live comments and markers

The discussion thread contained exactly three comments:

1. Alfaiz, 28 Jan 2026: the old [Ha] link loaded J. A. Haight's thesis

rather than L. Habsieger's paper; the comment is marked as addressed by a

site update.

2. Alfaiz, 25 Oct 2025: a summary of the two-factor literature—Caldwell

through \(C\le10^6\), Erdős's \(5\log\log C\) bound, Bhat–Ramachandra's

asymptotic constant, and Habsieger's \(B\le10^{3000}\) check and explicit

bound

\[ C-B\le\frac{\log\log(B+1)}{\log2}-0.8803. \]

This is also marked as addressed by a site update.

3. Dogmachine, 9 Aug 2025: a comment that the question is “essentially

equivalent” to a formulation involving Jordan–Pólya numbers. The site

explicitly warns that comments are unverified. I therefore classify this

equivalence assertion as (c) and do not use it.

The remaining markers were: likes by hediibl and Dogmachine; “looks

difficult” by Dogmachine; “looks tractable: None”; “results could be

formalisable: None”; and “working on formalising: None”. The external-data

field says the statement itself has been formalised.

Literature audit beyond the live page

I searched by the exact equation, title, author, and DOI, and checked the

primary paper/author abstract or full primary text before recording a claim.

A stronger unconditional bound omitted from the live page

The most important literature finding is that the live page omits a much

stronger arbitrary-\(k\) result.

product of consecutive positive integers, Journal of Number Theory* 159

(2016), 307–328, DOI

<https://doi.org/10.1016/j.jnt.2015.07.014>.

Its primary abstract says that Baker's explicit \(abc\) conjecture gives

the complete Hickerson result.

\(n!=a_1!a_2!\cdots a_t!\), Indagationes Mathematicae* 27 (2016),

634–642, DOI <https://doi.org/10.1016/j.indag.2015.12.002>.

applications, Hardy-Ramanujan Journal* 41 (2018), 143–156,

<https://hrj.episciences.org/5117/pdf>. Section 3.A explicitly states that

Nair–Shorey confirmed Hickerson for \(n\le e^{80}\), unconditionally, and

confirmed it completely under Baker's explicit \(abc\).

Shorey in their primary 2018 paper

<https://publi.math.unideb.hu/paper/2192/download/10_5486_PMD_2018_7582.pdf>.

Thus:

for

\[ n\le e^{80} = 55406223843935100525711733958316612.9248\ldots. \]

the complete list for all \(n\).

This is not an unconditional solution: \(e^{80}\) is a finite cutoff.

The same primary 2018 text records two further consequences of the 2016

papers:

\[ d:=n-a_1\le \left(\frac1{\log2}+0.2658\right)\log\log n. \]

\(P(n+1)\le79\).

It also states the Nair–Shorey consecutive-block theorem:

\[ P(x(x+1)\cdots(x+d-1))>4.42d, \]

apart from \(x=125,224,2400,4374\) for \(d=2\), and \(x=350\) for

\(d=3\).

Other primary checks

533–542, DOI <https://doi.org/10.1017/S0305004107000308>; its publisher

abstract confirms conditional finiteness and unconditional density zero.

<https://arxiv.org/abs/1903.08370>. Its own abstract says the two-factor

conjecture is checked for \(B\le10^{3000}\).

<https://arxiv.org/abs/2512.03188>, was updated in March 2026 and still

describes even the two-factor question as open; it proves sparsity results,

not finiteness.

<https://arxiv.org/abs/2602.23838>, studies a broader equation and obtains

an explicit-\(abc\) result only on a positive-density subset. It does not

settle problem 373.

I found no primary source claiming an unconditional proof of finiteness.

This agrees with the live OPEN status. Search misses are not evidence of

nonexistence, but the two recent 2026 primary sources explicitly continue to

treat the relevant questions as open.

Elementary exact reduction

Everything in this section is (a).

Let \(p(n)\) be the largest prime at most \(n\). Bertrand's postulate gives

\(p(n)>n/2\), and hence \(v_{p(n)}(n!)=1\). If

\[ n!=a_1!a_2!\cdots a_k!, \]

some \(a_i!\) must contain \(p(n)\), so

\[ a_1\ge p(n). \]

If \(p(n)\in\{n-1,n\}\), this contradicts \(a_1\le n-2\). Otherwise every

possible leading index lies in the finite interval

\[ p(n)\le a_1\le n-2. \tag{1} \]

Fix such an \(a_1\), and put

\[ x=a_1+1,\qquad d=n-a_1\ge2,\qquad R=\frac{n!}{a_1!}=x(x+1)\cdots(x+d-1). \tag{2} \]

The original equation for this \(a_1\) is equivalent to asking whether \(R\)

is a product of factorials \(b!\) with \(2\le b\le a_1\). Moreover, every

integer in the block in (2) must be composite: a prime in the block would be

larger than every \(a_i\), so it could not occur on the right side.

Largest-residual-prime lemma

Let \(q=P(R)\), and let \(q^+\) be the next prime after \(q\). In any

factorial decomposition of \(R\), choose its largest remaining factorial

index \(b\). Some factorial must supply \(q\), while no factorial may supply

\(q^+\). Therefore

\[ q\le bIn particular,

\[ q!\mid R. \tag{4} \]

Projecting (4) to the 2-adic valuation and using Legendre's formula gives the

especially cheap necessary test

\[ q-s_2(q)=v_2(q!)\le v_2(R), \tag{5} \]

where \(s_2(q)\) is the number of 1-bits of \(q\).

Complete recursion for one block

Represent \(R\) by its exact vector \((v_r(R))_{r\ {\rm prime}}\).

1. If the vector is zero, return the empty factorization.

2. Let \(q\) be its largest supported prime.

3. Try every integer

\[ q\le b\le\min\{B,q^+-1\} \]

in descending order, where \(B\) is the preceding chosen index.

4. Use Legendre's formula to retain exactly those \(b\) for which

\(b!\mid R\), subtract the valuation vector of \(b!\), and recurse with

new cap \(b\).

5. If only a power of 2 remains, the unique possible tail is \(2!\) repeated

\(v_2(R)\) times.

This recursion is exhaustive by induction on \(R\): (3) contains the largest

factorial index of every possible decomposition, every member of that

interval is tried, and division by \(b!\ge2\) strictly decreases the state.

The descending cap produces each non-increasing tuple once.

Consequently, (1), (2), and this recursion are an exact decision procedure

for every fixed finite range of \(n\); no logarithmic or floating-point

comparison is involved.

For context, counting multiples of \(2^j\) in a block also gives the

elementary estimate

\[ v_2(R)\le d+\lfloor\log_2 n\rfloor. \]

Together with (5), it yields

\[ q\le d+2\lfloor\log_2 n\rfloor+1. \tag{6} \]

This upper bound is far too weak to close the problem, but it makes the

remaining gap explicit.

Exact finite computation

The standalone verifier is:

runs/erdos373_wave8y_reverify.py

It uses only the Python standard library. Its two logically distinct checks

are:

1. the full prime-valuation recursion above through the requested limit; and

2. an independent big-integer divisibility recursion through \(n=24\), plus

direct math.factorial equality and an independently generated

trial-division prime list for every reported tuple.

Run:

python3 runs/erdos373_wave8y_reverify.py

The default limit is \(10^6\). On this VM with Python 3.12.3 it completed in

33.234 seconds and used 57,572 KiB maximum resident memory.

Exhaustion statistics

These are (d), with algorithmic completeness supplied by the proof (a).

  • Range: \(2\le n\le1,000,000\).
  • Values of \(n\) eliminated immediately by \(n\) or \(n-1\) prime:

156,995.

  • Remaining \(n\): 843,004.
  • Candidate leading indices \(a_1\) tested: 9,097,742.
  • Candidates rejected by the root \(v_2(q!)\le v_2(R)\) test: 9,097,730.
  • Root \(v_2\) survivors: exactly 12.
  • Of those 12, six fail the full necessary test \(q!\mid R\).
  • Only 13 factorial-fit branches were needed after those filters.
  • Certificate SHA-256 of the canonical JSON payload:

bc59468c663e5ced367279ae1d7b742a2a240b9fb61f97208ee8825c6f38f4f3.

Complete 12-row survivor table

Here \(R=(a_1+1)\cdots n\). “No (recursive)” means \(q!\mid R\) but the exact

factorial recursion rejects the quotient(s).

| \(n\) | \(a_1\) | \(R\) | \(q=P(R)\) | \(q!\mid R\)? | exact outcome |

|---:|---:|---:|---:|:---:|:---|

| 9 | 7 | \(72=2^3 3^2\) | 3 | yes | \(R=3!3!2!\) |

| 10 | 7 | \(720=2^4 3^2 5\) | 5 | yes | \(R=6!=5!3!\) |

| 16 | 14 | \(240=2^4 3\cdot5\) | 5 | yes | \(R=5!2!\) |

| 16 | 13 | \(3360=2^5 3\cdot5\cdot7\) | 7 | no | rejected |

| 25 | 23 | \(600=2^3 3\cdot5^2\) | 5 | yes | no (recursive) |

| 49 | 47 | \(2352=2^4 3\cdot7^2\) | 7 | no | rejected |

| 50 | 47 | \(117600=2^5 3\cdot5^2 7^2\) | 7 | no | rejected |

| 64 | 62 | \(4032=2^6 3^2 7\) | 7 | no | rejected |

| 81 | 79 | \(6480=2^4 3^4 5\) | 5 | yes | no (recursive) |

| 225 | 223 | \(50400=2^5 3^2 5^2 7\) | 7 | yes | no (recursive) |

| 2401 | 2399 | \(5762400=2^5 3\cdot5^2 7^4\) | 7 | no | rejected |

| 4096 | 4094 | \(16773120=2^{12}3^2 5\cdot7\cdot13\) | 13 | no | rejected |

The three recursive failures are transparent:

  • \(600/5!=5\), not a product of factorials \(\ge2!\).
  • \(6480/6!=9\), while \(6480/5!=54\); neither remainder is a product of

factorials.

  • The 2-adic exponent forces the largest factor of \(50400\) to be \(7!\),

and \(50400/7!=10\), not a factorial product.

Thus the exact solutions through \(10^6\) are (d):

\[ \begin{aligned} 9!&=7!3!3!2!,\\ 10!&=7!6!,\\ 10!&=7!5!3!,\\ 16!&=14!5!2!. \end{aligned} \]

The independent big-integer enumerator agrees on every \(n\le24\), and each

displayed identity is also rechecked directly as an integer equality and

prime by prime.

This computation is not advertised as a new numerical frontier: the

published Nair–Shorey \(e^{80}\) theorem is vastly stronger. Its value is a

short from-scratch audit, an exact small-case table, and a reproducible

valuation certificate.

Exactly what remains

Combining the elementary reduction with the named results, any unknown

solution must satisfy all of the following:

1. (b) \(n>e^{80}\) and \(P(n+1)>79\).

2. (a) \(p(n)\le a_1\le n-2\).

3. (b) With \(d=n-a_1\),

\[ 2\le d\le\left(\frac1{\log2}+0.2658\right)\log\log n. \]

4. (a) The block \(x=a_1+1,\ldots,x+d-1=n\) consists entirely of

composite integers.

5. (b) For \(q=P(R)\), the large-\(n\) case of the consecutive-block

theorem gives \(q>4.42d\).

6. (a) The entire vector inequality

\[ v_r(R)\ge v_r(q!)\quad\text{for every prime }r\le q \]

must hold; equivalently \(q!\mid R\). In particular (5) must hold.

7. (a) After choosing some

\(b\in[q,q^+-1]\), the residual valuation vector must pass the same test

recursively.

This is a clean exact reduction, but not a finiteness proof.

Precise wall for this route

The missing ingredient is a uniform valuation-deficit lemma, not another

finite search. A sufficient lemma would be:

> For all sufficiently large admissible composite blocks in the

> Nair–Shorey \(d=O(\log\log n)\) range, if \(q=P(R)\), then there is a prime

> \(r\le q\) with \(v_r(R)

That would force \(q!\nmid R\) and finish the problem up to a finite check.

The experiment shows why it is attractive: \(r=2\) rejects all but 12 of

9,097,742 candidates through \(10^6\), and \(r=3\) or \(5\) rejects another

six.

However, the displayed lemma is (c); I found no theorem proving it

uniformly. Existing machinery supplies a lower bound on the largest prime

\(q\), such as \(q>4.42d\), but it does not supply the required deficit in

one of the many small-prime valuations. There is no contradiction between

\(q>4.42d\) and (5), because the elementary upper bound for \(v_2(R)\)

contains a \(\log n\) term, much larger than \(d=O(\log\log n)\). This is the

specific place the route stalls.

Brute force cannot bridge the published cutoff. At the measured

\(\approx2.76\times10^5\) leading states per core-second, merely touching one

state for every \(n\le e^{80}\) would already take more than

\(2.0\times10^{29}\) core-seconds, about \(6.4\times10^{21}\) core-years, or

roughly \(2.2\times10^{24}\) USD at 0.04 USD/core-hour. The real cost is

higher because each \(n\) has multiple possible \(a_1\). No such computation

was attempted.

PARTIAL: The problem remains open; primary literature gives the omitted unconditional Hickerson classification through \(n\le e^{80}\), while an exact valuation recursion independently finds only the four known tuples through \(10^6\) and isolates the missing uniform small-prime valuation-deficit lemma.

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