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.

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.

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In this paper

An elementary polynomial bound for the primes

We derive the bound $p_{n} \leq 1250 (n + 1)^{4}$ by comparing the size and prime factorisation of a central binomial coefficient, following Erdős’s proof of Chebyshev’s theorem [5]. With $π (x)$ denoting the number of primes at most $x$ , we first prove, for $m \geq 4$ ,

\begin{matrix} (11) & 4^{m} < m (\binom{2 m}{m}) \leq m (2 m)^{π (2 m)} . \end{matrix}

For the first inequality, $m (\binom{2 m}{m}) / 4^{m}$ equals $70 / 64$ at $m = 4$ and its ratio at successive indices is $(2 m + 1) / (2 m) > 1$ . For the second, the exponent of a prime $ℓ$ in $(\binom{2 m}{m})$ is $\sum_{k \geq 1} (⌊ 2 m / ℓ^{k} ⌋ - 2 ⌊ m / ℓ^{k} ⌋)$ , each summand is $0$ or $1$ , and every summand with $ℓ^{k} > 2 m$ vanishes; the exponent is therefore at most $⌊ \log_{ℓ} (2 m) ⌋$ . Thus the full power of each prime appearing in $(\binom{2 m}{m})$ is at most $2 m$ . Multiplying over at most $π (2 m)$ distinct primes gives the second inequality.

Now fix $n \geq 0$ , put $x = n + 5$ and $m = x^{4} \geq 625$ , and suppose $π (2 m) \leq n$ . Since $x \leq 2^{x}$ and $n + 4 (n + 1) x = 4 x^{2} - 15 x - 5 \leq 2 x^{4}$ ,

m (2 m)^{n} = 2^{n} x^{4 (n + 1)} \leq 2^{n + 4 (n + 1) x} \leq 2^{2 x^{4}} = 4^{m},

contradicting (11). Hence $π (2 m) > n$ , so at least $n + 1$ primes lie below $2 m$ and therefore

p_{n} \leq 2 (n + 5)^{4} \leq 1250 (n + 1)^{4},

the last step because $(n + 5) \leq 5 (n + 1)$ for $n \geq 0$ . The coarser bound $p_{n} \leq 1250 (n + 1)^{4}$ is checked by the kernel at line 360.

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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