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ERDőS #929 · PARTIAL

Erdős problem #929 — wave w021

Date: 2026-07-28 UTC

This report does not solve the asymptotic problem. It gives:

  1. a complete live-page/status audit;
  2. an elementary exact reduction to the primorial Jacobsthal function;
  3. a stronger analytic bracket obtained by reading the cited FGKMT theorem in its native \(Y(x)\) formulation;
  4. a from-scratch exact computation of \(S(k)\) for every \(2\leq k\leq73\), with explicit positive-density residue classes and independent upper-bound checks; and
  5. the precise uniform lemma still missing.

Claim labels used below:

0. Mandatory live-page gate

I fetched the live page, its LaTeX-source view, its bibliography pop-up, and its discussion thread through the Bright Data cloud browser. Direct datacenter access was not used as the authority.

Live state observed on 2026-07-28:

Thus the mandatory stop condition was not triggered.

Verbatim current statement

Let \(k\geq 2\) be large and let \(S(k)\) be the minimal \(x\) such that there is a positive density set of \(n\) where \[ > n+1,n+2,\ldots,n+k > \] are all divisible by primes \(\leq x\).

Estimate \(S(k)\) - in particular, is it true that \(S(k)\geq k^{1-o(1)}\)?

Results and comment listed on the page

The page states:

\[ S(k)\ll k\, \frac{\log\log\log k} {\log\log k\,\log\log\log\log k}. \]

The sole comment, by Adenwalla on 15 October 2025, points out that the earlier example \(n\equiv1\pmod{k!}\) did not handle \(n+k\), and corrects it to \(n\equiv1\pmod{(k+1)!}\). The live statement has already incorporated that correction. The comment contains no claimed proof or current-worker marker.

The bibliography pop-up identifies the original source as:

P. Erdős, Problems and results on number theoretic properties of consecutive integers and related questions, Proceedings of the Fifth Manitoba Conference on Numerical Mathematics (1976), 25–44, MR 0422146. The source is available in the Rényi archive.

1. Exact reduction: this is the inverse primorial Jacobsthal function

For real \(x\geq2\), put

\[ P(x)=\prod_{p\leq x}p \]

and let

\[ Y(x)=\max\left\{m:\ \exists a\in\mathbb Z\ \forall 1\leq i\leq m,\ (a+i,P(x))>1\right\}. \]

[a] Periodicity removes the apparent density issue. For fixed \(x,k\), the condition

\[ \forall 1\leq i\leq k,\quad (n+i,P(x))>1 \]

depends only on \(n\bmod P(x)\). Hence its set of solutions is a union of residue classes modulo \(P(x)\). It has positive natural density if and only if it is nonempty; one working class alone has density \(1/P(x)\).

It follows immediately that

\[ \boxed{S(k)=\min\{x:Y(x)\geq k\}.}\tag{1} \]

Since \(Y(x)\) changes only when \(x\) passes a prime, the minimum in (1) is a prime.

Let \(j(m)\) denote the Jacobsthal function: the least \(L\) such that every \(L\) consecutive integers contain an integer coprime to \(m\). If \(p_r\) is the \(r\)-th prime and

\[ h(r)=j(p_r\#),\qquad p_r\#=\prod_{i=1}^r p_i, \]

then the longest non-coprime run has length \(h(r)-1\). Therefore

\[ \boxed{Y(p_r)=h(r)-1,\qquad S(k)=p_{\min\{r:h(r)-1\geq k\}}.}\tag{2} \]

This is also exactly equation (1.3), \(Y(x)=j(P(x))-1\), in the cited FGKMT paper.

Exact asymptotic target

[a] Generalized-inverse monotonicity in (1) gives the equivalence

\[ \boxed{S(k)\geq k^{1-o(1)} \quad\Longleftrightarrow\quad Y(x)\leq x^{1+o(1)} \quad\Longleftrightarrow\quad h(r)\leq p_r^{\,1+o(1)}.}\tag{3} \]

For example, if \(Y(x)\leq x^{1+\varepsilon}\), then

\[ k\leq Y(S(k))\leq S(k)^{1+\varepsilon}. \]

Conversely, put \(k=Y(x)\) and use \(S(Y(x))\leq x\). Letting \(\varepsilon\to0\) proves both directions.

Thus the exact missing lemma is not a density statement or a CRT construction. It is the uniform upper bound

\[ \boxed{Y(x)\leq x^{1+o(1)}.}\tag{4} \]

2. Literature audit: a sharper bracket than the page displays

2.1 Direct consequence of FGKMT

FGKMT define the same \(Y(x)\) by residue-class coverings and prove, in their equation (1.2),

\[ Y(x)\gg x\frac{\log x\,\log_3 x}{\log_2 x},\tag{5} \]

where \(\log_j\) denotes the \(j\)-fold iterated natural logarithm. They also explicitly record \(Y(x)=j(P(x))-1\).

[b, modulo FGKMT Theorem/equation (1.2)] Inverting (5) in the correct variable gives

\[ \boxed{ S(k)\ll k\frac{\log_2 k}{\log k\,\log_3 k}. }\tag{6} \]

Indeed, if

\[ x=A\,k\frac{\log_2 k}{\log k\,\log_3 k}, \]

then

\[ \log x\sim\log k,\qquad \log_2x\sim\log_2k,\qquad \log_3x\sim\log_3k, \]

so the right side of (5) is \((cA+o(1))k\). Choosing \(A>1/c\) makes \(Y(x)\geq k\), and (1) gives \(S(k)\leq x\).

This is asymptotically stronger than the upper bound printed on the live page. The ratio of (6) to the page’s displayed bound is

\[ \frac{(\log_2 k)^2\log_4 k} {\log k\,(\log_3 k)^2}\longrightarrow0. \]

This is a literature extraction and inversion, not a new proof of FGKMT.

2.2 Iwaniec’s upper bound for \(Y\)

Iwaniec’s On the problem of Jacobsthal proves the classical linear-sieve bound \(C(r)\ll(r\log r)^2\) for a covering by \(r\) prime residue classes. FGKMT restate its primorial consequence as

\[ Y(x)\ll x^2. \]

[b, modulo Iwaniec and the prime number theorem] Inverting this yields

\[ \boxed{S(k)\gg \sqrt{k}.}\tag{7} \]

This modestly sharpens the \(k^{1/2-o(1)}\) formulation printed on the page.

Combining (6) and (7), the presently verified analytic bracket found in the primary literature is

\[ \boxed{ \sqrt{k}\ll S(k)\ll k\frac{\log_2 k}{\log k\,\log_3 k}. }\tag{8} \]

The lower bound sought by Erdős would replace the exponent \(1/2\) in (7) by \(1-o(1)\).

2.3 Computational literature

Hagedorn’s Computation of Jacobsthal’s Function \(h(n)\) for \(n<50\) gives exact values through \(r=49\). Ziller and Morack’s arXiv:1611.03310 reports exhaustive values through \(p_r=251\), i.e. \(r=54\), and supplies residue/permutation data.

Those published tables are much longer than the independently verified table below. I do not use their entries as certificates: the accompanying Python program recomputes every finite value it claims.

2.4 Search miss, stated narrowly

Targeted searches for “primorial Jacobsthal function,” “Jacobsthal upper bound,” the Iwaniec paper, later citing work, and \(Y(x)\) located no primary paper improving the exponent \(2\) in Iwaniec’s uniform upper bound for this primorial case. Pascal Stumpf’s 2023 dissertation On coverings and reduced residues in combinatorial number theory also describes Iwaniec’s bound as the best known and identifies the linear-sieve limitation. This is an honest search report, not a proof that no unindexed result exists.

[c, explicitly conjectural] FGKMT mention the Maier–Pomerance conjecture

\[ Y(x)\ll x(\log x)^{2+o(1)}, \]

which would imply (4), but it is not a theorem and is not used as one here.

3. From-scratch finite computation

The standalone verifier is runs/erdos929_wavew021_reverify.py.

3.1 Removing the prime \(2\)

Let \(Q\) be odd. [a]

\[ j(2Q)=2j(Q).\tag{9} \]

For the lower bound, lift a maximal gap modulo \(Q\) by CRT to an odd left endpoint; odd offsets are then even, and even offsets reproduce the \(Q\)-gap after division by \(2\). For the upper bound, the \(j(Q)\) odd members of any \(2j(Q)\)-term interval map, after an affine division by \(2\bmod Q\), to \(j(Q)\) consecutive residue classes modulo \(Q\), one of which is coprime to \(Q\).

Let \(\omega(r)\) be the maximum \(m\) such that \([1,m]\) is covered by one nonzero residue class modulo each odd prime \(3,\ldots,p_r\). Equation (9) gives

\[ h(r)=2\omega(r)+2.\tag{10} \]

3.2 Pure-Python exhaustive search through \(p_r=31\)

[a] Correctness of the enumeration. Normalize a maximal odd-prime run so that position \(0\) is uncovered. All selected residues are then nonzero. Given any such cover, take its first uncovered position \(q\); some unused prime class covering \(q\) has forced residue \(q\bmod p\). Repeating this operation orders the prime classes canonically. Therefore it is exhaustive to branch only on the unused prime chosen at the current first uncovered position.

The verifier represents the first 128 positions by a Python integer bit mask and enumerates every such branch. It records the first uncovered position at every node. If a normalized cover of 128 positions existed, the corresponding canonical branch would fill the entire mask and trigger an overflow. No overflow occurs.

[d] The largest pure-Python case, \(p_r=31\), visits 5,736,589 nodes and obtains \(\omega=28\), hence \(h=58\). All smaller prefixes are recomputed rather than read from a table.

3.3 Independent full-period scan through \(p_r=19\)

[d] A structurally different checker marks every residue in the complete period \(0,\ldots,p_r\#-1\) having nontrivial gcd with the primorial and scans for the longest run. It independently reproduces

\[ 2,4,6,10,14,22,26,34 \]

for \(h(r)\), \(1\leq r\leq8\). The largest scanned period is

\[ 19\#=9{,}699{,}690. \]

3.4 SAT upper certificates through \(p_r=41\)

For each odd \(p\leq p_r\) and \(1\leq s<p\), introduce a Boolean variable \(z_{p,s}\). Impose:

  1. exactly one \(z_{p,s}\) is true for each \(p\);
  2. for every position \(q\leq m\),

\[ \bigvee_{\substack{3\leq p\leq p_r\\p\ \mathrm{prime}\\q\not\equiv0\pmod p}} z_{p,q\bmod p}. \]

[a] This CNF is satisfiable exactly when \([1,m]\) has a normalized cover by those odd primes. Thus an explicit cover at \(m=\omega\) and UNSAT at \(m=\omega+1\) prove the exact value.

[d] The script builds the CNF from scratch and uses CaDiCaL 1.9.5 through PySAT. For \(p_r=41\), the UNSAT instance has 435 variables and 672 clauses; the clean run used 714,302 conflicts and about 16 seconds. Cases through \(31\) are therefore checked both by the pure permutation enumerator and by the independent SAT formulation.

3.5 Explicit positive-density witnesses

Given condensed residues \(s_p\), the verifier constructs \(n_0\) by

\[ n_0\equiv1\pmod2,\qquad n_0\equiv-2s_p\pmod p \quad(3\leq p\leq p_r). \]

Then odd offsets are even, while at offset \(2q\) some odd prime divides \(n_0+2q\). The program directly checks every gcd and also checks that both endpoints \(n_0\) and \(n_0+h(r)\) are coprime to \(p_r\#\).

[d] Exact verified table:

| \(r\) | \(p_r\) | \(p_r\#\) | \(h(r)\) | max run \(h(r)-1\) | witness \(n_0\bmod p_r\#\) | |---:|---:|---:|---:|---:|---:| | 1 | 2 | 2 | 2 | 1 | 1 | | 2 | 3 | 6 | 4 | 3 | 1 | | 3 | 5 | 30 | 6 | 5 | 1 | | 4 | 7 | 210 | 10 | 9 | 1 | | 5 | 11 | 2,310 | 14 | 13 | 113 | | 6 | 13 | 30,030 | 22 | 21 | 9,439 | | 7 | 17 | 510,510 | 26 | 25 | 217,127 | | 8 | 19 | 9,699,690 | 34 | 33 | 9,639,613 | | 9 | 23 | 223,092,870 | 40 | 39 | 140,722,741 | | 10 | 29 | 6,469,693,230 | 46 | 45 | 417,086,647 | | 11 | 31 | 200,560,490,130 | 58 | 57 | 125,601,285,781 | | 12 | 37 | 7,420,738,134,810 | 66 | 65 | 740,266,380,323 | | 13 | 41 | 304,250,263,527,210 | 74 | 73 | 79,622,514,581,573 |

For every row and every integer \(t\),

\[ n=n_0+t(p_r\#) \]

has all of \(n+1,\ldots,n+h(r)-1\) divisible by primes at most \(p_r\). Hence the displayed class supplies positive density \(1/(p_r\#)\).

Using (2), this gives the following exact inverse table.

[d]

| range of \(k\) | exact \(S(k)\) | |---:|---:| | \(2\leq k\leq3\) | 3 | | \(4\leq k\leq5\) | 5 | | \(6\leq k\leq9\) | 7 | | \(10\leq k\leq13\) | 11 | | \(14\leq k\leq21\) | 13 | | \(22\leq k\leq25\) | 17 | | \(26\leq k\leq33\) | 19 | | \(34\leq k\leq39\) | 23 | | \(40\leq k\leq45\) | 29 | | \(46\leq k\leq57\) | 31 | | \(58\leq k\leq65\) | 37 | | \(66\leq k\leq73\) | 41 |

4. Reproduction

Run:

python runs/erdos929_wavew021_reverify.py

One clean full run on this VM took 25.249 seconds and ended:

PASS: exact h(r) through p_r=41; exact S(k) for 2 <= k <= 73; elapsed=25.249s

The full verifier uses python-sat (installed version 1.9.dev7) for the SAT upper bounds. A standard-library-only independent subset is available:

python runs/erdos929_wavew021_reverify.py --quick

It exhaustively certifies through \(p_r=31\), hence \(S(k)\) through \(k=57\), and still runs the complete-period cross-check through \(p_r=19\).

5. Precise wall

The finite computation does not address uniformity in \(r\). Published exact data through \(p_r=251\), or any other fixed cutoff, cannot imply (4).

The exact analytic gap is:

\[ \underbrace{Y(x)\ll x^2}_{\text{Iwaniec, proved}} \qquad\text{versus}\qquad \underbrace{Y(x)\leq x^{1+o(1)}}_{\text{needed for \#929}}. \]

Iwaniec’s exponent \(2\) is the linear-sieve threshold in this covering problem. The missing ingredient is a uniform argument exploiting correlations among the particular initial-prime residue classes strongly enough to save essentially a full power of \(x\). This is a missing theorem, not a computation of presently estimable core-hour cost. No bounded computation can supply the required \(x\to\infty\) step.

PARTIAL: reduced #929 exactly to the primorial Jacobsthal bound \(Y(x)\le x^{1+o(1)}\), extracted the sharper proved bracket \(\sqrt{k}\ll S(k)\ll k\log_2k/(\log k\log_3k)\), and independently certified exact \(S(k)\) for \(2\le k\le73\).

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