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

Will Cook

Long paper

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

Erdős #251 By Will Cook September 2026 Authorship, AI use and citation

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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 $T_{N} = \sum_{j \geq 1} g_{N + j} 2^{- j}$ satisfy $T_{N + 1} = 2 T_{N} - g_{N + 1}$ . 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 $2^{39998} > 10^{12040}$ . The bound does not establish irrationality.

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In this paper

A bounded sequence with recurring values and a rational sum

The following bounded example isolates the recurring-values argument without a growth estimate. Logarithmic growth was needed in Theorem 8.8 to obtain cumulative size $n \log n$ , but is unnecessary for the recurrence itself.

Proposition D.6 (recurring values with integral tails). Put $U_{0} = 4$ and, for $n \geq 1$ , $U_{n} = 6$ when $n = k!$ for some $k \geq 3$ and $U_{n} = 4$ otherwise, and set $a_{n} = 2 U_{n - 1} - U_{n}$ for $n \geq 1$ . Then $a_{n} \in {2, 4, 8}$ . For every $k \geq 3$ , the value $2$ occurs at index $k!$ and the value $4$ at index $2 k!$ , so both recur infinitely often at indices divisible by any fixed $t \geq 1$ . The series $\sum_{n \geq 1} a_{n} 2^{- n}$ equals $4$ and every tail $\sum_{j \geq 1} a_{N + j} 2^{- j}$ equals the integer $U_{N}$ .

Proof. For $k \geq 3$ the index $k!$ is even and at least $6$ , and $k! - 1$ is odd, so the preceding tail is $4$ ; hence $a_{k!} = 8 - 6 = 2$ , $a_{k! + 1} = 12 - 4 = 8$ , and $a_{n} = 8 - 4 = 4$ elsewhere. For $k \geq 2$ one has $k! < 2 k! < (k + 1)!$ , so $2 k!$ is not a factorial. Its odd predecessor $2 k! - 1 \geq 3$ is not a factorial either; therefore $a_{2 k!} = 4$ . Since $t ∣ k!$ for every large $k$ , both values recur in the residue class $0$ modulo $t$ . The identity $a_{n} 2^{- n} = U_{n - 1} 2^{- (n - 1)} - U_{n} 2^{- n}$ telescopes to $\sum_{j = 1}^{m} a_{N + j} 2^{- j} = U_{N} - U_{N + m} 2^{- m}$ , and $U$ is bounded, so the endpoint vanishes. ◻

The indices with coefficient $2$ are precisely the factorials $k!$ , $k \geq 3$ . They occur infinitely often with unbounded gaps, which is impossible for a value recurring in an eventually periodic sequence. Thus even this bounded coefficient sequence is nonperiodic, despite its rational dyadic sum. Its cumulative sum has linear growth. The logarithmic construction in Theorem 8.8 adds unboundedness and cumulative growth $N \log N$ without changing the telescoping mechanism. Neither construction guarantees prime cumulative positions.

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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