\(N(X)\), and \(N(X)\) assumes every sufficiently large integer value as

\(X\) increases. This equivalence is (a).

For the remainder, set

Thus \(R_N\) is exactly the live-page expression at \(x=N+1\).

Exact reduction to chains of two large gaps

Define

Proposition

For every \(N\ge2\),

Consequently,

Proof (a). Take a pair \(a=d_{n-1}\), \(b=d_n\). Since

\(\max(a,b)\le G_N\) and \(\min(a,b)\le H_N\),

Taking the maximum over pairs gives \(Q_N\le H_NG_N\). A pair attaining

\(H_N\) has both entries at least \(H_N\), so its product is at least

\(H_N^2\), and \(Q_N\ge H_N^2\). Divide by \(G_N^2\). The two implications

follow respectively from the upper bound and from

\(H_N/G_N\le\sqrt{R_N}\). \(\square\)

This isolates the question exactly: are the largest two consecutive gaps

negligible compared with the largest single gap?

There is also an exact match with the standard notation of

Ford–Maynard–Tao. They define

At \(X=p_{N+1}\),

Therefore problem #1137 is precisely

Relevant primary literature

Chains of large gaps

Ford, Maynard, and Tao, Chains of large gaps between primes,

arXiv:1511.04468, Theorem 1, prove that

for every fixed \(k\) and all sufficiently large \(X\),

with an absolute effective implied constant. This is exactly about \(H_N\)

when \(k=2\), not merely about isolated large gaps.

Keiju Sono, *An explicit lower bound for large gaps between some consecutive

primes*, arXiv:2404.06951, published in

Le Matematiche 80 (2025), 521–544, makes the constant explicit:

for sufficiently large \(X\).

Let

At \(X=p_{N+1}\), the exact reduction gives the direct corollary (b)

and hence

for all sufficiently large \(N\). The threshold implicit in “sufficiently

large” is not supplied numerically, so this is not a finite-range estimate.

Established upper bound for a single gap

Baker, Harman, and Pintz, The Difference Between Consecutive Primes, II,

Proc. London Math. Soc. 83 (2001), 532–562,

DOI 10.1112/plms/83.3.532,

prove that \([x-x^{0.525},x]\) contains a prime for all sufficiently large

\(x\). The standard consequence is (b)

This and the chain lower bound do not compare

\(\mathcal G_2(X)\) with the actual value of \(\mathcal G_1(X)\). Their

scales are separated by almost a power of \(X\), so inserting them in the

sandwich gives no nonzero or zero limiting bound.

A July 2026 claimed \(O(\log^2p)\) bound is not usable

Cheng-Ting Wang, On Maximal Prime Gaps,

arXiv:2605.14871v5, revised

23 July 2026, claims

for all \(n\), and consequently claims Oppermann’s conjecture. This would be

a major improvement over the established published bound. It has no journal

reference, and its supplied argument has a concrete invalid step.

In the proof of Lemma 2.3, lines 143–145 of the HTML version assert that

forces \(n\le4\). At \(n=195\), however,

so (*) holds although \(195>4\). More fundamentally, the right side grows

like \(n\log n\), while the left grows like \(\log^2n\). Thus this step cannot

produce the claimed contradiction, and the downstream lemma and theorem are

not proved by the posted argument. This diagnosis is (a); it does not

prove Wang’s stated bound false. I do not use the preprint as a theorem.

Search result

Exact-formula searches, searches for products of adjacent gaps, and searches

through the cited chain literature found no primary source deciding

\(\mathcal G_2(X)/\mathcal G_1(X)\). Baker–Freiberg’s normalized-gap results

and the Ford–Maynard–Tao chain theorem control prescribed polylogarithmic

normalizations, not normalization by the all-time maximal gap. This is an

honest literature miss, not evidence that no such paper can exist.

Exact computation through \(N=10^7\)

The standalone checker is

erdos1137_wave6v_verify.py. It uses only the

Python standard library and generates every prime itself with a segmented

Eratosthenes sieve.

For every cutoff it updates \(G_N,Q_N,H_N\) and checks

It independently audits every record gap, product, and two-gap chain by:

• checking the prime endpoints with deterministic 64-bit Miller–Rabin;
• finding a trial-division certificate for every integer strictly inside

each asserted prime gap;

• checking that all record values and indices increase as required.

The following is exact; fractions are not rounded in the calculation.

| \(N\) | \(p_{N+1}\) | \(G_N\) | \(H_N\) | \(Q_N\) | \(R_N=Q_N/G_N^2\) |

|---:|---:|---:|---:|---:|---:|

| 10 | 31 | 6 | 4 | 24 | \(2/3=0.666666666667\) |

| 100 | 547 | 18 | 12 | 144 | \(4/9=0.444444444444\) |

| 1,000 | 7,927 | 34 | 20 | 512 | \(128/289=0.442906574394\) |

| 10,000 | 104,743 | 72 | 42 | 2,436 | \(203/432=0.469907407407\) |

| 100,000 | 1,299,721 | 114 | 62 | 4,340 | \(1085/3249=0.333948907356\) |

| 1,000,000 | 15,485,867 | 154 | 98 | 9,800 | \(50/121=0.413223140496\) |

| 10,000,000 | 179,424,691 | 222 | 136 | 22,040 | \(5510/12321=0.447203960717\) |

The checker also determines the exact extrema over every inclusive index

decade:

| inclusive \(N\)-range | exact minimum (first \(N\)) | exact maximum (first \(N\)) |

|---:|---:|---:|

| \([10,100]\) | \(2/7\) at 30 | \(1\) at 16 |

| \([100,10^3]\) | \(3/17\) at 217 | \(4/9\) at 100 |

| \([10^3,10^4]\) | \(5/24\) at 3,385 | \(5/9\) at 1,663 |

| \([10^4,10^5]\) | \(345/1568\) at 31,545 | \(203/432\) at 10,000 |

| \([10^5,10^6]\) | \(1085/5476\) at 149,689 | \(50/121\) at 826,235 |

| \([10^6,10^7]\) | \(2/9\) at 1,319,945 | \(5510/12321\) at 8,040,878 |

In particular, the sharp finite statement (d) is

Finite data cannot imply that the limiting value is positive.

Endpoint record certificates

At \(N=10^7\):

• the maximal single gap is

\[ G_N=222,\qquad 122164969-122164747=222, \]

first occurring at gap index \(6,957,876\);

• the maximal adjacent product is

\[ Q_N=116\cdot190=22040 \]

at

\[ 142414553,\ 142414669,\ 142414859, \]

with product index \(8,040,878\);

• the maximal adjacent minimum is

\[ H_N=136 \]

at

\[ 163709971,\ 163710121,\ 163710257, \]

whose two gaps are \(150\) and \(136\), at product index \(9,170,830\).

The pair that maximizes the product need not be the pair that maximizes the

minimum; the checker keeps the three record processes separate.

Independent recomputation

The principal run used \(1,048,576\) odd integers per sieve segment. A second

full run used the deliberately incommensurate segment size \(100,003\).

Both produced:

• \(10,000,001\) primes ending at \(p_{10,000,001}=179,424,691\);
• 26 single-gap records, 39 product records, and 24 chain-minimum records;
• SHA-256 of \(d_1,\ldots,d_{10^7}\), each encoded as little-endian unsigned

32-bit:

33d7700aeb812181cd82a6222d27a2876a9ab0fe6afac0ce7f7a28ef83cfaab5;

• the same snapshots and all the same decade extrema.

The expected hash and exact tables are frozen as assertions in the checker.

Each Python run took about 35 seconds on this VM.

Finally, PARI/GP’s independent forprime iterator was run through

179,424,691. It returned, independently of the Python sieve:

10       31         6    24     4     2/3
100      547        18   144    12    4/9
1000     7927       34   512    20    128/289
10000    104743     72   2436   42    203/432
100000   1299721    114  4340   62    1085/3249
1000000  15485867   154  9800   98    50/121
10000000 179424691  222  22040  136   5510/12321
PARI_OK

Columns after \(N,p_{N+1}\) are \(G_N,Q_N,H_N,R_N\).

An additional PARI pass maintained exact cross-products at every one of the

\(10^7\) cutoffs and independently returned all six minimum/maximum fractions

and first-attainment indices in the decade table, ending

PARI_EXTREMA_OK.

Reproduction

From the repository root:

python3 runs/erdos1137_wave6v_verify.py --summary-only

To force the alternative segmentation and save every record:

python3 runs/erdos1137_wave6v_verify.py \
  --segment-odds 100003 \
  --summary-only \
  --json-out /tmp/erdos1137_alt.json

No downloaded prime list, OEIS b-file, or probabilistic prime generator is

used.

The \(1/4\) heuristic is a theorem in the independent-exponential model

This does not prove anything about primes, but it makes the page comment

mathematically precise.

Let \(X_1,X_2,\ldots\) be independent \({\rm Exp}(1)\) variables and define

Then (a), for this random model only,

Indeed, standard union bounds give

in probability. For \(a>1/2\), AM–GM and the exact

\(\Gamma(2,1)\) tail give

For \(a<1/2\), use the \(\lfloor N/2\rfloor\) disjoint pairs. Each pair has

both entries at least \(a\log N\) with probability \(N^{-2a}\), so the

probability that none does is at most

Hence \(\sqrt{P_N}/\log N\to1/2\), proving the claim.

Transferring this conclusion to prime gaps requires uniform extreme-value

independence far beyond anything proved. That transfer is (c).

Exact remaining wall

The reduction shows that there are only two possible asymptotic routes:

• A yes proof needs

\[ \mathcal G_2(X)=o(\mathcal G_1(X)), \]

i.e. an isolation theorem saying that every pair of adjacent large gaps is

negligible compared with the record single gap.

• A no proof needs a constant \(c>0\) and arbitrarily large \(X\) for

which

\[ \mathcal G_2(X)\ge c\,\mathcal G_1(X). \]

Current large-gap machinery constructs two consecutive gaps on the

Erdős–Rankin scale \(L(X)\). It supplies no upper control on all other gaps

below the same \(X\), which is exactly what comparison with

\(\mathcal G_1(X)\) requires. Conversely, short-interval prime theorems bound

each individual gap but do not show that two adjacent gaps enjoy a smaller

bound.

A fixed-offset Hardy–Littlewood prime-tuples conjecture is also insufficient:

the offsets here grow with \(X\), the two intervals between the three primes

must contain no additional primes, and the constructed pair must be compared

uniformly with every gap below \(X\). A quantitative, growing-offset,

prime-free, uniform version would be needed.

Extending the finite scan cannot resolve that uniformity. At the measured

Python rate, \(N=10^8\) would cost about 350 seconds, roughly \(0.10\)

single-core hours, and would reach \(p_N\) near \(2.0\times10^9\). An

optimized compiled segmented sieve would likely reduce that to a few minutes,

but any finite endpoint still leaves the same asymptotic lemma missing, so I

did not spend the box’s few-minute compute budget on it.

PARTIAL: proved the exact equivalence \(R_N\to0\iff\mathcal G_2(p_{N+1})/\mathcal G_1(p_{N+1})\to0\), derived a rigorous chain-theorem lower bound for the numerator, and exactly verified \(2/9\le R_N\le5510/12321\) for every \(10^6\le N\le10^7\); the required asymptotic comparison remains open.