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

Integer recurrences and rational binary series

For coefficients $0 \leq c (n) \leq n$ , rationality is equivalent to the existence of an integer carry with $u (N) / 2^{N} \to 0$ . The recurrence alone is not restrictive: every integer initial value produces an integer sequence. The limit condition selects the initial value that represents the infinite sum.

Theorem 6.125 (Integer recurrence criterion for a binary series). Let $c : N \to N$ satisfy $c (n) \leq n$ . Then $\sum_{n \geq 1} c (n) 2^{- n}$ is rational if and only if there are an integer $v > 0$ and an integer sequence $u$ such that

u (N + 1) = 2 u (N) - v c (N + 1), \frac{u (N)}{2^{N}} ⟶ 0.

For each fixed $v$ , such a sequence is unique and equals

u (N) = v R_{N}^{c}, R_{N}^{c} = \sum_{j \geq 1} \frac{c (N + j)}{2^{j}} .

Proof. Iterating the recurrence gives the finite identity

\frac{u (N)}{2^{N}} = u (0) - v \sum_{n = 1}^{N} \frac{c (n)}{2^{n}} .

Taking $N \to \infty$ gives $S_{c} = u (0) / v$ , where $S_{c} = \sum_{n \geq 1} c (n) 2^{- n}$ . Subtracting the finite prefix then gives $u (N) = v R_{N}^{c}$ .

Conversely, suppose $S_{c}$ is rational. Choose an integer $v > 0$ with $v S_{c} \in Z$ , set $u (0) = v S_{c}$ , and iterate the recurrence. The terms are integers by construction; the displayed identity gives $u (N) = v R_{N}^{c}$ . Finally $0 \leq R_{N}^{c} \leq N + 2$ proves the required limit. ◻

Without the boundary condition, any integer $u (0)$ gives an integer solution of the recurrence, whether the series is rational or irrational. Two solutions with the same $v$ differ by $2^{N}$ times their initial difference. Thus the limit, not the recurrence alone, determines the initial value. Changing $v$ rescales the canonical solution; uniqueness is asserted only for a fixed multiplier.

Proposition 6.126 (Uniqueness under the growth condition). A real sequence $d$ with $d (N + 1) = 2 d (N)$ and $d (N) = o (2^{N})$ is identically zero.

Observation 6.127 (The specified finite-state decoder is impossible). Fix $m \geq 2$ , put $R = ⌊ (m + 1) / 2 ⌋$ , and for $0 \leq r \leq R$ define

c_{m, r} (n) = {\begin{cases} R - r, & n = m, \\ 2 r, & n = m + 1, \\ 0, & otherwise . \end{cases}

The two nonzero terms have total contribution $(R - r) 2^{- m} + 2 r 2^{- (m + 1)} = R 2^{- m}$ , independent of $r$ . All coefficients and scaled tails before $m$ are therefore the same, whereas the scaled tail at $m$ is $r$ . No map from a state common to this whole family can recover that tail. More generally, exact recovery of all its possible values requires at least $R + 1$ distinct labels. Erdos249257.balancedPulse_no_autonomous_decoder Erdos249257.balancedPulse_label_card_lower_bound The sequences in this family have identical histories before the isolated change but different subsequent tails. Therefore that finite history alone cannot determine the exact tail in this class of sequences. This is not a conclusion about the totient sequence.

There is also an infinite family with a common rational value. Let $a : N \to {0, 1}$ satisfy $a (0) = 0$ , and put $c (2 k) = k - a (k)$ and $c (2 k + 1) = 2 a (k)$ . The adjacent terms contribute $k 4^{- k}$ , so

\sum_{n \geq 1} \frac{c (n)}{2^{n}} = \sum_{k \geq 1} \frac{k}{4^{k}} = \frac{4}{9},

regardless of the binary sequence $a$ . The cited result also gives an integer recurrence satisfying the growth condition (Erdos249257.globalBalancedCoeff_value). The assumptions $a (0) = 0$ and $a (k) \in {0, 1}$ are part of that formal statement, not an arbitrary bounded-input hypothesis. For comparison, the identity coefficients $c (n) = n$ have the explicit multiplier-one recurrence $u (N) = N + 2$ (Erdos249257.idCoeff_temperedOrbit).

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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