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 local certificate, and one actual pair

To rule out an integral shift, we can bound two consecutive complete differences in $(- 1, 1)$ and check that the corresponding gaps are unequal. Indeed, simultaneous integrality would force both differences to vanish, and their recurrence would then give $g_{N + h + 1} = g_{N + 1}$ . For even gaps we can state these conditions as a single signed interval for the first difference.

Theorem 6.1 (two consecutive small tail differences). Let $T$ satisfy the dyadic tail recurrence with integer coefficients $g$ , and fix $h$ and $N$ . If

- 1 < σ_{h} (N) < 1, - 1 < σ_{h} (N + 1) < 1, g_{N + h + 1} \neq g_{N + 1},

then $σ_{h} (N)$ and $σ_{h} (N + 1)$ cannot both be integers. Consequently, if such a pair occurs beyond every threshold, the $h$ -shift is not eventually integral.

Proof. An integer strictly between $- 1$ and $1$ is zero. If both shifts were integers, both would vanish, and (7) would give $g_{N + h + 1} = g_{N + 1}$ . The final assertion chooses one such adjacent pair after the alleged onset of integrality. ◻

Corollary 6.2 (real form and the sufficient condition). The same statement holds for a real orbit, with the same proof. If for every $h \geq 1$ and every cutoff some later $N$ satisfies the three displayed conditions for the actual prime gaps, then $Π$ is irrational.

The unequal-gap condition alone is available: at $h = 1$ , $N = 2$ it reads $g_{4} = 2 \neq 4 = g_{3}$ , and Proposition 6.3 gives such inequalities at arbitrarily large indices for each fixed $h$ . The missing assertion is that both small-tail inequalities and the unequal-gap condition hold at the same arbitrarily late indices. Separate infinite sets of witnesses for the three requirements need not intersect.

Proposition 6.3 (prime gaps do not become periodic). For every positive $h$ , the actual consecutive-prime-gap sequence is not eventually periodic with period $h$ .

Proof. For $m \geq 2$ the integers $(m + 1)! + 2, \dots, (m + 1)! + m + 1$ are composite, so the gaps are unbounded. An eventually periodic sequence of natural numbers has finite range after its preperiod and a bounded initial segment, hence is bounded. ◻

Far stronger lower bounds for large gaps are known [8]; the unboundedness of the gaps is all that is needed.

With even gap differences, the three conditions in Theorem 6.1 reduce to two possible signed intervals, each accompanied by a prescribed gap difference. For fixed $h$ write $D_{N} = σ_{h} (N)$ and $δ_{N} = g_{N + h + 1} - g_{N + 1}$ , so that $D_{N + 1} = 2 D_{N} - δ_{N}$ .

Theorem 6.4 (an equivalent signed interval test). Assume $δ_{N}$ is even. The conjunction $- 1 < D_{N} < 1$ , $- 1 < D_{N + 1} < 1$ , $δ_{N} \neq 0$ is equivalent to

(δ_{N} = 2 and \frac{1}{2} < D_{N} < 1) or (δ_{N} = - 2 and - 1 < D_{N} < - \frac{1}{2}) .

Proof. Since $δ_{N} = 2 D_{N} - D_{N + 1}$ , the bounds $| D_{N} | < 1$ and $| D_{N + 1} | < 1$ give $- 3 < δ_{N} < 3$ . A nonzero even integer in this interval is $2$ or $- 2$ . If $δ_{N} = 2$ , the inequality $- 1 < 2 D_{N} - 2 < 1$ , together with $| D_{N} | < 1$ , is equivalent to $1 / 2 < D_{N} < 1$ . Negating both differences gives the case $δ_{N} = - 2$ . Each substitution is reversible. ◻

For the actual gaps, both indices $N + 1$ and $N + h + 1$ are positive, so $δ_{N}$ is even for every $N \geq 0$ . Theorem 6.4 therefore applies to the actual tails. Its proof uses order, the recurrence and the evenness of $δ_{N}$ ; it does not require rational tail values. This is the real version used in Theorem 8.7.

An explicit remainder and a certified actual pair

Here $N$ locates the two tails, $h$ separates them, and $L$ is the number of terms retained from each. A finite block supplies the centre of an enclosure, not the full difference. The polynomial prime bound supplies its radius; clearing denominators then makes each test an integer comparison.

Proposition 6.5 (explicit remainder). For integers $h, N \geq 0$ and $L \geq 1$ put

F_{h, N, L} = \sum_{j = 1}^{L} \frac{g_{N + h + j} - g_{N + j}}{2^{j}}, P (x) = x^{4} + 8 x^{3} + 36 x^{2} + 104 x + 150,

E_{h, N, L} = \frac{1250}{2^{L}} (P (N + h + L + 2) + P (N + L + 2)) .

Then $| σ_{h} (N) - F_{h, N, L} | \leq E_{h, N, L}$ . In particular $| F_{h, N, L} | + E_{h, N, L} < 1$ certifies $| σ_{h} (N) | < 1$ , and $dist (F_{h, N, L}, Z) > E_{h, N, L}$ certifies $σ_{h} (N) \notin Z$ .

Proof. Splitting the complete tails after $L$ terms gives

σ_{h} (N) = F_{h, N, L} + 2^{- L} σ_{h} (N + L) .

The last term is the part not determined by the finite gap blocks. The checked bound $p_{n} \leq 1250 (n + 1)^{4}$ gives $0 \leq g_{m} \leq p_{m + 1} \leq 1250 (m + 2)^{4}$ , so its absolute value is at most

1250 \sum_{j > L} \frac{(N + h + j + 2)^{4} + (N + j + 2)^{4}}{2^{j}} .

The polynomial identity $2 P (x) = (x + 1)^{4} + P (x + 1)$ telescopes to $\sum_{k = 1}^{J} (m + k)^{4} 2^{- k} = P (m) - P (m + J) 2^{- J}$ , whose last term tends to zero. Apply it with $m = N + h + L + 2$ and $m = N + L + 2$ after writing $j = L + k$ . The two tests follow from the triangle inequality and the definition of distance to $Z$ . ◻

Proposition 6.6 (a certified adjacent pair). For the actual prime gaps, $h = 1$ and $N = 2$ satisfy the three hypotheses of Theorem 6.1: both $σ_{1} (2)$ and $σ_{1} (3)$ lie in $(- 1, 1)$ and are nonintegral, and $g_{4} = 2 \neq 4 = g_{3}$ .

Proof. Take $L = 40$ and $Q = 2^{40} = 1099511627776$ . Exact integer arithmetic over the first $46$ primes gives

\begin{array}{crr} N & Q F_{1, N, 40} & Q E_{1, N, 40} \\ 2 & - 662838684750 & 11764181250 \\ 3 & 873345886050 & 12805761250 \end{array}

In each row $| Q F | + Q E < Q$ , which puts the whole certified interval inside $(- 1, 1)$ , and $Q E$ is smaller than the distance from $Q F$ to the nearest multiple of $Q$ , which excludes every integer. The margins are $424908761776$ and $213359980476$ respectively. These are strict integer comparisons rather than inferences from rounded numerical tails. Appendix B replays them. ◻

This certifies one actual adjacent pair, not a recurrence theorem for such pairs. Irrationality still requires a witness for every positive $h$ beyond every cutoff; the successful indices may depend on $h$ .

Proposition 6.7 (certifying two small differences from one enclosure). Let $D_{N + 1} = 2 D_{N} - δ_{N}$ with real $D_{N}$ , and suppose $δ_{N} = 2 s$ for $s \in {- 1, 1}$ . If integers $A, B, Q$ satisfy $Q > 0$ , $B \geq 0$ , $| Q D_{N} - A | \leq B$ , and

2 s A - Q > 2 B, Q - s A > B,

then $1 / 2 < s D_{N} < 1$ , $| D_{N + 1} | < 1$ , and both $D_{N}, D_{N + 1}$ are nonintegral. In particular, the two small-difference conditions are certified from just one tail enclosure.

Proof. Since $| s | = 1$ , the enclosure gives

\frac{s A - B}{Q} \leq s D_{N} \leq \frac{s A + B}{Q} .

The first strict inequality says that the lower endpoint exceeds $1 / 2$ ; the second says that the upper endpoint is less than $1$ . Thus $s D_{N} \in (1 / 2, 1)$ . The recurrence and $δ_{N} = 2 s$ now give $s D_{N + 1} = 2 s D_{N} - 2 \in (- 1, 0)$ . Both intervals exclude integers, and the second also gives $| D_{N + 1} | < 1$ . ◻

For $h = 1, N = 2$ , the exact data already used above give $Q = 2^{40}$ , $A = - 662838684750$ , $B = 11764181250$ and $s = - 1$ . The two positive integer margins are respectively $202637379224$ and $424908761776$ . Proposition 6.6 encloses each difference separately. Here the first enclosure and the exact gap difference give the second interval through the recurrence. The second enclosure is therefore an independent cross-check, not a necessary input.

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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