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.

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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Recurring gap values differing by two do not suffice

Theorem 6.4 needs values differing by $2$ at a prescribed distance $h$ and a complete tail difference in a signed interval. Recurrence of two values in a residue class supplies neither requirement. We demonstrate this with an integer sequence whose dyadic sum is rational and whose complete tail differences are integral, while retaining the recurrence and growth properties stated below. This is not a sequence of actual prime gaps.

We reverse the usual order: choose integer $U_{n}$ first, then set $a_{n} = 2 U_{n - 1} - U_{n}$ . The construction must still make the coefficients positive and ensure $2^{- n} U_{n} \to 0$ , so that these are actual tails. Constant tails $U_{n} = 4$ give constant coefficients $a_{n} = 4$ . Raising one isolated tail to $6$ changes the two affected coefficients from $4, 4$ to $2, 8$ , with the same combined dyadic contribution. A slowly growing even background will supply cumulative growth $n \log n$ ; factorial indices will impose the recurrence in residue classes while keeping these changes sparse.

Theorem 8.8 (recurring values are not enough). There is a sequence $(a_{n})_{n \geq 1}$ of positive even integers with the following properties. The values $2$ and $4$ each occur infinitely often at indices divisible by every fixed $t \geq 1$ . The sequence is unbounded, not eventually periodic, and satisfies $a_{n} = O (\log n)$ . The series $\sum_{n \geq 1} a_{n} 2^{- n}$ equals $6$ , and every scaled tail $\sum_{j \geq 1} a_{N + j} 2^{- j}$ is an integer, so every tail shift is integral. The increasing odd sequence $P_{n} = 3 + \sum_{j \leq n} a_{j}$ satisfies $P_{n} \sim n \log n$ .

Proof. Put $b_{n} = 2 ⌈ \frac{1}{2} \log (n + 64) ⌉$ , so that $b_{n}$ is even, $b_{n} \geq 6$ , and $b_{n} = \log (n + 64) + O (1)$ . Since $\frac{1}{2} \log (n + 65) - \frac{1}{2} \log (n + 63) < 1$ for every $n \geq 1$ , consecutive increments of $b$ lie in ${0, 2}$ and $b_{n + 1} - b_{n - 1} \leq 2$ . Call an index $n$ special when $n = k!$ or $n = 2 k!$ for some $k \geq 5$ , and define

U_{n} = {\begin{cases} 2 b_{n - 1} - 2, & n = k!, k \geq 5, \\ 2 b_{n - 1} - 4, & n = 2 k!, k \geq 5, \\ b_{n}, & otherwise, \end{cases} a_{n} = 2 U_{n - 1} - U_{n} (n \geq 1) .

For $k \geq 5$ one has $k! < 2 k! < (k + 1)!$ , so all special indices are distinct. They are even and therefore never adjacent. Each $U_{n}$ is an even integer by construction, so each $a_{n}$ is an even integer.

At an ordinary index whose predecessor is also ordinary, $a_{n} = 2 b_{n - 1} - b_{n} = b_{n - 1} - (b_{n} - b_{n - 1}) \geq 6 - 2 = 4$ . At a special index the predecessor is ordinary and $a_{n} = 2 b_{n - 1} - (2 b_{n - 1} - r) = r$ , which is $2$ at $n = k!$ and $4$ at $n = 2 k!$ . Immediately after a special index carrying $r$ ,

a_{n + 1} = 2 (2 b_{n - 1} - r) - b_{n + 1} \geq 4 b_{n - 1} - 2 r - (b_{n - 1} + 2) = 3 b_{n - 1} - 2 r - 2 \geq 8 .

Every $a_{n}$ is therefore a positive even integer, and $U_{n} = O (\log n)$ gives $a_{n} = O (\log n)$ . For $k \geq 5$ , the index $n = k! + 2$ and its predecessor are ordinary: both lie strictly between $k!$ and $2 k!$ , where there are no special indices. Along these indices $a_{n} \geq b_{n - 1} - 2 \to \infty$ . Thus $(a_{n})$ is unbounded and hence not eventually periodic.

The definition $a_{n} = 2 U_{n - 1} - U_{n}$ is exactly $a_{n} 2^{- n} = U_{n - 1} 2^{- (n - 1)} - U_{n} 2^{- n}$ , so for every $N \geq 0$ and $m \geq 1$

\sum_{j = 1}^{m} \frac{a_{N + j}}{2^{j}} = U_{N} - \frac{U_{N + m}}{2^{m}} .

Since $U_{n} = O (\log n)$ the endpoint tends to zero, so the tail at $N$ equals the integer $U_{N}$ ; at $N = 0$ the series equals $U_{0} = b_{0} = 6$ . Every difference $U_{N + h} - U_{N}$ is an integer, so every tail shift is integral.

For each $t$ and every sufficiently large $k$ we have $t ∣ k!$ , hence $t ∣ 2 k!$ , and $a_{k!} = 2$ while $a_{2 k!} = 4$ : both values recur infinitely often at indices congruent to $0$ modulo $t$ .

Finally, summing $a_{j} = 2 U_{j - 1} - U_{j}$ gives $\sum_{j = 1}^{n} a_{j} = 2 U_{0} + \sum_{j = 1}^{n - 1} U_{j} - U_{n}$ . The baseline $\sum_{j < n} b_{j} = n \log n + O (n)$ . The special indices up to $n$ number $O (\log n / \log \log n)$ and each changes the summand by $O (\log n)$ , so their total contribution is $O ((\log n)^{2}) = o (n)$ . Hence $P_{n} = 3 + \sum_{j \leq n} a_{j} \sim n \log n$ , and $P_{n}$ is odd and increasing. ◻

The factorial indices used above give recurrence in the zero residue class. The following separate construction strengthens this to every residue class.

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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