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

Approximations to the greedy orbit

Finite quotient procedures approximate the real greedy sequence for $1 / 2$ . Agreement on a fixed prefix does not give agreement at a cutoff that grows with the scale. We track the cutoff in each of the four calculations below and state the uniform estimate that would be needed to pass to unbounded depths.

Four reported calculations

Fix the weights $x_{n} := 1 / (2^{n} - 1)$ , $n \geq 2$ , and the target $1 / 2$ . The single object under study throughout this programme is the greedy orbit for this target: a residual $ρ$ starts at $1 / 2$ , and at rank $k$ the orbit takes iff $ρ \geq x_{k}$ . A skip at rank $k$ is safe iff $ρ \leq T_{k + 1} := \sum_{j > k} x_{j}$ and fatal iff $ρ \in (T_{k + 1}, x_{k})$ ; the fatal interval is exactly the gap no later tail can repair. The four calculations below concern this target, but their finite parameters and certification records differ. Agreement with the real greedy prefix requires a convergence argument for the approximants used in each calculation; agreement over a tested range is not that argument.

Truncation rung $J$ . Replace the full weight $x_{n}$ by the $J$ -term truncation $w_{n}^{(J)} := \sum_{q = 1}^{J} 2^{- q n}$ and ask whether some Boolean support hits $1 / 2$ exactly under the truncated weights. Certified for $3 \leq J \leq 22$ (Table tab:scale-ladder, finite ranges; Theorem 8.6–8.9 below).
Seam row $s$ . The base-4 seam recursion tracking the deviation $Δ_{s} := rem (s) - 2^{s}$ of the integer greedy orbit from its dyadic target. Certified rows $6$ through $2500$ for the reset-crossing classification, and the underlying orbit separately certified to row $200,000$ , with exactly one violation of the target inequality at reset rows $6 \leq r \leq 200,000$ , at $r = 7$ (saved runs to row $2500$ and to row $200,000$ ).
Integer margin $m_{c}$ . An exploratory, unverified scan of the integer-greedy margin out to $5 \times 10^{5}$ , tracking how far the certified region sits from the danger threshold.
Sharp tail margin. The rank-by-rank take/skip margin against the exact tail, checked for ranks $2$ through $3000$ : $1497$ takes against $1502$ skips, zero fatal skips anywhere, minimum safe-skip margin $+ 2.9922$ bits at rank $5$ , and the smallest reported take-margin $+ 0.0340$ bits at rank $7$ , where the remainder $1 / 126$ exceeds the weight $1 / 127$ . Here the margins are the base-two logarithms of the corresponding positive slack multiplied by $4^{k}$ (saved run; working note, section 8).

Proposition 8.1 (Stability of each fixed greedy prefix). Let $t_{j} \to 1 / 2$ and, for each fixed $n \geq 2$ , let $v_{n}^{(j)} \to x_{n} = (2^{n} - 1)^{- 1}$ , with $v_{n}^{(j)} > 0$ . Apply the greedy rule with target $t_{j}$ and weights $v_{n}^{(j)}$ in increasing order of $n$ , through depths $m_{j} \to \infty$ . For every fixed depth $K$ , the decisions at ranks $2, \dots, K$ eventually agree with those of the real half-greedy rule. This assertion concerns finite prefixes, not survival at all ranks.

Proof. At rank $2$ , the target and the weight converge to their real counterparts. Their comparison is strict by Lemma 8.2, so its sign is eventually unchanged. Inductively, once all earlier decisions agree, the approximate remainder is the target minus a fixed finite sum of convergent weights. It therefore converges to the real remainder. The next comparison is again strict, so the next decision eventually agrees. A finite induction through $K$ proves the claim. ◻

Lemma 8.2 (No-ties lemma). At every rank $k$ of the full greedy orbit for target $1 / 2$ , both defining comparisons are strict: $ρ \neq x_{k}$ and $ρ \neq T_{k + 1}$ .

Proof. The two strict comparisons have different reasons: parity excludes equality with a weight, and irrationality excludes equality with the complete tail.

Take boundary excluded (parity). Write a finite Mersenne sum as $p / q = \sum_{d \in F} x_{d}$ in lowest terms. Its denominator $q$ is odd, because every $2^{d} - 1$ is odd (equivalently, halfGreedyPrefixDenominator_odd, itself traced to finiteErdosSum_den_odd). Hence the residual after any finite prefix is $ρ = \frac{1}{2} - \frac{p}{q} = \frac{q - 2 p}{2 q}$ : its reduced denominator is even. Every weight $x_{k} = 1 / (2^{k} - 1)$ has odd denominator. An even-denominator rational can never equal an odd-denominator rational, so $ρ = x_{k}$ is impossible at any rank: the take comparison is never a tie.

Skip-safety boundary excluded (irrationality). The tail past rank $k$ is $T_{k + 1} = E - 1 - \sum_{2 \leq j \leq k} x_{j}$ , where $E = \sum_{n \geq 1} 1 / (2^{n} - 1)$ is the Erdős–Borwein constant: a rational shift of $E$ . Erdős’s own 1948 theorem gives $E$ irrational (irrational_erdosBorwein_series, instantiated at $b = 2$ ), so $T_{k + 1}$ is irrational for every $k$ . The residual $ρ$ arising from any finite Boolean prefix is rational. An irrational number never equals a rational one, so $ρ = T_{k + 1}$ is impossible at any rank; the skip-safety comparison is never a tie either.

Both boundaries of the fatal interval $(T_{k + 1}, x_{k})$ are therefore approached only strictly, at every rank, unconditionally. ◻

Remark 8.3. The hypotheses of Proposition 8.1 can be checked for two concrete approximations. The truncated geometric weights $\sum_{q = 1}^{J} 2^{- q n}$ increase to $x_{n}$ . For the integer quotient rows, division by $4^{s}$ gives weights $4^{- s} ⌊ 4^{s} / (2^{n} - 1) ⌋ \to x_{n}$ and target $4^{- s} (2^{2 s - 1} - 2^{s}) = 1 / 2 - 2^{- s} \to 1 / 2$ . Hence both greedy rules agree with each fixed real prefix once their parameters are large enough. Convergent upper and lower bounds for $R_{k}$ likewise determine the strict tail comparison at a fixed rank. No conclusion about the exploratory margin scan follows without specifying its approximants and proving their convergence. Fixed-prefix agreement is not uniform agreement through a growing depth. For the quotient rows, however, summability bounds the unmatched tail by $R_{K}$ and does give convergence of the coded values, as proved in Theorem 6.2. It identifies their limit as $X_{G} (2)$ ; it does not establish that this value is $1 / 2$ .

Successive tail truncations

Finite truncations of the geometric expansion give rational weights and permit a finite decision procedure for each fixed truncation order $J$ . The bounds below explain why only finitely many ranks need to be tested. Fix $J \geq 2$ and set

\begin{aligned} w_{n}^{(J)} & := \sum_{q = 1}^{J} 2^{- q n}, & T_{n + 1}^{(J)} & := \sum_{q = 1}^{J} \frac{2^{- q n}}{2^{q} - 1}, \\ HalfRung (J) & :⟺ \exists A \subseteq {2, 3, \dots}, & \sum_{n \in A} w_{n}^{(J)} & = \frac{1}{2} . \end{aligned}

The arguments below are ordinary mathematics. The accompanying exact integer computation rechecks the finite instances $3 \leq J \leq 22$ ; neither the arguments nor that computation are presented as a Lean verification.

Lemma 8.4 (Forced greedy). For every $J \geq 2$ : $w_{n}^{(J)} > T_{n + 1}^{(J)}$ for all $n$ (the $q = 1$ terms agree exactly, and every $q \geq 2$ tail term is strictly smaller than the corresponding weight term). Consequently the greedy support is the unique candidate support, and $HalfRung (J)$ holds iff the greedy orbit for $1 / 2$ under weights $w_{n}^{(J)}$ never lands in a fatal interval $(T_{n + 1}^{(J)}, w_{n}^{(J)})$ . Rank $1$ is always a safe skip; ranks $2$ and $3$ are always takes.

Lemma 8.5 (Parity forces infinite support). For every $J \geq 2$ , no finite $A \subseteq {2, 3, \dots}$ attains $HalfRung (J)$ .

Proof. The empty set has value zero. For a nonempty finite set, put $m = max A$ . After multiplication by $2^{J m}$ , the $m$ th summand is odd and every earlier summand is even, whereas the target $2^{J m - 1}$ is even. This parity contradiction excludes finite support. ◻

Theorem 8.6 (Witness exclusion). Define the misalignment mass

μ_{J} (M) := \sum_{q = 2}^{J} \frac{2^{M mod q}}{2^{q} - 1} .

Let $J \geq 3$ , $n \geq 4$ , and suppose some $M \in [n, 2 n - 2]$ has $μ_{J} (M) \leq \frac{11}{15}$ . Then no Boolean prefix $D \subseteq {2, \dots, n - 1}$ satisfies

T_{n + 1}^{(J)} < \frac{1}{2} - \sum_{d \in D} w_{d}^{(J)} < w_{n}^{(J)};

This excludes the specified gap for every Boolean prefix, not only the greedy prefix.

Proof. Suppose such a prefix exists and write $ρ = 1 / 2 - \sum_{d \in D} w_{d}^{(J)}$ . Since $T_{n + 1}^{(J)} \geq 2^{- n}$ and $n \leq M \leq 2 n - 2$ ,

0 < 2^{M} ρ - 2^{M - n} < \sum_{q = 2}^{J} 2^{M - q n} \leq \frac{2^{M - 2 n}}{1 - 2^{- n}} \leq \frac{4}{15} .

Here $n \geq 4$ gives the last inequality. Split each scaled summand $2^{M - q d}$ according to whether $q d \leq M$ . The terms with $q d \leq M$ are integers. The other terms have total

F = \sum_{d \in D} \sum_{\begin{matrix} 1 \leq q \leq J \\ q d > M \end{matrix}} 2^{M - q d} \leq \sum_{q = 2}^{J} \sum_{d > ⌊ M / q ⌋} 2^{M - q d} = μ_{J} (M) \leq \frac{11}{15} .

There is no $q = 1$ contribution to $F$ , since $d < n \leq M$ . Thus $2^{M} ρ + F - 2^{M - n}$ is an integer strictly between zero and one, a contradiction. The same calculation also applies when $J = 2$ . ◻

Corollary 8.7 (Half-LCM horizon). Let $J \geq 2$ and $L_{J} := lcm (2, 3, \dots, J)$ . For every $n \geq max {4, L_{J} / 2 + 1}$ there is an $M \in [n, 2 n - 2]$ with $μ_{J} (M) < 11 / 15$ . Thus witness exclusion leaves only the finite window $[4, L_{J} / 2]$ to check.

Proof. Let $M$ be the least multiple of $L_{J}$ with $M \geq n$ . If $M = L_{J}$ , then $M \leq 2 n - 2$ is the assumed inequality. If $M = k L_{J}$ with $k \geq 2$ , then $n \geq (k - 1) L_{J} + 1$ gives $2 n - 2 \geq 2 (k - 1) L_{J} \geq M$ . At this witness,

μ_{J} (M) = \sum_{q = 2}^{J} \frac{1}{2^{q} - 1} < \sum_{q = 2}^{\infty} \frac{4}{3 \cdot 2^{q}} = \frac{2}{3} < \frac{11}{15} .

The bound uses $2^{q} - 1 \geq 3 \cdot 2^{q - 2}$ , with strict inequality for $q > 2$ , so no numerical approximation to the full series is needed. Apply Theorem 8.6; its proof also covers $J = 2$ . Ranks $2$ and $3$ are takes, as already noted. ◻

Lemma 8.8 (Mod-12 filter, $J \geq 7$ ). For $J \geq 7$ , the inequality $μ_{J} (M) \leq 11 / 15$ implies $12 ∣ M$ . Thus only multiples of $12$ need be tested as witnesses in each interval $[n, 2 n - 2]$ ; no potentially uncovered rank is discarded.

Proof. If $M$ is odd, the $q = 2, 3$ terms give at least $2 / 3 + 1 / 7 > 11 / 15$ . If $M \equiv 2 (\mod 4)$ , the $q = 2, 3, 4$ terms give at least $1 / 3 + 1 / 7 + 4 / 15 = 26 / 35 > 11 / 15$ . Finally, if $4 ∣ M$ but $3 ∤ M$ , the $q = 2, 3, 4, 6$ terms give at least $1 / 3 + 2 / 7 + 1 / 15 + 4 / 63 = 236 / 315 > 11 / 15$ . Each case contradicts the assumed upper bound. ◻

Theorem 8.9 (Finite decision procedure). Call $n \in [4, L_{J} / 2]$ bad if no $M \in [n, 2 n - 2]$ has $μ_{J} (M) \leq \frac{11}{15}$ , and set $B (J) := max (bad \cup {3})$ . Then $HalfRung (J)$ holds iff the greedy orbit for $1 / 2$ under weights $w_{n}^{(J)}$ survives every rank from $2$ through $B (J)$ . This is a finite exact decision procedure.

Proof. Survival at all ranks plainly implies survival through $B (J)$ . Conversely, every later rank at most $L_{J} / 2$ has a witness by the definition of $B (J)$ , and every larger rank has one by Corollary 8.7. Witness exclusion rules out a fatal gap at all these ranks. The strict-tail criterion then gives membership. For $J = 2$ , the same proof works because $μ_{2} (M) \leq 2 / 3$ for every $M$ ; the window is empty and $B (2) = 3$ . ◻

Remark 8.10 (Verified truncations and the unbounded-index question). The finite decision procedure verifies $HalfRung (J)$ for every $3 \leq J \leq 22$ . The checking program uses exact integers: it verifies witness intervals covering the ranks beyond $B (J)$ , proves for $J \leq 16$ that the listed bad ranks have no witness, and tests the greedy remainders through $B (J)$ . In particular, it reproduces $B (19) = 1008$ and $B (20) = B (21) = B (22) = 1530$ . The checking program and its saved outputs for $3 \leq J \leq 16$ and $17 \leq J \leq 22$ are published. For $J \geq 17$ a floating-point filter proposes witnesses and each accepted witness is rechecked with integers, so a missed witness could only lengthen the exact greedy check. These are finite computer-assisted proofs using the ordinary lemmas above, not Lean proofs and not a test of the full weights through all ranks.

If $A$ represents $1 / 2$ with the $J$ -truncated weights, then

0 \leq X_{A} (2) - \frac{1}{2} \leq \sum_{q > J} \frac{2^{- 2 q}}{1 - 2^{- q}} \leq 2 \sum_{q > J} 4^{- q} = \frac{2}{3} 4^{- J} .

Consequently, such representations for an unbounded sequence of $J$ would put $1 / 2$ in the closed set $A$ . Its representing support would be infinite, since finite Mersenne sums have odd denominator (Lemma 8.2); Lemma 8.5 separately excludes finite support for each truncation. The unbounded-index assertion is still unproved. Twenty verified values of $J$ do not establish 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.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

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