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

The tail recurrence and the exact criteria

We rescale each tail so that its first denominator is again $2$ . For a convergent series $\sum_{i \geq 0} a_{i} 2^{- i - 1}$ with $a_{i} \in Z$ , put

T_{N} = 2^{N + 1} \sum_{i > N} a_{i} 2^{- i - 1} = \frac{a_{N + 1}}{2} + \frac{a_{N + 2}}{4} + \dots .

Thus $T_{N}$ starts with coefficient $a_{N + 1}$ , not $a_{N}$ . Doubling and removing that first coefficient gives $T_{N + 1} = 2 T_{N} - a_{N + 1}$ . Modulo integers, a step is simply doubling; the detailed values of the integer coefficients disappear. This is why the recurrence gives a rationality criterion for arbitrary integer coefficients.

At $N = 0$ we have $T_{0} = 2 \sum_{i \geq 0} a_{i} 2^{- i - 1} - a_{0}$ . In particular, for the prime gaps $T_{0} = 2 S - 1 = 2 Π - 5$ , so $T_{0}$ , $S$ and $Π$ have the same rationality status. We first analyse the recurrence abstractly. We then impose the vanishing boundary condition that singles out complete tails and apply the criterion to these prime-gap tails.

Definition 5.1. Let $g : N \to Z$ and $T : N \to Q$ or $T : N \to R$ . Say $T$ satisfies the dyadic tail recurrence with coefficients $g$ when $T_{N + 1} = 2 T_{N} - g_{N + 1}$ for every $N$ . Write $σ_{h} (N) = T_{N + h} - T_{N}$ for the shift of length $h$ at $N$ , and call a real number integral when it belongs to $Z$ .

For a complete-tail example, take $g_{n} = 2$ for odd $n \geq 1$ and $g_{n} = 4$ for even $n \geq 1$ . Its complete tail at index $0$ is $T_{0} = (2 / 2 + 4 / 4) / (1 - 1 / 4) = 8 / 3$ , and the tails alternate between $8 / 3$ and $10 / 3$ . Thus $σ_{1} (N)$ is always $2 / 3$ or $- 2 / 3$ , whereas $σ_{2} (N) = 0$ for every $N$ . Nonintegrality at one prescribed shift length does not imply irrationality, even for positive even coefficients and genuine tails.

An integral shift returns to the same fractional part. To track the integer accumulated in $h$ steps, define $B_{0, N} = 0$ and $B_{h + 1, N} = 2 B_{h, N} + g_{N + h + 1}$ , so that $B_{h, N} = g_{N + 1} 2^{h - 1} + \dots + g_{N + h}$ is the weighted integer sum accumulated in those $h$ steps.

Theorem 5.2 (block identity). For every $N$ and $h$ ,

T_{N + h} = 2^{h} T_{N} - B_{h, N}, hence σ_{h} (N) = (2^{h} - 1) T_{N} - B_{h, N} .

Proof. The case $h = 0$ follows from $B_{0, N} = 0$ . If the first identity holds at $h$ , then

T_{N + h + 1} = 2 (2^{h} T_{N} - B_{h, N}) - g_{N + h + 1} = 2^{h + 1} T_{N} - B_{h + 1, N} .

Subtracting $T_{N}$ gives the second identity. ◻

Subtracting the tail recurrences at $N + h$ and $N$ gives

\begin{matrix} (7) & σ_{h} (N + 1) = 2 σ_{h} (N) - (g_{N + h + 1} - g_{N + 1}), \end{matrix}

which is the recurrence for the tail differences. Since $B_{h, N}$ is an integer, Theorem 5.2 also gives the integrality criterion: $σ_{h} (N)$ is integral if and only if $(2^{h} - 1) T_{N}$ is integral. For fixed $h$ in this general recurrence, $T_{N}$ therefore determines whether the shift is integral, regardless of the later integer coefficients. For a complete tail, however, changing those coefficients also changes $T_{N}$ ; the observation does not permit arbitrary changes to prime gaps.

For rational $T_{N} = u / v$ in lowest terms, the recurrence gives $T_{N + 1} = (2 u - g_{N + 1} v) / v$ . The numerator has greatest common divisor $gcd (2, v)$ with $v$ , because $gcd (u, v) = 1$ . Thus

\begin{matrix} (8) & den (T_{N + 1}) = \frac{den (T_{N})}{gcd (2, den (T_{N}))}, σ_{h} (N) \in Z ⟺ den (T_{N}) ∣ 2^{h} - 1 . \end{matrix}

The power of two in the denominator determines how many initial steps must pass before a return is possible; the remaining odd factor $d$ determines the possible return lengths. Once that factor is reached, Euler’s congruence supplies $d ∣ 2^{φ (d)} - 1$ . The least positive return length is the multiplicative order of $2$ modulo $d$ , taken to be $1$ when $d = 1$ . Before the power of two is removed there is no return: for example, $T_{N} = 1 / 2$ gives no integral shift at $N$ , since $2^{h} - 1$ is always odd. This periodicity concerns fractional parts, not full tail values or coefficients.

Theorem 5.3 (exact rationality classification). Let $T : N \to R$ satisfy $T_{N + 1} = 2 T_{N} - g_{N + 1}$ with integer coefficients $g$ . The following are equivalent:

$T_{0}$ is rational;
$σ_{h} (N)$ is an integer for some $h \geq 1$ and some $N$ ;
for some fixed $h \geq 1$ , $σ_{h} (N)$ is an integer at every sufficiently large $N$ .

Consequently $T_{0}$ is irrational if and only if every positive-length shift is nonintegral at every index, equivalently if and only if for every $h \geq 1$ and every cutoff some later $N$ has $σ_{h} (N) \notin Z$ .

Proof. For (i) $\Rightarrow$ (iii), the block identity makes every $T_{N}$ rational. By (8), its denominator is a fixed odd integer $d$ for all sufficiently large $N$ . Choose $h = φ (d)$ ; Euler’s congruence and the same formula show that every subsequent $σ_{h} (N)$ is integral. The implication (iii) $\Rightarrow$ (ii) is immediate. For (ii) $\Rightarrow$ (i), the block identity gives $σ_{h} (N) = (2^{h} - 1) T_{N} - B_{h, N}$ with $B_{h, N}$ and $σ_{h} (N)$ integers and $2^{h} - 1 \neq 0$ , so $T_{N}$ is rational, and $T_{N} = 2^{N} T_{0} - B_{N, 0}$ then makes $T_{0}$ rational. Negating the three conditions gives the two irrationality formulations. ◻

Lemma 5.4 (the boundary condition for the tail recurrence). Let $a_{1}, a_{2}, \dots$ be real numbers with $\sum_{j \geq 1} | a_{j} | 2^{- j} < \infty$ , and let $U_{N + 1} = 2 U_{N} - a_{N + 1}$ . Then

U_{N} = \sum_{j \geq 1} a_{N + j} 2^{- j} for every N ⟺ 2^{- N} U_{N} ⟶ 0.

Proof. Iteration gives $2^{- N} U_{N} = U_{0} - \sum_{j = 1}^{N} a_{j} 2^{- j}$ . The limit is zero exactly when $U_{0}$ equals the full series; subtracting its first $N$ terms then gives the claimed tail formula. Without the boundary condition, one may add $C 2^{N}$ to every $U_{N}$ without changing the recurrence. ◻

For instance, zero coefficients and initial value $U_{0} = 1 / 12$ give $\frac{1}{12}, \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \dots$ , with $U_{N + h} - U_{N} = (2^{h} - 1) 2^{N} / 12$ . Shifts of length $2$ become integral at index $2$ ; shifts of length $1$ never do. These are not complete tails of the zero sequence, since $2^{- N} U_{N} = 1 / 12$ does not tend to zero. The denominator calculation remains valid, but the boundary condition excludes this extra homogeneous term.

The actual prime-gap tails.

Put

T_{N} = \sum_{j \geq 1} \frac{g_{N + j}}{2^{j}},

which converges by Theorem 4.3. Reindexing identifies this sum with the rescaled complete tail of $S$ , as in the shifted-gap series identity. Splitting off the first term gives $T_{N + 1} = 2 T_{N} - g_{N + 1}$ without a rationality hypothesis, as formalised in the real tail recurrence. Since $T_{0} = 2 S - 1 = 2.349287932 \dots$ , the numbers $Π$ , $S$ and $T_{0}$ have the same rationality status, and Theorem 5.3 specialises to the actual series: irrationality of $Π$ is equivalent to nonintegrality arbitrarily far out for each positive shift of $T$ . The formal statement is the irrationality equivalence for the actual prime tails. Thus the passage from the series to its complete tails is proved. What remains is to establish the required nonintegrality for these prime-gap tails.

Pairs of congruent indices and least-common-multiple shifts

Instead of prescribing the offset $M - N$ , we may prescribe a positive modulus $t$ and allow any pair with $M \equiv N (\mod t)$ . This gives more freedom in finding a witness, but the test must still work for every $t$ and beyond every cutoff.

Theorem 5.5 (pairs of congruent indices). $S$ is irrational if and only if for every $t \geq 1$ and every $N_{0}$ there are $N, M \geq N_{0}$ with $M \equiv N (\mod t)$ and $T_{M} - T_{N} \notin Z$ .

Proof. If $S$ is irrational, take $M = N + t$ for the $N$ supplied by Theorem 5.3 at shift length $t$ . Conversely, suppose $S$ is rational. By (8) the orbit reaches an index $N_{0}$ beyond which the reduced denominator is a fixed odd $d$ ; let $t$ be the multiplicative order of $2$ modulo $d$ , taking $t = 1$ when $d = 1$ . For $N, M \geq N_{0}$ with $M \equiv N$ modulo $t$ , the difference $T_{M} - T_{N}$ is then an integer, contradicting the stated property at this $t$ and this cutoff. ◻

The obstruction under rationality is therefore one particular modulus: the order of $2$ modulo the eventual odd denominator. A successful pair for some other modulus does not exclude it.

Equal tail values do not help this criterion, since their difference is zero. Section D.4 shows that rationality and the prime number theorem would in fact force many equal tail values. It also explains why, beyond the rational denominator cutoff, an interval condition excluding integers already contradicts rationality at indices congruent modulo the corresponding multiplicative order. These are ordinary deductions, not additional formalised statements.

Alternatively, one can fix the pairs in advance. Let $L_{0} = 1$ and $L_{j} = lcm (1, \dots, j)$ . These numbers eventually exceed any starting cutoff and are eventually divisible by any proposed period.

Theorem 5.6 (criterion using least common multiples). Let $g : N \to Z$ and $T : N \to R$ satisfy $T_{N + 1} = 2 T_{N} - g_{N + 1}$ . Then $T_{0}$ is irrational if and only if $T_{2 L_{j}} - T_{L_{j}} \notin Z$ for every $j \geq 0$ .

Proof. Theorem 5.3 gives the forward implication. Conversely, if $T_{0}$ is rational then some positive shift length $h$ is integral at every basepoint beyond an index $N_{0}$ . Choose $j$ so large that $N_{0} \leq L_{j}$ and $h ∣ L_{j}$ . The telescoping identity $σ_{a + b} (N) = σ_{a} (N) + σ_{b} (N + a)$ shows by induction that every positive multiple of $h$ is integral at every such basepoint, so $T_{2 L_{j}} - T_{L_{j}}$ is an integer. ◻

One could use factorials instead of least common multiples, since they have the same eventual divisibility property. The least common multiples are no larger at each index. Theorems 5.5 and 5.6 change which tail differences one must test, but remain exact equivalences for integer-coefficient recurrences. Neither proves the needed nonintegrality for the actual prime gaps.

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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