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ERDőS #386 · PARTIAL

Erdős problem #386 — wave 9b

Access date: 2026-07-28 (UTC).

0. Mandatory live-page check

I fetched the live page through the Bright Data browser path before doing any mathematics. The page was not taken from the stale tracker.

A280992.

Thus neither of the mandatory stop conditions applies.

Verbatim live statement

Let \(2\leq k\leq n-2\). Can \(\binom{n}{k}\) be the product of consecutive primes infinitely often? For example \[ > \binom{21}{2}=2\cdot3\cdot5\cdot7. > \]

The page attributes this to Erdős and Graham, Old and New Problems and Results in Combinatorial Number Theory (1980), p. 74. The additional page-grounded facts are:

hopeless, and guessed there were no examples for \(3\leq k\leq n-3\).

\[ \binom73=5\cdot7,\quad \binom{10}4=2\cdot3\cdot5\cdot7,\quad \binom{14}4=7\cdot11\cdot13,\quad \binom{15}6=5\cdot7\cdot11\cdot13. \]

All 13 live comments read

The forum itself warns that comments are not verified. I therefore record them for collision/state awareness but do not promote a comment to a theorem.

  1. Terence Tao (2025-08-10) suggested adapting the methods of his

Singmaster paper to exclude the interior of Pascal's triangle.

  1. Desmond Weisenberg (2025-08-13) pointed to Sander's work on nonsquarefree

interior coefficients.

  1. Weisenberg (2025-08-21) gave an argument for excluding

\(n/3<k\leq n/2\) for large \(n\).

  1. Thomas Bloom (2025-08-21) suggested strengthening this to

\(k\in[\epsilon n,n/2]\).

  1. Weisenberg (2025-08-21) gave a follow-up argument that a solution must

have \(k=o(n)\).

  1. StijnC (2025-08-22) sketched restrictions using primes in

\((n-k,n]\) and \(((n-k)/2,n/2]\), with a conditional Cramér heuristic.

  1. StijnC (2025-08-24) gave a counting heuristic for fixed \(k\), reported

an independent search with no new example, and explicitly caveated a claimed \(k=2\) lower range near \(10^{500}\) by possible machine and estimate errors. I do not use that number.

  1. Dogmachine (2025-08-24) opined that an unconditional fixed-\(k\) result

may require ideas comparable to \(abc\); this is only an opinion.

  1. Tao (2025-08-26) sketched how Proposition 1.12 of arXiv:2106.03335

could force a squared prime when \(k\geq\exp((\log n)^{2/3+\epsilon})\).

  1. Vjeko Kovač (2025-08-27) observed that Granville–Ramaré Theorem 2

already supplies the relevant squarefreeness bound.

  1. StijnC (2025-08-27) commented on Granville–Ramaré's examples and

conjectural polylogarithmic edge scale.

  1. Tao (2025-08-27) noted that Granville–Ramaré also treated the methods

relevant to Erdős problem #175.

  1. Dogmachine (2026-03-31) speculated that “product of consecutive

primes” might originally have meant a primorial, explaining the missed small counterexamples.

There is no claimed proof and no current worker among these comments.

1. Claim labels

I use the requested labels throughout:

2. Primary-source literature check

Published squarefreeness results

[b] Granville and Ramaré, “Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients,” Mathematika 43 (1996), 73–107, DOI 10.1112/S0025579300011608, was inspected in the authors' PDF. Its Theorem 2 says exactly that there is a constant \(\tau_1>0\) such that, for sufficiently large \(n\), squarefreeness of \(\binom nk\) implies

\[ \min(k,n-k) <\exp\!\left(\tau_1(\log n)^{2/3}(\log\log n)^{1/3}\right). \tag{GR} \]

The downloaded 45-page PDF had SHA-256 2fc57e71c90e5246a661464b82a3bd7e33514660e3cf1bb0d6bdadb2ff3fad63. The theorem does not give a numerical \(\tau_1,n_0\) in its displayed statement, so I do not turn it into an explicit finite cutoff.

[b] J. W. Sander, “Prime power divisors of binomial coefficients,” J. reine angew. Math. 430 (1992), 1–20, DOI 10.1515/crll.1992.430.1, was verified to exist in the digitised primary text. Its Theorem 1 gives large prime-power divisors when a coefficient is sufficiently close to the middle. This supports the older forum remark, but (GR) is the sharper result used here.

The Tao et al. paper cited in the comments

[b] arXiv 2106.03335 really is Matomäki, Radziwiłł, Shao, Tao, and Teräväinen, “Singmaster's conjecture in the interior of Pascal's triangle,” Q. J. Math. 73 (2022), 1137–1177. Proposition 1.12 is indeed the stated prime equidistribution estimate for expressions involving \(N/p\) and \(M/p^j\). The downloaded 33-page arXiv PDF had SHA-256 f2058a35ab3cb6f9f1ca6d5a78cdd9416eeb43f218c4277118650be47b63feba. However, the paper does not state Erdős #386's desired result; the squarefreeness application is a forum sketch. I therefore do not cite that sketch as proved.

Current adjacent work and search miss

The May 2026 preprint of Bui, Pratt, and Zaharescu, arXiv:2605.21221, was also checked. It is about divisors of \(\binom nk\) near \(n\), i.e. the adjacent Erdős problem #387, not consecutive-prime products. Searches by the exact phrase, the Erdős–Graham citation, and citations around squarefree binomial coefficients found no primary paper that resolves #386 itself. This is an honest search miss, not a claim that no such paper exists.

I found bibliographic confirmation of the 1980 Erdős–Graham monograph but not a readable primary scan of its p. 74 during this run. Consequently, the exact historical wording above is taken only from the authoritative live page, as the task permits.

3. An elementary reduction

By symmetry, assume throughout that

\[ 2\leq k\leq n/2. \]

Two-interval lemma

Lemma [a]. Suppose

\[ n^{\lfloor k/2\rfloor}>k^k. \tag{1} \]

If \(\binom nk\) is a product of consecutive primes, then at least one of

\[ (n-k,n]\quad\text{and}\quad(n/2,n-k] \tag{2} \]

contains no prime.

Proof. For every prime \(r\in(n-k,n]\), Legendre's formula gives

\[ v_r\binom nk=1, \]

because \(r\) occurs once in \(n!\), and not in \(k!\) or \((n-k)!\). For every prime \(q\in(n/2,n-k]\), the same formula gives

\[ v_q\binom nk=1-0-1=0. \tag{3} \]

Assume both intervals in (2) contain primes. The consecutive prime support contains a prime \(r>n-k\), omits a prime \(q\leq n-k\), and cannot cross that omitted prime. Hence it has no prime factor below \(q\). Equations (3) and the fact that every prime in \((n-k,n]\) divides the coefficient then force the exact identity

\[ \binom nk=\prod_{n-k<p\leq n}p. \tag{4} \]

There are at most \(\lceil k/2\rceil\) primes in the \(k\) integers \((n-k,n]\), since all of them are odd. Also, term by term,

\[ \binom nk =\prod_{i=0}^{k-1}\frac{n-i}{k-i} \geq (n/k)^k. \]

Thus (1) implies

\[ \binom nk\geq(n/k)^k >n^{\lceil k/2\rceil} \geq\prod_{n-k<p\leq n}p, \]

contradicting (4). \(\square\)

For even \(k\), condition (1) is simply \(n>k^2\); for odd \(k\), it is \(n>k^{2k/(k-1)}\). The standalone checker separately audited the Legendre valuations and size comparison for all 6,241 half-triangle pairs through \(n=160\) (70,722 individual valuation checks and 678 activated size obstructions).

What every sufficiently large solution must look like

Corollary [b]. There is a constant \(\tau>0\) such that every sufficiently large solution, represented with \(k\leq n/2\), satisfies

\[ \begin{aligned} k&<\exp\!\left(\tau(\log n)^{2/3}(\log\log n)^{1/3}\right),\\ (n-k,n]\cap\mathbb P&=\varnothing. \tag{5} \end{aligned} \]

Indeed, a consecutive-prime product is squarefree, so the first line is Granville–Ramaré Theorem 2. In particular \(k=n^{o(1)}=o(n)\). The prime number theorem then puts a prime in \((n/2,n-k]\) for all sufficiently large \(n\). Uniformly over the Granville–Ramaré range, \(\log k=o(\log n)\), so (1) also holds. The lemma forces the other interval, \((n-k,n]\), to be prime-free.

This is a clean reduction, not a finiteness proof. It says that a large solution must simultaneously be:

  1. extremely close to an edge of Pascal's triangle;
  2. immediately preceded by a prime gap of length at least \(k\);
  3. squarefree; and
  4. supported on one uninterrupted interval of prime indices.

4. Exact computation

The standalone verifier is erdos386_wave9b_reverify.py. It uses only the Python standard library.

Pascal-triangle scan

[d] The scanner maintains the exact prime-exponent vector via

\[ \binom nk=\binom n{k-1}\frac{n-k+1}{k}. \]

It does not accept its own state as the sole reason for a rejection. For every tested pair it independently checks, using the factorial form of Legendre's formula, one of:

\(q\nmid\binom nk\).

The first certificate violates squarefreeness; the second violates consecutive support. A survivor is recomputed with math.comb, multiplied from the claimed prime interval, and factored again by literal trial division. A completely literal trial-division oracle differentially checks the full prefix through \(n=160\).

Two exact regions were scanned:

| Region (using \(k\leq n/2\)) | Pairs | Square witnesses | Gap witnesses | Survivors | |---|---:|---:|---:|---:| | all \(n\leq5{,}000\), all \(2\leq k\leq n/2\) | 6,245,001 | 6,225,435 | 19,557 | 9 | | all \(n\leq500{,}000\), \(2\leq k\leq32\) | 15,498,977 | 14,014,314 | 1,484,654 | 9 |

After removing the overlap, this is 21,590,001 distinct half-triangle pairs. By symmetry it proves the corresponding statements for \(n-k\) as well.

The nine half-triangle survivors were exactly:

| \((n,k)\) | Exact value and consecutive-prime factorisation | |---|---| | \((4,2)\) | \(6=2\cdot3\) | | \((6,2)\) | \(15=3\cdot5\) | | \((7,3)\) | \(35=5\cdot7\) | | \((10,4)\) | \(210=2\cdot3\cdot5\cdot7\) | | \((14,4)\) | \(1001=7\cdot11\cdot13\) | | \((15,2)\) | \(105=3\cdot5\cdot7\) | | \((15,6)\) | \(5005=5\cdot7\cdot11\cdot13\) | | \((21,2)\) | \(210=2\cdot3\cdot5\cdot7\) | | \((715,2)\) | \(255255=3\cdot5\cdot7\cdot11\cdot13\cdot17\) |

No new example occurred.

Independent \(k=2\) computation

[d] A second algorithm does not scan Pascal rows or use their exponent states. If a consecutive-prime interval has product \(P=\binom n2\), then

\[ 8P+1=(2n-1)^2. \tag{6} \]

Conversely, an odd square in (6) recovers \(n\) exactly. For \(n\leq50{,}000{,}000\), every prime factor is at most \(n\), so the program sieves all endpoint primes and enumerates every consecutive-prime interval whose product is at most \(\binom{50{,}000{,}000}{2}\).

It tested 5,179,663 intervals and found precisely

\[ n=4,6,15,21,715. \]

This independently extends the OEIS/page computation from \(n\leq5{,}000{,}000\) to \(n\leq50{,}000{,}000\), with no additional term.

The sorted survivor-pair SHA-256 is 3231326f2ca05ebb2e43ee40650646ce5c5da3f7115a51b062f916ab0ad045d1.

Reproduction

Run:

python3 runs/erdos386_wave9b_reverify.py

The report-scale run printed VERIFIED, used 123 MB peak resident memory, and took 104.8 CPU-seconds on this VM. A short smoke test is:

python3 runs/erdos386_wave9b_reverify.py --quick

The final verifier source has SHA-256 db8515a13d646c783ad30980e3bafc3a202b146f2e263de99d67cfc2b7d8bf81.

5. Exact wall

Nothing above supplies the uniform finiteness step.

consecutive-prime intervals \(P\) make \(8P+1\) a square. Neither Granville–Ramaré nor the prime number theorem supplies such a Diophantine statement.

interior squarefreeness machinery no longer applies. Prime gaps of growing size genuinely occur, so “\((n-k,n]\) is prime-free” cannot by itself be contradicted.

prime-gap condition in (5) forces either a squared prime divisor of \(\binom nk\) or a missing prime inside its support, uniformly in \(n\). This is the exact unsupported step; no such theorem was found.

The certified Pascal scan processes about \(2.0\)–\(2.9\times10^5\) pairs per core-second. A full half-triangle scan through \(n=100{,}000\) would contain about \(2.5\times10^9\) pairs and cost roughly 3.5 core-hours with this independently checked implementation; through \(10^6\) it would cost hundreds of core-hours. I did not run those computations. More finite search would strengthen a cutoff but would not address the missing uniformity lemma.

PARTIAL: Proved an elementary prime-gap reduction and, modulo Granville–Ramaré plus PNT, isolated all sufficiently large solutions to a thin prime-free edge regime; exact certified searches found only the nine known half-triangle examples through \(n\le5000\) globally, through \(n\le500000\) for \(k\le32\), and through \(n\le50000000\) for \(k=2\), but no uniform finiteness proof.

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