Irrationality criteria for Lambert subseries

Will Cook

Short paper

Irrationality criteria for Lambert subseries

Erdős #257 By Will Cook 30 September 2026 Authorship, AI use and citation

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Summary. The series $\sum_{a \in A} (b^{a} - 1)^{- 1}$ is irrational for every integer $b \geq 2$ and infinite A contained in E∪V when E satisfies a finite-prime weighted summability condition and V has a summable positive divisor cover. The two classes are incomparable, and their union reaches supports in neither class. A common average over dyadic lengths produces one index with both displacements small. Arbitrary infinite supports at base two remain unresolved.

This paper covers the weighted, divisor-cover and mixed support criteria with complete proofs and method limits; the long paper treats finite denominators, the achievement set and rational membership.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, which remains open.

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In this paper

Positive divisor majorants

The weighted criterion treats exponents separately. A divisor cover instead uses their joint divisibility. To describe it, write the series in terms of its divisor counts:

\begin{matrix} (6) & c_{A} (n) = # {a \in A : a ∣ n}, X_{A} (b) = \sum_{n \geq 1} c_{A} (n) b^{- n} . \end{matrix}

Expand $(b^{a} - 1)^{- 1} = \sum_{j \geq 1} b^{- a j}$ and interchange the nonnegative sums. The selector $1_{A}$ is zero or one, whereas $0 \leq c_{A} (n) \leq τ (n)$ , with $τ (n)$ the number of positive divisors of $n$ . These are power-series coefficients before any carrying; they need not be base- $b$ digits.

Kaneko–Suzuki–Tachiya [2] require sparsity of the nonzero power-series coefficients. Sparsity of the selected Lambert denominators does not suffice: for any nonempty $A$ , $c_{A} (n)$ is positive on every multiple of $min A$ . Its nonzero positions therefore have positive lower density, violating those support-counting hypotheses. The criteria do not apply directly to (6). The density calculation and the distinction between their remote-tail average and our displacement appear in Section 1.2 of the companion paper.

A finite divisor set and the cover condition

For finite $F \subseteq N_{> 0}$ , write $f_{F} (n) = # {a \in F : a ∣ n}$ . Consider the four exponents $F = {2, 6, 10, 30}$ . If $2 ∤ n$ , the count is zero. Otherwise each of the primes $3$ and $5$ that divides $n$ doubles the count. Thus

f_{F} (n) = 1_{2 ∣ n} (1 + 1_{3 ∣ n}) (1 + 1_{5 ∣ n}) .

For $0 < α \leq 1$ , put $z = 2^{α} - 1$ . Taking the fractional power and expanding gives a positive divisor sum:

f_{F} (n)^{α} = 1_{2 ∣ n} + z 1_{6 ∣ n} + z 1_{10 ∣ n} + z^{2} 1_{30 ∣ n}, C = \frac{1}{2} (1 + \frac{z}{3}) (1 + \frac{z}{5}) .

Since $1_{d ∣ n}$ has mean $1 / d$ , $C$ is the mean of this expansion. The same factorisation applies to ${q d : d ∣ \prod_{p \in P} p}$ when no prime in $P$ divides $q$ . Small $α$ reduces the coefficients, but the tail estimate costs $(2^{α} - 1)^{- 1} = 1 / z$ . The condition below balances these effects over a sequence of finite sets.

Theorem 3.1 (a summable divisor-cover criterion). 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} (V) & \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 $b \geq 2$ .

For example, take $F_{j} = {4^{j}}$ , $α_{j} = 1$ and $c_{j, 4^{j}} = 1$ , with all other coefficients zero. Then $C_{j} = 4^{- j}$ , and the cost in (V) is $\sum_{j} 2^{- j}$ . Every finite set has the majorant $α_{j} = 1$ , $c_{j, d} = 1_{F_{j}} (d)$ ; the restriction is summability of the full cost. Covering sets may overlap, and their majorants may use divisors outside them. Section 5 gives a necessary condition independent of these choices.

A finite mean for shifted divisor tails

A divisor majorant becomes a sum of geometric tails, one for each divisor. Indeed, the positive offsets $r$ with $d ∣ N + r$ are $d - (N mod d), 2 d - (N mod d), \dots$ , so

\sum_{\begin{matrix} r \geq 1 \\ d ∣ N + r \end{matrix}} B^{- r} = \frac{B^{N mod d}}{B^{d} - 1} .

Fractional powers introduce $B = 2^{α}$ , possibly close to $1$ . The estimate must remain uniform after multiplication by $B - 1$ and allow the modulus to grow: the mixed proof fixes the first covering sets before enlarging that modulus.

For $1 < B \leq 2$ , positive integers $L, d, M$ , and an integer $R \geq 0$ , put

w_{B, d} (n) = \frac{B^{n mod d}}{B^{d} - 1}, D_{L; R, M} F = \frac{1}{M} \sum_{k = R}^{R + M - 1} \frac{1}{2^{k}} \sum_{m = 1}^{2^{k}} F (L m) .

Choose $k$ uniformly from $R, \dots, R + M - 1$ , then $m$ uniformly from $1, \dots, 2^{k}$ and observe $N = L m$ . Scales, not individual observations, have equal weight; repeated observations of $N$ add their weights.

The finite estimate

\begin{matrix} (S) & D_{L; R, M} w_{B, d} \leq \frac{1 + 4 L / M}{d (B - 1)} \end{matrix}

is uniform as $B ↓ 1$ after multiplication by $B - 1$ : the normalised bound is $(1 + 4 L / M) / d$ . The unnormalised factor $(B - 1)^{- 1}$ becomes the cost $(2^{α} - 1)^{- 1}$ in (V). The bound is independent of the starting scale $R$ and tracks the ratio $L / M$ , which tends to zero on the weighted proof’s scales.

Fix $T = 2^{k}$ and compare $d$ with the largest sampled multiple $L T$ . When $d \leq L T$ , the complete orbit has length $d / g$ , with $g = (L, d)$ , and total weight $1 / (B^{g} - 1)$ . Split the sample into complete cycles (possibly none) and one remaining piece to obtain

\frac{1}{T} \sum_{m = 1}^{T} w_{B, d} (L m) \leq \frac{g}{d (B^{g} - 1)} + \frac{1}{T (B^{g} - 1)} \leq \frac{1}{d (B - 1)} + \frac{1}{T (B - 1)} .

Across dyadic lengths satisfying $d \leq L 2^{k}$ , the reciprocal-length errors sum to at most $2 L / [d (B - 1)]$ . When $d > 2 L T$ , we have $L m mod d = L m$ and $2 L m \leq d - 1$ ; hence $\sum_{i = 0}^{d - 1} B^{i} \geq d B^{(d - 1) / 2} \geq d B^{L m}$ . Each term $w_{B, d} (L m)$ is then at most $1 / [d (B - 1)]$ . For fixed $d$ , the remaining transition range $L T < d \leq 2 L T$ occurs at most once as $T$ doubles. At that scale, geometric summation gives

\frac{1}{T} \sum_{m = 1}^{T} w_{B, d} (L m) = \frac{B^{L}}{T (B^{L} - 1)} \frac{B^{L T} - 1}{B^{d} - 1} \leq \frac{L B^{L}}{d (B^{L} - 1)} \leq \frac{2 L}{d (B - 1)} .

The first inequality uses the convexity of $x \mapsto B^{x} - 1$ and its value $0$ at the origin to bound the ratio by $L T / d$ . For the last inequality we used $(B^{L} - 1) / (B - 1) = \sum_{i = 0}^{L - 1} B^{i} \geq B^{L - 1}$ and $B \leq 2$ . The incomplete periods and the possible transition scale together cost at most $4 L / [d (B - 1)]$ before division by $M$ . Adding the mean term $1 / [d (B - 1)]$ proves (S). The corresponding source estimate uses the same finite average. Nonnegative interchange permits summation against any coefficients $c_{d}$ with $\sum_{d} c_{d} / d < \infty$ .

Proof of the cover criterion

Proof of Theorem 3.1. 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 $U_{j}$ are the tails to be made small; the $V_{j}$ have means controlled by (S). Subadditivity of the power $α_{j}$ and the divisor majorant connect them:

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) .

For the displacement itself, the geometric-series identity gives

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

For $j \leq J$ , divisibility will cancel the displacement, so $U_{j} (N) = X_{F_{j}} (2)$ . Only the tails with $j > J$ must be made small.

Fix $ε > 0$ and allot $t_{j} = ε 2^{- j}$ to the $j$ th tail. These allowances sum to $ε$ . The sufficient condition $t_{j}^{- α_{j}} V_{j} < 1$ for $U_{j} < t_{j}$ explains the factor $2^{j α_{j}}$ in the cover cost. 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 $J$ finite sets. For every $R \geq 0$ and $M \geq 4 L$ , (S) bounds the $D_{L; R, M}$ -mean of $S_{J} (N) := \sum_{j > J} t_{j}^{- α_{j}} V_{j} (N)$ by $(1 + 4 L / M) K_{J} < 1 / 2$ . Choose a sample with $S_{J} (N) < 1$ . This controls all countably many tails at once: each nonnegative summand is below $1$ , hence $U_{j} (N) < t_{j}$ for every $j > J$ . The first $J$ sets contribute no displacement; later sets cover the rest of $A$ , with overlaps only enlarging the upper bound. Thus,

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

Equation (D) now excludes rationality at base two. For $0 \leq r < d$ , the function $(b^{r} - 1) / (b^{d} - 1)$ is nonincreasing on $b > 1$ . For $r > 0$ , cancel $b - 1$ and write it as $A (b) / (A (b) + C (b))$ , where $A (b) = \sum_{i < r} b^{i}$ 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 every real $b \geq 2$ ; when $b$ is an integer, this contradicts the lower bound obtained from (D) under rationality. Any prescribed positive integer can be included as a divisor of $L$ , so the chosen 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 (V) by $η_{j}^{- α_{j}}$ and take $t_{j} = ε η_{j}$ . These weights allocate error allowances; they do not reweight the Lambert subseries. The stronger condition $\sum_{j} C_{j} 2^{j α_{j}} 2^{α_{j}} / (2^{α_{j}} - 1)^{2} < \infty$ implies (V), since each of its summands is the corresponding summand of (V) multiplied by $2^{α_{j}} / (2^{α_{j}} - 1) > 1$ . This proves inclusion of the hypotheses for a given cover; after optimisation over all covers, the inclusion of support classes is strict by a corresponding source 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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