Plectis

Long paper

Prime-Gap Dyadic Series: Perturbations, Exact Criteria and Certificates

Summary. A convergent dyadic series with nonnegative integer coefficients can be made rational by sparse nonnegative corrections of upper Banach density zero, eventually bounded by any f(n)→∞, while preserving every fixed modulus and, for a logarithmic allowance, short-block empirical distributions. For prime gaps the cumulative positions remain asymptotic to n log n; no primality conclusion is asserted. Summation by parts reduces the prime series to the gap series, and the tails TN=∑j≥1gN+j2−j satisfy TN+1=2TN−gN+1. That recurrence characterises rationality and yields finite tests with explicit remainder bounds. An exact finite certificate shows that a rational value of the prime series would have denominator at least 239998>1012040. The bound does not establish irrationality.

This paper covers the complete problem-specific reasoning surface for Erdős #251, including all registered result families and their boundaries.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #251, which remains open.

In this paper

Density of indices with two small tail differences

When the two tail differences have absolute value less than one and the gaps are unequal, the calculation above restricts the gap difference to 2 or −2. For each fixed shift, Schlage-Puchta’s lemma gives density zero for both events. We use this to bound the set of possible witnesses, without excluding witnesses at arbitrarily large indices.

Theorem 8.7 (density zero for the small-difference condition). Fix h≥1. The set of N≥1 at which the three hypotheses of Theorem 6.1 hold for the actual prime gaps has density zero. For the same h, the set of N with gN+h+1=gN+1 also has density zero.

Proof. For N≥1 every gap involved is even, so δN=gN+h+1−gN+1 is even. If |σh(N)|<1, |σh(N+1)|<1 and δN≠0, then δN=2σh(N)−σh(N+1) lies in (−3,3), so δN=±2. Apply Schlage-Puchta’s lemma [18] at index n=N+1 with k=h to the two polynomials xh−x0−2 and xh−x0+2, neither of which vanishes identically: each of {N:δN=2} and {N:δN=−2} has density zero, and the event is contained in their union. The second assertion is the same lemma applied to xh−x0. ◻

Unequal gaps occur at almost every index, a stronger conclusion than Proposition 6.3. But combining unequal gaps with the two small tail differences forces their difference to be exactly 2 or −2, which occurs only on a density-zero set. Thus a proof of Problem 10.2 must find arbitrarily large successful indices in that sparse set. A positive-density lower bound for the same event is impossible. An averaging argument that detects a sparse set is not ruled out.

The proof of Schlage-Puchta’s lemma gives a quantitative upper bound as well. For fixed h, it bounds the number of eligible indices in [X,2X) by Oh(X/log⁡log⁡X). To see the two contributions, discard starts whose block of h+1 gaps contains a gap exceeding log⁡Xlog⁡log⁡X. The sum of these block sums is Oh(Xlog⁡X) by the prime number theorem, so there are at most Oh(X/log⁡log⁡X) discarded starts. For the remaining starts, the sieve calculation in the cited proof bounds the zeros of each of xh−x0−2 and xh−x0+2 by

Oh(X(log⁡log⁡X)2h+2log⁡X)=oh(Xlog⁡log⁡X).

The shift from n=N+1 changes only the band endpoints. Thus the same upper bound applies to the indices satisfying the three conditions of Theorem 6.1. This is an ordinary consequence of the cited sieve argument, not a new lower bound or an asymptotic formula for the number of successful indices.

The observed proportions in Section 9 decrease across the sampled bands. These finite observations establish neither an asymptotic rate nor the required existence beyond every cutoff. Counting by prime size and counting by gap index also give different cutoffs and must not be compared without converting between them.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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