The Binary Totient Series

Will Cook

Long paper

The Binary Totient Series

Erdős #249 By Will Cook July 2026; revised 1 October 2026 Authorship, AI use and citation

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Summary. The dyadic totient sections through level $e \geq 1$ have an explicit rational basis of size $2^{e} + 1$ . Truncating a scaled tail difference produces an integer whose residue certifies nonintegrality when its distances from both endpoints of the residue interval exceed an explicit error bound. The record develops the resulting finite denominator exclusions, equivalent residue criteria, and sufficient exponential-sum estimates. Rational comparison sequences identify the limitations of particular size, parity, rank, and finite-prefix arguments. The short paper supplies the all-base basis theorem; here each conditional result carries the arithmetic hypothesis still needed to apply it. Those hypotheses are not established in the form required to prove irrationality of S=∑ $φ (n) 2^{- n}$ .

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #249: selected checked premises, finite certificates, 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 #249, which remains open.

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In this paper

Exact dyadic comparisons and tail bounds

A finite prefix locates a dyadic rational relative to the tail. The following two formulas give the exact comparison and an upper bound for the omitted terms, so that the residue margin can be compared with its error.

Definition 6.138. For $p \in Z$ , $n \geq 0$ and an integer $L > 0$ , the difference between $p / (2 L)$ and the dyadic point $2^{- (n + 1)}$ is

\frac{p}{2 L} - 2^{- (n + 1)} = \frac{2^{n} p - L}{2^{n + 1} L} .

Its sign is therefore the sign of the integer numerator $2^{n} p - L$ . For fixed $p, L$ , the next numerator satisfies $2^{n + 1} p - L = 2 (2^{n} p - L) + L$ . Erdos249257.nextDyadicExcessIntNumerator Erdos249257.nextDyadicExcessIntNumerator_succ

Proposition 6.139 (The geometric remainder bound). The remainder after the first two geometric terms of $1 / (2^{n} - 1)$ is

\frac{1}{2^{n} - 1} - 2^{- n} - 4^{- n} = \frac{8^{- n}}{1 - 2^{- n}} \leq \frac{4}{3} 8^{- n} (n \geq 2) .

Indeed, $1 - 2^{- n} \geq 3 / 4$ . Summing over $n > m$ , for an integer $m \geq 1$ , gives

\sum_{n > m} (\frac{1}{2^{n} - 1} - 2^{- n} - 4^{- n}) \leq \frac{4}{3} \sum_{n > m} 8^{- n} = \frac{4}{21} 8^{- m} .

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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