Erdős problem 218 — wave 5m report
Access and computation date: 2026-07-26 UTC
Claim labels
- (a) elementary-rigorous: proved directly here.
- (b) rigorous-modulo-named-theorem: rigorous after invoking the named, standard theorem.
- (c) plausible/structural-unverified: heuristic, interpretation, or a search-limited literature conclusion.
- (d) computational-only: exact output of the independently cross-checked finite program described below, within its software trust boundary.
The problem remains open. The concrete outputs here are:
1. (b) an explicit quantitative version of the already-known zero-density result for ties,
\[ \#\{n\leq N:d_{n+1}=d_n\}\ll \frac{N}{\sqrt{\log N}}; \]
2. (a) an exact reduction of both density assertions to the single cancellation estimate
\[ \sum_{n\leq N}\operatorname{sgn}(d_{n+1}-d_n)=o(N); \]
3. (a)/(b) a fixed admissible triple \(210t+47,210t+53,210t+59\) whose simultaneous primality would give consecutive equal gaps of size \(6\);
4. (d) a two-sieve exact table through \(p_n\leq10^9\).
0. Mandatory live-page check
I used the Bright Data browser path, rather than datacenter curl, to fetch all three relevant live views:
- <https://www.erdosproblems.com/218>
- <https://www.erdosproblems.com/latex/218>
- <https://www.erdosproblems.com/forum/discuss/218>
(d; live-page observation) The page was marked OPEN, was last edited 29 January 2026, displayed 0 claimed proofs, and listed None for both “Currently working on this problem” and “Interested in collaborating.” Thus the mandatory stop condition did not apply.
(d; live-page observation) The header source badges were [Er55c] [Er57] [Er61] [Er65b] [Er85c]. The remaining activity markers were: likes—None; looks difficult—TerenceTao and banrovegrie; looks tractable—None; results could be formalisable—None; working on formalisation—None. The page also listed the formalised-statement flag as Yes and OEIS sequences A333230, A333231, and A064113.
Verbatim current statement
> Let \(d_n=p_{n+1}-p_n\). The set of \(n\) such that \(d_{n+1}\geq d_n\) has density \(1/2\), and similarly for \(d_{n+1}\leq d_n\). Furthermore, there are infinitely many \(n\) such that \(d_{n+1}=d_n\).
Results listed on the page
(d; page transcription) The page additionally says:
- In Erdős [Er85c], the stronger equation
\[ d_n=d_{n+1}=\cdots=d_{n+k} \]
is conjectured to be solvable for every \(k\), equivalently giving consecutive primes in arithmetic progression (the page points to problem 141).
- Banks [Ba23] gives a heuristic based on a quantitative prime-tuples conjecture. The page prints
\[ \#\{n:p_n\leq x,\ d_{n+1}\geq c d_n\} =\frac{\pi(x)}{c+1} +O\!\left(\frac{x}{(\log x)^{3/2+\epsilon}}\right). \]
The sign of \(\epsilon\) in that displayed error does not agree with the primary Banks paper; see the literature audit below.
Complete comment/activity inventory
(d; live-thread observation) The thread had 14 comments and no proof claim registered by the site:
- On 3–4 September 2025, Terence Tao said existing sieve methods should make adjacent ties density zero, Dogmachine agreed that Brun's method should do this, and Tao said an appropriately uniform prime-tuples conjecture should imply the full assertions.
- On 26 November 2025, John N. Dvorak posted a claimed proof only of the partial result that ties have density zero, splitting at \(y=(\log x)^{4/3}\). Replies discussed Aristotle, Boris Alexeev's forum comments, the absence of a ready Brun sieve in Mathlib, the PNT+ repository, the strong-PNT repository, and Lean-version compatibility. This was not a claimed solution of problem 218.
- On 28–29 January 2026, Dvorak posted a uniform-Hardy–Littlewood/Poisson-process conditional argument. Tao pointed out that this implication is already in Banks's paper. Dvorak asked whether it counted as a conditional resolution; Tao said he had added it to the AI-contributions wiki and recommended literature review; Dvorak then summarized the zero-density split and the Banks heuristic.
The PDF linked from the November comment currently returns HTTP 404 at its GitHub raw URL. The mathematical outline in the comment is nevertheless enough to identify the claimed partial result, and the result already appears in Erdős's 1985 paper.
1. Primary-source literature audit
Verified sources
1. (b) P. Erdős, On some of my problems in number theory I would most like to see solved, Lecture Notes in Mathematics 1122 (1985), 74–84. In §5, Erdős states that Brun's method easily gives density zero to the indices where two gaps in a fixed block are equal. On the same page he says that even infinitely many solutions of \(d_n=d_{n+1}\) seemed beyond reach. Thus zero density of ties is old, not a new 2025 result.
2. (c; the paper itself labels the argument heuristic) William D. Banks, On ratios of consecutive prime gaps, Integers 23 (2023), A50, DOI 10.5281/zenodo.8174512. Its Conjecture 1 is
\[ \pi_c(x)=\frac{\pi(x)}{c+1} +O\!\left(x(\log x)^{-3/2+\epsilon}\right) =\frac{\pi(x)}{c+1} +O\!\left(\frac{x}{(\log x)^{3/2-\epsilon}}\right), \]
for every \(c\geq0\) and \(\epsilon>0\). This is weaker than the live page's denominator exponent \(3/2+\epsilon\). (c) The most likely explanation is a transcription sign error on the website. This discrepancy does not affect the predicted density: choosing any \(0<\epsilon<1/2\) still gives an \(o(\pi(x))\) error.
3. (b) W. D. Banks, T. Freiberg, and C. L. Turnage-Butterbaugh, Consecutive primes in tuples, arXiv:1311.7003v3, published in Acta Arithmetica 167 (2015), 261–266, DOI 10.4064/aa167-3-4. Corollary 1 proves that for every fixed \(m\) there are infinitely many runs of \(m\) strictly increasing consecutive prime gaps and infinitely many strictly decreasing runs. This proves infinitude of both strict signs, but gives no density.
4. (b) János Pintz, On a conjecture of Erdős, Pólya and Turán on consecutive gaps between primes, arXiv:1504.06860. Its Theorem 1 proves the stated sign-change criterion for fixed linear combinations of consecutive prime gaps; Theorem 2 gives particularly large isolated gaps relative to finitely many neighboring gaps. Again, this is an infinitude/limsup theorem, not a density theorem.
5. (b) D. K. L. Shiu, Increasing and decreasing prime gaps, arXiv:1604.01761, proves infinitely many increasing and decreasing runs whose length grows at least on the order of \(\log\log\log n\). This strengthens the existence result but does not control the proportion of rises and falls.
Search scope, rejected lead, and honest miss
I searched exact-title, exact-problem-number, “equal consecutive prime gaps,” “balanced primes,” ratios of consecutive prime gaps, and increasing/decreasing gap queries over arXiv, publisher pages, the Rényi Erdős archive, and the live discussion.
One search result claimed that a “Theo Johnson” had solved “Erdős #218” and cited arXiv:2410.05678. The primary arXiv record is actually Fern Gossow's unrelated paper Polynomial and combinatorial analogues of Gauss congruence. The news/SEO claim also states a different combinatorial problem, so it was rejected.
(c; search-limited) I found no primary source proving either density \(1/2\) assertion or infinitely many adjacent equal prime gaps. This is a report of the search, not a theorem that no such source exists; it agrees with the authoritative live status.
2. Quantitative zero-density theorem for equal adjacent gaps
Define
\[ E(x)=\#\{n:p_n\leq x,\ d_{n+1}=d_n\}, \qquad E_N=\#\{n\leq N:d_{n+1}=d_n\}. \]Theorem
(b; Selberg upper-bound sieve plus the prime number theorem)
\[ E(x)\ll \frac{x}{(\log x)^{3/2}}, \qquad E_N\ll \frac{N}{\sqrt{\log N}}. \]This makes Erdős's qualitative Brun-sieve remark quantitative. I do not claim that the exponent or observation is novel.
Proof
2.1 Congruence restriction
(a) Suppose \(p,p+h,p+2h\) are odd primes with \(p>3\). Then \(2\mid h\). If \(3\nmid h\), the three terms occupy all residue classes modulo \(3\), so one is divisible by \(3\), impossible. Hence every adjacent equal gap other than \(3,5,7\) has
\[ 6\mid h. \]2.2 Uniform upper-bound sieve for small \(h\)
For \(6\mid h\), let
\[ C_h(x)=\#\{m\leq x:m,\ m+h,\ m+2h\text{ are prime}\}. \](b; standard Selberg upper-bound sieve for a prime triple) Uniformly for
\[ h\leq y=(\log x)^{3/2}, \]one has
\[ C_h(x)\ll \mathfrak S(h)\frac{x}{(\log x)^3}, \]where
\[ \mathfrak S(h)= \prod_q\left(1-\frac{\nu_q(h)}q\right) \left(1-\frac1q\right)^{-3}, \quad \nu_q(h)=|\{0,h,2h\}\bmod q|. \]Uniformity here is the routine polylogarithmic-shift form of the sieve: one may sieve to a fixed power of \(x\), while all shifts are \(x^{o(1)}\).
(a) For \(6\mid h\), the local factors at \(2,3\) are fixed. For every prime \(q\geq5\), the factor has \(\nu_q=1\) if \(q\mid h\), and \(\nu_q=3\) otherwise. Therefore
\[ \mathfrak S(h) =C_0\prod_{\substack{q\mid h\\q\geq5}}\frac{q-1}{q-3} =C_0\prod_{\substack{q\mid h\\q\geq5}} \left(1+\frac2{q-3}\right), \]where
\[ C_0=9\prod_{q\geq5} \left(1-\frac3q\right)\left(1-\frac1q\right)^{-3} \]is a finite positive absolute constant.
(a) Expanding the last product over squarefree divisors gives
\[ \begin{aligned} \sum_{h\leq y}\prod_{\substack{q\mid h\\q\geq5}} \left(1+\frac2{q-3}\right) &\leq y\sum_{\substack{d\geq1\ {\rm squarefree}\\q\mid d\Rightarrow q\geq5}} \frac1d\prod_{q\mid d}\frac2{q-3}\\ &= y\prod_{q\geq5}\left(1+\frac{2}{q(q-3)}\right) \ll y. \end{aligned} \]Thus
\[ \sum_{\substack{h\leq y\\6\mid h}}\mathfrak S(h)\ll y. \](b) Every small tie is among the triples counted above, so
\[ E_{\rm small}(x) \ll \frac{x}{(\log x)^3} \sum_{\substack{h\leq y\\6\mid h}}\mathfrak S(h) \ll \frac{xy}{(\log x)^3}. \]Consecutiveness is not needed for this upper bound.
2.3 Large \(h\) by telescoping
(a) Let \(M=\pi(x)\). Then
\[ \sum_{n\leq M}d_n=p_{M+1}-2<2x \]by telescoping and Bertrand's postulate. Hence
\[ E_{\rm large}(x) \leq\#\{n\leq M:d_n>y\} \leq\frac{2x}{y}. \]2.4 Balance and translate to the index variable
Taking \(y=(\log x)^{3/2}\) balances the two bounds:
\[ E(x)\ll \frac{xy}{(\log x)^3}+\frac{x}{y} \ll\frac{x}{(\log x)^{3/2}}. \]This part is (b) only because of the invoked upper-bound sieve; the balancing and telescoping are (a).
(b; prime number theorem) Set \(x=p_N\). Since \(p_N\sim N\log N\) and \(\log p_N\sim\log N\),
\[ E_N=E(p_N)\ll \frac{N}{\sqrt{\log N}}. \]In particular \(E_N=o(N)\).
3. Exact reduction of the density part
Let
\[ \begin{aligned} G_N&=\#\{n\leq N:d_{n+1}>d_n\},\\ L_N&=\#\{n\leq N:d_{n+1}(a) The identities
\[ G_N+L_N+E_N=N,\qquad G_N-L_N=S_N \]give
\[ G_N=\frac{N-E_N+S_N}{2}, \qquad L_N=\frac{N-E_N-S_N}{2}. \]Consequently,
\[ \#\{n\leq N:d_{n+1}\geq d_n\} =\frac{N+E_N+S_N}{2}, \] \[ \#\{n\leq N:d_{n+1}\leq d_n\} =\frac{N+E_N-S_N}{2}. \]Since \(E_N=o(N)\) is already known, (a) both density assertions on the page are equivalent to the one remaining estimate
\[ \boxed{S_N=o(N).} \]Indeed, either one of the two asserted densities already implies this cancellation and then implies the other density.
This reduction does not address the separate infinitude assertion for ties.
4. A fixed admissible construction for equal consecutive gaps
Put
\[ P_0(t)=210t+47,\quad P_1(t)=210t+53,\quad P_2(t)=210t+59. \](a) For every integer \(t\geq0\), all ten integers strictly between \(P_0(t)\) and \(P_2(t)\), except \(P_1(t)\), are composite:
- offsets \(1,3,5,7,9,11\) from \(P_0(t)\) are even;
- offset \(2\) is divisible by \(7\);
- offsets \(4,10\) are divisible by \(3\);
- offset \(8\) is divisible by \(5\).
Therefore, whenever all three displayed forms are prime, they are three consecutive primes and give
\[ d_n=d_{n+1}=6. \](a) The three linear forms are admissible. For \(q\mid210\), their constant terms \(47,53,59\) are all nonzero modulo \(q\). If \(q\nmid210\), each form excludes at most one residue class of \(t\bmod q\), so at most three classes are excluded; necessarily \(q\geq11\), hence not all classes are excluded.
(b; Hardy–Littlewood prime \(3\)-tuple conjecture) That conjecture would imply infinitely many \(t\) for which all three forms are prime, and hence would prove the tie-infinitude assertion in the particularly strong fixed-gap form \(d_n=d_{n+1}=6\) infinitely often.
This is a sufficient conditional route, not an unconditional solution and not an equivalence: the original infinitude assertion could conceivably be proved by another method.
5. Exact computation through \(10^9\)
For a cutoff \(x\), the table counts every \(n\) with \(p_n\leq x\); it sieves far enough past \(x\) to determine \(p_{n+1}\) and \(p_{n+2}\).
(d) The exact results are:
| \(x\) | \(\pi(x)\) | \(d_{n+1}>d_n\) | \(=\) | \(<\) | \(\geq/\pi(x)\) | \(\leq/\pi(x)\) | \((G-L)/\pi(x)\) |
|---:|---:|---:|---:|---:|---:|---:|---:|
| \(10\) | 4 | 2 | 1 | 1 | 0.750000000000 | 0.500000000000 | +0.250000000000 |
| \(10^2\) | 25 | 12 | 2 | 11 | 0.560000000000 | 0.520000000000 | +0.040000000000 |
| \(10^3\) | 168 | 79 | 15 | 74 | 0.559523809524 | 0.529761904762 | +0.029761904762 |
| \(10^4\) | 1,229 | 589 | 65 | 575 | 0.532139951180 | 0.520748576078 | +0.011391375102 |
| \(10^5\) | 9,592 | 4,615 | 434 | 4,543 | 0.526376146789 | 0.518869891576 | +0.007506255213 |
| \(10^6\) | 78,498 | 37,781 | 2,994 | 37,723 | 0.519439985732 | 0.518701113404 | +0.000738872328 |
| \(10^7\) | 664,579 | 321,751 | 21,837 | 320,991 | 0.517000988596 | 0.515857407471 | +0.001143581124 |
| \(10^8\) | 5,761,455 | 2,797,477 | 167,032 | 2,796,946 | 0.514541726005 | 0.514449561786 | +0.000092164219 |
| \(10^9\) | 50,847,534 | 24,760,598 | 1,328,401 | 24,758,535 | 0.513082876350 | 0.513042304077 | +0.000040572272 |
At \(10^9\), (d) the exact strict imbalance is only
\[ G-L=2,063, \]but this finite smallness is not an asymptotic theorem. The tie fraction is
\[ \frac{1,328,401}{50,847,534}=0.026125180427\ldots. \]Further exact checks:
- (d) The ten most frequent tied gap sizes were
\[ \begin{array}{c|rrrrrrrrrr} h&6&12&18&24&30&36&42&48&54&60\\ \hline \#&595279&330221&181903&90678&84523&19444&15333&4899&2577&2329. \end{array} \]
- (d) The largest tied gap with starting prime at most \(10^9\) was
\[ (383204539,\ 383204683,\ 383204827), \qquad h=144. \]
- (d) The fixed family from §4 supplied 47,836 triples with first term at most \(10^9\); the last was
\[ (999994427,\ 999994433,\ 999994439). \]
- (d) The SHA-256 of the complete \(G/E/L\) sign stream through \(10^9\) was
2b5445797be0b0abba03ddc8020743f872256891748cdd179f3f402ea43ae256.
Standalone verifier and trust boundary
The required standalone verifier is:
runs/erdos218_wave5m_verify.py
(d) Its SHA-256 in the final run was
499b8136b91e5bbb2f35cc92952371b18e1ae2861a2ce57c515528f32a0a750f.
Run it from the repository root with:
python runs/erdos218_wave5m_verify.py
It uses only the Python standard library and performs two full, independently implemented prime enumerations:
1. a monolithic odd-only Eratosthenes sieve through \(1,001,000,000\);
2. a segmented sieve whose base primes are generated independently by trial division.
The two implementations must agree on every table row and on the complete sign-stream hash. The script also:
- compares the monolithic sieve with trial division on a 101,000 prefix;
- verifies \(\pi(x)=G+E+L\) at every cutoff;
- checks every computed nonexceptional tie has \(6\mid h\);
- verifies the covering congruences and admissibility logic for \(210t+47,53,59\);
- checks the singular-series local-factor algebra using exact rational arithmetic;
- asserts all frozen counts, witnesses, and hashes.
The final default run reported:
sieve_limit=1001000000 sieve_seconds=12.279
independent_segmented_audit_seconds=25.679
largest_tie=(gap,p0,p1,p2)=(144, 383204539, 383204683, 383204827)
fixed_210t_plus_47_count=47836
fixed_210t_plus_47_last=(999994427, 999994433, 999994439)
total_seconds=75.542
all_assertions_passed=True
The finite claims depend on Python's integer and bytearray behavior and on the correctness of the two small implementations. Agreement of structurally different sieves, trial division on a prefix, identities, frozen hashes, and exact witnesses gives a strong reproducibility check, but it is not a formally verified certificate; the claims therefore remain labeled (d).
6. Exact wall
Density half
(a) Knowledge of the one-gap marginal distribution cannot by itself force equal frequencies of rises and falls: a stationary periodic real sequence can have identical time-shifted marginals yet unequal ascent/descent counts. The missing input must control the joint ordering of adjacent prime gaps.
(c) The specific absent lemma is
\[ \sum_{n\leq N}\operatorname{sgn}(d_{n+1}-d_n)=o(N). \]Present sieve methods give upper bounds for specified prime configurations and can prove \(E_N=o(N)\), but they do not supply cancellation for this discontinuous signed statistic. Banks obtains the expected answer only heuristically by using quantitative Hardy–Littlewood correlations and conjecturally truncating the high-order inclusion-exclusion terms that enforce the absence of intervening primes.
Infinitely many ties
(c) A concrete sufficient missing lemma is that the admissible triple
\[ 210t+47,\quad210t+53,\quad210t+59 \]is simultaneously prime for infinitely many \(t\). This is a fixed instance of the prime \(3\)-tuple conjecture. An upper-bound sieve cannot give the required lower bound because of the sieve parity obstruction; Maynard–Tao methods guarantee several primes among a much larger admissible tuple, not these three prescribed forms.
Why more computation does not close either assertion
(a) No finite cutoff proves a natural-density limit or infinitude. The \(10^9\) audit is therefore evidence and an exact regression target only.
(c; measured extrapolation) The default double audit used about 0.5 GB for its monolithic sieve and 75.5 seconds on this VM. A direct \(10^{10}\) repetition of the same design would need about 5 GB and roughly 12–20 single-core minutes (about 0.2–0.35 core-hours), outside the requested few-minute run. A segmented-only optimized C/C++ computation would reduce memory and time, but would still not address the required uniform asymptotic step.
PARTIAL: proved the quantitative tie bound \(E_N\ll N/\sqrt{\log N}\) modulo the standard Selberg sieve, reduced both density claims exactly to \(S_N=o(N)\), isolated a fixed admissible triple sufficient for infinite ties, and independently verified all gap comparisons through \(10^9\); neither open assertion is closed.