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ERDőS #683 · PARTIAL

Erdős problem 683 — live-page audit, reduction, and verified finite progress

Accessed and computed on 2026-07-27 UTC.

Claim labels used throughout:

unproved input.

only on the accurately quoted published theorem.

search conclusion, not a theorem.

standalone checker, not a statement about all integers.

0. Mandatory live-page gate

(d) I fetched the live problem page

through the Bright Data browser route (direct datacenter access is

Cloudflare-walled) and visually inspected a full-page screenshot. The page

showed OPEN, 0 claimed proofs, **Currently working on this problem:

None, and Interested in collaborating: None**. It also showed None for

likes, “looks difficult,” “looks tractable,” “results could be formalisable,”

and “working on formalising.” Thus the mandatory stop condition did not

trigger.

The exact live statement was:

> Is it true that for every \(1\leq k\leq n\) the largest prime divisor of

> \(\binom{n}{k}\), say \(P(\binom{n}{k})\), satisfies

> \[ > P\left(\binom{n}{k}\right)\geq \min(n-k+1, k^{1+c}) > \]

> for some constant \(c>0\)?

(a) There is a literal endpoint defect: at \(k=n\),

\(\binom{n}{n}=1\), whose largest prime divisor is normally undefined. All

results below either restrict to \(1\leq k\leq n-1\), or use the common

convention \(P(1)=1\), under which the endpoint is true.

Results and remarks listed on the live page

(b) The page lists the Sylvester–Schur theorem

\[ P\!\left(\binom nk\right)>k\qquad(k\leq n/2), \]

citing Erdős's 1934 paper.

(b) It lists Erdős's 1955 bound

\[ P\!\left(\binom nk\right)\gg k\log k\qquad(k\leq n/2). \]

(c) It reports Erdős's view that the displayed problem should hold for

every fixed \(c>0\), with only finitely many exceptions depending on \(c\),

and the prime-gap heuristic

\[ P\!\left(\binom nk\right)>\exp(c\sqrt{k}) \]

in the lower half. These are explicitly conjectural, not known results.

(a) The page says the problem is “essentially equivalent” to

problem 961. Section 2 below makes that

equivalence precise, including the quantifiers and constants.

All six live comments

(d) I also read the complete

discussion thread:

1. On 3 December 2025, Alfaiz observed that the then-strict version was

contradicted by \(n=2^t,\ k=n-1\), because both sides equal \(2\).

2. Terence Tao replied that the intended form was apparently

\(\geq\min(n-k+1,\ldots)\), rather than the old strict inequality.

3. Thomas Bloom confirmed the typo, explained that \(n-k+1\) only handles the

upper half, and said the substantive problem is the \(k\leq n/2\) case,

equivalent to 961.

4. On 31 December 2025, seewoo5 noted that \(P(1)\) is undefined at \(k=n\);

the thread says the site was updated, although the current displayed

quantifier still includes \(k=n\).

5. On 7 January 2026, seewoo5 asked about a stray \(c\) in the description of

the 1955 result and could not locate the cited 1979 passage.

6. Bloom replied that the relevant source is Er79d, page 74, and that the

extra \(c\) was a typo.

(d) None of those comments claims a proof of the present non-strict

statement, and none is a current-worker marker.

1. Primary-source and current-literature audit

(b) The following sources were opened and checked, rather than inferred

from titles or database snippets.

“A theorem of Sylvester and Schur”,

J. London Math. Soc. 9 (1934), 282–288.

“On consecutive integers”,

Nieuw Arch. Wisk. (3) 3 (1955), 124–128. Its Theorem 1 defines the same

gap function used below and proves \(f(y)\ll y/\log y\).

[“Problems and results on number theoretic properties of consecutive

integers and related questions”](https://users.renyi.hu/~p_erdos/1976-39.pdf),

Proc. Fifth Manitoba Conference on Numerical Mathematics (1975), 25–44.

Its formulas (11)–(13) contain the \(k\log k\) result, the proposed

\(k^{1+c}\) replacement, and the stronger prime-gap heuristic.

[“Some unconventional problems in number

theory”](https://users.renyi.hu/~p_erdos/1979-23.pdf),

Acta Math. Acad. Sci. Hungar. 33 (1979), 71–80. Formula (6) on printed

page 74 is the conjecture on the live page and is followed by the

every-\(c\), finitely-many-exceptions remark.

[“On gaps between numbers with a large prime

factor”](https://doi.org/10.4064/aa-24-1-99-111),

Acta Arith. 24 (1973), 99–111.

[“Lower bounds for the greatest prime factor of product of consecutive

positive integers”](https://doi.org/10.1016/j.jnt.2015.07.014),

J. Number Theory 159 (2016), 307–328. The exact theorem and its 39-pair

exception set are also reproduced as Lemma 3.3 of T. N. Shorey and

S. B. Sinha,

[“Extension of Laguerre polynomials with negative

arguments”](https://arxiv.org/abs/2103.02353).

[“On the interval containing at least one prime

number”](https://doi.org/10.3792/pja/1195570997),

Proc. Japan Acad. 28 (1952), 177–181.

[“Arithmetic properties of blocks of consecutive

integers”](https://arxiv.org/abs/1612.05438), a survey that records the

piecewise state of the art and makes clear where the uniform estimates

revert to logarithmic improvements.

[“Integers without large prime

factors”](https://jtnb.centre-mersenne.org/item/JTNB_1993__5_2_411_0/),

J. Théorie des Nombres de Bordeaux 5 (1993), 411–484, which records the

gap-function estimate used in Section 2. Its bibliography verifies

M. Jutila, “On numbers with a large prime factor II,” *J. Indian Math.

Soc.* (N.S.) 38 (1974), 125–130; I did not locate an open primary scan of

that particular paper.

(c) Searches through 27 July 2026 did not locate a paper claiming the

fixed-power saving for the worst-case gap function isolated in Section 2.

This is a search result, not proof of absence. In particular, I screened two

very recent adjacent papers: van Doorn–Tang,

arXiv:2606.19863, controls blocks avoiding

primes specifically in \((k,2k)\), and Yang,

arXiv:2607.16032, proves density results

about the relative order of \(P^+(n)\) and \(P^+(n+1)\). Neither supplies a

worst-case, all-block power saving.

2. Exact reduction to blocks and to problem 961

Let \(P(m)\) denote the largest prime factor of \(m>1\), and put \(P(1)=1\)

only when discussing the endpoint. In the lower half, set

\[ x=n-k+1,\qquad M(x,k)=\max_{0\leq iThen \(x>k\).

Lemma 1: the binomial coefficient loses no relevant prime

(a) For \(1\leq k\leq n/2\),

\[ \boxed{\quad P\!\left(\binom nk\right)=M(n-k+1,k).\quad} \tag{1} \]

Proof. Write

\[ \binom nk=\frac{x(x+1)\cdots(x+k-1)}{k!}. \]

Sylvester–Schur gives a prime \(q>k\) in the numerator block. No prime

larger than \(k\) divides \(k!\), so every numerator prime larger than \(k\)

survives cancellation. In particular, the largest numerator prime

\(M(x,k)>k\) survives. Conversely every prime of the binomial coefficient

comes from the numerator. This proves (1). \(\square\)

Lemma 2: the upper half is automatic

(a) If \(n/2 \[ P\!\left(\binom nk\right) =P\!\left(\binom nj\right)>j. \]

The left side is an integer, so it is at least \(j+1=n-k+1\), which is at

least the required minimum. Thus only \(k\leq n/2\), equivalently blocks

\(x,x+1,\ldots,x+k-1\) with \(x>k\), contain any substance.

Exact gap-function equivalence

Define

\[ f(y)=\min\left\{L:\ \text{every length-\(L\) block beginning at an integer \(x>y\) has a prime factor \(>y\)}\right\}. \tag{2} \]

(a) If problem 683 holds with an exponent \(c>0\), then, for all

sufficiently large \(y\),

\[ \boxed{\quad f(y)\leq \left\lfloor y^{1/(1+c)}\right\rfloor+1.\quad} \tag{3} \]

Indeed, take \(L=\lfloor y^{1/(1+c)}\rfloor+1\). For a block starting at

\(x>y\), both \(x\) and \(L^{1+c}\) exceed \(y\), so the block form of 683

forces a prime factor \(>y\).

(a) Conversely, suppose that for some \(C>0,\delta>0\),

\[ f(y)\leq C y^{1-\delta} \tag{4} \]

for all sufficiently large \(y\). Choose any \(c>0\) satisfying

\[ (1+c)(1-\delta)<1 \quad\left(\text{equivalently }c<\frac{\delta}{1-\delta}\right). \]

For a lower-half block, put \(Y=\min(x,k^{1+c})\). If \(M(x,k)

with \(y=\lceil Y\rceil-1\) we have \(M(x,k)\leq y \[ f(y)\leq C y^{1-\delta} \leq Ck^{(1+c)(1-\delta)}for large \(k\). The definition of \(f\) then forces a prime \(>y\) into

the block, a contradiction. Finitely many smaller \(k\) are absorbed by

decreasing \(c\): Sylvester gives \(M(x,k)\geq k+1\), and

\(k^{1+c}\leq k+1\) can be imposed simultaneously for a finite set of

\(k\)'s.

(a) Consequently, the existence assertion in problem 683 is equivalent,

up to harmless finite exceptions and endpoint conventions, to the existence

of a fixed \(\delta>0\) such that

\[ \boxed{\quad f(y)=O(y^{1-\delta}).\quad} \tag{5} \]

This is the precise missing lemma.

(b) The best general estimate cited on problem 961, due to Jutila and

Ramachandra–Shorey, is

\[ f(y)\ll \frac{\log\log\log y}{\log\log y}\frac{y}{\log y}. \tag{6} \]

This is \(y^{1-o(1)}\), not \(O(y^{1-\delta})\) for a fixed \(\delta\).

(a) Substituting \(y=k^{1+c}\) into (6) does not reach \(f(y)\leq k\):

apart from iterated logarithms, its ratio to \(k\) is

\(k^c/\log k\), which tends to infinity. This identifies exactly why the

standard general machinery stalls.

3. A uniform theorem for every \(k\leq 2,845,920\)

Proposition

(b) Let

\[ K=2,845,920. \]

For every \(n\) and every \(1\leq k\leq\min(n,K)\), problem 683 holds with

\[ \boxed{c=\frac1{10}},\qquad P\!\left(\binom nk\right)\geq \min(n-k+1,k^{11/10}), \tag{7} \]

where either \(k

prime-interval theorem and Nair–Shorey's greatest-prime-factor theorem; all

their finite exceptional cases are checked independently by the verifier.

Proof

(a) The upper half and \(k=1\) follow from Lemma 2 and directly,

respectively. It remains to consider \(2\leq k\leq n/2\). Set

\[ x=n-k+1,\qquad m=x+k-1=n. \]

The relevant product is

\[ {}_k(m)=m(m-1)\cdots(m-k+1)=x(x+1)\cdots(x+k-1). \]

(b) Nair–Shorey prove that if \(m>4k\), \(k\geq2\), and

\((m,k)\notin T\), then

\[ P({}_k(m))>4.42k=\frac{221}{50}k, \tag{8} \]

where \(T\) is their explicit 39-pair set. The integer \(K\) is certified

exactly by

\[ 50^{10}K\leq221^{10}<50^{10}(K+1). \]

Therefore, for every \(k\leq K\),

\[ k^{1/10}\leq\frac{221}{50}, \qquad P({}_k(m))>\frac{221}{50}k\geq k^{11/10}. \tag{9} \]

(d) The verifier trial-factors every block belonging to \(T\). Here is

the complete certificate, written as \(m\mapsto M(m-k+1,k)\):

| \(k\) | exceptional endpoints \(m\mapsto M\) |

|---:|:---|

| 2 | \(9\mapsto3,\ 14\mapsto13,\ 15\mapsto7,\ 20\mapsto19,\ 24\mapsto23,\ 27\mapsto13,\ 35\mapsto17,\ 48\mapsto47,\ 49\mapsto7,\ 63\mapsto31,\ 80\mapsto79,\ 125\mapsto31,\ 224\mapsto223,\ 2400\mapsto2399,\ 4374\mapsto4373\) |

| 3 | \(13\mapsto13,\ 14\mapsto13,\ 20\mapsto19,\ 24\mapsto23,\ 25\mapsto23,\ 26\mapsto13,\ 48\mapsto47,\ 54\mapsto53,\ 63\mapsto61,\ 64\mapsto31,\ 98\mapsto97,\ 350\mapsto349\) |

| 4 | \(24\mapsto23,\ 25\mapsto23,\ 32\mapsto31,\ 33\mapsto31,\ 48\mapsto47,\ 49\mapsto47,\ 63\mapsto61\) |

| 5 | \(24\mapsto23,\ 32\mapsto31,\ 48\mapsto47\) |

| 7 | \(29\mapsto29,\ 30\mapsto29\) |

(d) Direct exact tests show

\[ M(m-k+1,k)\geq \min(m-k+1,k^{11/10}) \]

for all 39 rows. The comparison is performed as

\(M^{10}\geq k^{11}\), never with floating point.

(b) Now suppose \(m\leq4k\). Then

\[ x=m-k+1\leq3k+1. \]

If \(x\geq25\), Nagura gives a prime \(p\) with

\[ xSince

\[ p-x<\frac{x}{5}\leq\frac{3k+1}{5}\leq k, \]

the integer prime \(p\) lies in \(x,\ldots,x+k-1\). Hence

\(M(x,k)\geq p>x\), making (7) automatic.

(d) The only remaining branch is

\[ 3\leq x<25,\qquad 2\leq kThere are exactly 196 such pairs. The standalone verifier trial-factors all

of them and checks \(M\geq x\) or \(M^{10}\geq k^{11}\). There are no

failures.

(b) These cases exhaust all \(n,k\), proving (7) modulo the two named

published theorems. \(\square\)

4. Exact finite computation through \(n=5,000,000\)

Result

(d) For every

\[ 1\leq n\leq5,000,000,\qquad 1\leq k\leq n, \]

with \(P(1)=1\) at \(k=n\), the stronger bound

\[ \boxed{\quad P\!\left(\binom nk\right)\geq \min\left(n-k+1,k^{\,1+4649/10000}\right) \quad} \tag{10} \]

holds. Thus the verified finite exponent is

\[ c=\frac{4649}{10000}=0.4649. \]

(a) A universal exponent cannot exceed

\[ c_*=\frac{\log5}{\log3}-1 =0.464973520717927\ldots \tag{11} \]

because

\[ \binom{10}{3}=120,\qquad P(120)=5,\qquad n-k+1=8. \]

For every \(c>c_*\), \(3^{1+c}>5\), so the required lower bound fails at

\((n,k)=(10,3)\). The finite computation's exponent is only

\[ c_*-0.4649=0.000073520717927\ldots \]

below this unavoidable global ceiling.

Exhaustive algorithm

(a) By Lemmas 1 and 2 it suffices to scan all integer pairs

\[ x>k\geq2,\qquad x+k-1\leq N. \]

For each fixed \(x\), maintain

\[ M_k=\max_{0\leq iOnce \(M_k\geq x\), every longer block with the same start is automatic,

because its required minimum is at most \(x\). Breaking at that point skips

no possible failure.

(d) Largest prime factors on \(1,\ldots,N\) are made by a sieve. Before

the main scan, the verifier independently:

  • compares the sieve with trial division for every integer through 10,000

and at 1,000 deterministic points spread through the remaining range;

  • compares (1) with direct Legendre valuations of \(\binom nk\) for every

lower-half pair with \(n\leq200\);

  • compares the early-break scan with a literal no-break scan through

\(n=300\);

  • verifies the exact Nair–Shorey cutoff, all 39 exceptional blocks, and the

196 small-start blocks.

(a) Floating point is used only to seed an integer root search. Every

actual decision in (10) is certified by

\[ M^{10000}\geq k^{14649}. \]

The root routine adjusts its seed until the exact certificate

\[ (r-1)^{10000}holds.

The core scan in the standalone code is:

for start in range(3, limit):
    max_length = min(start - 1, limit - start + 1)
    running_lpf = 1
    for length in range(1, max_length + 1):
        running_lpf = max(running_lpf, lpf[start + length - 1])
        if running_lpf >= start:
            break
        if length == 1:
            continue
        threshold = thresholds[length]
        if threshold == 0:
            threshold = ceil_rational_power(length, 14649, 10000)
            thresholds[length] = threshold
        if running_lpf < threshold:
            n = start + length - 1
            return threatening_pairs, largest_threatening_k, (n, length, running_lpf)

(d) Full reproducible source:

erdos683_wave6j_reverify.py, 334 lines,

SHA-256

ddf1688b25f3a4263daab2fecb17c51169546ff2bc3d17700ebe65101d3d8aa1.

Run it with:

python3 runs/erdos683_wave6j_reverify.py --limit 5000000

The completed clean run printed:

PASS reduction: direct Legendre valuations agree for every n <= 200
PASS theorem finite inputs: cutoff=2,845,920, Nair-Shorey exceptions=39, small-start pairs=196
PASS sieve: independent trial-factor checks; built in 1.80s
PASS scan logic: optimized and literal scans agree through n=300
PASS exhaustive: c=4649/10000 for every n <= 5,000,000; 53,393,869 nonautomatic lower-half pairs; largest such k=153; scan=31.13s
PASS obstruction: (n,k)=(10,3), P(C(10,3))=5, so universal c <= log(5)/log(3)-1 = 0.464973520717927
CERTIFIED: finite checks supporting c=1/10 through k=2,845,920, and exhaustive c=4649/10000 through n=5,000,000
elapsed=34.50s

(d) Peak resident memory was 42,640 KiB and measured user CPU time was

34.50 seconds. The scan examined 53,393,869 nonautomatic pairs

\(M(x,k)

observation is empirical and must not be extrapolated as a theorem.

5. What remains, precisely

(a) The open uniform step is exactly (5): prove

\[ f(y)=O(y^{1-\delta}) \]

for one fixed \(\delta>0\). Equivalently, prove that every block of

\(k\) consecutive integers exceeding \(k\) contains a prime factor

\(\geq k^{1+c}\) for one fixed \(c>0\), after finitely many \(k\).

(b) The finite-\(k\) theorem stops at \(K\) for a transparent reason.

Nair–Shorey's general complete estimate supplies the fixed linear factor

\(4.42k\); it implies \(k^{11/10}\) exactly while

\(k^{1/10}\leq4.42\), and ceases to do so after

\(k=2,845,920\). Nagura covers the complementary near-diagonal blocks but

does not repair arbitrarily remote blocks.

(a) No scan bounded by \(n\leq N\) can prove the conjecture: it leaves

every block start above \(N\) unchecked, while the conjecture requires one

exponent uniform in both the block start and its length. Even determining

\(f(y)\) exactly for finitely many \(y\)'s would not prove a fixed

\(\delta>0\) for all \(y\). The computation therefore supplies a checked

finite regime and a sharp numerical benchmark, not the missing uniformity

step.

PARTIAL: Reduced #683 exactly to a fixed-power saving for the gap function f(y), proved c=1/10 for every k<=2,845,920 modulo Nagura and Nair--Shorey with all finite exceptions checked, and exhaustively verified c=0.4649 for every n<=5,000,000.

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