Squarefull kernel truncation

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Theorem: proved

For a positive integer $q$, write

$$\lambda(q)=\#\{p:v_p(q)=1\}.$$

Consider a family of distinct moduli $q$, with one arbitrary normalized residue $a_q$ at each modulus, and let

$$C_q=[a_q]_q,\qquad U=\bigcup_q C_q\subseteq\widehat{\mathbb Z},\qquad B=\{n\ge1:\exists q\le n,\ n\equiv a_q\pmod q\}.$$

If $\lambda(q)\le r$ throughout the family, for some fixed integer $r\ge0$, then $B$ has finite-periodic approximation in upper natural density. Its natural density is $\mu(U)$, and the survivor set has natural and logarithmic density $1-\mu(U)$.

In particular, arbitrary singleton residues with bounded $\omega(q)$ have natural density, without any restriction on exponents or coherence. The theorem is stronger: the number of distinct prime factors occurring to exponent at least two need not be bounded.

The proof uses the finite counting inequality from bounded prime rank, which permits multiple residues at a modulus:

$$\Omega(d_i)\le r \quad\Longrightarrow\quad \left|\bigcup_i\{k\ge d_i:k\equiv c_i\pmod{d_i}\} \cap[1,T]\right| \le 2^r T\, \mu\left(\bigcup_i[c_i]_{d_i}\right). \tag{1}$$

This is a finite inequality, uniform in the moduli and residues. It does not require the already established density theorem for bounded $\Omega$.

Two elementary estimates

Call $m$ powerful, or squarefull, if every prime divisor of $m$ occurs to exponent at least two; include $1$. Let $\mathcal P$ denote the set of powerful integers. Every $m\in\mathcal P$ has a unique representation $m=a^2b^3$ with $b$ squarefree. Therefore

$$\#\{m\in\mathcal P:m\le T\} \le \sqrt T\sum_{b\ge1}b^{-3/2}\ll\sqrt T. \tag{2}$$

Dyadic summation gives

$$S:=\sum_{m\in\mathcal P}\frac1m<\infty,\qquad \sum_{\substack{m\in\mathcal P\\m>M}}\frac1m\ll M^{-1/2}. \tag{3}$$

If

$$L_0(M)=\operatorname{lcm}\{m\le M:m\in\mathcal P\},$$

then the product bound for an lcm and (2) give

$$\log L_0(M) \le\sum_{\substack{m\in\mathcal P\\m\le M}}\log m \ll\sqrt M\log(2M). \tag{4}$$

Write $\mathcal T_r$ for the squarefree integers with at most $r$ prime factors, including $1$. The elementary prime reciprocal bound $\sum_{p\le T}1/p\ll\log\log(3T)$ implies

$$R_r(T):=\sum_{\substack{t\le T\\t\in\mathcal T_r}}\frac1t \ll_r\bigl(\log\log(3T)\bigr)^r. \tag{5}$$

Indeed, for each number of factors, enlarge the sum to ordered tuples of primes at most $T$.

We will also use the following consequence, proved here to avoid invoking an almost-prime asymptotic. For $r\ge1$, put $A_r(T)=\#(\mathcal T_r\cap[1,T])$. Chebyshev's elementary estimate $\vartheta(y)=\sum_{p\le y}\log p\ll y$ gives

$$\begin{aligned} \sum_{\substack{t\le T\\t\in\mathcal T_r}}\log t &=\sum_{\substack{t\le T\\t\in\mathcal T_r}}\sum_{p\mid t}\log p\\ &\le\sum_{\substack{m\le T\\m\in\mathcal T_{r-1}}} \vartheta(T/m) \ll T R_{r-1}(T). \end{aligned}$$

The terms $t>\sqrt T$ each have $\log t>\frac12\log T$; there are at most $\sqrt T$ smaller integers. Thus, for large $T$,

$$A_r(T)\ll_r \frac{T}{\log T}\bigl(\log\log(3T)\bigr)^{r-1}. \tag{6}$$

For $X^{1/2}\le Y<X$, partial summation consequently yields

$$\sum_{\substack{Y<t\le X\\t\in\mathcal T_r}}\frac1t \ll_r \bigl(\log\log(3X)\bigr)^{r-1} \left(\frac1{\log X}+ \frac{\log(X/Y)}{\log Y}\right). \tag{7}$$

Here the negative lower endpoint in partial summation was discarded. For $r=0$, the left side is zero as soon as $Y\ge1$.

A conditioned asymptotic counting bound

We prove the following stronger intermediate assertion. Let $W$ be any periodic set, viewed both as a union of integer residue classes and as a clopen subset of $\widehat{\mathbb Z}$. Then

$$\overline d(B\cap W)\le 2^r\mu(U\cap W). \tag{8}$$

The conditioning period of $W$ is fixed when the natural-density limit is taken.

Let $K$ be such a period. Every modulus in the family has a unique factorization

$$q=mt,\qquad m\in\mathcal P,\quad t\in\mathcal T_r,\quad (m,t)=1: \tag{9}$$

put in $m$ precisely the prime powers with exponent at least two, and in $t$ the primes with exponent exactly one.

For a sufficiently large counting cutoff $X$, choose

$$M=\lfloor\log X\rfloor,\qquad L=\operatorname{lcm}(K,L_0(M)),\qquad \eta=\frac1{\log X}. \tag{10}$$

By (4),

$$\log L\le\log K+O(\sqrt{\log X}\log\log X)=o_K(\log X). \tag{11}$$

Although $L$ grows with $X$, all ensuing finite bounds are uniform in $L$.

Large powerful kernels

Since a single delayed class modulo $q$ has at most $X/q$ points below $X$, the union coming from $m>M$ has at most

$$\begin{aligned} X\sum_{\substack{m\in\mathcal P\\m>M}}\frac1m \sum_{\substack{t\le X\\t\in\mathcal T_r}}\frac1t &\ll_r X\frac{(\log\log(3X))^r}{\sqrt{\log X}} =o_r(X) \end{aligned} \tag{12}$$

points. We enlarged the set of coprime pairs in (9), which only increases the positive sum. Distinctness of the original moduli is used here.

Refining the small kernels

For $m\le M$, we have $m\mid L$. As $t$ is squarefree and coprime to $m$, set

$$u=\frac{t}{(t,L)},\qquad D=\operatorname{lcm}(L,q)=Lu. \tag{13}$$

In particular, $\Omega(u)\le r$, $q\mid D$, and $D\le Lt$.

Refine $C_q$ into residue classes modulo $D$, but activate these refined classes only at $n\ge D$. Call the union of the resulting delayed sets $B_X^*$; this construction uses only the classes with small powerful kernel $m\le M$. It is a subset of $B$.

On a fiber $n=b+Lk$, $0\le b<L$, a nonempty refinement is one residue class for $k$ modulo $u$. The activation condition is exactly

$$b+Lk\ge Lu\quad\Longleftrightarrow\quad k\ge u. \tag{14}$$

The quotient family on that fiber can have several residues at a modulus, which is permitted in (1).

Because $K\mid L$, each fiber lies wholly in $W$ or wholly outside it. Let $\beta_b$ be the conditional Haar measure, on the fiber $[b]_L$, of the full union of all the small-kernel classes. Apply (1) to the finitely many refined classes active below $X$. Their Haar union is no larger than $\beta_b$; the quotient cutoff is at most $X/L$. Summing over the fibers belonging to $W$ gives

$$\begin{aligned} |B_X^*\cap W\cap[1,X]| &\le 2^r\frac XL \sum_{\substack{0\le b<L\\ {}[b]_L\subseteq W}}\beta_b\\ &=2^r X\,\mu(U_{\le M}\cap W) \le2^r X\,\mu(U\cap W), \end{aligned} \tag{15}$$

where $U_{\le M}$ is the full small-kernel union. Classes with $u=1$, including those with $q\mid L$, present no exception: their quotient activation is $k\ge1$.

The extra activation delay costs $o(X)$

Any point counted by a small-kernel class at the original activation but not at the stronger one satisfies, for some such class,

$$q\le n<D\le Lt. \tag{16}$$

There are at most $\eta X$ positive integers with $n\le\eta X$. For every other point in (16),

$$t>Y:=\frac{\eta X}{L}. \tag{17}$$

Since $q\le n\le X$, also $t\le X$. Another application of the individual delayed progression bound therefore gives at most

$$\eta X+ X\left(\sum_{m\in\mathcal P}\frac1m\right) \sum_{\substack{Y<t\le X\\t\in\mathcal T_r}}\frac1t \tag{18}$$

extra points. This is the second place where singleton distinctness is used.

By (10)--(11),

$$\log(X/Y)=\log L+\log\log X =O_K(\sqrt{\log X}\log\log X), \qquad \log Y\sim\log X. \tag{19}$$

In particular $X^{1/2}\le Y<X$ eventually. Equations (3), (7), and (19) show that (18) is $o_{r,K}(X)$. Combining (12), (15), and (18) proves (8).

The proof in fact supplies a family-independent error for each fixed $r$ and $K$; no rate of convergence for the full Haar union is assumed.

Finite-periodic approximation

Let $F$ be a finite subfamily, and write $U_F$ for its full periodic union and $B_F$ for its delayed union. They differ on only finitely many positive integers. Thus

$$B\setminus B_F \quad\hbox{and}\quad B\cap(\widehat{\mathbb Z}\setminus U_F)\cap\mathbb N$$

differ on only finitely many integers. Apply (8) with $W=\widehat{\mathbb Z}\setminus U_F$:

$$\overline d(B\setminus B_F) \le 2^r\mu(U\setminus U_F). \tag{20}$$

As finite subfamilies exhaust the family, the right side tends to zero by continuity of Haar measure from below. This is the claimed finite-periodic approximation. Since $d(B_F)=\mu(U_F)$, it also proves $d(B)=\mu(U)$. Complementation and partial summation yield the corresponding natural and logarithmic survivor densities.

The case $r=0$ is included; it also follows directly from (3), because all original moduli are powerful.

Scope and the earlier product obstruction

The argument establishes arbitrary bounded-$\omega$ density, but does not settle unrestricted Problem 25: $\lambda(q)$ can be unbounded. Its two elementary tail estimates rely on fixed $r$; the finite bound (1) also has the factor $2^r$.

The three-prime product-capacity obstruction does not contradict this result. Its abstract masses satisfy individual divisor-face capacities, but the present proof uses singleton arithmetic union bounds, a cutoff-dependent lcm refinement, and an asymptotically negligible activation discrepancy. It does not deduce a uniform finite bound solely from those capacities.

The same proof permits a uniformly bounded number of residues per original modulus: only (12) and (18) change, by that fixed multiplicity factor, and their errors still tend to zero. No unbounded-multiplicity version is claimed.

A growing-rank cutoff and its limitation

For every fixed real $z>1$,

$$\sum_{q\le X}\frac{z^{\lambda(q)}}q \le\prod_{p\le X}\left(1+\frac zp+\frac1{p(p-1)}\right) \ll_z(\log(3X))^z. \tag{21}$$

Each local factor is $(1-1/p)^{-z}(1+O_z(p^{-2}))$, so the previously used Euler-product bound proves the last estimate. Unlike the analogous generating function for $\Omega(q)$, this one has no restriction $z<2$: exponents at least two contribute the unweighted geometric tail $1/(p(p-1))$.

Write $B_{>r}$ for the delayed union using moduli with $\lambda(q)>r$. Singleton distinctness and the exact individual counting and harmonic bounds give, uniformly in the residues,

$$\frac{|B_{>r}\cap[1,X]|}{X},\quad \frac{H_{B_{>r}}(X)}{H_{\lfloor X\rfloor}} \ll_z z^{-r}(\log(3X))^z. \tag{22}$$

Taking $z=e$ and $r=\lceil(e+\varepsilon)\log\log X\rceil$ makes both omitted proportions $O_\varepsilon((\log X)^{-\varepsilon})$. This does not extend the fixed-rank density theorem: the retained rank grows, and its comparison constant $2^r$ is unbounded.

There is also a precise obstruction to using a fixed rank and only reciprocal counting. The powerful-kernel decomposition (9) gives

$$\sum_{\substack{q\le X\\\lambda(q)\le r}}\frac1q \ll_r(\log\log(3X))^r=o(\log X).$$

Thus high-rank moduli carry asymptotically all unrestricted reciprocal mass. They also carry its leading raw harmonic-progression budget:

$$\sum_{\substack{q\le X\\\lambda(q)>r}} \frac{H_{\lfloor X/q\rfloor}}q =\left(\frac12+o(1)\right)(\log X)^2 \qquad(r\text{ fixed}). \tag{23}$$

The unrestricted double sum is $\frac12(\log X)^2+O(\log X)$, by summing $q^{-1}(\log(X/q)+O(1))$; the low-rank part is $O_r(\log X(\log\log(3X))^r)$. A footprint- or first-kill-sensitive comparison remains necessary.