Reciprocal Mersenne Subseries

Will Cook

Long paper

Reciprocal Mersenne Subseries

Erdős #257 By Will Cook 30 September 2026 Authorship, AI use and citation

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Summary. A weighted summability criterion proves irrationality of reciprocal Mersenne subseries: if h is the part of a supported on a finite nonempty prime set P, every infinite A with ∑ $h (a) / (a (b^{h (a)} - 1))$ finite has irrational ∑ $(b^{a} - 1)^{- 1}$ . The base-two condition gives the conclusion at every integer base for every infinite subset, and permits an example with divergent reciprocal sum, while excluding full support and all odd exponents. The proof averages over a finite part of the support and over dyadic observation lengths. Erdős's weaker reciprocal-summable theorem is proved in full, and a common window extends both arguments to unions of weighted and positive-cover supports. Finite-denominator and achievement-set calculations formulate tests for the targets $1 / 2$ and $1 / 21$ ; those tests do not decide either target or the universal question. Prime support at base 2 and squarefree support at power-of-two bases are attributed as existing work.

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #257: selected checked premises, finite computations, closed routes, surviving obligations, and their exact evidence bands.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, whose universal and half-value questions remain open.

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In this paper

A sufficient test for a safe skip

Definition 6.119 (Three geometric terms in the tail lower bound). For $k \geq 1$ put $ℓ_{3} (k) = 1 / 2^{k} + 1 / (3 \cdot 4^{k}) + 1 / (7 \cdot 8^{k})$ . Then $ℓ_{3} (k) < R (k)$ : this is a strict rational lower bound obtained by truncating the Lambert-type series at three terms.

mersenneTailLB3 three_div_lt_mersenneTailLB3 mersenneTailLB3_lt_mersenneTail

Theorem 6.120 (A sufficient inequality for a safe skip). Let $k, u, L \geq 1$ be integers, set $a = 2 L - (2^{k} - 1) u$ , and suppose $a > 0$ . For the skipped rational remainder $ρ = u / (2 L)$ , $ρ \leq 2^{- k}$ is equivalent to $u \leq a$ , whereas

2 u \leq 3 a ⟹ ρ < R_{k} .

The latter implication uses the three-term lower bound above. More explicitly, with $t = 2^{k} \geq 2$ ,

(\frac{1}{t} + \frac{1}{3 t^{2}} + \frac{1}{7 t^{3}}) - \frac{3}{3 t - 1} = \frac{2 t - 3}{21 t^{3} (3 t - 1)} > 0.

The inequality $2 u \leq 3 a$ gives $ρ \leq 3 / (3 t - 1)$ , proving the claim. The first two terms alone would not prove this comparison. This is a sufficient exclusion of the current tail-mass deficit, not an exact test for membership in the remaining achievement set.

The sufficient condition is weaker than $u \leq a$ . A realizable example is $(k, u, L, a) = (2, 7, 13, 5)$ ; it passes $2 u \leq 3 a$ but not $u \leq a$ . The linked scalar example $(u, a) = (3, 2)$ proves strict containment of the inequalities, but is not integral rational data for this substitution. For comparison, the exact mass threshold is $a / u \geq R_{k}^{- 1} - (2^{k} - 1)$ ; its right-hand side lies strictly between $0$ and $2 / 3$ . None of these comparisons asserts that the half-greedy orbit satisfies the sufficient inequality at every skipped rank.

Proposition 6.121 (Two unconditional safety cases). Under the positive-integer and skipped-step hypotheses of Theorem 6.120, $u = 1$ implies $a \geq 1$ , so the current remainder is less than $R_{k}$ . Conversely, a fatal tail-mass deficit forces $3 a < 2 u$ , hence $u \geq 2$ , or $u \geq 3$ when $u$ is odd. These are statements about the current step, not infinite survival.

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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