---

We prove L65. All manuscript results invoked below are conditional premises from the unrefereed source. This proof checks their composition, not their underlying number-theoretic proofs.

Notation and source interface

Write $\mathcal V$ for the positive totients and $\ell(v)=\min\{n:\varphi(n)=v\}$. Then

$$V(x)=\#(\mathcal V\cap[1,x]),\qquad V'(x)=\#\{v\in\mathcal V:\ell(v)\le x\}.$$

The latter set is contained in the former, since $v\le\ell(v)$. Use natural logarithms throughout this proof. In the manuscript's notation,

$$B=\log\log x,\quad m=\left\lfloor\frac{\log B-\log\log B}{\lambda}\right\rfloor, \quad \theta(x)=\frac{\log B-\log\log B}{\lambda}-m, \quad Z(x)=\frac{xG_m}{\log x},$$

where $\lambda>0$ and $G_m$ are defined in its Equations (2.1)–(2.4). Its Theorem 2.1 supplies

$$V(x)=Z(x)\bigl(A(1;\theta(x))+o(1)\bigr),\qquad 0<c_-\le A(1;s)\le c_+<\infty, \tag{1}$$

and $V(cx)/V(x)\to c$ for every fixed $c>0$.

For each sufficiently large fixed cut $H$, the manuscript counts distinct data $(d,p_1,\ldots,p_R)$, where $R=m-H$, rather than their possibly many tail witnesses. Its Equation (6.1) defines

$$M_f(x;H)=\sum_{(d,p_1,\ldots,p_R)} \frac{f(\ell(d)/d)}{d\prod_{i=1}^R(p_i-1)}.$$

The needed source estimates are these. Proposition 5.3 gives, outside $o_H(1)Z(x)+o_{x;H}(Z(x))$ exceptional values and tuples, a unique tuple $(p_0,\ldots,p_R,d)$ for each value $v$, with

$$v=d\prod_{i=0}^R(p_i-1),\qquad \ell(v)=\ell(d)\prod_{i=0}^R p_i. \tag{2}$$

The identity for $\ell(v)$ uses its assertion about every preimage. Sections 6.1 and 7.1 give, uniformly in the remaining data,

$$\begin{gathered} D=d\prod_{i=1}^R(p_i-1)\le\exp((\log x)^{0.8}),\qquad M_1(x;H)\ll K_HG_m,\quad K_H=H^{o(1)},\\ \sum_{i=1}^R\frac1{p_i}\ll\exp(-cH\rho^{-H}), \end{gathered} \tag{3}$$

where $c>0$, $0<\rho<1$. Each fixed $H$ bounds the tail witnesses, hence $\ell(d)/d$, independently of $x$. The admissible remaining data do not change when the largest prime $p_0\ge x^{0.9}$ varies. Proposition 6.5 approximates $M_f/G_m$ by the finite coefficient $A_H(f;\theta)$ with an error tending to zero as $H\to\infty$, for every $0\le f\le1$.

Count the preimages at most x

Set $f_0(r)=1/r$. For fixed remaining data put

$$r=\ell(d)/d,\quad q=\prod_{i=1}^R(1-1/p_i),\quad T=x/D,\quad U=\frac{x}{\ell(d)\prod_{i=1}^R p_i} =Tq/r.$$

The conditions $\ell(v)\le x$ and $v\le x$ are exactly $p_0\le U$ and $p_0\le T+1$. Since $r\ge1$ and $q\le1$, the first already implies the second. We must therefore count

$$x^{0.9}\le p_0\le U. \tag{4}$$

For each fixed $H$, (3) and the boundedness of $r$ give $T,U=x^{1-o(1)}$ uniformly in the data; $q$ is bounded away from zero. The prime number theorem applied to $U$ gives $\pi(U)=(U/\log x)(1+o_{x;H}(1))$. The lower cutoff in (4) costs at most $x^{0.9}$ primes, which is $o(T/\log x)$ uniformly. Also

$$\left|\frac q r-\frac1r\right| \le1-q\le\sum_{i=1}^R\frac1{p_i}.$$

Thus (4) has

$$\frac{T}{\log x} \left(f_0(r)+O\left(\sum_{i=1}^R\frac1{p_i}\right) +o_{x;H}(1)\right) \tag{5}$$

prime solutions. This is an additive estimate, and requires no prime theorem for short intervals. Summing (5), using (3), and removing the exceptional values and tuples in Proposition 5.3 yields

$$\limsup_{x\to\infty} \left|\frac{V'(x)}{Z(x)}-\frac{M_{f_0}(x;H)}{G_m}\right| \le\eta_H,\qquad \eta_H\longrightarrow0. \tag{6}$$

In particular the extra error in (5) is $O(K_H\exp(-cH\rho^{-H}))$, which tends to zero.

Apply the manuscript's Lemma 6.6 to the comparison quantity $C_{f_0}(x)=V'(x)/Z(x)$, independent of $H$. It proves uniform existence of $A(f_0;s)$ and

$$V'(x)=Z(x)\bigl(A(f_0;\theta(x))+o(1)\bigr). \tag{7}$$

For clarity, uniformity does not require continuity of $A_H$ in $s$. For any fixed $s\in[0,1)$ the increasing continuous function $\psi(B)=(\log B-\log\log B)/\lambda$ assumes $n+s$ at some $B_n\to\infty$. Then $x_n=\exp(\exp B_n)$ has $\theta(x_n)=s$ exactly. Comparing any two cut indices on this sequence bounds their coefficients' difference by their uniform comparison errors. This is the uniform Cauchy argument of Lemma 6.6.

Positivity and the tail of the least-preimage intervals

Lemma 7.3 of the manuscript says that $\ell(v)/v$ is tight when distinct totients $v\le x$ receive equal weight: for every $\varepsilon>0$ some constant $C$ bounds the ratio outside at most $\varepsilon V(x)$ values, for all sufficiently large $x$. Take $\varepsilon=1/2$. At least $V(x/C)/2$ values $v\le x/C$ satisfy $\ell(v)\le Cv\le x$. Hence

$$V'(x)\ge\tfrac12 V(x/C),\qquad \liminf_{x\to\infty}\frac{V'(x)}{V(x)}\ge\frac1{2C}>0, \tag{8}$$

using fixed dilation from (1). Equations (1), (7), (8), and the exact-phase sequences bound $A(f_0;s)$ away from zero uniformly in $s$. The upper bound follows from $V'\le V$.

For $k\ge1$ let $N_k(x)$ and $f_k$ be as in the manuscript's Equation (2.9). The intervals partition the values with $\ell(v)>x$, so

$$V(x)-V'(x)=\sum_{k\ge1}N_k(x),\qquad \sum_{k>K}N_k(x) \le\#\{v\le x:v\in\mathcal V,\ \ell(v)/v>K+1\}. \tag{9}$$

Lemma 7.3 makes the right-hand side at most $\varepsilon V(x)$ for all large $x$ once $K$ is sufficiently large. Now assume the manuscript's Theorem 2.2 as well. For a fixed finite $K$, substitute its asymptotics and (1), (7) in (9), and take an exact-phase sequence. For every $s$ this gives

$$0\le A(1;s)-A(f_0;s)-\sum_{k=1}^K A(f_k;s) \le\varepsilon A(1;s)\le\varepsilon c_+.$$

The choice of $K$ is independent of $s$, proving the stated identity and uniform convergence of the series. No interchange of the cut limit and an uncontrolled infinite sum has been used.

The remaining phase question

Write $\Phi(s)=A(1;s)/A(f_0;s)$. Its denominator has a uniform positive lower bound, so (1) and (7) imply

$$\frac{V(x)}{V'(x)}=\Phi(\theta(x))+o(1). \tag{10}$$

If $\Phi$ is constant, (10) proves convergence. Conversely, every phase has an exact sequence tending to infinity, so convergence forces the same value at every $s$. This is equivalent to constancy of its reciprocal.

For completeness the set of subsequential limits is an interval. For integer $n$, each of $V(n)$ and $V'(n)$ increases by at most one at the next integer, $V(n)\le n$, and $V'(n)\ge\pi(n)$, since the distinct values $p-1$ have preimages $p\le n$. Consequently

$$\left|\frac{V(n+1)}{V'(n+1)}-\frac{V(n)}{V'(n)}\right| \le\frac1{\pi(n)}+\frac n{\pi(n)^2}\longrightarrow0.$$

The ratio is bounded by (8). A bounded real sequence with successive differences tending to zero has as its cluster set the closed interval between its liminf and limsup: every intermediate value is approached during arbitrarily late crossings. The real-variable ratio is constant between consecutive integers. If $\Phi$ is nonconstant, two exact-phase sequences give different limits, so this interval is nondegenerate and has continuum many points. This also recovers the relevant consequence of Proposition 417.2.

The remaining question is a property of the finite arithmetic coefficient, not a missing tail interchange. With $p_{H,s}(d)$ denoting the union probability in the manuscript's Equation (2.7), its common prefactor cancels:

$$\frac{A_H(f_0;s)}{A_H(1;s)} =\frac{\sum_d p_{H,s}(d)/\ell(d)}{\sum_d p_{H,s}(d)/d}. \tag{11}$$

The sums range over distinct tail totients having a witness. For large $H$ the denominator is uniformly positive, and (11) converges uniformly to $A(f_0;s)/A(1;s)$. The witness sets and their union probabilities depend on $s$; $\ell(d)$ is the least preimage among all preimages of $d$. Neither deduplication nor minimization can be replaced by a sum over independent witnesses. No phase independence or phase dependence of (11) is proved here.