The Binary Totient Series

The quantified prime condition.

This condition is unproved; it implies irrationality (ErdosProblems.Erdos249.DTWNaturalPrimeTailOrbitStrictGap, ErdosProblems.Erdos249.irrational_totient_series_of_naturalPrimeTailOrbitStrictGap).

Its meaning for the doubling map.

$\mathrm{Re}e(2^{p-h-1}{\alpha }_h)<9/10$ says $\Vert 2^{p-h-1}{\alpha }_h{\Vert }_{\mathbb{R}/\mathbb{Z}}>\arccos (9/10)/2\pi =0.0717831\dots$, so the requirement is exactly: for each $h$, the indicated prime-indexed $\times 2$ orbit point stays more than $0.0717831\dots$ from the nearest integer, for cofinally many primes $p$. A short-run condition on the binary digits can be a sufficient proxy for this circle-distance inequality, but it is not equivalent to it; a run-length description alone does not determine the residual position inside the dyadic cylinder.

What the prime condition adds.

The preceding condition asks for a separated point at a prime-indexed position. The full-block average does not by itself locate such a point on a prescribed sparse set. Conversely, the prime condition does not assert cancellation over the whole interval. They are different sufficient hypotheses; no general comparison of their difficulty is proved here.

Finite comparison.

Among the $501$ primes in $[1000,5000)$, the strict inequality holds at $432$, $426$ and $430$ primes for $h=1,2,5$, respectively. The corresponding proportions are $0.8623$, $0.8503$ and $0.8583$ to four decimal places, and each sample has a cosine value below $-0.9999$. These counts were recomputed using the same interval for $S$ as in Table 1, with no interval meeting the threshold $9/10$. They do not establish separation at arbitrarily large primes.

An average over a prescribed prime sample.

The subset implication, Theorem 9.5, allows a nonempty finite set $T$. If $T$ is chosen after a separated point is known, the subset formulation may add no arithmetic content: the equivalence proved using selected pairs, Proposition 2.7(b), illustrates this issue. To avoid that particular choice, fix the sample in advance by

For $X\ge 4$ and $L\ge h$, this is precisely the assigned group with cofactor $m_N=1$ of §3.2, taken with $s=L-h$ (Erdos249257.TotientTailPeriodKiller.mem_pivotFiber_one_overlap_iff).

A sufficient, unproved condition is

The subset theorem then gives a certificate at some $N\ge X\ge X_0$. Nonemptiness matters: for the empty set the displayed sum inequality would hold automatically and would produce no certificate. The bounds $X\ge 4$ and $L\ge h$ are also the hypotheses of the identification with that cofactor group.

On this sample the selected cofactor is $1$, so its coefficient is the odd integer $2^h-1$ and its modulus is $2^{h+1}$, fixed once $h$ is fixed. The unweighted prime average has zero Ramanujan main term, since $c_{2^{h+1}}(2^h-1)=\mu (2^{h+1})=0$. That observation does not establish (5): the prime phase is multiplied by a residual phase built from the other totient values. Their correlation is the missing estimate. Restricting the sample removes the complement and unwanted cofactor terms, but not this residual correlation.