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ERDőS #1073 · PARTIAL

Erdős problem #1073 — wave 8c

Date of audit and computation: 2026-07-28 (UTC).

Claim labels used below:

0. Mandatory live-page audit

I fetched both the rendered problem page and its LaTeX view through the Bright

Data browser, rather than relying on the supplied tracker YAML.

The exact current statement is:

> Let \(A(x)\) count the number of composite \(u

Source: live page and

LaTeX view.

(d) Page status. The page says OPEN. It displayed:

Thus none of the mandated stop conditions was present.

(d) Context actually listed on the page. It gives the initial values

\(25,121,169,437,\ldots\), links OEIS A256519, recalls Wilson's theorem, cites

Hardy–Subbarao, and says the question occurs in Guy's problem A2. It lists no

partial bound.

1. Source and literature audit

1. (d) G. E. Hardy and M. V. Subbarao,

“A Modified Problem of Pillai and Some Related Questions”,

American Mathematical Monthly 109 (2002), 554–559,

DOI 10.1080/00029890.2002.11919885,

exists and was read. Its Problem F on p. 557 is the source of this question.

It gives \(25,121,721\) as examples but supplies no bound for \(A(x)\).

2. (d) Richard Guy's third-edition

Unsolved Problems in Number Theory

exists; “Prime Numbers” is the chapter containing A2. The live Erdős page

identifies A2 as a mention of this question. I did not have the chapter text

needed to attribute any stronger statement to Guy.

3. (d) OEIS A256519 has the equivalent

definition and a 719-entry b-file

ending at \(99\,795\,337\). It is sequence data, not a proof of an

asymptotic assertion.

4. (d) Exact-title, exact-congruence, DOI-citation, and sequence searches

found no later primary paper specifically proving a bound for this \(A(x)\).

OpenAlex's citation record for the 2002 paper showed only Guy's 2004

“Prime Numbers” chapter. This is an honest search result, not a theorem that

no such literature exists.

5. (b) The adjacent paper of Klurman and Munsch,

“Distribution of factorials modulo \(p\)”

(journal version here),

concerns the number of residue classes occupied or missed by \(n!\bmod p\).

Its one-modulus set-size results do not give an upper bound for occurrences

of the particular residue \(-1\), much less correlations of that occurrence

for two different primes. I therefore do not use it as if it solved the

required counting problem.

2. Elementary exact reductions

Write \(P^-(u)\) for the least prime divisor of \(u>1\).

Lemma 1

(a) If \(u\mid n!+1\), then every prime divisor of \(u\) is greater than

\(n\). In particular,

\[ nProof. If a prime \(p\mid u\) had \(p\leq n\), then \(p\mid n!\) and

\(p\mid n!+1\), which is impossible. \(\square\)

Lemma 2: finite search identity

(a) If \(u \[ nConsequently, with \(s=\lfloor\sqrt{X-1}\rfloor\),

\[ \{uThis identity is exact, not heuristic. The apparently unbounded witness search

therefore terminates at \(n=s-1\).

Lemma 3: how to enumerate without factoring a huge factorial

Fix \(n

dividing \(N\).

(a) If no such \(p\) exists, \(N\) has no composite divisor below \(X\).

If it exists, every prime \(q\) occurring in any composite divisor \(d

\(N\) obeys

\[ q\leq B:=\min\left(\left\lfloor\frac{X-1}{p}\right\rfloor, \left\lfloor\frac{N}{p}\right\rfloor\right). \tag{2} \]

Proof. A composite \(d

Removing one prime factor from \(d\) shows every other prime factor is at most

\((X-1)/r\leq(X-1)/p\); the same argument inside \(N\), which is also

divisible by \(p\), gives \(q\leq N/p\). The least factor itself also satisfies

these bounds whenever a composite divisor is possible. \(\square\)

Now

\[ \gcd\!\left(N,\prod_{q\leq B,\ q\ {\rm prime}}q\right) \]

is precisely the squarefree product of all relevant prime divisors. Exact

division recovers their valuations, after which ordinary divisor generation

recovers every \(d

most 4096 primes so it never has to factor \(N\) or hold one enormous global

primorial.

Core executable logic:

common = gcd(factorial + 1, block_product)
for q in primes_in_block:
    if common % q == 0:
        relevant_primes.append(q)

# Recover v_q(n!+1), then form every product below X.
for q in relevant_primes:
    while residual % q == 0:
        residual //= q
        exponent[q] += 1

The full standard-library-only implementation is

erdos1073_wave8c_reverify.py.

3. A global density-zero bound

This does not reach the requested subpolynomial bound, but it is a rigorous

uniform estimate.

Theorem

(a) As \(x\to\infty\),

\[ A(x)\leq (1+o(1))\frac{x}{\log\log x}. \tag{3} \]

In particular, \(A(x)=o(x)\).

Proof

Let \(L=\log x\), \(\ell=\log\log x\), and, for sufficiently large \(x\), put

\[ y=\left\lfloor\frac{L}{8\ell}\right\rfloor. \]

First discard the fewer than \(\sqrt x\) possible \(u<\sqrt x\).

Suppose \(\sqrt x\leq u \[ \log(n!+1)\leq n\log n+\log 2 \leq y\log y+\log 2 <\tfrac12\log x \]

for all sufficiently large \(x\). This contradicts \(u\leq n!+1\) and

\(u\geq\sqrt x\). Hence \(n\geq y\). Lemma 1 now says that \(u\) has no prime

divisor at most \(y\).

Let \(P(y)=\prod_{p\leq y}p\). Full inclusion–exclusion gives

\[ \begin{aligned} R(x,y) &:=\#\{mNo sieve theorem is hidden here: the error is at most one for each of the

\(2^{\pi(y)}\) squarefree divisors of \(P(y)\).

The finite Euler product contains \(1/k\) for every \(1\leq k\leq y\), so

\[ \prod_{p\leq y}\left(1-\frac1p\right)^{-1} \geq\sum_{k\leq y}\frac1k\geq\log(y+1). \tag{5} \]

Also \(2^{\pi(y)}\leq2^y=x^{o(1)}\). Equations (4)–(5) therefore yield the

fully explicit intermediate bound

\[ A(x)\leq \sqrt x+\frac{x}{\log(y+1)}+2^y. \]

Since \(\log(y+1)=(1+o(1))\log\log x\), (3) follows. \(\square\)

A matching-scale elementary construction only at the logarithmic end

(a) For every prime \(p\geq5\),

\[ u_p=(p-1)!+1 \]

is a distinct composite member of the set, witnessed by \(n=p-1\): Wilson's

theorem gives \(p\mid u_p\), while \(u_p>p\).

(b) Stirling's formula and the prime number theorem therefore give

\[ A(x)\geq (1-o(1))\frac{\log x}{(\log\log x)^2}. \tag{6} \]

This is consistent with \(x^{o(1)}\) and is not a counterexample.

4. Exact computation through \(2\cdot10^8\)

All counts here use the problem's strict convention \(u

Certified count table

(d) The complete enumeration gives:

| \(X\) | exact \(A(X)\) |

|---:|---:|

| \(10\) | 0 |

| \(10^2\) | 1 |

| \(10^3\) | 7 |

| \(10^4\) | 28 |

| \(10^5\) | 60 |

| \(10^6\) | 131 |

| \(10^7\) | 290 |

| \(10^8\) | 719 |

| \(1.1\cdot10^8\) | 754 |

| \(1.2\cdot10^8\) | 779 |

| \(1.3\cdot10^8\) | 808 |

| \(1.4\cdot10^8\) | 825 |

| \(1.5\cdot10^8\) | 850 |

| \(1.6\cdot10^8\) | 879 |

| \(1.7\cdot10^8\) | 902 |

| \(1.8\cdot10^8\) | 921 |

| \(1.9\cdot10^8\) | 941 |

| \(2\cdot10^8\) | 960 |

Thus there are **241 certified values in

\([10^8,2\cdot10^8)\)** beyond the range of the current OEIS b-file. The first

20 are

100227431 100347833 100413697 100839833 100966727
101150461 101671667 101700509 101920531 101960623
102182161 102190469 102607441 102703933 103430519
103480999 103580573 104505013 104729117 104771231

and the final 20 below \(2\cdot10^8\) are

189172153 190002377 190341313 190397173 190460383
191151277 192423271 192734741 194490823 194570603
195611149 196309783 196779521 197160037 197535997
198660673 198674293 199595467 199886243 199899121

For example, direct independent scans give

\[ \begin{array}{c|c|c} u&\text{prime factorization}&\text{unique witness }n\\ \hline 100227431&7541\cdot13291&6645\\ 143067073&211\cdot239\cdot2837&210\\ 199899121&2087\cdot95783&2086. \end{array} \]

Structural finite data

(d) Among the 960 values:

| prime-exponent pattern | count |

|---|---:|

| \(1+1\) | 922 |

| \(1+1+1\) | 23 |

| \(1+1+1+1\) | 6 |

| \(1+1+1+1+1\) | 1 |

| \(2\) | 7 |

| \(2+1\) | 1 |

Exactly eight are nonsquarefree:

\[ 25,121,169,2209,5041,174511,316969,375769. \]

All 960 have exactly one witness \(n\). This last observation is finite data

only; it is not asserted uniformly.

The five-prime example is

\[ 51794551=19\cdot23\cdot29\cdot61\cdot67\mid18!+1. \]

Indeed, a separate multiplication/primality check verified

\[ 18!+1=19\cdot23\cdot29\cdot61\cdot67\cdot123610951. \]

Independent checks and reproducibility

(d) The final run performed all of the following:

  • reconstructed every candidate using (1)–(2);
  • factored every output integer independently by trial division;
  • recomputed \(n!\bmod u\) from 1 for all 960 displayed witness pairs;
  • ran a completely different literal modulus-by-modulus scan through

\(10^6\), where both methods returned the same 131 integers and witnesses;

  • independently matched all 719 OEIS values below \(10^8\), including its last

value \(99\,795\,337\).

Canonical lines u:n1,n2,... for the 960-value certificate have SHA-256

5d32210f884e15884d55629db48eaf11bfded8a5ce88b41fd0e872a695f7a1bf

Run:

python runs/erdos1073_wave8c_reverify.py

The default run uses only the Python standard library. On this VM it used

68.52 seconds wall time and 77,920 KiB maximum RSS. Internally it checked

\(n\leq14141\), sieved 993,933 primes through 15,384,615, processed 3,218

relevant \(n\), and made 9,980 block gcds.

5. What remains, precisely

For a prime \(p\), define

\[ W(p)=\{\,n:1\leq n(a) For distinct primes \(p

if and only if

\[ W(p)\cap W(q)\ne\varnothing. \tag{7} \]

Prime squares require the stronger lift modulo \(p^2\), and products with

three or more prime factors require a common witness for every prime-power

factor.

(d) The finite data show why (7) is the main numerical obstruction:

922 of the first 960 values are squarefree semiprimes.

(c) A random-residue model predicts that these cross-prime coincidences

are very sparse, but that model supplies no uniform error term and is not used

as proof.

The exact missing analytic lemma is already visible in the restricted

semiprime question:

\[ \#\{(p,q):pThe full conjecture would imply (8), yet the literature located above gives no

estimate of this strength. Bounds for how many distinct residues the

factorial sequence occupies modulo one prime do not control the specified

residue \(-1\) simultaneously modulo \(p\) and \(q\). Even proving (8) would

still leave uniform treatment of prime powers and common intersections of

three or more prime-power witness sets.

(d) Computation cost wall. Extending the same exact Python computation to

\(10^9\) would require \(n<31\,623\) and primes through at least \(58\,823\,529\)

(already forced by \(n=16,p=17\)). Extrapolating the measured \(10^8\) and

\(2\cdot10^8\) runs gives roughly 5–15 CPU-minutes

(0.08–0.25 core-hours) and about 0.1–0.3 GB RAM. I did not run it because it

would exceed the requested few-minute budget, and a longer finite table would

not supply the uniform correlation estimate (8).

PARTIAL: Proved the elementary bound \(A(x)\leq(1+o(1))x/\log\log x\) and exactly certified \(A(200000000)=960\), including 241 values beyond the OEIS \(10^8\) table; the \(x^{o(1)}\) claim remains open at the cross-prime factorial-correlation step.

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