The Binary Totient Series

Approach

Complete terminal divisibility pattern

Under $a\ge 8$, $m\le K$, $J+K+a+6<2\cdot 2^a$ and $2H+J+K+2<2^m$, the full set of divisibilities in Observation 6.107 is impossible: the last difference would be a positive multiple of $2^m$ smaller than $2^m$. not_actualLcmTerminalDyadicStaircase_of_room.

The obstruction is specific to that pattern and those bounds. If the last divisibility is omitted, the additional terminal bound of Proposition 6.86 forces the penultimate difference to be $2^{m-1}$. It does not prove that such partial patterns exist. puncturedDyadicStaircase_penultimate_eq_half.

The explicit sequence in Observation 6.115 has $a_{(q-1)H-1}=\phi (H)$ for $2\le q<t$ and a bounded state satisfying $a_i=2c_i-c_{i+1}$. Every finite integer combination of its shifts retains a recurrence of this form, including products of differences along multiples of $H$. Its bounds acquire the factor $\sum _j|{\lambda }_j|$; they are not uniform in the coefficients of the linear combination. The explicit theorem and ordinary proof are given in Observation 6.115.

The constructed coefficients are not asserted to be actual totient differences. Preserving the displayed recurrence under shifts therefore does not distinguish them from the coefficients required in the proposed argument. The full totient divisor identity totient_eq_sum_mobiusTotientChannel retains information absent from the recurrence alone. The counterexample does not require every future proof to use that particular identity.

For $H\ge 0$ and positive integers $p,q$, requiring $R_{2kH}-R_{kH}\in \mathbb{Z}$ for every $k\in \{1,p,q,pq\}$ is equivalent to requiring only $R_{2H}-R_H\in \mathbb{Z}$. Neither multiplier need be prime. The affine transport formula along $H\mapsto kH$ carries integrality at the first scale to each of the other three. Erdos249257.FullTargetPrimeAdjunctionNoGo.fullTarget_primeAdjunction_diamond_iff_root.

Adding these three integrality assertions does not strengthen the first one. A different test involving part of the tail would need an estimate for the part it discards; this equivalence provides no such estimate. The result concerns the displayed four conditions, not every test using several scales.

For the valuations ${\nu }_j$ and odd residues $a_j$ of Proposition 2.11(iii), the free integers $z_j$ can always be chosen so that $e_{j+1}=2e_j+2^{{\nu }_j}(a_j+2^u{z}_j)$ lies in the stated centred interval at every step. Erdos249257.TotientTailPeriodKiller.fixedPrecisionTropicalNoGo.

The specified residues do not exclude all finite carry sequences while the higher quotients $z_j$ remain unrestricted. An argument would need further constraints on those quotients or on the actual coefficients. The lemma does not assert that its freely chosen coefficients are totient differences.

For an integer sequence $a$ and integer initial values $u_0,v_0$, let $u_{n+1}=2u_n-a(n+1)$ and $v_{n+1}=2v_n-a(n+1)$. Induction gives $u_L-v_L=2^L(u_0-v_0)$ for every $L\ge 0$. Erdos249257.affineBinaryOrbit_mod_twoPow_eq.

The endpoints agree modulo $2^L$, but they are distinct integers whenever $u_0\ne v_0$. Indeed, if the forcing terms are known, then

Only reduction modulo $2^L$ discards the initial value. The identity does not exclude an argument that retains the full endpoint or other information about the sequence.

For the family in Proposition 2.11(i), with $m\ge 2$, all histories before $m$ agree but the next tail takes $\lfloor (m+1)/2\rfloor +1$ distinct values. A common state cannot determine that tail; a set of labels permitting exact recovery must have at least that many elements. Erdos249257.balancedPulse_no_autonomous_decoder.

This counterexample concerns states determined only by the common history of the displayed family. It does not exclude a state using additional information specific to the totient sequence. The bound counts labels, not bits; it is not a complexity theorem about the binary digits of $S$.

Let ${\chi }_4(n)$ be $0$ for even $n$, $1$ for $n\equiv 1(\mathrm{mod}4)$, and $-1$ for $n\equiv 3(\mathrm{mod}4)$, and put $c(n)=\sum _{d\mid n}{\chi }_4(d)$ for $n\ge 1$. Divisor pairing gives $c(n)=0$ when $n\equiv 3(\mathrm{mod}4)$. Erdos249257.intWeightedCoeff_periodFourSignWeight_eq_zero_of_mod_four_eq_three. Nevertheless $c(2^k)=1$ for every $k\ge 0$, since $1$ is the only odd divisor of $2^k$.

This rules out nonvanishing on an arbitrarily prescribed progression, not nonzero coefficients arbitrarily far out. In fact $c$ is multiplicative. Its prime-power values are $1$ at powers of $2$, $a+1$ at $p^a$ with $p\equiv 1(\mathrm{mod}4)$, and $1$ or $0$ at $p^a$ with $p\equiv 3(\mathrm{mod}4)$ according as $a$ is even or odd. Thus $c(n)\ge 0$ for every $n\ge 1$. These elementary calculations verify both coefficient hypotheses of irrational_intWeightedErdosSeries_periodic_of_coeff_nonneg_of_frequently_ne_zero: $\sum _{a\ge 1}{\chi }_4(a)/(b^a-1)$ is irrational for every integer $b\ge 2$. This application concerns the displayed weighted Lambert series, not the totient series $S$.

$\text{finite Mersenne-weight exclusion}(x,\mathrm{level},\mathrm{lookahead})$ is a decidable, finite-depth certificate proving non-membership only (greedyMersenneDeath_not_mem); a finite computation excludes $3/4$ at level 1 (three_fourths_certifiedGreedyMersenneDeath).

The cited test has a sound exclusion implication. The analogous soundness statement for the residue test of this record, Proposition 6.18, also does not make a finite unsuccessful search a proof of integrality. This is a limit of what follows from those tests, not an impossibility theorem for rationality or membership proofs using additional identities.

For any finite $w:\iota \to \mathbb{Q}$, $x:\iota \to \mathbb{N}$ with $\sum w_i\phi (x_i)=1$ (exact isolation), the crude two-tail cost $\sum |w_i|(2(x_i+1)+(x_i+2))\ge 3$, using only $\phi (x)\le x$. Erdos249257.three_le_totientAdjugateTailCost (see also not_totientAdjugateTailCost_lt_one).

The displayed termwise absolute-value estimate cannot be less than $1$. The same inequality holds for any coefficients $0\le c(n)\le n$ with $\sum _i{w}_{i}c(x_i)=1$. It does not rule out a smaller error bound that uses cancellation between the tails, or a different approximation identity.

For the completely multiplicative sequence $c(n)=n$, if $0<K<q$ and $q\mid 2^K-1$, then the scaled-tail difference is the integer $K$ at every starting index, but $q\nmid K$. idCoeff_mersenneModulus_does_not_annihilate_tailShift.

A factor of $2^K-1$ does not, from the recurrence alone, have to divide the inhomogeneous tail difference. The example $K=3$, $q=7$ already has a primitive prime factor. An application to another series would need a correspondence between its coefficients and this recurrence; Mersenne denominators alone do not provide one.

For $p$ prime and $e\ge 1$, if every candidate integer in the interval at denominator $p^e$ is a nonunit modulo $p^e$, then there is at most one candidate. Equivalently, the notation of Observation 6.114 satisfies $L=0$, or $L=1$ with $p\mid J+1$. reducedDenominatorCandidateCount_prime_pow_le_one (for the equivalent characterisation, reducedDenominatorUnitGapCert_prime_pow_iff).

This candidate-count bound does not apply to every odd denominator; the example modulo $15$ in Observation 6.114 has two candidates. It also does not bound the denominator cutoff obtainable by changing the window or approximation. It supplies no proof that the fixed exclusion in Section 10.4 extends to arbitrarily large denominators.

At $H=2^{n+2}$, $j=3\cdot 2^{n+1}$, $\phi (3H+j)-2\phi (2H+j)+\phi (H+j)=-2^{n+1}$ for every $n\ge 0$. fixedRankSecondDifference_dyadic_fixture_exact.

The exact valuation concerns this scaled family, not the LCM family with $j^2\le 2^a$; see Observation 6.108. The independent coefficient calculation says that integer solutions of $c_1+c_2+c_3=c_1+2c_2+3c_3=0$ are exactly the multiples of $(1,-2,1)$ (affine_rank_three_kernel_classification). It classifies these three coefficients; it does not determine the valuation of their totient combination on every restricted family.