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

A rational series with recurring values in every residue class

There is a synthetic integer sequence $(a_{n})_{n \geq 1}$ with all of the following properties. Each $a_{n}$ is positive and divisible by $2$ , and

a_{n} \leq 4 \log (n + 1) + 24.

For every $t \geq 1$ , every $0 \leq r < t$ , and every cutoff $N$ , there are $i, j \geq N$ such that

i \equiv j \equiv r (\mod t), a_{i} = 2, a_{j} = 4.

For every integer $B$ and cutoff $N$ , some $n \geq N$ satisfies $a_{n} > B$ . For every $h \geq 1$ there is no cutoff after which $a_{n + h} = a_{n}$ for all $n$ . Nevertheless, all the complete tails

U_{N} = \sum_{j \geq 1} \frac{a_{N + j}}{2^{j}} (N \geq 0)

are integers, $U_{0} = 6$ , and $U_{N + h} - U_{N} \in Z$ for every $N, h \geq 0$ . The sequence

P_{N} = 3 + \sum_{j = 1}^{N} a_{j}

is strictly increasing and odd, with $P_{N} / (N \log N) \to 1$ . These are synthetic positions; no $P_{N}$ is asserted to be prime.

Enumerate the triples $(t, r, m)$ with $t \geq 1$ , $0 \leq r < t$ and $m \geq 0$ , each once. The extra index $m$ repeats each residue-class requirement infinitely often; its parity will select the value $2$ or $4$ . At the $k$ th step choose a centre $c_{k} \equiv r (\mod t)$ , subject to $c_{0} \geq 100$ , $c_{k + 1} \geq c_{k} + 3$ and $c_{k} \geq 2^{k^{2}}$ for $k \geq 1$ . Each arithmetic progression is unbounded, so these lower bounds can be met recursively. Set $v_{k} = 2$ for even $m$ and $v_{k} = 4$ for odd $m$ . Thus each value is assigned beyond every cutoff in each residue class. Use the even baseline

b_{n} = 6 + 2 ⌊ \frac{\log (n + 1)}{2} ⌋, U_{n} = {\begin{cases} 2 b_{n - 1} - v_{k}, & n = c_{k}, \\ b_{n}, & n \notin {c_{k} : k \geq 0}, \end{cases}

and define $a_{n} = 2 U_{n - 1} - U_{n}$ . The separation of the centres will ensure that $a_{c_{k}} = v_{k}$ .

The baseline is even and at least $6$ . For $n \geq 1$ , the increase of $\log (n + 1)$ over two consecutive steps is less than $2$ , so the floor in the definition gives $b_{n} - b_{n - 1} \in {0, 2}$ and $b_{n + 1} - b_{n - 1} \leq 2$ . Away from a centre and its successor, $a_{n} = 2 b_{n - 1} - b_{n} \geq 4$ . At a centre, $a_{c_{k}} = v_{k}$ . Immediately afterwards,

a_{c_{k} + 1} = 4 b_{c_{k} - 1} - 2 v_{k} - b_{c_{k} + 1} \geq 3 b_{c_{k} - 1} - 2 v_{k} - 2 \geq 8.

Since the centres are at least three apart, these cases exhaust all indices. Also $0 < U_{n} \leq 2 b_{n}$ , so $a_{n} \leq 4 b_{n - 1} \leq 4 \log (n + 1) + 24$ . The indices $n = c_{k} + 2$ lie outside the centres and their successors, because consecutive centres are at least three apart. Along these indices $a_{n} \geq b_{n - 1} - 2 \to \infty$ . Thus the sequence is unbounded and cannot be eventually periodic.

Finite telescoping gives

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

The bound $U_{n} = O (\log (n + 1))$ makes the last term tend to zero. Hence the complete tail equals the integer $U_{N}$ , and its initial value is $6$ . For the cumulative growth, $b_{n} = \log (n + 1) + O (1)$ and there are at most $1 + \sqrt{\log_{2} N}$ centres up to $N$ . Replacing $b_{n}$ by $U_{n}$ changes its partial sum by $O ((\log N)^{3 / 2})$ . Therefore

\sum_{j = 1}^{N} a_{j} = 2 U_{0} + \sum_{j = 1}^{N - 1} U_{j} - U_{N} = N \log N + O (N),

which gives the stated asymptotic for $P_{N}$ .

The full statement is kernel-checked in Lean.

Even with all the displayed growth and recurrence properties, every tail difference is integral. Requiring the values $2$ and $4$ in every residue class therefore does not supply the signed tail interval. The construction is not a prime-gap sequence: its logarithmic bound excludes the extreme large gaps of the primes, so hypotheses using those gaps are not tested by this example.

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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