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

Combining the two support criteria

Separate sequences of small weighted and covered displacements would not give a small sum: their small values could occur on disjoint sets of indices. Instead, the weighted proof chooses one finite distribution and the explicit modulus dependence in (13) lets the cover estimate use that same distribution. We add the two nonnegative errors there and choose an index only afterwards.

Theorem 1.4 (mixed weighted and cover supports). Let $E, V \subseteq N_{> 0}$ . Suppose $E$ has finite weighted mass (1) at $b = 2$ for a finite nonempty prime set $P$ . Suppose also that $V \subseteq ⋃_{j} F_{j}$ for sets and majorants satisfying Theorem 1.3, with (14) or its positive-weight variant. Then $X_{A} (b)$ is irrational for every infinite $A \subseteq E \cup V$ and every integer $b \geq 2$ .

Proof. Fix $ε > 0$ and an integer $N_{0} \geq 1$ , and put $ρ = ε / 3$ . Use the cover notation $B_{j}, U_{j}, V_{j}$ above, with $η_{j} = 2^{- j}$ in the case (14). Set $t_{j} = ρ η_{j}$ and choose $J$ so that

K_{J} = \sum_{j > J} \frac{C_{j} t_{j}^{- α_{j}}}{B_{j} - 1} < \frac{1}{16} .

Choose a finite $F \subseteq E$ whose complementary weighted mass is $κ < ρ / 16$ . Let $L$ be a positive common multiple of $N_{0}$ , all members of $F$ , and all members of the first $J$ covering sets. The order is deliberate: first the finite pieces determine $L$ ; then the weighted argument enlarges $L$ to $Q$ and chooses the dyadic scales; only after both estimates are on that distribution is a sample selected. With the finite sets and $L$ fixed, choose the base-two weighted parameters

Q = L \prod_{p \in P} p^{⌊ \log_{p} H ⌋}, G = ⌊ H / max P ⌋, M = ⌊ 2^{G / 2} ⌋ .

We use $D_{Q; M, M}$ throughout the following estimates. Its scales run from $M$ to $2 M - 1$ , and both quantities are evaluated at the same sampled multiple $N = Q m$ . The finite estimate (11), with complementary weighted mass $κ$ , remains valid for this $L$ : its proof requires only that every member of $F$ divide $L$ . It also applies to finite or empty $E$ , since positivity was used only after the averaging estimate. Thus

D_{Q; M, M} (Δ_{2, E} / ρ) < \frac{1}{8}

for sufficiently large $H$ . For the same finite distribution, (13) and nonnegative interchange give

D_{Q; M, M} S_{J} \leq (1 + 4 Q / M) K_{J} < \frac{1}{8}, S_{J} = \sum_{j > J} t_{j}^{- α_{j}} V_{j},

because $Q / M \to 0$ . Here the cover estimate is applied with modulus $Q$ , not merely with the fixed divisor $L$ ; this is why its explicit modulus dependence was needed. On this one distribution,

D_{Q; M, M} (Δ_{2, E} / ρ + S_{J}) < \frac{1}{4} .

Choose a sampled $N = Q m$ at which the nonnegative sum is less than $1$ . Nonnegativity of both quantities and of every summand of $S_{J}$ gives $Δ_{2, E} (N) < ρ$ and $U_{j} (N) < t_{j}$ for every $j > J$ . For $j \leq J$ , it is the displacement $Δ_{2, F_{j}} (N)$ that vanishes, since every exponent in $F_{j}$ divides $L$ and hence divides $N$ . Therefore

Δ_{2, A} (N) \leq Δ_{2, E} (N) + Δ_{2, V} (N) < 2 ρ < ε, N \geq Q \geq L \geq N_{0} .

The inequality also holds when $E$ and $V$ overlap. Since $A$ is infinite, $Δ_{b, A} (N) > 0$ . The comparison $Δ_{b, A} (N) \leq Δ_{2, A} (N)$ from the cover proof makes these values arbitrarily small at every integer base, contradicting the positive lower bound imposed by rationality. ◻

Why the mixed criterion is strict.

Take an infinite support $E$ satisfying the weighted condition but admitting no strengthened positive cover, and an infinite support $V$ admitting such a cover but failing the weighted condition for every finite prime set at every integer base. The two constructions in the short paper’s section on comparison of support classes supply these sets. Put $U = E \cup V$ . A strengthened cover of $U$ restricts to one of $E$ , since each divisor count decreases on a subset. Convergence of the weighted sum over $U$ would imply convergence over $V$ with the same prime set, since every summand is nonnegative. Hence $U$ belongs to neither individual class. For every infinite $A \subseteq U$ , Theorem 1.4 applies with these same sets $E, V$ and gives irrationality at every integer base. Irrationality of the two summands separately would not suffice. This proves the short paper’s strictness corollary from the two constructions, with the ordinary-proof qualification stated below.

The positive-weight variant of Theorem 1.3, with its hereditary all-base conclusion, also has a kernel-checked Lean proof. Theorem 1.4 alone does not exhibit a support outside both individual classes; the preceding strictness argument also uses the two separating constructions. Existence of two sets with those properties has a Lean proof whose second set is formed from squarefree divisor sets with a different choice of parameters. The printed fresh-prime construction and the strictness corollary have ordinary proofs, and none of these constructions has been independently reviewed. The second set in the Lean proof also admits no cover with $\sum_{j} C_{j} 2^{j α_{j}} 2^{α_{j}} / (2^{α_{j}} - 1)^{2} < \infty$ , so this stronger cost defines a strictly smaller support class. The mixed implication itself is kernel-checked. Universal irrationality and the quantitative and prime-power thinning claims remain open.

Supports satisfying the mixed criterion are closed under finite unions and finite changes, even with the fixed dyadic cover cost in (14). For two covers, interleave their $j$ th finite sets at positions $2 j - 1, 2 j$ and halve each exponent. Halving compensates for the increased index in $2^{j α_{j}}$ . The same coefficients still majorise the smaller powers, since integer divisor counts satisfy $f^{α / 2} \leq f^{α}$ . The denominator $2^{α} - 1$ is the only remaining cost: replacing $α$ by $α / 2$ increases the full cost of each finite set by at most $1 + \sqrt{2}$ . For $r$ covers, positions at most $r j$ and exponents divided by $r$ give a cost ratio below $2 r$ . For weighted supports, take the union of the finite prime sets: the prime part can only increase, while $h / (2^{h} - 1)$ decreases. Each condition is inherited by subsets, and the weighted sum is finite on a finite support. Lean proves these closures with the constant $3$ in place of $1 + \sqrt{2}$ . For a countable union the cover costs must still be summable. Every prime singleton is admitted, but for the full prime support $P$ one has $Δ_{2, P} (N) > 1 / 3$ for every $N \geq 1$ : indeed $\sum_{r \geq 1} 2^{- r} ω (N + r) \geq 1$ , whereas $X_{P} (2) \leq \sum_{a \geq 2} (2^{a} - 1)^{- 1} < 2 / 3$ . Here $ω (n)$ is the number of distinct prime divisors of $n$ .

The elementary independence boundary.

For the divisor counts $c_{A}$ and a fixed $a \in A$ , the identity $c_{A} (a n) - c_{A} (n) = 1 - 1_{a ∣ n}$ holds for every $n \geq 1$ exactly when $a$ is coprime to every other member of $A$ . Every summand $1_{t ∣ a n} - 1_{t ∣ n}$ is nonnegative; $t = a$ supplies the right side, and coprimality makes all other terms zero. If $g = (a, t) > 1$ for some $t \neq a$ , take $n = t / g$ to obtain an additional positive term. For distinct odd primes $p, q$ , the covariance of $1_{2 p ∣ n}$ and $1_{2 q ∣ n}$ over a common period is $1 / (2 p q) - 1 / (4 p q) = 1 / (4 p q)$ . Thus dilating a prime support already destroys the independence used in the prime-incidence argument. This brief ordinary calculation marks a boundary of that argument, not an irrationality criterion.

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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