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.

Start here, by the editorial route

In this paper

Positive divisor majorants

For a finite set $F \subseteq N_{> 0}$ , put $f_{F} (n) = # {a \in F : a ∣ n}$ . Consider the finite set $F = {2, 6, 10, 30}$ , obtained by multiplying each divisor of $15$ by $2$ . For $0 < α \leq 1$ and $z = 2^{α} - 1$ , its count has the exact expansion

f_{F} (n)^{α} = 1_{2 ∣ n} + z 1_{6 ∣ n} + z 1_{10 ∣ n} + z^{2} 1_{30 ∣ n} .

Indeed, conditional on $2 ∣ n$ , each of $3$ and $5$ dividing $n$ doubles $f_{F} (n)$ and multiplies its fractional power by $1 + z$ . The coefficient sum weighted by reciprocal divisors is $(1 + z / 3) (1 + z / 5) / 2$ . This illustrates the general positive expansion used in the separating constructions. The extra factor $(2^{α} - 1)^{- 1}$ in the averaging bound remains part of the cover cost, so small $α$ alone does not make that cost small. The theorem permits a different fractional power for each finite set.

Theorem 1.3 (A summable family of divisor majorants). For each $j \geq 1$ , let $F_{j} \subseteq N_{> 0}$ be finite, let $0 < α_{j} \leq 1$ , and let $c_{j, d} \geq 0$ satisfy

f_{F_{j}} (n)^{α_{j}} \leq \sum_{d ∣ n} c_{j, d} (n \geq 1) .

Set $C_{j} = \sum_{d \geq 1} c_{j, d} / d$ . If

\begin{matrix} (14) & \sum_{j \geq 1} \frac{C_{j} 2^{j α_{j}}}{2^{α_{j}} - 1} < \infty, \end{matrix}

then $X_{A} (b)$ is irrational for every infinite $A \subseteq ⋃_{j} F_{j}$ and every integer $b \geq 2$ .

Every finite set admits such a majorant: take $α_{j} = 1$ and $c_{j, d} = 1_{F_{j}} (d)$ . The restrictive part is summability over the whole family, not the existence of a majorant for an individual set. For example, $F_{j} = {4^{j}}$ with these choices contributes $2^{- j}$ to (14), so the theorem applies. Full support cannot admit a cover satisfying the theorem, because the proof gives arbitrarily small positive displacements and the full-support displacement is at least $1 / 2$ .

Proof. Write $B_{j} = 2^{α_{j}}$ and

U_{j} (N) = \sum_{r \geq 1} 2^{- r} f_{F_{j}} (N + r), V_{j} (N) = \sum_{d \geq 1} c_{j, d} w_{B_{j}, d} (N) .

The choice $B_{j} = 2^{α_{j}}$ matches the decay after taking a fractional power. Subadditivity and the divisor majorant give

U_{j} (N)^{α_{j}} \leq \sum_{r \geq 1} B_{j}^{- r} f_{F_{j}} (N + r)^{α_{j}} \leq \sum_{d \geq 1} c_{j, d} \sum_{\begin{matrix} r \geq 1 \\ d ∣ N + r \end{matrix}} B_{j}^{- r} = V_{j} (N) .

The last identity sums a geometric progression in each residue class. The geometric-series identity also gives

Δ_{2, F_{j}} (N) = U_{j} (N) - X_{F_{j}} (2) \leq U_{j} (N) .

For the first $J$ covering sets we will impose divisibility, making $Δ_{2, F_{j}} (N)$ vanish. At those indices the identity above gives $U_{j} (N) = X_{F_{j}} (2)$ , which need not be zero. We bound only the remaining displacements by the corresponding tails $U_{j} (N)$ .

Fix $ε > 0$ and allot $t_{j} = ε 2^{- j}$ to the $j$ th tail; these allowances sum to $ε$ . Since $U_{j}^{α_{j}} \leq V_{j}$ , it suffices to make $t_{j}^{- α_{j}} V_{j} < 1$ . This normalisation explains the factor $2^{j α_{j}}$ in the cover condition. Choose $J$ with

K_{J} := \sum_{j > J} \frac{C_{j} t_{j}^{- α_{j}}}{B_{j} - 1} < \frac{1}{4} .

This is possible since $ε^{- α_{j}} \leq max (1, ε^{- 1})$ . Choose $L$ divisible by every member of the first $J$ covering sets. Equation (13) bounds the finite mean of $S_{J} (N) := \sum_{j > J} t_{j}^{- α_{j}} V_{j} (N)$ by $(1 + 4 L / M) K_{J} < 1 / 2$ whenever $M \geq 4 L$ . One sample therefore has $S_{J} (N) < 1$ , forcing $U_{j} (N) < t_{j}$ for every $j > J$ . Every exponent in the first $J$ finite sets divides $L$ , so those terms have zero displacement. Consequently,

0 < Δ_{2, A} (N) \leq \sum_{j > J} U_{j} (N) \leq ε .

The positive lower bound under rationality excludes a rational value at base two. For $0 \leq r < d$ , the summand $(b^{r} - 1) / (b^{d} - 1)$ is nonincreasing in $b > 1$ . For $r > 0$ , cancellation of $b - 1$ writes it as $A (b) / (A (b) + C (b))$ , with $A (b) = \sum_{i < r} b^{i}$ and $C (b) = \sum_{r \leq j < d} b^{j}$ . Its derivative has the sign of $A^{'} C - A C^{'} = \sum_{i < r \leq j < d} (i - j) b^{i + j - 1} < 0$ . Thus $Δ_{b, A} (N) \leq Δ_{2, A} (N)$ for real $b \geq 2$ ; for integer $b$ the same positive-integer argument gives irrationality. This ordinary argument is sharper than the factor-two bound in the Lean source’s scalar comparison between bases. Any prescribed positive integer can be included as a divisor of $L$ , so the selected indices can also be required to be arbitrarily large. ◻

The same proof permits any positive weights $η_{j}$ with $\sum_{j} η_{j} = 1$ : replace $2^{j α_{j}}$ in (14) by $η_{j}^{- α_{j}}$ and take $t_{j} = ε η_{j}$ .

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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