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

Conditional membership tests

The finite tests below use $g_{n} = w_{n} - R_{n}$ . When written as functions, $R (n) = R_{n}$ , $g (n) = g_{n}$ , $r (x, n) = r_{n} (x)$ and $C (A, N) = C_{A} (N)$ . For a finite set $D$ , $V (D) = X_{D} (2)$ . For $n \geq 2$ , the integer remainder $s (n) = rem (n)$ uses weights $q (2 n, d)$ , $2 \leq d < n$ , and target $T_{n} = 2^{2 n - 1} - 2^{n}$ ; it is not the real remainder $r_{n} (1 / 2)$ . In the branch conditions below, $D_{s}, B_{s}, o_{s}, p_{s}^{-}, p_{s}^{+}$ denote the lower and upper supports, overshoot and correction terms of Theorem 5.16, with the row added as a subscript. The labels $U, M, R$ mean precisely its three arithmetic cases. Finally, $F (k, J) = F_{k} (J)$ denotes the expression in Theorem 6.13.

Theorem 6.1 (Infinitely many greedy skips force membership). Fix $x \geq 0$ . If $r_{n} (x) > R_{n}$ at some rank $n$ , every later rank is selected and $r_{n + k} (x) > R_{n + k}$ for every $k \geq 0$ . Consequently, infinitely many omitted positive ranks imply $x \in A$ .

Proof. The identity $R_{n} = w_{n + 1} + R_{n + 1}$ gives $r_{n} (x) > w_{n + 1}$ , so the rule takes rank $n + 1$ and leaves $r_{n + 1} (x) = r_{n} (x) - w_{n + 1} > R_{n + 1}$ . Induction proves persistence. Infinitely many skips therefore force $0 \leq r_{n} (x) \leq R_{n}$ at every rank; since $R_{n} \to 0$ , the greedy sum equals $x$ . ◻

Theorem 6.2 (The quotient remainders converge to the greedy deficit). Let $G$ be the real greedy support for $1 / 2$ , and let $D_{s}$ be the integer-greedy support at row $s$ . Then

X_{D_{s}} (2) ⟶ X_{G} (2), \frac{rem (s)}{4^{s}} ⟶ δ := \frac{1}{2} - X_{G} (2) \geq 0.

In particular, the following conditions are equivalent: $1 / 2 \in A$ ; the full sequence $rem (s) / 4^{s}$ tends to zero; and there exist $s_{j} \to \infty$ with $rem (s_{j}) / 4^{s_{j}} \to 0$ .

Proof. For each fixed $d$ , the scaled weight $4^{- s} q (2 s, d)$ tends to $w_{d}$ , and the scaled target tends to $1 / 2$ . The finite-prefix stability argument of Proposition 8.1 therefore gives $D_{s} \cap {2, \dots, K} = G \cap {2, \dots, K}$ for all sufficiently large $s$ , for each fixed $K$ . At such rows,

| X_{D_{s}} (2) - X_{G} (2) | \leq R_{K} .

First letting $s \to \infty$ and then $K \to \infty$ proves convergence of the values. This uses summability, not an unproved uniform rate of prefix agreement.

For $s \geq 5$ , put $Φ_{s} = \sum_{d \in D_{s}} {4^{s} / (2^{d} - 1)}$ , so $0 \leq Φ_{s} < s - 2$ . The quotient identity gives exactly

X_{D_{s}} (2) = \frac{1}{2} - 2^{- s} - \frac{rem (s)}{4^{s}} + \frac{Φ_{s}}{4^{s}} .

The two error terms tend to zero, proving the asserted limit. The real greedy sums never exceed $1 / 2$ , so $δ \geq 0$ . By strict tail separation, a support representing $1 / 2$ must agree with the greedy rule, as in Theorem 6.35. Thus membership is equivalent to $δ = 0$ , which is also equivalent to a zero limit along any one sequence $s_{j} \to \infty$ . ◻

Convergence itself is automatic; proving that its limit is zero is exactly the unresolved membership question.

Theorem 6.3 (Two-channel and dyadic cap sufficiency). Suppose that for every $n \geq 0$ at which the next weight is skipped,

w_{n + 1} > r_{n} (1 / 2) ⟹ r_{n} (1 / 2) \leq 2^{- (n + 1)} + \frac{1}{3} 4^{- (n + 1)} .

Then $1 / 2 \in A$ . The stronger bound $r_{n} (1 / 2) \leq 2^{- (n + 1)}$ at the same skipped ranks also suffices. The first bound retains two positive geometric terms of the full tail; it is larger, and hence less restrictive, than the second. Both are sufficient tests on the specified greedy orbit. The result does not assert that either bound holds at all its skipped ranks.

Theorem 6.4 (Second-channel phase separation sufficiency). Define $P_{n} = 4^{n} (2 r_{n} (1 / 2) - 2^{- n})$ . If

\frac{1}{6} + \frac{37}{56} 2^{- n} \leq | P_{n} - \frac{1}{3} | for every n \geq 1,

then $1 / 2 \in A$ . These are exact rational inequalities because the target and every finite greedy remainder are rational. The cited finite calculation establishes them for $1 \leq n \leq 6$ ; it therefore suffices to prove them for every $n \geq 7$ .

Unlike Theorem 6.3, this hypothesis concerns every positive rank, not only skipped ranks. It excludes an explicit interval around $1 / 3$ . For example, $P_{n} \notin (0, 1)$ implies the displayed inequality when $n \geq 2$ , but no such avoidance theorem for the whole orbit is proved here.

Theorem 6.5 (Straddle-prefix closed-set criterion). Suppose that for every $d \geq 0$ there is a finite support $D_{d} \subseteq {1, \dots, d}$ such that

X_{D_{d}} (2) \leq t \leq X_{D_{d}} (2) + R_{d} .

Then $t \in A$ . The supports need not agree at different depths: compactness, together with $R_{d} \to 0$ , supplies a limiting support. Conversely, truncating a support representing $t$ supplies such finite supports. Thus this condition is an exact finite-approximation formulation of membership, not a criterion already known to hold for an unresolved target.

Theorem 6.6 (Keeping the largest skipped rank beyond two thirds). Let $D_{s} \subseteq {2, \dots, s - 1}$ be the support of the integer-greedy row defined in Section 5.4. Assume the following implication for every $s \geq 14$ and every largest omitted rank $d = max ({2, \dots, s - 1} ∖ D_{s})$ :

2 s < 3 d ⟹ 2 (s + 1) < 3 d or s \notin D_{s + 1} .

Then $1 / 2 \in A$ . The hypothesis says that either the same omitted rank remains beyond two thirds of the next row, or the next row omits its terminal rank. The latter alternative is exactly an upper or middle transition in the preceding classification.

The cited finite calculation establishes the initial inequality at row $14$ . The displayed implication then preserves the inequality by induction and supplies omitted ranks tending to infinity. It is an unproved condition on every later row; the verified base case alone does not establish it.

Theorem 6.7 (Two sufficient bounds at middle transitions). Suppose that at every middle transition with row $s \geq 13$ ,

| D_{s} | + p_{s}^{-} + 5 < 4 rem (s) .

Then $1 / 2 \in A$ . The stronger condition $rem (s) \geq s$ at all such rows also suffices.

Indeed, $| D_{s} | \leq s - 2$ and $p_{s}^{-} \leq 2 (s - 2)$ give $| D_{s} | + p_{s}^{-} + 5 \leq 3 s - 1 < 4 s$ . This explains the implication between the two hypotheses. The implication from the first bound to membership is proved in Lean; no ordinary proof is printed here. Neither all-row hypothesis is established by the finite computations.

Theorem 6.8 (The tail inequality at a final middle transition). Suppose row $D \geq 13$ is middle and all rows after it are right. Let $F_{D} = D_{D} \cup {D}$ and put

C_{D} = 4 rem (D) - p_{D}^{-} - 4, Θ_{D} = \sum_{j \geq 1} c_{F_{D}} (2 D + 2 + j) 2^{- j} .

Then $C_{D} < Θ_{D}$ . Since $0 \leq Θ_{D} \leq | F_{D} |$ , an estimate in the opposite direction would exclude this scenario. No such reverse estimate is assumed or proved here. Remark 10.8 states the corresponding sufficient hypothesis with the value $- 3$ excepted.

Theorem 6.9 (A conditional two-sided dyadic bound). Assume the following two conditions for every row $s \geq 5$ :

\begin{aligned} M at s & ⟹ 4 rem (s) - p_{s}^{-} - 4 \notin {- 3, - 2, - 1}, \\ R at s and o_{s} \leq 2^{s} & ⟹ 4 o_{s} + p_{s}^{+} \leq 2^{s + 2} . \end{aligned}

Then at every row $s \geq 5$ ,

rem (s) \leq 2^{s} or o_{s} \leq 2^{s} .

The first hypothesis concerns only middle transitions. The second concerns only right transitions at which the old overshoot is at most $2^{s}$ . Neither condition is asserted here for the whole sequence.

Proof. At row $5$ , $D_{5} = {2, 4}$ and $rem (5) = 71$ , whereas $B_{5} = {2, 3}$ and $o_{5} = 7 \leq 32$ . For the induction step use Theorem 5.16. The upper branch gives $rem (s + 1) \leq 2^{s + 1}$ . On a middle branch put $c = 4 rem (s) - p_{s}^{-} - 4$ . If $c \leq - 4$ , the new remainder is $2^{s + 1} + c + 4 \leq 2^{s + 1}$ . Otherwise the hypothesis forces $c \geq 0$ . Adding the terminal weight to the corrected lower word then exceeds the target by $2^{s + 1} - c > 0$ , so the new minimal overshoot is at most $2^{s + 1}$ .

On a right branch, an old remainder at most $2^{s}$ gives a new remainder at most $2^{s + 1}$ by the recurrence. In the other case the inductive hypothesis gives $o_{s} \leq 2^{s}$ ; the corrected upper word exceeds the new target by $4 o_{s} + p_{s}^{+} - 2^{s + 1}$ , which is positive and at most $2^{s + 1}$ by the second assumption. ◻

This conclusion is a disjunction. It does not by itself give an upper bound for the remainder on rows where only the overshoot is small, and therefore is not by itself a membership criterion.

Theorem 6.10 (Upper-reset dyadic-band escape, checked for $13 \leq d \leq 30$ ). The upper-reset dyadic-band escape condition requires the following at every actual upper reset $d \geq 13$ : for every $0 \leq j \leq d$ , the reset charge avoids a linear-width band immediately below the dyadic power $2^{d - j + 1}$ ( $2^{d - j + 1} < resetCharge$ or $resetCharge + 2 (d + j) \leq 2^{d - j + 1}$ ). Granted this, $1 / 2 \in A$ . The linked proof verifies this condition for $13 \leq d \leq 30$ , using exact successor remainders at rows $14$ – $31$ (for example, $rem (14) = 392$ and $rem (31) = 4187487147$ ). This finite verification does not supply the hypothesis for every $d \geq 13$ .

Theorem 6.11 (Möbius-centred carry nonnegativity below $1 / 2$ ). If $1 \notin A$ and $X_{A} (2) < 1 / 2$ , then $C_{A} (N) \geq 0$ for every $N$ . Indeed,

ihc (A, N) = 2^{N + 1} (1 / 2 - X_{A} (2)) + \sum_{j > N + 1} c_{A} (j) 2^{N + 1 - j} > 0.

The carry is an integer, so it is at least $1$ ; subtracting $1$ gives the asserted nonnegativity of the centred carry. The strict inequality and integrality are both needed in this argument. The statement is about the divisor counts $c_{A}$ , not a new irrationality criterion.

Theorem 6.12 (A square-root bound on the greedy carry). For the greedy support $G = G_{1 / 2}$ , the centred carry is nonnegative. If in addition

C_{G} (N) \leq 2 \sqrt{N} + 4 for every N \geq 0,

then $G$ has infinitely many skipped indices and $X_{G} (2) = 1 / 2$ . The unresolved part is the upper bound along this particular orbit. The carry identity explains its strength: a positive gap $1 / 2 - X_{G} (2)$ would contribute a term of order $2^{N}$ , which cannot satisfy a square-root bound. This is distinct from the reset-deviation hypothesis of Theorem 6.10; no equivalence between the two hypotheses is asserted.

Theorem 6.13 (Equivalent sign and vanishing conditions). For the prefix $D = G \cap {1, \dots, k}$ define

F_{k} (J) = \sum_{i = 1}^{J} c_{D} (k + 1 + i) 2^{J - i} - 2^{J} C_{D} (k) .

This is a finite integer expression: it compares the next $J$ divisor counts of the fixed prefix with its centred carry. Let $s_{n}$ denote the integer greedy remainder at depth $2 n$ , with weights $q (2 n, d)$ for $2 \leq d < n$ and target $2^{2 n - 1} - 2^{n}$ . The following conditions each imply $1 / 2 \in A$ :

For every skipped rank $n \geq 3$ , $F_{n - 1} (n) \geq 0$ .
For every skipped rank $n \geq 3$ at which the real and integer greedy words agree on ${2, \dots, n - 1}$ , $s_{n} = 0$ .
For every skipped rank $n \geq 3$ , $B (2 n) < s_{n}$ .

Conditions (i) and (ii) are equivalent. Condition (iii) implies them; no converse or strict separation is established here. These hypotheses must hold at every indicated skipped rank, not merely through a finite computed range.

Theorem 6.14 (Erdős–Borwein full-support irrationality, unconditional). For every integer $b \geq 2$ , the sum $\sum_{n \geq 1} (b^{n} - 1)^{- 1}$ is irrational. This is Erdős’s theorem [1]; at base $2$ the sum is the Erdős–Borwein constant. The result concerns full support, not all its infinite subsets.

Theorem 6.15 (Pairwise-coprime support irrationality, Erdős 1968). For every integer $b \geq 2$ and every infinite pairwise-coprime support $A \subseteq N_{> 0}$ with summable reciprocals, $\sum_{a \in A} 1 / (b^{a} - 1)$ is irrational.

Theorem 6.16 (Irrationality from divisible coefficient blocks). Let $b \geq 2$ be an integer and let $c : N \to N$ satisfy $c (n) \leq n$ for every $n$ . Suppose that for every integer $q \geq 1$ there are nonnegative integers $N, K, L, C$ , with $K \leq L$ , such that

\begin{aligned} b^{r} & ∣ c (N + r) & (1 \leq r \leq K), \\ \sum_{r = K + 1}^{L} c (N + r) b^{L - r} & \leq C, \\ c (N + L + 1 + t) & > 0 & for some integer t \geq 0, \\ q (C + N + L + 2) & < b^{L} . \end{aligned}

Then $\sum_{n \geq 1} c (n) b^{- n}$ is irrational.

Proof. Write $x = \sum_{n \geq 1} c (n) b^{- n}$ . Multiplication by $b^{N}$ makes the terms through $N$ integral. The first condition also makes the terms at $N + 1, \dots, N + K$ integral. Subtracting their sum leaves, for some integer $z$ ,

0 < b^{N} x - z = \sum_{r > K} \frac{c (N + r)}{b^{r}} \leq \frac{C + N + L + 2}{b^{L}} < \frac{1}{q} .

Positivity follows from the specified nonzero coefficient beyond $L$ . The middle sum contributes at most $C / b^{L}$ , while $c (N + r) \leq N + r$ gives

\sum_{r > L} \frac{c (N + r)}{b^{r}} \leq b^{- L} (\frac{N + L}{b - 1} + \frac{b}{(b - 1)^{2}}) \leq \frac{N + L + 2}{b^{L}} .

If $x = p / q$ were rational, the positive difference $b^{N} x - z$ could not be smaller than $1 / q$ . ◻

A certificate at one precision is not enough: the same coefficient sequence must satisfy these conditions for every $q$ . The positive-tail condition is essential; without it the zero sequence would satisfy the other conditions at every precision. The full-support theorem supplies such certificates for $c = τ$ . Their existence for $c = φ$ is not proved here. This is a limitation of this particular sufficient criterion, not a comparison of every approach to Problem 249.

Theorem 6.17 (Irrationality from a gap beyond the preceding least common multiple). For an integer base $b \geq 2$ and strictly increasing support $a : N \to N$ with $a (0) \geq 1$ : if $a (k) - lcm (a (0), \dots, a (k - 1)) \to \infty$ , then $\sum_{k}^{'} 1 / (b^{a (k)} - 1)$ is irrational. The base may vary over the integers, but the denominators remain $b^{a (k)} - 1$ . Applying the argument to another denominator sequence would require its own divisibility and tail estimates.

Theorem 6.18 (Factorial-support and $2^{k}$ -support instances). For every integer $b \geq 2$ , both

\sum_{k \geq 0} \frac{1}{b^{(k + 1)!} - 1} and \sum_{k \geq 0} \frac{1}{b^{2^{k}} - 1}

are irrational. These are instances of the preceding theorem. For the factorial support the preceding least common multiple is $k!$ when $k \geq 1$ , so the gap is $k k!$ . For the powers of two it is $2^{k - 1}$ , so the gap is $2^{k - 1}$ . Both tend to infinity. The conclusion concerns these two supports, not every infinite support.

Theorem 6.19 (Multiples-support irrationality via dilation). For integers $b \geq 2$ and $d \geq 1$ ,

X_{d N_{> 0}} (b) = \sum_{k \geq 1} \frac{1}{(b^{d})^{k} - 1}

is irrational. This is the full-support result (Theorem 6.14) at the integer base $b^{d}$ ; no new support argument is required.

Luca and Tachiya proved irrationality for every purely periodic integer weight that is not identically zero [3]. The periodic, eventually periodic, residue-class and odd-support statements below follow from their theorem; their Lean proofs are separate and use the certificate method described here.

Theorem 6.20 (Periodic-support irrationality). Let $b \geq 2$ and $m \geq 1$ be integers. Suppose $A \subseteq N_{> 0}$ is nonempty and satisfies $n \in A$ if and only if $n + m \in A$ for every $n \geq 1$ . Then $X_{A} (b)$ is irrational.

The indicator of $A$ is a nonzero purely periodic integer weight, so this follows from Luca and Tachiya’s theorem stated above. The linked declaration supplies a separate formal proof. At $m = 1$ the only nonempty periodic support is full support; residue classes and unions of residue classes give the other immediate examples.

Theorem 6.21 (Eventually-periodic support irrationality). Let $b \geq 2$ and $m \geq 1$ be integers. An infinite support whose membership is $m$ -periodic from some threshold $N_{0}$ onward has irrational $X_{A} (b)$ .

Theorem 6.22 (Residue-class and odd-support irrationality). For integers $b \geq 2$ , $m \geq 1$ and a residue $c$ , $\sum_{\begin{matrix} n \geq 1 \\ n \equiv c (\mod m) \end{matrix}} (b^{n} - 1)^{- 1}$ is irrational. Specializing $m = 2, c = 1$ : $\sum_{n odd} 1 / (b^{n} - 1)$ is irrational for every $b \geq 2$ , the case treated explicitly in [3].

comparison-with-the-known-theorem.

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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