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

Examples and limits of the hypothesis.

Squares, powers of two and factorials satisfy reciprocal summability and therefore satisfy the base-two weighted condition. Full support does not: for any fixed finite prime set $P$ , its subset coprime to $\prod_{p \in P} p$ has weight $1 / [a (b - 1)]$ and divergent reciprocal sum. The set of all odd exponents also fails, by restricting to integers coprime to $2 \prod_{p \in P} p$ . Inclusion–exclusion in the harmonic sum proves both divergences. This is a limitation of the sufficient condition, not an argument for rationality; full support is already known to be irrational.

Writing $a = h m$ with $h$ its $P$ -part gives the exact decomposition

W_{b, P} (A) = \sum_{h \in H_{P}} \frac{1}{b^{h} - 1} \sum_{\begin{matrix} m \geq 1, h m \in A \\ (m, \prod_{p \in P} p) = 1 \end{matrix}} \frac{1}{m},

where $H_{P}$ consists of $1$ and the positive integers with no prime factor outside $P$ . Each inner reciprocal sum must be finite; the displayed outer sum imposes the additional summability across different $P$ -parts. For example, at base two with $P = {2}$ , choose in each layer $2^{k}$ , $k \geq 1$ , a finite set of odd factors whose reciprocal sum is at least $2^{2^{k}}$ . Every inner sum is finite, but the outer weighted sum diverges. Such finite sets exist by divergence of the harmonic sum over odd integers. Prime support also fails this criterion for every finite $P$ : its contribution from primes outside $P$ is a constant multiple of their divergent reciprocal sum. These failures concern the sufficient hypothesis, not rationality of the resulting values. The following construction shows, nonetheless, that this condition is strictly weaker than reciprocal summability. At base two take

A_{⋆} = {2^{k} m : k \geq 1, m odd, m \leq 2^{2^{k}}} .

Write $H_{N} = \sum_{m = 1}^{N} 1 / m$ and $N_{k} = 2^{2^{k}}$ . The sum of reciprocals of the exponents with $v_{2} (a) = k$ is

ρ_{k} = 2^{- k} \sum_{\begin{matrix} m \leq N_{k} \\ m odd \end{matrix}} \frac{1}{m} = 2^{- k} (H_{N_{k}} - \frac{1}{2} H_{N_{k} / 2}) ⟶ \frac{\log 2}{2} .

Thus $\sum_{a \in A_{⋆}} 1 / a$ diverges. For $P = {2}$ , the weighted contribution of the same layer is

\frac{2^{k}}{2^{2^{k}} - 1} ρ_{k} .

Since $H_{N} \leq 1 + \log N$ , these terms are at most $(1 + 2^{k} \log 2) 2^{1 - 2^{k}}$ , which is summable. The theorem therefore proves irrationality at every integer base for every infinite subset of this set, although its reciprocal sum diverges.

Proof. We first derive the lower bound implied by rationality and then estimate a finite average of the same quantities. Division of $m$ by $a$ , say $m = q a + r$ with $0 \leq r < a$ , gives

\frac{b^{m} - 1}{b^{a} - 1} - \frac{b^{r} - 1}{b^{a} - 1} = \sum_{j = 1}^{q} b^{m - j a} .

Consequently

\begin{matrix} (2) & Δ_{b, A} (m) = (b^{m} - 1) X_{A} (b) - J_{b, A} (m), J_{b, A} (m) = \sum_{\begin{matrix} a \in A \\ a \leq m \end{matrix}} \sum_{j = 1}^{⌊ m / a ⌋} b^{m - j a} \in Z . \end{matrix}

The integer sum is finite. For $m \geq 1$ , moreover, $Δ_{b, A} (m) > 0$ : since $A$ is infinite, some $a \in A$ exceeds $m$ , and its summand is positive. At $m = 0$ the displacement is zero. Thus, if $X_{A} (b) = u / v$ with $v > 0$ , equation (2) gives $v Δ_{b, A} (m) \in Z$ , whence

\begin{matrix} (3) & Δ_{b, A} (m) \geq \frac{1}{v} (m \geq 1) . \end{matrix}

We therefore seek positive indices at which the displacement is arbitrarily small.

Put

ϕ_{a} (m) = \frac{b^{m mod a}}{b^{a} - 1}, d_{a} (m) = ϕ_{a} (m) - \frac{1}{b^{a} - 1} .

Then $d_{a} (m) \geq 0$ and $Δ_{b, A} (m) = \sum_{a \in A} d_{a} (m)$ . For $Q, a, T \geq 1$ , put $g = (Q, a)$ . The residues $t Q mod a$ have period $a / g$ and visit $0, g, \dots, a - g$ once in a period, so

\sum_{t = 1}^{a / g} ϕ_{a} (t Q) = \frac{1 + b^{g} + \dots + b^{a - g}}{b^{a} - 1} = \frac{1}{b^{g} - 1} .

Complete periods and at most one incomplete period therefore give

\begin{matrix} (4) & \frac{1}{T} \sum_{t = 1}^{T} d_{a} (t Q) \leq \frac{g}{a (b^{g} - 1)} + \frac{1}{T (b^{g} - 1)} . \end{matrix}

The complete-period term explains the weight in (1): replacing $g$ by the $P$ -part $h (a)$ gives its summand. The modulus below will permit this replacement for $h (a) \leq H$ ; larger prime parts require a separate estimate. For this comparison we use that $n / (b^{n} - 1)$ decreases for positive integers $n$ ; indeed

n (b^{n + 1} - 1) - (n + 1) (b^{n} - 1) = b^{n} (n (b - 1) - 1) + 1 > 0.

The part of the support beyond the observation range has a uniform bound. If $Y = Q T$ and $a > Y \geq t Q$ , then $t Q mod a = t Q$ and $d_{a} (t Q) \leq 2 2^{t Q - a}$ . Hence

\sum_{a > Y} d_{a} (t Q) \leq 2 2^{t Q - Y}, \sum_{t = 1}^{T} 2^{t Q - Y} = \sum_{j = 0}^{T - 1} 2^{- j Q} \leq 2,

and therefore

\begin{matrix} (5) & \frac{1}{T} \sum_{t = 1}^{T} \sum_{\begin{matrix} a \in A \\ a > Q T \end{matrix}} d_{a} (t Q) \leq \frac{4}{T} . \end{matrix}

The incomplete-period term in (4) has no factor $1 / a$ . Although it tends to zero for a fixed exponent, there are more observed exponents as the window grows. To recover a summable bound, average over lengths and sum in the other order. An exponent $a$ first enters when the window reaches $a / Q$ ; choosing geometrically spaced lengths makes the sum of their later reciprocals at most $2 Q / a$ . Let $α_{a} \geq 0$ with $\sum_{a} α_{a} / a < \infty$ . For $Q, M \geq 1$ , finite rearrangement gives

\begin{matrix} (6) & \sum_{j = M}^{2 M - 1} \frac{1}{2^{j}} \sum_{a \leq Q 2^{j}} α_{a} \leq 2 Q \sum_{a \geq 1} \frac{α_{a}}{a} . \end{matrix}

Indeed, for each fixed $a$ the admissible indices satisfy $2^{j} \geq a / Q$ and

\sum_{\begin{matrix} M \leq j < 2 M \\ a \leq Q 2^{j} \end{matrix}} 2^{- j} \leq \frac{2 Q}{a} .

Only finitely many terms occur on the left, so no infinite interchange is hidden here.

We now choose the modulus to be divisible by a fixed finite subset of the support and by all powers of primes in $P$ up to a growing threshold. Fix $ε > 0$ and a lower bound $N$ . Select a finite $F \subseteq A$ such that

\begin{matrix} (7) & \sum_{a \in A \ F} \frac{h (a)}{a (b^{h (a)} - 1)} < ε, \end{matrix}

and put $L = lcm (F)$ , with $L = 1$ if $F$ is empty. Write $p_{*} = max P$ and $r = | P |$ . For a large integer $H \geq 2 p_{*}$ , set

Q_{0} (H) = \prod_{p \in P} p^{⌊ \log_{p} H ⌋}, Q = L Q_{0} (H), G = ⌊ \frac{H}{p_{*}} ⌋ .

Then $Q \leq L H^{r}$ , $G \geq 1$ , and every $a \in F$ divides $Q$ , so its $d_{a} (t Q)$ contribution vanishes.

Consider first $a \in A \ F$ with $a \leq Q T$ and $h (a) \leq H$ . Every prime-power factor of $h (a)$ occurs in $Q_{0} (H)$ , hence $h (a) ∣ Q_{0} (H)$ and $(Q, a) \geq h (a)$ . Applying (4) and the monotonicity above, the total contribution of these $a$ is at most

\begin{matrix} (8) & ε + \frac{1}{T} \sum_{\begin{matrix} a \in A, a \leq Q T \\ h (a) \leq H \end{matrix}} \frac{1}{b^{h (a)} - 1} . \end{matrix}

The first term is bounded using (7).

For $h (a) > H$ , one has $(Q, a) \geq G$ . If every $P$ -prime-power component of $h (a)$ is at most $H$ , then $h (a) ∣ Q_{0} (H)$ and the gcd is larger than $H$ . Otherwise some $p^{v_{p} (a)} > H$ , and the gcd contains $p^{⌊ \log_{p} H ⌋} > H / p \geq H / p_{*}$ . Thus (4), enlarged from the support to all $a \leq Q T$ , bounds the large- $h (a)$ part by

\begin{matrix} (9) & \frac{G (1 + \log (Q T)) + Q}{b^{G} - 1}; \end{matrix}

we used $\sum_{a \leq Q T} a^{- 1} \leq 1 + \log (Q T)$ and $# {a \leq Q T} / T = Q$ .

Let

M (Q, T) = \frac{1}{T} \sum_{t = 1}^{T} Δ_{b, A} (t Q) .

Combining (5), (8), and (9) gives

\begin{matrix} (10) & M (Q, T) \leq ε + \frac{1}{T} \sum_{\begin{matrix} a \in A, a \leq Q T \\ h (a) \leq H \end{matrix}} \frac{1}{b^{h (a)} - 1} + \frac{G (1 + \log (Q T)) + Q}{b^{G} - 1} + \frac{4}{T} . \end{matrix}

Increasing the number $M$ of dyadic lengths reduces the incomplete-period error to order $Q / M$ . It also makes $\log T$ of order $M$ on the largest window, increasing the large-gcd contribution to order $G M / b^{G}$ . Thus the two quantities must be made small together. With $L$ fixed, $Q$ grows polynomially in $H$ and $G$ grows linearly, so an intermediate exponential choice of $M$ has both properties. Choose

M = ⌊ b^{G / 2} ⌋

and average (10) over $T = 2^{j}$ , $M \leq j < 2 M$ . Apply (6) with $α_{a} = 1_{A} (a) / (b^{h (a)} - 1)$ . Its hypothesis holds because

\sum_{a} \frac{α_{a}}{a} = \sum_{a \in A} \frac{1}{a (b^{h (a)} - 1)} \leq W_{b, P} (A) .

The resulting finite average satisfies

\begin{matrix} (11) & \frac{1}{M} \sum_{j = M}^{2 M - 1} M (Q, 2^{j}) \leq ε + \frac{2 Q W_{b, P} (A)}{M} + \frac{G (1 + \log Q + 2 M \log 2) + Q}{b^{G} - 1} + 4 2^{- M} . \end{matrix}

As $H \to \infty$ ,

Q \leq L H^{r}, G = H / p_{*} + O (1), M ≍ b^{G / 2} .

Consequently $Q / M \to 0$ , $G M / b^{G} \to 0$ , $(G \log Q + Q) / b^{G} \to 0$ , and $2^{- M} \to 0$ . For sufficiently large $H$ , we may also arrange $Q \geq N$ and make the right side of (11) smaller than $2 ε$ . The left side is a finite average of positive numbers $Δ_{b, A} (t Q)$ . One of them therefore satisfies

0 < Δ_{b, A} (t Q) < 2 ε, t Q \geq Q \geq N .

The construction is valid for every positive $ε$ ; applying it with $ε / 2$ proves the stated cofinal bound with $Δ_{b, A} (m) < ε$ . This contradicts (3) if $X_{A} (b)$ is rational. ◻

Since $b^{h (a)} - 1 \geq 2^{h (a)} - 1$ for every $b \geq 2$ , one has $W_{b, P} (A) \leq W_{2, P} (A)$ . Thus the base-two weighted hypothesis gives the conclusion simultaneously at every integer base. The hypothesis is also hereditary under passage to infinite subsets. The theorem does not cover all infinite supports and gives no quantitative irrationality measure; no priority is claimed for it.

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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