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

Will Cook

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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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Repeated tail values and congruent indices

For the complete actual tails $T_{N} = \sum_{j \geq 1} g_{N + j} 2^{- j}$ , the prime number theorem controls the mean size without assuming rationality. Rationality would then restrict these tails to a fixed rational lattice, forcing many repetitions by pigeonhole. Repetition gives difference zero, not the nonintegral difference required at congruent indices. We make both steps explicit. These are ordinary deductions; the linked formal results establish the recurrence and the congruent-index criterion.

Repeated tail values.

For an integer $X \geq 0$ , summing $g_{N + 1} = 2 T_{N} - T_{N + 1}$ over $0 \leq N < X$ gives the exact identity

\sum_{N = 0}^{X} T_{N} = p_{X + 1} - p_{1} - T_{0} + 2 T_{X} \leq (p_{X + 1} - p_{0}) + 2 T_{X} .

The inequality uses $T_{0} \geq 0$ and $p_{1} \geq p_{0}$ . The prime number theorem [17] also gives

\begin{matrix} (12) & \frac{T_{X}}{p_{X}} = \sum_{j \geq 1} \frac{g_{X + j}}{2^{j} p_{X}} ⟶ 0. \end{matrix}

Indeed, for each fixed $j$ , both $p_{X + j + 1} / p_{X}$ and $p_{X + j} / p_{X}$ tend to $1$ , so their difference $g_{X + j} / p_{X}$ tends to zero. For the uniform bound needed to pass the limit through the sum, the two-sided estimates $p_{n} ≍ n \log n$ give one constant $K$ such that, for $X \geq 2$ and $j \geq 1$ ,

\begin{aligned} \frac{g_{X + j}}{p_{X}} & \leq K \frac{(X + j + 1) \log (X + j + 1)}{X \log X} \\ \leq K (1 + j) (1 + \frac{\log (1 + j)}{\log X}) \leq K (1 + j)^{2} . \end{aligned}

The second line uses $X + j + 1 \leq X (1 + j)$ and $\log (1 + j) \leq j \log 2 \leq j \log X$ . Since $\sum_{j \geq 1} (1 + j)^{2} 2^{- j} < \infty$ , dominated convergence applies. Subtracting the exact summation identities at $2 X$ and $X$ now gives

\sum_{X < N \leq 2 X} T_{N} = p_{2 X + 1} - p_{X + 1} + 2 (T_{2 X} - T_{X}) \sim X \log X .

Indeed, the prime difference is asymptotic to $X \log X$ , and (12) makes both tail terms $o (X \log X)$ . Thus the mean complete tail in this band is asymptotic to $\log X$ , without a rationality assumption. This is a mean, not a pointwise estimate. For the counting argument we only need the upper bound $\sum_{X < N \leq 2 X} T_{N} \leq K_{0} X \log X$ , valid for any fixed $K_{0} > 1$ and all sufficiently large $X$ . For each fixed $C > 1$ , Markov’s inequality then leaves at least $(1 - 1 / C) X$ indices with $T_{N} \leq K_{0} C \log X$ .

Under rationality, take $X$ beyond the point where the reduced denominator is the fixed odd integer $d$ . Among the indices just selected by Markov’s inequality, the values lie in $d^{- 1} Z_{\geq 0} \cap [0, K_{0} C \log X]$ . There are at most $d K_{0} C \log X + 1$ such values. Pigeonhole therefore gives one value occurring at least

\frac{(1 - 1 / C) X}{d K_{0} C \log X + 1}

times. Taking, for example, $K_{0} = 2$ gives a lower bound of order $≫_{C, d} X / \log X$ . The value may depend on $X$ ; no single value recurring in every band is established.

The growth input is essential: the rational quadratic example in Proposition 8.9 has strictly increasing tails $T_{N} = 2 (N + 4)^{2}$ . For the primes, however, repeated values give zero difference, not the required nonintegral one.

The terminal term in the exact identity cannot simply be omitted. The factorial-gap argument in the proof of Proposition 6.3 shows that $T_{X} \geq g_{X + 1} / 2$ is unbounded, whereas $p_{1} + T_{0}$ is fixed. The exact identity therefore gives $\sum_{N = 0}^{X} T_{N} > p_{X + 1}$ for arbitrarily large $X$ . It does not establish that this strict inequality holds eventually.

The interval condition at congruent indices.

Suppose the sum is rational, let $d$ be the eventual odd reduced denominator, and let $t$ be the order of $2$ modulo $d$ , with $t = 1$ when $d = 1$ . Beyond the denominator cutoff, $M \equiv N (\mod t)$ forces $T_{M} - T_{N} \in Z$ . Neither this difference nor its negative can then lie in $(\frac{1}{2}, 1)$ . Thus a witness in that interval, at two such congruent indices, would contradict rationality without a further condition on the intervening gaps. This uses only the interval occurring in Theorem 6.4; it does not assert that its other hypotheses hold for arbitrary congruent pairs.

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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