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ERDőS #679 · PARTIAL

Erdős problem 679 — wave w009

Date: 2026-07-28 (UTC)

Claim labels

Every substantive conclusion is marked as requested:

theorem;

diagnosis rather than a theorem;

computation, not an asymptotic theorem.

0. Mandatory live-page audit

(a, direct source audit) I accessed the live problem page through the Bright Data browser path on 2026-07-28, before doing any mathematics. I also opened the full discussion thread and the live LaTeX source.

The page showed:

Thus neither mandatory stop condition applied.

Live statement

The first 22 words of the live statement are, verbatim:

Let \(\epsilon>0\) and \(\omega(n)\) count the number of distinct prime factors of \(n\). Are there infinitely many values of \(n\) such that

Here is an exact symbolic transcription of the complete question. For a fixed \(\epsilon>0\), does there exist a cutoff \(K_\epsilon\), depending on \(\epsilon\) only, and infinitely many \(n\) such that

\[ \omega(n-k)<(1+\epsilon)\frac{\log k}{\log\log k} \qquad(K_\epsilon\leq k<n)? \tag{P} \]

The page then asks whether the stronger assertion

\[ \omega(n-k)<\frac{\log k}{\log\log k}+O(1) \qquad(K\leq k<n) \tag{P+} \]

can be shown to be false.

This transcription preserves the page's quantifiers: in particular, the lower cutoff is uniform in \(n\), and “sufficiently large” depends on \(\epsilon\) only. This uniformity is essential.

Results and related problems listed on the live page

(a, direct source audit) The page lists the following.

  1. The analogous \(\Omega\) question replaces

\(\log k/\log\log k\) by \(\log_2 k\), where \(\Omega\) counts prime factors with multiplicity.

  1. DottedCalculator's comment disproves (P+). The page records the stronger

consequence that, for all sufficiently large \(n\), some \(k<n\) obeys \[ \omega(n-k)\geq \frac{\log k}{\log\log k} +c\frac{\log k}{(\log\log k)^2} \] for an absolute \(c>0\).

  1. Cheuk Fung Lau [La26] proves that an absolute \(C>0\) exists such that

infinitely many \(n\) satisfy \[ \omega(n-k)\leq\Omega(n-k)\leq C\log k \qquad(1<k<n). \] The page says Lau conjectures that, for \(\omega\), only the constant can be improved, which would make the answer to (P) negative.

  1. The page links related Erdős problems 248, 413, and 1203.

All four comments

(a, direct source audit) I read all comments rather than relying on the page summary.

  1. DottedCalculator, 11 January 2026. For each \(n\), choose the largest

primorial below \(n\), say \(Q_{m-1}=p_1\cdots p_{m-1}\), and put \(k=n-Q_{m-1}<Q_m\). Then \(\omega(n-k)=m-1\). The asymptotics for \(p_m\) and the Chebyshev function show that this exceeds \(\log k/\log\log k+C\) for every fixed \(C\), once \(m\) is large. The comment says the site was updated to incorporate it.

  1. Terence Tao, 11 January 2026. Tao noted the apparent use of the third

term in the asymptotic for \(p_m\), and suggested that an error on the order of \(\log k/(\log\log k)^2\) might still have been possible.

  1. llllvvuu, 12 January 2026. The implication was formalised with

Aristotle. An edit reports that the formalisation was weakened to use an asymptotic already available in the PNT+ project.

  1. Terence Tao, 12 January 2026. Tao explained why weak PNT already

suffices: \(p_m=(1+o(1))m\log m\) gives \(\log p_m=\log m+\log\log m+o(1)\), from which the needed expansion for \(\vartheta(p_m)\) follows. He reports moving the formalisation record accordingly.

None of these comments claims to solve (P), and the claimed-proof count remains zero.

1. Result of this run

Put

\[ g(k):=\frac{\log k}{\log\log k}. \]

For fixed \(c>1\) and \(K\geq16\), call \(n>K\) \((c,K)\)-admissible when

\[ \omega(n-k)<c\,g(k)\qquad(K\leq k<n). \tag{1} \]

Let \(p_t\) be the \(t\)-th prime and

\[ Q_t:=\prod_{i=1}^{t}p_i. \]

The output is the following exact finite result.

Finite exclusion theorem (d). Let \(c=11/10\) and \(K=16\). Among \(17\leq n\leq Q_{101}+15\), exactly 94 integers are \((11/10,16)\)-admissible, and every one is at most \(225\). Equivalently, no \(n\) in \[ > 226\leq n\leq > 2577426147548683169379880845613862450845254401055092509543183257270179188707233718999293223417932941038924189941484105421516996015467418326179536384362799440729804187886824533414953001905801090622787969540076319408964006245 > \tag{2} > \] satisfies (1).

The upper endpoint is \(Q_{101}+15\), where \(p_{101}=547\); it has 223 decimal digits and is approximately \(2.5774\times10^{222}\).

(d) The complete list below \(226\), compressed into consecutive runs, is

\[ \begin{split} &[17,45], [50,57], [62,75], [80,81], [90,93], [98,99],\\ &[104,105], [110,117], \{125\}, [134,135], [140,141],\\ &[152,153], [160,165], [176,180], [194,195], \{210\},\\ &[218,219], [224,225]. \end{split} \tag{3} \]

This is a concrete finite exclusion, not a solution of (P). The original problem allows a different, possibly enormous, cutoff \(K_\epsilon\), and asks for infinitely many \(n\).

2. Exact interval-cover reduction

Monotonicity

(a) For real \(x>e^e\),

\[ g'(x)=\frac{\log\log x-1} {x(\log\log x)^2}>0. \tag{4} \]

Thus \(g\) is strictly increasing on the integers \(k\geq16\).

For integers \(t\geq0\), define the exact inverse endpoint

\[ B_{c,K}(t):= \max\bigl(\{k\in\mathbb Z:k\geq K,\;c\,g(k)\leq t\} \cup\{K-1\}\bigr) \tag{5} \]

and the level set

\[ A_t:=\{m\geq1:\omega(m)\geq t\}. \tag{6} \]

Covering lemma

Lemma (a). An integer \(n>K\) fails (1) if and only if \[ > n\in\bigcup_{t\geq0} > \bigl(A_t+[K,B_{c,K}(t)]\bigr), > \tag{7} > \] where sums and intervals are over integers.

Proof. If \(m\in A_t\) and \(k=n-m\in[K,B_{c,K}(t)]\), then

\[ \omega(n-k)=\omega(m)\geq t\geq c\,g(k), \]

so (1) fails. Conversely, if (1) fails at \(k\), put \(m=n-k\) and \(t=\omega(m)\). Then \(m\in A_t\), and monotonicity plus (5) gives \(k\leq B_{c,K}(t)\). \(\square\)

This reformulates (P) as a simultaneous gap question for all the nested sets \(A_t\). It also gives an \(O(N\log\log N)\) finite scan: sieve \(\omega(m)\) up to \(N\), emit one interval from each \(m\), and merge the intervals in increasing order of their left endpoints.

Primorial-block lemma

Lemma (a). If \[ > t\geq c\,g(Q_{t+1}-Q_t+K-1), > \tag{8} > \] then every integer in \[ > [Q_t+K,\;Q_{t+1}+K-1] > \tag{9} > \] fails (1).

Proof. Given \(n\) in (9), take \(m=Q_t\) and \(k=n-Q_t\). Then

\[ K\leq k\leq Q_{t+1}-Q_t+K-1. \]

Since \(\omega(Q_t)=t\), (4) and (8) imply

\[ \omega(n-k)=t\geq c\,g(k). \]

This contradicts the strict inequality in (1). \(\square\)

3. Certificate for the 223-digit exclusion

Exact base scan

(d) It is enough to scan through

\[ Q_5+15=2\cdot3\cdot5\cdot7\cdot11+15=2325. \]

Directed 80-digit interval arithmetic certifies the only threshold transitions needed in this range:

\[ \left\lceil\frac{11}{10}g(k)\right\rceil= \begin{cases} 3,&16\leq k\leq19,\\ 4,&20\leq k\leq1282,\\ 5,&1283\leq k\leq2324. \end{cases} \tag{10} \]

For orientation only, the ordinary decimal approximations at the transition points are

\[ \begin{array}{c|rrrrrr} k&16&19&20&1282&1283&2324\\ \hline \frac{11}{10}g(k)& 2.990687&2.999193&3.003408&3.999945&4.000159&4.163509. \end{array} \]

The checker does not use these rounded display values. It encloses each logarithm with an outward-rounded interval.

(d) Two independent implementations then give the same set (3):

  1. direct evaluation of (1) for every \(17\leq n\leq2325\);
  2. the interval-union scan from Lemma (7).

The \(\omega\)-sieve is separately checked against trial division for every \(1\leq m\leq2325\). Both scans find 94 admissible integers, with last value \(225\).

Primorial chain

For \(5\leq t\leq100\), put

\[ L_t:=Q_{t+1}-Q_t+15. \]

(d) The checker certifies, with outward-rounded intervals,

\[ \frac{11}{10}g(L_t)<t \qquad(5\leq t\leq100). \tag{11} \]

The weakest of these 96 inequalities is \(t=5\), where the certified positive margin is greater than

\[ 0.160554146312. \]

By the primorial-block lemma, (11) covers the consecutive blocks

\[ [Q_t+16,Q_{t+1}+15]\qquad(5\leq t\leq100). \]

They begin at \(Q_5+16=2326\), immediately after the base scan, and end at \(Q_{101}+15\). Together with the absence of a base-scan candidate after 225, this proves the finite exclusion (2).

4. Primary-source and literature audit

Erdős's source

(a, direct source audit) The original source corresponding to the live question is P. Erdős, Some unconventional problems in number theory, Acta Mathematica Academiae Scientiarum Hungaricae 33 (1979), 71–80. On page 72, equation (3) asks for exactly the two \(\omega\) and \(\Omega\) inequalities with an \(\epsilon\)-dependent cutoff, and equation (4) asks the stronger additive-constant version. Thus the live page's quantifiers agree with the primary source.

The inspected PDF's SHA-256 was

a4703a104adcbbde0ed605c9ff81379adee8b7ceda3714ea455b5b68ae720dfd

Tao–Teräväinen

(b) T. Tao and J. Teräväinen, Quantitative correlations and some problems on prime factors of consecutive integers, arXiv:2512.01739v2 (25 April 2026), prove in Theorem 1.1 that infinitely many \(n\) satisfy \(\omega(n+k)\leq\Omega(n+k)\leq Ck\) for all positive \(k\). Their Remark 1.2 states problem 679 with the \((1+\epsilon)\log k/\log_2 k\) threshold and says it appears beyond their methods. A v2 footnote points to Lau's subsequent \(C\log k\) refinement.

The inspected PDF's SHA-256 was

ce10e83b10c6544e1dbff037a5e4efa0e387892e0fc596ae09dce76025d7b41e

Lau

(b) C. F. Lau, On the Number of Prime Factors of Consecutive Integers, arXiv:2604.15042v2 (24 June 2026), is the source cited on the live page. Theorem 1.3 proves that an absolute \(C>0\) exists such that infinitely many \(n\) obey

\[ \omega(n-k)\leq\Omega(n-k)\leq C\log k \qquad(1<k<n). \]

Conjecture 6 says this is best possible up to a constant. Section 7 gives a more targeted conditional route. In Conjecture 8, Lau asks for a polylogarithmic every-interval bound for gaps in

\[ \left\{m:\omega(m)\geq C_0\frac{\log_2 m}{\log_3 m}\right\}. \]

Theorem 7.3 proves that Conjecture 8 would imply the existence of \(\delta>0\) such that every sufficiently large \(n\) has some \(1\ll k<n\) with

\[ \omega(n-k)>(1+\delta)\frac{\log k}{\log\log k}. \]

This would refute (P) for small enough \(\epsilon\).

The inspected PDF's SHA-256 was

90443d4ebbdfe9e05052e1a8115b8acd9d29d3ce7269d7e9c71cb09716c718d1

Short-interval results

(b) É. Goudout, Lois locales de la fonction \(\omega\) dans presque tous les petits intervalles, arXiv:1607.08666, Proc. London Math. Soc. 115 (2017), 599–637, proves local laws for integers with prescribed \(\omega\) in almost all sufficiently short intervals, in ranges including \(k\asymp\log_2 x\). This supports the random-model heuristic but does not give Lau's required statement in every interval. An exceptional set of interval starting points is fatal here, because the negation of (P) needs an obstruction for every sufficiently large \(n\).

(c) Exact-title, arXiv-ID, exact-phrase, Erdős-number, and forward searches through 2026-07-28 found no primary source after Lau claiming an unconditional resolution of (P) or the every-interval estimate of his Conjecture 8. This is a documented search miss, not a proof that no such source exists. It agrees with the live page's OPEN status and zero proof-claim count.

5. Exactly what remains

The interval-cover lemma gives a precise two-sided formulation.

(a) To refute (P) for a fixed \(c=1+\epsilon\), it is enough—and for each fixed cutoff \(K\), necessary—to prove that every sufficiently large \(n\) has some \(t\) and some

\[ m\in A_t\cap[n-B_{c,K}(t),\,n-K]. \tag{12} \]

To prove (P), one must instead find a cutoff \(K\) and infinitely many simultaneous holes:

\[ A_t\cap[n-B_{c,K}(t),\,n-K]=\varnothing \qquad\text{for every }t. \tag{13} \]

Thus the exact missing input is a uniform multiscale maximal-gap theorem, not a mean-value estimate for \(\omega\).

Why the primorial chain cannot close the problem

(b, modulo the prime number theorem) For fixed \(c>1\), inversion of \(t=c\,g(k)\) gives

\[ \log B_{c,K}(t)=\left(\frac1c+o(1)\right)t\log t. \tag{14} \]

The PNT gives

\[ \log Q_t=(1+o(1))t\log t \quad\text{and}\quad \log(Q_{t+1}-Q_t)=(1+o(1))t\log t. \tag{15} \]

Consequently

\[ B_{c,K}(t)\ll Q_{t+1}-Q_t \]

on the logarithmic scale for all sufficiently large \(t\). The consecutive primorial blocks used here must eventually stop overlapping for every \(c>1\). The 223-digit result is therefore a genuine finite certificate, not the beginning of an unproved induction to infinity.

Why the known analytic machinery stalls

(c) Global Sathe–Selberg estimates and Goudout's local laws control the number of members of \(A_t\) globally or in almost all intervals. Condition (12) needs a member in a specifically positioned interval for every \(n\), at a level \(t\) that changes with the scale. No cited theorem removes the exceptional intervals with the uniformity needed here.

(c) The Tao–Teräväinen/Lau weighted-sieve construction goes in the opposite direction: it constructs special \(n\) whose shifts have few prime factors. Lau reaches \(C\log k\) by exponential concentration and a summable union bound, and explicitly conjectures that this logarithmic scale is optimal up to its constant. Replacing \(C\log k\) by \(\log k/\log\log k\) requires a qualitatively stronger lower-tail phenomenon, not a constant optimisation in the existing argument.

(c) Brute force cannot address the remaining quantifier. A literal one-byte sieve through the endpoint in (2) would require about \(2.6\times10^{222}\) bytes and the same order of operations. Even at an unrealistic \(10^9\) tested integers per core-second, this is about \(8\times10^{205}\) core-years. The primorial certificate avoids that scan, but no finite extension can prove an infinitude statement. The needed advance is the every-interval lemma (12), essentially the role played by Lau's Conjecture 8.

6. Standalone re-verification

The standalone checker is erdos679_wavew009_reverify.py. It uses only the Python standard library and recomputes:

base range;

Run:

python3 runs/erdos679_wavew009_reverify.py

Observed output on the final run:

PASS: directed 80-digit log intervals and two base scans agree
base admissible count: 94
base admissible runs: [(17, 45), (50, 57), (62, 75), (80, 81), (90, 93), (98, 99), (104, 105), (110, 117), (125, 125), (134, 135), (140, 141), (152, 153), (160, 165), (176, 180), (194, 195), (210, 210), (218, 219), (224, 225)]
weakest primorial-chain inequality: t=5
certified positive margin > 0.160554146312
Q_101: 2577426147548683169379880845613862450845254401055092509543183257270179188707233718999293223417932941038924189941484105421516996015467418326179536384362799440729804187886824533414953001905801090622787969540076319408964006230
excluded n interval: [226, 2577426147548683169379880845613862450845254401055092509543183257270179188707233718999293223417932941038924189941484105421516996015467418326179536384362799440729804187886824533414953001905801090622787969540076319408964006245]
certificate sha256: 938d3bcd76dd1452172777d591bb98a9acafd3cd8b2243a86b086c43a34026bc

Runtime was 0.10 seconds with peak RSS 16512 KB on this VM. The final checker SHA-256 is

5eb19cab8569b7452a6ac9560d1d52ea640aa9b1f44f29f0e37783d99808e6aa

PARTIAL: For epsilon=1/10 and cutoff K=16, exactly 94 admissible n occur through Q_101+15 and all are at most 225, while the infinite problem reduces to an every-interval multiscale gap lemma not supplied by known results.

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