Geometric Moments and Rational Lambert Values: Proofs and Further Results

Will Cook

Long paper

Geometric Moments and Rational Lambert Values: Proofs and Further Results

Erdős #1049 By Will Cook September 2026 Authorship, AI use and citation

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Summary. The record proves the fixed-base Hankel asymptotic for Zudilin's normalised moments and the rational-base irrationality criterion log b/log a<0.4056830213840605…. The determinant proof separates geometric node configurations from positive weights; the irrationality proof instead specialises Zudilin's 2004 polynomial forms after cancellation. Coefficient moments, evaluated lattices and congruence conditions distinguish integer divisibility from nonzero remainder decay. A broader ratio-limit argument and a rational-value model remain unformalised. None of these arguments settles $F (3 / 2)$ .

This paper covers the sharp fixed-base Hankel asymptotic and rational-base irrationality proofs, coefficient moments and arithmetic obstructions, with ordinary ratio-limit and rational-value models distinguished from checked results.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #1049, which remains open.

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Failure of the stated clearing conditions at $3 / 2$

At an integer base, congruences for divisor coefficients can produce long blocks of zero digits. We recall that argument before comparing it with the coordinatewise clearing conditions below.

The integer-base argument.

For an integer $b \geq 2$ , Erdős uses the expansion $F (b) = \sum_{n \geq 1} τ (n) b^{- n}$ and the Chinese remainder theorem to produce arbitrarily long blocks of divisor coefficients with prescribed powers of $b$ . After controlling carries from the tail, he obtains arbitrarily long blocks of zero digits and proves that the expansion does not terminate. An eventually periodic expansion with arbitrarily long zero blocks must terminate, so these two facts imply irrationality [1]. Positivity of the original summands alone would not prove nontermination; for example, $\sum_{n \geq 1} (b - 1) b^{- n} = 1$ .

Direct clearing of a partial sum.

A more naive attempt is to suppose $F (b) = p / q$ , cut at $N$ , and multiply the remaining identity by $q b^{N}$ . This does not by itself trap a positive integer below $1$ : the first uncleared tail contribution is $q τ (N + 1) / b$ . The clearing conditions below describe one bounded-window version of this attempt. They fail at $3 / 2$ .

Write $β = r / s$ for the base and $c (n)$ for the coefficient of $β^{- n}$ , so that $c = τ$ in the case at hand. The term $c (n) β^{- n}$ is $c (n) s^{n} / r^{n}$ . Clearing the power of $r$ leaves the numerator factor $s^{n}$ in place. That factor is invisible when $s = 1$ and grows geometrically when $s \geq 2$ . At $3 / 2$ it is $2^{n}$ .

We return to the attempt to clear successive terms of the series. We retain just two necessary demands: the last coordinate in the window must clear, and the coefficient-free lower bound from the first omitted positive term must be smaller than $1$ . Even these weaker tests are incompatible at $3 / 2$ . The argument below uses six natural-number parameters; it excludes the stated clearing scheme, not other approximations to $F (3 / 2)$ .

Definition 6.1 (clearing conditions). For natural numbers $a, b, N, K, Q, D$ , the clearing conditions are

a > 0, Q > 0, D > 0, D \leq N + K, a^{K} ∣ Q D, Q b^{N + K + 1} < a^{K + 1} .

The reading is: $a$ and $b$ are the numerator and denominator of the base, playing the roles of $r$ and $s$ in the partial-sum calculation above, so that $(a, b) = (3, 2)$ is the case of interest; $N$ is the shift, $K$ is the width of the cleared window, $Q$ is the accumulated clearing factor, and $D$ is the final coefficient being cleared. The bound $D \leq N + K$ is the only property of the coefficient used; for the divisor-counting coefficient it holds because $τ (n) \leq n$ . The divisibility $a^{K} ∣ Q D$ tests the final coordinate of the cleared window; it does not by itself clear every earlier coordinate. For a reduced base $a / b$ , the denominator of $Q a^{N} D (b / a)^{N + K}$ is cleared exactly when $a^{K} ∣ Q D$ , since $a$ and $b$ are coprime. For example, at $a = 2$ , $b = 1$ , $N = 21$ , $K = 3$ and $Q = 1$ , the final term clears because $2^{3} ∣ τ (24) = 8$ , but the preceding scaled term is $τ (23) / 2^{2} = 1 / 2$ and is not integral.

The final inequality is a necessary smallness test, not a sufficient bound for the entire tail. For $a > b \geq 1$ and $Q > 0$ , positivity and $τ (m) \geq 1$ give

Q a^{N} \sum_{m \geq N + K + 1} τ (m) (b / a)^{m} > \frac{Q b^{N + K + 1}}{a^{K + 1}} .

A tail smaller than $1$ must therefore satisfy the final inequality. The divisibility gives $a^{K} \leq Q (N + K)$ , while the smallness test requires $Q b^{N + K + 1} < a^{K + 1}$ . The same $Q$ must satisfy both. These conditions are defined in Lean by the six clearing conditions.

Theorem 6.2 (a necessary inequality for clearing). If $(a, b, N, K, Q, D)$ satisfies the clearing conditions, then

b^{N + K + 1} < a (N + K) .

Proof. Since $Q D > 0$ and $a^{K} ∣ Q D$ , we have $a^{K} \leq Q D$ , and $D \leq N + K$ gives $a^{K} \leq Q (N + K)$ . Hence

Q b^{N + K + 1} < a^{K + 1} = a^{K} \cdot a \leq Q (N + K) \cdot a,

and cancelling the positive factor $Q$ gives the claim. ◻

The necessary inequality in Theorem 6.2 shows the difference between these clearing tests at integer and noninteger rational bases. At $b = 1$ it reads $1 < a (N + K)$ , which holds for every $a \geq 2$ and every nonempty window; this necessary inequality imposes no obstruction. The other clearing conditions still have to be checked. At $b \geq 2$ the left side is exponential in $N + K$ and the right side is linear, so the conditions can hold only for bounded $N + K$ at each fixed base. At $(a, b) = (3, 2)$ the necessary inequality already fails when $N = K = 1$ .

Example 6.3. Take the smallest window with $N, K \geq 1$ , namely $N = K = 1$ , and numerator $a = 3$ . At $b = 1$ the tuple $(3, 1, 1, 1, 3, 1)$ satisfies the conditions: $D = 1 \leq 2$ , the divisibility reads $3 ∣ 3$ , and the last inequality reads $3 \cdot 1^{3} = 3 < 3^{2} = 9$ . At $b = 2$ no choice works. The last inequality becomes $Q \cdot 2^{3} < 3^{2}$ , which forces $Q = 1$ ; the divisibility then reads $3 ∣ D$ , and the only candidates are $D = 1$ and $D = 2$ , neither divisible by $3$ . This is the proof of Theorem 6.2 in miniature: it derives $a^{K} \leq Q (N + K)$ , here $3 \leq 2$ .

Feasibility of these necessary conditions is not a successful tail estimate. For $(a, b, N, K, Q, D) = (2, 1, 1, 1, 1, 2)$ all six conditions hold, and $D = τ (N + K)$ . Nevertheless,

2 \sum_{m \geq 3} \frac{τ (m)}{2^{m}} > 2 (\frac{2}{2^{3}} + \frac{3}{2^{4}} + \frac{2}{2^{5}}) = 1.

The failure at $3 / 2$ is useful because it already occurs at the weaker, necessary tests; their feasibility at an integer base proves no bound for the full remainder.

Proposition 6.4. For every natural number $x \geq 2$ we have $3 x < 2^{x + 1}$ .

Proof. Induct on $x$ , starting from $6 < 8$ at $x = 2$ . For the step, $2^{x + 1} \geq 2^{2} > 3$ when $x \geq 1$ , so $3 (x + 1) = 3 x + 3 < 2^{x + 1} + 2^{x + 1} = 2^{x + 2}$ . ◻

Theorem 6.5 (failure of the stated clearing conditions at $3 / 2$ ). For all $N \geq 1$ and $K \geq 1$ and all natural $Q, D$ , the tuple $(3, 2, N, K, Q, D)$ does not satisfy the clearing conditions.

Proof. The clearing conditions would give $2^{N + K + 1} < 3 (N + K)$ by Theorem 6.2, contradicting Proposition 6.4 applied to $x = N + K \geq 2$ . ◻

Theorem 6.5 excludes the divisibility pattern and necessary tail test in Definition 6.1 at $3 / 2$ . It gives no conclusion about the rationality of $F (3 / 2)$ or about clearing schemes with different conditions.

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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Failure of the stated clearing conditions at 3/23/2

The integer-base argument.

Direct clearing of a partial sum.

About this paper

Failure of the stated clearing conditions at $3 / 2$