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

Dependence on the chosen primes

The weighted condition need not single out one required prime. We construct a support for which convergence depends on a prescribed upward-closed condition on the chosen primes. This description holds at every integer base. We also allow the empty prime set, with $h (a) = 1$ ; it gives a divergent weighted sum because the constructed support has divergent reciprocal sum.

Proposition 1.2 (realising finite monotone witness rules). Let $E$ be a finite set of primes, and let $U$ be an upward-closed family of subsets of $E$ with $E \in U$ and $\emptyset \notin U$ . There is a set $A_{U} \subseteq N_{> 0}$ with $\sum_{a \in A_{U}} 1 / a = \infty$ such that, for every integer $b \geq 2$ and every finite set $P$ of primes,

\begin{matrix} (12) & W_{b, P} (A_{U}) < \infty ⟺ P \cap E \in U . \end{matrix}

Moreover $X_{B} (b)$ is irrational for every infinite $B \subseteq A_{U}$ and every integer $b \geq 2$ .

Proof. For each nonempty $C \subseteq E$ put $D_{C} = \prod_{p \in C} p$ and

B_{C} = {D_{C}^{k} m : k \geq 1, 1 \leq m \leq 2^{D_{C}^{k}}, gcd (m, D_{C}) = 1},

and let $g = \prod_{p \in C \cap P} p$ . The contribution to the weighted sum from the exponents with this value of $k$ lies between positive multiples of $g^{k} / (b^{g^{k}} - 1)$ , with constants depending only on $C \cup P$ . For the upper bound, an element $D_{C}^{k} m$ has $P$ -part $g^{k} u$ , and raising the $P$ -part from $g^{k}$ to $g^{k} u$ multiplies the weight by $u / (1 + b^{g^{k}} + \dots + b^{(u - 1) g^{k}}) \leq 1$ , while $\sum_{m \leq 2^{r}} 1 / m \leq 2 r$ . For the lower bound, the cofactors $m \equiv 1 (\mod Q)$ , with $Q = \prod_{p \in C \cup P} p$ , have $P$ -part exactly $g^{k}$ , and grouping $1, \dots, 2^{r}$ into blocks of at most $Q$ integers gives reciprocal sum at least $H_{2^{r}} / Q \geq r / (2 Q)$ , where $r = D_{C}^{k}$ and $H_{n} = \sum_{j \leq n} 1 / j$ . The comparison terms equal $1 / (b - 1)$ when $g = 1$ and are summable when $g \geq 2$ , since consecutive ratios are at most $g / b^{g^{k}} \to 0$ . Hence the weighted sum over $B_{C}$ converges exactly when $P$ meets $C$ , and the case $Q = D_{C}$ shows that each layer adds at least $1 / (2 D_{C})$ to the ordinary reciprocal sum.

Let $A_{U}$ be the union of the blocks $B_{E ∖ M}$ over the forbidden sets $M \subseteq E$ , $M \notin U$ ; each clause $E ∖ M$ is nonempty because $E \in U$ . A subset of $E$ lies in $U$ exactly when it meets every clause: an allowed set cannot lie inside a forbidden one, and a forbidden set misses its own clause. This finite correspondence is standard [5]. Nonnegative weights make summability on a finite union equivalent to summability on each block, which gives (12) and reciprocal divergence. Since $E \in U$ , (12) with $P = E$ gives $W_{b, E} (B) \leq W_{b, E} (A_{U}) < \infty$ for every infinite $B \subseteq A_{U}$ and every $b \geq 2$ , so Theorem 1.1 makes $X_{B} (b)$ irrational. ◻

For example, require the chosen prime set to contain at least two of $2, 3, 5$ . No individual prime is then mandatory, and one prime alone does not suffice. The maximal forbidden sets give the three clauses ${2, 3}$ , ${2, 5}$ and ${3, 5}$ . Using only these clauses gives the same witness classification as the formal statement, which uses every forbidden set. Historical novelty of this arithmetic realisation has not been assessed.

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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