Reciprocal Mersenne Subseries

Will Cook

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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

Bounds for a general coefficient sequence

Rationality of a binary coefficient series gives an integer recurrence. The necessary decay condition is on the scaled recurrence, and allows unbounded integer values. We first bound the coefficient tail, then recover the series value by telescoping. For support series, Möbius inversion imposes an additional condition on the coefficients.

Theorem 6.104 (An upper bound for a binary coefficient tail). Let $c : N \to N$ satisfy $c (n) \leq n$ for every $n$ . Then

\sum_{r \geq 1} c (N + r) 2^{- r} \leq N + 2 (N \geq 0) .

Indeed, $c (N + r) \leq N + r$ , while $\sum_{r \geq 1} 2^{- r} = 1$ and $\sum_{r \geq 1} r 2^{- r} = 2$ . Thus the scaled tail is $O (N)$ , and in particular $o (2^{N})$ , as required in Theorem 6.105. The coefficient bound also holds for Euler’s totient function.

Theorem 6.105 (Rationality through an integer recurrence). Let $c : N \to N$ satisfy $c (n) \leq n$ for every $n$ . Then $\sum_{n \geq 1} c (n) 2^{- n}$ is rational if and only if there exist a positive integer $v$ and a sequence $u : N \to Z$ such that

u (N + 1) = 2 u (N) - v c (N + 1) (N \geq 0), \frac{u (N)}{2^{N}} ⟶ 0.

The limit condition is $u (N) = o (2^{N})$ ; it does not assume that $u$ is bounded. No divisor-count hypothesis is imposed on $c$ . Thus the ordinary statement also applies to $c = φ$ ; the identification with the separate formal #249 development, including its indexing conventions, is not asserted here.

To prove the forward implication from a recurrence, we telescope:

\frac{u (N)}{2^{N}} = u (0) - v \sum_{n = 1}^{N} \frac{c (n)}{2^{n}} .

The limit condition gives the rational value $u (0) / v$ . Conversely, if the series equals $p / v$ , define $u (N) = 2^{N} p - v \sum_{n = 1}^{N} 2^{N - n} c (n)$ . This is an integer and equals $v \sum_{r \geq 1} c (N + r) 2^{- r}$ , so the preceding tail bound proves the limit. This argument also explains why an arbitrary integer solution of the recurrence is insufficient: its homogeneous part can grow like $2^{N}$ .

For example, the finite support $A = {2}$ has $c_{A} (n) = 1_{2 ∣ n}$ and $X_{A} (2) = 1 / 3$ . With $v = 3$ , the sequence $u (N) = 1$ for even $N$ and $u (N) = 2$ for odd $N$ satisfies the recurrence. The criterion therefore includes finite supports; it does not provide the infinite support needed to contradict Problem 257.

Theorem 6.106 (Möbius inversion of the divisor counts). For a set $A \subseteq N$ and each positive integer $n$ ,

\sum_{d ∣ n} μ (d) c_{A} (n / d) = 1_{A} (n),

where $μ$ is the Möbius function and $c_{A} (n)$ counts the positive elements of $A$ dividing $n$ . Thus the divisor counts determine the positive support exactly. In particular their Möbius transform takes only the values $0$ and $1$ . The identity does not depend on a base.

Theorem 6.107 (The support series as a coefficient series). For $A \subseteq N_{> 0}$ ,

X_{A} (2) = \sum_{a \in A} \frac{1}{2^{a} - 1} = \sum_{n \geq 1} \frac{c_{A} (n)}{2^{n}} .

Expanding each denominator as a geometric series and interchanging nonnegative sums gives the identity. Since $c_{A} (n) \leq τ (n) \leq n$ , Theorem 6.105 applies. Together with Theorem 6.106, it expresses rationality through an integer recurrence whose coefficients all come from the same support.

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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