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.

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In this paper

Totient-length shifts and persistence of integrality

Once the denominator is odd, Euler’s congruence specifies an integral shift of totient length. We record this choice and then prove that an integral shift persists under each subsequent recurrence step.

Proposition D.1 (a shift of totient length). Let $T : N \to Q$ satisfy the dyadic tail recurrence with integer coefficients. If the reduced denominator $d$ of $T_{N}$ is odd, then $σ_{φ (d)} (N)$ is an integer.

Proof. Since $d$ is odd, $2$ and $d$ are coprime, so Euler’s congruence gives $d ∣ 2^{φ (d)} - 1$ . Writing $2^{φ (d)} - 1 = d k$ and $T_{N} = u / d$ in lowest terms, $(2^{φ (d)} - 1) T_{N} = k u$ is an integer, and the integral-shift criterion transfers this to the shift. ◻

Proposition D.2 (propagation). Let $T : N \to R$ satisfy $T_{n + 1} = 2 T_{n} - a_{n + 1}$ with integer coefficients, and define $σ_{h} (N) = T_{N + h} - T_{N}$ . For fixed $h, N \geq 0$ , if $σ_{h} (N)$ is an integer, then $σ_{h} (N + k)$ is an integer for every $k \geq 0$ .

Proof. The shift step identity gives $σ_{h} (N + 1) = 2 σ_{h} (N) - (a_{N + h + 1} - a_{N + 1})$ , an integer combination of an integer and two coefficients. Induct on $k$ . ◻

These are the totient shift and the propagation theorems cited in the note. For example, if $den T_{N} = 3$ then $φ (3) = 2$ and $3 T_{N}$ is an integer, so $σ_{2} (N)$ is integral while $σ_{1} (N)$ is not; if $den T_{N} = 5$ then $σ_{4} (N)$ is integral. In the orbit with every coefficient zero and $T_{0} = 1 / 12$ , the denominator is $12 = 2^{2} \cdot 3$ , the orbit reaches $T_{2} = 1 / 3$ with odd denominator at $s = 2$ , and $h = φ (3) = 2$ is exactly the shift length seen to be integral from index $2$ onwards. The totient supplies one length, while the exact criterion is $den (T_{N}) ∣ 2^{h} - 1$ . At an index with odd denominator $d > 1$ , the least positive length is the multiplicative order of $2$ modulo $d$ . When $d = 1$ , every positive length works; before the denominator becomes odd, no positive length works.

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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