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This is the complete author-recorded proof of L220. Fix $y,M,\varepsilon$ as in its statement. Let $\mathcal H$ consist of the positive integers all of whose prime factors are at most $y$, including $1$, and put $P=\prod_{p\le y}p$. Every positive integer has a unique factorization $n=sk$ with $s\in\mathcal H$ and $\gcd(k,P)=1$.

Choose distinct primes $r_0,t_0>y$ and put $Q=r_0t_0P$. The progression $e\equiv r_0t_0\pmod Q$ consists of integers coprime to $P$ and divisible by both $r_0,t_0$. In particular none is $1$ or a prime power. Write $g(e)=\lfloor\tfrac14\log_2e\rfloor$. Choose $e_1$ in this progression sufficiently large that, for every $e\ge e_1$,

$$g(e)\ge2Q,\qquad \frac{3M}{2}e^{1/4}\le\varepsilon\sqrt{e/2}. \tag{1}$$

Set $e_{j+1}=e_j+Q\lfloor g(e_j)/Q\rfloor$ and $E=\{e_1,e_2,\ldots\}$. All these integers remain in the progression, and their gaps satisfy $g(e_j)/2\le e_{j+1}-e_j\le g(e_j)$. Thus

$$E(X):=\#(E\cap[1,X])=O\bigl(X/\log(X+2)\bigr). \tag{2}$$

For completeness, elements up to $\sqrt X$ contribute at most $\sqrt X$; above $\sqrt X$ their gaps are at least a fixed positive multiple of $\log X$, once $X$ is sufficiently large. Enlarging the constant covers the bounded range.

The entire binary weight of the orbit with rough part $e$ is

$$A_e=\sum_{s\in\mathcal H}\varphi(s)2^{-se}.$$

For $e\ge2$, positivity, the term $s=1$ and $\varphi(s)\le s$ give

$$2^{-e}\le A_e\le\frac{2^{-e}}{(1-2^{-e})^2}<2^{1-e}. \tag{3}$$

Put $d=MA_{e_1}/2$. For any real $r$ with $|r-S|<d$, start with $\eta_0=r-S$, and define, with nearest-integer rounding,

$$\delta_j=M\left\lfloor\frac{\eta_{j-1}}{MA_{e_j}}\right\rceil, \qquad \eta_j=\eta_{j-1}-\delta_j A_{e_j}. \tag{4}$$

Then $\delta_1=0$, every $\delta_j$ is divisible by $M$, and $|\eta_j|\le MA_{e_j}/2$. For $j\ge2$, (3) and the gap bound give

$$\begin{aligned} |\delta_j| &\le\frac M2\left(\frac{A_{e_{j-1}}}{A_{e_j}}+1\right)\\ &\le M2^{e_j-e_{j-1}}+\frac M2 \le\frac{3M}{2}e_j^{1/4} \le\varepsilon\sqrt{e_j/2} \le\varepsilon\varphi(e_j). \end{aligned} \tag{5}$$

The last elementary bound follows by taking the product over $p^a\Vert n$ in $\varphi(n)^2/n$: the factor $p^{a-2}(p-1)^2$ is at least one for odd $p$, and the factor for $p=2$, when present, is at least $1/2$.

Define $h(e_j)=\varphi(e_j)+\delta_j$ and $h(k)=\varphi(k)$ at all other integers coprime to $P$. For the unique decomposition $n=sk$ above, set

$$f(n)=\varphi(s)h(k). \tag{6}$$

Equation (5), including $\delta_1=0$, proves positivity, integrality and the asserted relative bounds. It also gives $f(n)\equiv\varphi(n)\pmod M$. Every prime power has rough part either $1$ or a prime power, hence outside $E$, so its value is unchanged. Multiplication by a prime $p\le y$ changes only the smooth part $s$. The two required laws for $f(pn)$ therefore follow from the corresponding exact laws for $\varphi(s)$.

The relative bound implies absolute convergence. The orbits with distinct rough parts are disjoint, so absolute convergence permits regrouping:

$$\sum_{n\ge1}(f(n)-\varphi(n))2^{-n} =\sum_{j\ge1}\delta_j A_{e_j} =r-S. \tag{7}$$

The last equality follows by telescoping (4), since (3) implies $\eta_j\to0$. All orbit terms were included in $A_e$; there is no omitted contribution from multiples of the changed integers.

Finally the changed set is contained in $\{se:s\in\mathcal H,e\in E\}$. For $s\le\sqrt X$, (2) bounds its count by $O(X\sum_{s\in\mathcal H}s^{-1}/\log X)$. For $s>\sqrt X$, the crude bound $E(X/s)\le X/s$ gives at most

$$X\sum_{\substack{s\in\mathcal H\\s>\sqrt X}}\frac1s \le X^{3/4}\sum_{s\in\mathcal H}s^{-1/2}=O(X^{3/4}).$$

Both sums over $\mathcal H$ converge, being finite Euler products $\prod_{p\le y}(1-p^{-\alpha})^{-1}$ for $\alpha=1,1/2$. This proves the claimed $O(X/\log X)$ bound and completes the proof. $\square$