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We prove L44. All manuscript results below are explicit conditional premises from the September 2026 unrefereed manuscript. We use the notation and conclusions of L65: $Z(x)=xG_m/\log x$, $q(s)=A(f_0;s)/A(1;s)$ and $f_0(r)=1/r$. The measure $\mu_{H,s}$ always uses the union probability for each distinct tail totient and the least preimage among all preimages.

Encoding a fixed set of primes

The map from a finite set of primes $S$ to $r(S)=\prod_{p\in S}p/(p-1)$ is injective. Indeed, cancel the primes common to two sets with equal products and take the largest prime $p$ remaining. After cross multiplication, $p$ divides the side containing its numerator and cannot divide the other side: every remaining prime and every factor of the form $q-1$ there is smaller than $p$. This is a contradiction.

For a squarefree integer $k$, define $h_k(r(S))$ to be $1$ if every prime dividing $k$ belongs to $S$, and $0$ otherwise. Extend $h_k$ arbitrarily with values in $[0,1]$ off these arguments. For every totient $d$,

$$h_k(\ell(d)/d)=\mathbf 1_{k\mid\ell(d)}. \tag{1}$$

This uses $d=\varphi(\ell(d))$, not the prime factors of a chosen witness. In particular $b_{H,k}(s)=A_H(h_k;s)/A_H(1;s)$ for all sufficiently large $H$.

Counting the divisibility condition

Fix $k$, and choose $H$ large enough that the lower bound for every prefix prime in the manuscript is greater than every prime dividing $k$. This choice is possible because that lower bound tends to infinity with $H$, uniformly in phase. Keep $H$ fixed while $x\to\infty$.

Outside the exceptional sets of Proposition 5.3, a value has a unique tuple and

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

Thus $k\mid\ell(v)$ if and only if $k\mid\ell(d)$. For fixed remaining data, counting $p_0$ as in Proposition 6.1 gives

$$\limsup_{x\to\infty} \left|\frac{C_k(x)}{Z(x)}-\frac{M_{h_k}(x;H)}{G_m}\right| \le e_H,\qquad e_H\longrightarrow0. \tag{2}$$

The count for each datum is its unweighted largest-prime count multiplied by (1). Its weight lies in $[0,1]$, so the errors from exceptional values and tuples are at most those in Proposition 5.3, and the prime-sum error is controlled by $M_1\ll K_HG_m$. No independence of prime divisibility and phase is used. The restriction on $H$ depends on $k$; once it holds, the error bounds do not.

Apply Proposition 6.5 and Lemma 6.6 to the comparison quantity $C_k/Z$ in (2). They give uniform convergence to $B_k(s)=A(h_k;s)$ and $C_k(x)/Z(x)=B_k(\theta(x))+o(1)$. Since $A(1;s)$ is bounded away from zero, division gives

$$b_{H,k}\longrightarrow b_k:=\frac{B_k}{A(1;\cdot)} \quad\hbox{uniformly},\qquad \frac{C_k(x)}{V(x)}=b_k(\theta(x))+o(1). \tag{3}$$

The bounds $0\le b_k\le1$ and $b_1=1$ follow directly.

By the finite-coefficient argument of L43, $B_k$ has one-sided limits; the positive denominator gives the same for $b_k$. The count $C_k$ changes by at most one on an interval of length one. Since $0\le C_k\le V$,

$$\left|\frac{C_k(x+1)}{V(x+1)}-\frac{C_k(x)}{V(x)}\right| \le\frac2{V(x)}\longrightarrow0. \tag{4}$$

Use (4) and (3) at the exact-phase arguments and their neighbors in L43's proof. At an interior phase both one-sided limits equal the value at that phase; at the period boundary the limit from $1$ equals the value at $0$. This proves continuity on $\mathbb R/\mathbb Z$ without a positive lower bound for $B_k$.

Prime deletion, uniformly in the prime

We give the estimates used in the unrefereed, non-arXiv October note An explicit bound for Erdős problem 417, Equations (2)–(4), at the uniformity needed here. The note's numerical evaluation is not used.

Fixed dilation of $V$ implies that for some absolute constant $D$,

$$\frac{V(x/t)}{V(x)}\le Dt^{-1/2}\qquad(t\ge1, x\ge x_0). \tag{5}$$

To see the uniformity, choose $x_0$ so that $V(y/2)/V(y)\le2^{-1/2}$ for $y\ge x_0$. Iterate down dyadic scales while the arguments remain above $x_0$, and use monotonicity for an intermediate $t$. The same iteration gives $V(x)\gg\sqrt{x}$; below $x_0$ the numerator is bounded, and when $x/t\ge1$ one has $t\le x$. When $x/t<1$ the numerator vanishes. These cases prove (5) after increasing $D$.

If $p^a\Vert\ell(v)$, deleting the full $p$-part leaves an integer with totient $v/[p^{a-1}(p-1)]$. For each fixed $a$ this is an injection on the values $v$. Therefore

$$\frac{C_p(x)}{V(x)} \le\sum_{a\ge1}\frac{V(x/[p^{a-1}(p-1)])}{V(x)} \le\frac{D}{\sqrt{p-1}(1-p^{-1/2})} \le Kp^{-1/2}, \tag{6}$$

where $K$ is independent of $p$ and $x\ge x_0$. For fixed $p$, (5) dominates the series in $a$, so fixed dilation also gives $\limsup C_p(x)/V(x)\le p/(p-1)^2$. If $2\mid\ell(v)$, minimality forces $4\mid\ell(v)$, since an integer with exactly one factor of $2$ has the same totient after that factor is removed. Consequently $v/2$ is a totient and $C_2(x)\le V(x/2)$. Taking the exact-phase sequences in (3) proves the bounds for $b_p$ and $b_2$ on the card.

Truncating the single weighted expectation

For a totient $v$ put $g(v)=v/\ell(v)$ and $w_Y(v)=\prod_{p\le Y,\ p\mid\ell(v)}(1-1/p)$. Elementary finite-product inequalities give

$$0\le w_Y(v)-g(v) \le\sum_{p>Y,\ p\mid\ell(v)}\frac1p.$$

Summing and applying (6) proves

$$0\le\frac1{V(x)}\sum_{v\le x,\ v\in\mathcal V} (w_Y(v)-g(v)) \le K\sum_{p>Y}p^{-3/2}\le\frac{2K}{\sqrt Y}. \tag{7}$$

The last bound compares the sum over primes with $\sum_{n=Y+1}^\infty n^{-3/2}\le\int_Y^\infty t^{-3/2}\,dt$.

We also need the comparison

$$\frac1{V(x)}\sum_{v\le x,\ v\in\mathcal V}g(v) =\frac{V'(x)}{V(x)}+o(1)=q(\theta(x))+o(1). \tag{8}$$

This is Equation (1) of the October note. It follows directly from the same premises as L65: on a good tuple, $g(v)=f_0(\ell(d)/d)\prod_{i=0}^R(1-1/p_i)$, and the prefix product is $1+o_H(1)+o_{x;H}(1)$ uniformly by the prime-band bounds. Counting the largest prime over its full range therefore gives $(x/\log x)M_{f_0}$ for the sum in (8), with normalized error tending to zero as $H\to\infty$. The discarded values and tuples have weights at most one. L65's asymptotic for $V'$ has the same limiting coefficient, which proves (8). This argument retains the full minimum in $\ell(v)$.

Expand the finite product $w_Y$ and use (3):

$$\frac1{V(x)}\sum_{v\le x,\ v\in\mathcal V}w_Y(v) =\sum_{k\mid P_Y}\frac{\operatorname{mobius}(k)}k \frac{C_k(x)}{V(x)} =Q_Y(\theta(x))+o_Y(1). \tag{9}$$

Keep $Y$ fixed and take an exact-phase sequence in (7)–(9). The bound in (7) is independent of phase and of $Y$, so

$$0\le Q_Y(s)-q(s)\le K\sum_{p>Y}p^{-3/2}\quad(0\le s<1). \tag{10}$$

There is no exchange of the limits in $x$ and $Y$. In particular $Q_Y\to q$ uniformly. This initial bound can now be sharpened to one with an explicit constant. For fixed integers $Z>Y\ge2$, the product inequality gives $w_Y-w_Z\le\sum_{Y<p\le Z,\ p\mid\ell(v)}1/p$. Take an exact-phase sequence and use (3), (9) and the bound for $b_p$ to obtain

$$0\le Q_Y(s)-Q_Z(s) \le\sum_{Y<p\le Z}\frac{b_p(s)}p \le\sum_{Y<p\le Z}\frac1{(p-1)^2}.$$

Let $Z\to\infty$ using (10). This proves the stronger bound

$$0\le Q_Y(s)-q(s)\le T_Y:=\sum_{p>Y}\frac1{(p-1)^2} \le\sum_{n=Y}^\infty\frac1{n^2} \le\frac1{Y-1}. \tag{11}$$

For fixed $Y$, the finite product expansion also gives

$$\sum_d\mu_{H,s}(d)w_Y(d) =\sum_{k\mid P_Y}\frac{\operatorname{mobius}(k)}k b_{H,k}(s) \longrightarrow Q_Y(s)$$

uniformly in $s$. Subtract L65's uniformly convergent $q_H$ and use (11) to obtain the claimed finite-coefficient truncation bound. Here $Y$ is fixed before $H\to\infty$; no error uniform over arbitrary $Y=Y(H)$ is claimed.

Finally (11) implies uniform convergence $Q_Y\to q$ as $Y\to\infty$. The oscillation of two continuous functions on the circle differs by at most twice their uniform distance. Thus $\operatorname{osc} Q_Y\to\operatorname{osc} q$, which is zero precisely when $q$ is constant. L65 makes this equivalent to the existence of the limit in Problem 417. No constancy of any $b_k$ or $Q_Y$ has been assumed or proved.