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This is an author-recorded argument supporting the binary totient sketch. No independent review or novelty claim is attached. Its mechanism is related to the progression-mean step in the identified Problem 252 manuscript, but the self-contained core proof is in the native claim linked below. It shows that finite support is unnecessary for that step. The later error bounds and applications are proved here. None makes rationality of the binary series imply the required limiting identity.

Write $\rho(n)=\varphi(n)/n$ for positive integers $n$, so $0<\rho(n)\le1$ and $\rho(n)=\sum_{d\mid n}\mu(d)/d$.

The mean and separation theorem

The canonical statement and complete proof are L16, at tier 0: author-recorded, without independent acceptance. The labels below retain the notation used by the research applications.

Lemma 1 (the mean calculation in L16). For fixed $Q\ge1$ and integer $a$, the mean of $\rho$ over positive members of the progression is

$$\kappa_Q(a) =\sum_{\substack{d\ge1\\(d,Q)\mid a}} \frac{\mu(d)(d,Q)}{d^2} =\frac1{\zeta(2)} \prod_{p\mid Q}(1-p^{-2})^{-1} \prod_{p\mid(Q,a)}(1-p^{-1})>0.$$

For a real absolutely summable kernel $c$, let $F(N)=\sum_{s\ge1}c_s\rho(N+s)$. Its mean is

$$\mathcal M_Q(a)=\sum_{s\ge1}c_s\kappa_Q(a+s).$$

For a prime $\ell\nmid Q$, let $a_b\equiv a\pmod Q$ and $a_b\equiv b\pmod\ell$. The refinement formula is

$$\kappa_{Q\ell}(a_b+s) =\frac{\kappa_Q(a+s)}{1-\ell^{-2}} \left(1-\frac{\boldsymbol1_{\ell\mid b+s}}\ell\right).$$

The canonical proof derives these formulas by divisor-congruence counting and absolute convergence; it uses no pointwise asymptotic for $\rho$.

Theorem 2 (L16). If $c$ is nonzero, then for every fixed $Q\ge1$ and integer $a$,

$$\limsup_{\substack{N\to\infty\\N\equiv a\ (Q)}}|F(N)|>0.$$

More precisely, put $d_s=c_s\kappa_Q(a+s)$. Choose any $s_*$ with $d_{s_*}\ne0$, any $K\ge s_*$ with $\sum_{s>K}|d_s|<|d_{s_*}|/4$, and any prime $\ell>K$ not dividing $Q$. The means on the refinements $N\equiv0$ and $N\equiv-s_*$ modulo $\ell$ differ; one has modulus greater than $\eta=|d_{s_*}|/[4\ell(1-\ell^{-2})]$. That refinement contains infinitely many $N$ with $|F(N)|>\eta/2$.

First fix the kernel and original progression, then choose the anchor, cutoff and prime, and finally take the limit. For growing data the anchor may shrink and $K,\ell,Q$ may grow. No uniform separation follows without controlling those quantities. The finite-window estimate below makes one such control available; Corollary 5 gives a particular family where its growing-shift error can actually be bounded.

An explicit error bound for the means

The limiting argument has the following quantitative form. This makes its dependence on a growing modulus or kernel available for comparison; it does not assert that a proposed growing family satisfies the resulting bounds. Logarithms in this subsection are natural.

Lemma 2a. Let $R,X,H\ge1$ and $T\ge0$ be integers, $0\le b<R$, and let $c$ be an absolutely summable real kernel. Write

$$B_H=\sum_{s\le H}|c_s|,\qquad E_H=\sum_{s>H}|c_s|, \qquad F(N)=\sum_{s\ge1}c_s\rho(N+s).$$

Then

$$\left|\frac1X\sum_{t=1}^X F\bigl(b+R(T+t)\bigr) -\mathcal M_R(b)\right| \le \frac{B_H\bigl(2+\log(R(T+X+1)+H)\bigr)}X+2E_H.$$

Proof. For a single $s\le H$, put $Y=b+R(T+X)+s$. As in Lemma 1, counting a soluble congruence for $t$ in this interval differs from its mean by at most one. After division by $X$, the divisor-sum error is at most $X^{-1}\sum_{d\le Y}1/d\le(1+\log Y)/X$. Completing the main series contributes at most $R\sum_{d>Y}d^{-2}\le R/Y\le1/X$. Since $Y\le R(T+X+1)+H$, multiplying by $|c_s|$ and summing gives the first term. The omitted kernel contributes at most $E_H$ to each of the sample average and the limiting mean, because $\rho,\kappa_R\le1$. $\square$

For the choices $s_*,K,\ell$ in Theorem 2, put

$$R=Q\ell,\qquad \eta=\frac{|d_{s_*}|}{4\ell(1-\ell^{-2})}>0.$$

One of the two refined means has magnitude greater than $\eta$. Choose any $T\ge0$, then $H$ and $X$ satisfying

$$E_H\le\eta/8,\qquad \frac{B_H\bigl(2+\log(R(T+X+1)+H)\bigr)}X\le\eta/4.$$

On at least one refinement there is a member $N=b+R(T+t)$, $1\le t\le X$, for which $|F(N)|>\eta/2$. Indeed the error from Lemma 2a is at most $\eta/2$, so the signed sample mean has magnitude greater than $\eta/2$. Such $H$ and $X$ exist for each fixed kernel, modulus and $T$. In particular the same refinement contains infinitely many such members, and the limsup in Theorem 2 is at least $\eta/2$.

This is an explicit finite-window separation criterion, including the infinite kernel tail. For a parameter family, its anchor, refinement modulus, tail mass and sample length must satisfy these very inequalities. A family may use a different kernel on each sample window, provided it is held fixed throughout that window and the displayed inequalities are verified with that window's data. The two candidate refinements for a given kernel use that same kernel. Substituting a different kernel at each $N$, or retaining only a selected member that need not be the separating one, is not licensed by the criterion. No implication from rationality to smallness on any such window has been supplied.

A precise obstruction for a fixed dilation grid

Retain $T_n=\sum_{j\ge1}\varphi(n+j)2^{-j}$ from the main sketch. Fix a nonempty finite set of pairwise distinct integers $p_e\ge2$, offsets $0\le t_e<p_e$, and integer weights $w_e$ not all zero. Suppose a fixed progression $N\equiv a\pmod Q$ satisfies all $N\equiv t_e\pmod{p_e}$, and define

$$\Lambda(N)=\sum_e w_e\varphi(p_e)T_{(N-t_e)/p_e}.$$

Only sufficiently large positive $N$ are considered.

Corollary 3. For every fixed finitely supported real sequence $(h_s)_{s\ge1}$, the expression

$$\Lambda(N)-\sum_{s\ge1}h_s\varphi(N+s)$$

is not $o(N)$ as $N\to\infty$ through the progression $N\equiv a\pmod Q$.

In particular a fixed finite correction cannot give an error $o(1)$ uniformly on the entire progression. This is a conclusion about one fixed grid and one fixed correction; a cutoff or grid depending on $N$ is outside its hypotheses. The same conclusion holds on every fixed subprogression, by applying the statement with that subprogression as the original one. A selected subsequence containing no full subprogression remains outside the claim.

Proof. Refine the progression to one whose modulus $Q'$ is divisible by $Q$ and every $p_e^2$. Such a refinement can be obtained by choosing any one member and its residue modulo the least common multiple. From $\varphi(p)\varphi(m)=\varphi(pm)\varphi((p,m))/(p,m)$, the exact expansion is

$$\Lambda(N)=\sum_{s\ge1}A_N(s)\varphi(N+s),\qquad A_N(s)=\sum_{e:\ p_e\mid s+t_e} w_e\frac{\varphi(g_{e,N,s})}{g_{e,N,s}}2^{-(s+t_e)/p_e}, \quad g_{e,N,s}=\gcd\!\left(p_e,\frac{N+s}{p_e}\right).$$

On the refined progression the coefficient $A_N(s)=A(s)$ is fixed, since the gcd depends only on $N\pmod{p_e^2}$. Also $\sum_s s|A(s)|\le\sum_e2p_e|w_e|<\infty$.

Let $p$ be the largest dilation with a nonzero weight. On the residue class $s\equiv-t\pmod p$ belonging to that dilation, its summand has magnitude at least $(\varphi(p)/p)2^{-(s+t)/p}$. Every other nonzero summand decays at least as fast as a constant times $2^{-s/(p-1)}$. Consequently $A(s)$ is nonzero for all sufficiently large $s$ in that class. This is the coefficient calculation of Proposition 4 of the main sketch, with the dependence on $N$ correctly retained before refinement.

Set $c_s=A(s)-h_s$. This is a nonzero sequence with an absolutely summable first moment. Uniformly over the refined progression,

$$\frac{\Lambda(N)-\sum_s h_s\varphi(N+s)}N =\sum_s c_s\rho(N+s)+O\!\left(\frac{\sum_s s|c_s|}{N}\right).$$

If the numerator were $o(N)$ on the original progression, the kernel on the right would tend to zero on the refinement. Theorem 2 forbids that. $\square$

The signed tail after a growing common cutoff

For the same fixed grid, a direct truncation of its exact coefficient expansion has a quantitative limitation even when the cutoff grows. This concerns the particular correction using the first $H$ coefficients $A_N(s)$, not an arbitrary correction with $H$ terms.

Lemma 4. Let $p=p_{e_*}$ be the largest dilation with nonzero weight, $t=t_{e_*}$, and set

$$V=\sum_e|w_e|,\qquad W=V-|w_{e_*}|,\qquad \alpha=|w_{e_*}|\frac{\varphi(p)}p\,2^{-t/p}>0.$$

For every compatible $N>2\max_ep_e$ and integer $H\ge0$, define the exact remainder

$$\mathcal R_H(N)=\Lambda(N)-\sum_{s\le H}A_N(s)\varphi(N+s).$$

It satisfies

$$|\mathcal R_H(N)|\le 2V(N+H+2p)2^{-H/p}.$$

If in addition

$$\frac{2W}{\alpha}(1+\log_2N) \left(1+\frac{H+2p}{N}\right) 2^{-H/[p(p-1)]}\le\frac12,$$

then $\mathcal R_H(N)$ has the sign of $w_{e_*}$ and

$$|\mathcal R_H(N)|\ge \frac{\alpha N}{2(1+\log_2N)}\,2^{-H/p}.$$

Proof. Put $J_e=\lfloor(H+t_e)/p_e\rfloor$. The contribution of dilation $e$ to the remainder is

$$w_e\sum_{j>J_e} \frac{\varphi(\gcd(p_e,n_e+j))}{\gcd(p_e,n_e+j)} \varphi(N+jp_e-t_e)2^{-j},\qquad n_e=(N-t_e)/p_e.$$

Every argument $N+jp_e-t_e$ exceeds $N$. The elementary bound on the main sketch, and the monotonicity of $x/(1+\log_2x)$ for $x\ge2$, give a lower bound $N/(1+\log_2N)$ for each totient. The contribution of $e_*$ therefore has its weight's sign and magnitude at least

$$|w_{e_*}|\frac{\varphi(p)}p\frac{N}{1+\log_2N}2^{-J_{e_*}} \ge \frac{\alpha N}{1+\log_2N}2^{-H/p}.$$

For any active dilation, the bounds on its gcd factor and $\varphi(n)\le n$ give an absolute contribution at most

$$|w_e|\bigl(N-t_e+p_e(J_e+2)\bigr)2^{-J_e} \le 2|w_e|(N+H+2p)2^{-H/p_e}.$$

Summing proves the upper bound. Every other active dilation has $p_e\le p-1$, so their total absolute contribution is at most $2W(N+H+2p)2^{-H/(p-1)}$. The displayed hypothesis makes this at most half the lower bound for the $e_*$ contribution, proving the sign and lower bound. The case $W=0$ is included. $\square$

In particular fix a real $c>0$ and use $H=\lceil c\log_2N\rceil$. If $0<c<p$, the dominance hypothesis eventually holds and $|\mathcal R_H(N)|\to\infty$, uniformly over compatible $N$. If $c>p$, the upper bound gives $\mathcal R_H(N)\to0$ uniformly. At $c=p$ these estimates give only a lower bound of order $1/\log N$ and an upper bound of order one; they do not decide convergence to zero. Thus the direct common-shift truncation needs the logarithmic scale set by its largest dilation. The calculation controls the entire infinite remainder and does not exchange a limit with a kernel whose parameters vary. It does not supply the arithmetic fractional-part separation needed under rationality.

The finite-window mean estimate settles uniform convergence at the critical scale, by keeping the varying shift fixed on each sampled block.

Corollary 5. On every fixed compatible progression $N\equiv a\pmod Q$, let $H_N=\lceil p\log_2N\rceil$. Then

$$\limsup_{\substack{N\to\infty\\N\equiv a\ (Q)}} |\mathcal R_{H_N}(N)| \ge \frac{|w_{e_*}|\varphi(p)}{8p}\,\kappa_Q(0)>0.$$

Proof. Write $D=|w_{e_*}|\varphi(p)/p$ and $\gamma=\kappa_Q(0)>0$. The product in Lemma 1 gives $\kappa_Q(r)\ge\gamma$ for every integer $r$. For each large integer $k$, consider all progression members in the interval

$$2^k<N\le2^{k+1/p}.$$

Throughout this interval $H_N=pk+1$, so the leading dilation's $J_{e_*}=\lfloor(H_N+t)/p\rfloor=k+\epsilon$, where $\epsilon=\lfloor(1+t)/p\rfloor\in\{0,1\}$. Its first retained term has the fixed shift $s_k=p(k+\epsilon+1)-t$.

The interval contains $X_k=(2^{1/p}-1)2^k/Q+O(1)$ consecutive members of the progression. Apply Lemma 2a with $R=Q$, its starting index $T=O_Q(2^k)$, and the kernel consisting of the single coefficient $c_{s_k}=1$, with cutoff $s_k$. Here $B_{s_k}=1$, $E_{s_k}=0$ and $s_k=O_p(k)$, so the error is $O_{p,Q}(k2^{-k})$. This tends to zero uniformly in the phase $a+s_k\pmod Q$. Thus for each sufficiently large $k$ at least one member $N_k$ in this interval satisfies $\rho(N_k+s_k)\ge\gamma/2$.

At that member, the leading dilation's whole tail has its weight's sign and magnitude at least its first retained term, namely

$$D\,\varphi(N_k+s_k)2^{-(k+\epsilon+1)} \ge \frac{D\gamma}{2}(N_k+s_k)2^{-(k+\epsilon+1)} > \frac{D\gamma}{2^{\epsilon+2}} \ge \frac{D\gamma}{8}.$$

The other dilations contribute in total at most $2W(N_k+H_{N_k}+2p)2^{-H_{N_k}/(p-1)} =O_{p,w}(N_k^{-1/(p-1)})=o(1)$, by the proof of Lemma 4. Subtracting this bound along $N_k\to\infty$ proves the result. $\square$

Consequently the direct truncation at $H_N=\lceil c\log_2N\rceil$ has a remainder tending to zero on a whole fixed compatible progression exactly when $c>p$, among fixed real $c>0$. For $c=p$ the conclusion is a positive limsup, not a positive lower bound at every member or a claim about every selected subsequence. The quantitative mean bound is what permits this particular family of growing shifts; a bare exchange of fixed-kernel limits would not do so.

The unsupplied irrationality implication

Rationality $S=a/b$ makes $b\Lambda(N)$ integral for integer $w_e$. It does not make $\Lambda(N)$ bounded or its normalized kernel tend to zero. Corollary 3 explains why subtracting a fixed number of shifted values cannot supply such a uniform reduction on a whole progression. Theorem 2 remains usable for a different construction that really does produce a null kernel; that construction is open.

Possible changes of mechanism include growing grids, corrections more general than the direct truncation of Lemma 4, coefficients with an additional parameter, or a carefully selected proper subsequence of a progression. They require their own error bounds and nonvanishing arguments. Even where Lemma 4 makes the tail small, the needed separation of the finite prefix from the rationality lattice remains unproved.

The same finite-window estimate has a separate quantitative application to [[research/leads/totient_binary_series/integer_interpolation/geometry|shifted-totient interpolation matrices]]. Averaging the squared limiting means over every class of one prime $\ell>2k$ gives a lower bound uniform over unit coefficient vectors on at most $2k$ shifts. Lemma 2a transfers that bound to common finite samples with explicit growing parameters. The resulting singular-value estimate bounds the prescribed real solution's contribution to the weighted interpolation cost. It supplies neither a quantitative integral-image endpoint nor a small rounding error in the integer coset, so the irrationality implication remains conditional on that additional arithmetic control.