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

Positivity and the remaining residue inequality

Definition 9.9. The actual LCM tail orbit at exponent $a$ is $Ω_{a} = R_{2 H} - R_{H}$ , where $H = H (2^{a})$ (actualLcmTailOrbit_pos).

Theorem 9.10. For every $a \geq 8$ and every $J$ with $J + (a + 6) < 2 \cdot 2^{a}$ ,

0 < R (2 H + J) - R (H + J), H = H (2^{a}) .

Hence $0 < Ω_{a}$ for every $a \geq 8$ . The proof is genuinely uniform in $a$ : it rests on two facts proved for all $a \geq 8$ and $0 < j < 2^{a + 1}$ , namely $δ_{2^{a}} (j) > 0$ and $H < 8 \cdot 2^{a} δ_{2^{a}} (j)$ , by a two-case structural split (divisor letter vs. foreign prime power), with absolute constants $4, 8, 32$ and no lookup table.

Remark 9.11 (Which endpoint interval must be excluded). Under the hypotheses of Proposition 6.85, integrality would force the residue to be $2^{K} - e$ , where $e = R_{2 H + J + K} - R_{H + J + K}$ is a positive integer smaller than $2 H + J + K + 2$ . Thus the residue lies near the upper endpoint, not near zero in its least nonnegative representation. The representative $- e$ is negative; the carry $e$ is positive.

The one-sided test below excludes precisely that upper interval. This uses the bound $J + K + (a + 6) < 2 \cdot 2^{a}$ , as well as the modulus bound. It does not follow from positivity that the residue already satisfies half of a nonintegrality certificate without those assumptions.

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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