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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Arithmetic-progression reformulations of integrality

A longer observed block can appear to provide a stronger arithmetic test. Here it does not: the recurrence inserts the same block sum into the quantity tested and into the comparison progression, so that sum cancels. The moving progression tests even integrality of the original difference. Under the stated growth bound, separation from the fixed progression $2^{r} Z$ tests nonintegrality. We record the two reductions separately.

Fix $h$ and abbreviate $D_{N} = σ_{h} (N)$ and $δ_{N} = g_{N + h + 1} - g_{N + 1}$ . Iterating over a longer block gives

B_{h, N, r} = \sum_{i = 0}^{r - 1} 2^{r - 1 - i} δ_{N + i}; D_{N + r} = 2^{r} D_{N} - B_{h, N, r} .

Both the progression $- B_{h, N, r} + 2^{r + 1} Z$ and the value $D_{N + r}$ contain the same integer offset. Substituting the recurrence cancels it, leaving a condition on $D_{N}$ alone. Thus increasing the block length in this particular test adds no arithmetic information.

Theorem E.1 (equivalent arithmetic-progression tests). For every rational dyadic tail recurrence and all $h, N, r \geq 0$ ,

\begin{matrix} (13) & D_{N + r} \in - B_{h, N, r} + 2^{r + 1} Z ⟺ D_{N} \in 2 Z . \end{matrix}

Consequently, if every $δ_{N}$ is even, then

\begin{matrix} (14) & \begin{aligned} (\forall N_{0} \exists N, r : N_{0} < N and D_{N + r} \notin - B_{h, N, r} + 2^{r + 1} Z) \\ ⟺ D_{N} \notin Z for arbitrarily large N . \end{aligned} \end{matrix}

There is a second equivalence. Let $b : N \to Q$ satisfy $| D_{N} | \leq b (N)$ for every $N$ , and suppose that for every $N$ and every positive integer $q$ there is an $r$ with

\begin{matrix} (15) & 2 b (N + r) q < 2^{r} . \end{matrix}

Then

\begin{matrix} (16) & \begin{aligned} \forall N_{0} \exists N, r : N_{0} < N and \forall z \in Z, b (N + r) < | B_{h, N, r} - 2^{r} z | \\ ⟺ D_{N} \notin Z for arbitrarily large N . \end{aligned} \end{matrix}

Proof. Substituting $D_{N + r} = 2^{r} D_{N} - B_{h, N, r}$ into the first membership condition cancels the observed block from both sides and leaves $D_{N} = 2 z$ . This proves (13). When every $δ_{N}$ is even, the recurrence also gives $D_{N + 1} \in 2 Z$ if and only if $D_{N} \in Z$ . Thus eventual integrality would force eventual even integrality and contradict the left side of (14). This proves the forward implication. Conversely, given a cutoff, choose a nonintegral $D_{N}$ beyond it and take $r = 0$ in (13).

For (16), eventual integrality and the block identity give some $z \in Z$ with

B_{h, N, r} - 2^{r} z = - D_{N + r},

so the displayed strict separation contradicts $| D_{N + r} | \leq b (N + r)$ . Conversely, if $D_{N}$ is nonintegral, choose a positive integer $q$ with $1 / q \leq dist (D_{N}, Z)$ . Such a $q$ exists because $Z$ is closed and $D_{N} \notin Z$ . Choose $r$ from (15). The block identity and the triangle inequality give, for every $z \in Z$ ,

| B_{h, N, r} - 2^{r} z | \geq 2^{r} | D_{N} - z | - | D_{N + r} | \geq \frac{2^{r}}{q} - b (N + r) > b (N + r) .

The same $r$ works for all integers $z$ , as required. ◻

Evenness is essential in (14). Without it, the recurrence $T_{N} = N + 1$ , $g_{n} = n - 1$ for $n \geq 1$ , has $D_{N} = σ_{1} (N) = 1$ and $δ_{N} = 1$ at every index. Thus $D_{N}$ is never an even integer but is always an integer. By (13), the left side of (14) holds while the right side fails.

The growth hypothesis (15) is equivalent to

\underset{n \to \infty}{lim inf} 2^{- n} b (n) = 0.

Indeed, writing $n = N + r$ , its inequality becomes $2^{- n} b (n) < 1 / (2^{N + 1} q)$ . Since $b (n) \geq 0$ , finding such an $n \geq N$ for every $N$ and every positive integer $q$ is exactly the stated liminf condition. In particular it holds for polynomial bounds and for $C c^{n}$ with fixed $C > 0$ and $1 < c < 2$ , but fails for $b (n) = 2^{n}$ . A limit of zero is not necessary: the bound $b (n) = 1$ at odd indices and $b (n) = 2^{n}$ at even indices also satisfies the hypothesis. This example concerns the growth condition alone, not a bound for the prime tails. Only one suitable scale is needed to magnify the positive distance to $Z$ beyond the error; no comparison at every sufficiently large scale is required.

The proof in fact uses no rationality of $D_{N}$ : the integer $q$ is chosen from its positive distance to $Z$ , not from a reduced denominator. Thus (16) holds for real integer-coefficient recurrences under the same bound and growth hypothesis. The same is true of (13), and of (14) when the coefficient differences are even. These extensions follow from the printed argument; the Lean proof covers the rational case stated in the theorem.

The first test is exactly even integrality of $D_{N}$ , regardless of $r$ . The fixed-progression test compares $2^{r} dist (D_{N}, Z)$ with the bound on the remaining tail. When $D_{N}$ is nonintegral, the growth hypothesis makes the former larger than twice the latter at a suitable scale. Rewriting the test does not provide an estimate about the distribution of consecutive primes.

A signed comparison with more analytic data

A signed variant appears in the totient companion, The Binary Totient Series, under “Signed interpolation with entire Dirichlet defects”. We recall its different hypotheses to delimit the comparison. Around the totient sequence, signed integer changes bounded by any positive $f (n) \to \infty$ fill an interval at every reciprocal integer base. They can preserve the odd coefficients, every fixed eventual congruence, all arithmetic-progression Dirichlet singularities, and prescribed finite root-of-unity jets. The proof takes finite differences of increasing order on separated finite sets of indices. The support of the changes has upper Banach density zero. The union of the intervals spanned by those sets has ordinary density zero and upper Banach density one.

This theorem allows the bound on the corrections to grow arbitrarily slowly and retains more analytic observations. It uses both signs, however, and does not subsume the one-sided prime-gap perturbation theorem proved here. In particular, it constructs comparison integer sequences, not consecutive prime gaps, and neither theorem supplies an irrationality proof for the actual prime series. The companion contains a full ordinary proof and the precise block estimate in terms of the prescribed bound; no infinite formal theorem is claimed.

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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