Geometric boundary

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Write $G=\widehat{\mathbb Z}$, $C_q=[a_q]_q$, $U=\bigcup_q C_q$, and $K=G\setminus U$. The Haar measures are $\beta=\mu(U)$ and $\delta=\mu(K)$. The actual excluded set is

$$B=\{n\ge1:\exists q\le n,\ n\equiv a_q\pmod q\}.$$

The boundary model below does not alone settle the problem. It does give a sufficient condition for natural density, uniform under common profinite affine transformations of the residues.

Location of a possible logarithmic defect

The ambient measures

$$\lambda_X=\frac1{H_{\lfloor X\rfloor}} \sum_{n\le X}\frac1n \delta_{(\log n/\log X,\,n)}$$

on $[0,1]\times G$ converge weakly to $dt\,d\mu$. For a fixed residue class modulo $L$ and $0\le s<t\le1$, harmonic summation between $X^s$ and $X^t$ gives the required mass $(t-s)/L+o(1)$. Finite unions of such product sets, or approximation of continuous functions by them, prove the assertion.

Replace the sum by its restriction to $n\in B$, obtaining $\lambda_X^B$. Every subsequential weak limit $\lambda^B$ has the form

$$d\lambda^B =\mathbf1_U(z)\,dt\,d\mu(z) +\rho(t,z)\mathbf1_{\partial U}(z)\,dt\,d\mu(z), \qquad0\le\rho\le1. \tag{1}$$

Indeed $0\le\lambda_X^B\le\lambda_X$ implies domination of the limit by $dt\,d\mu$. Every compact subset of $U$ is covered by a fixed finite set of cylinders. Away from $t=0$, these are all active for large $X$, so the limit agrees with the ambient measure on $[0,1]\times U$. No deleted integer lies in $G\setminus\overline U$; open-set approximation makes the limit zero there. The slice $t=0$ has zero ambient measure. These observations prove (1).

In particular, a positive discrepancy from $\beta$ requires $\mu(\partial U)>0$. If $\mu(\partial U)=0$, the analogous ordinary counting argument already gives natural density. Formula (1) locates the possible defect; it does not prove that $\rho=0$.

A local reciprocal criterion

Define the closed set

$$D=\left\{x\in G:\ \text{for every }L\ge1, \sum_{\substack{q\,: a_q\equiv x\pmod{(q,L)}}}\frac1q=\infty\right\}. \tag{2}$$

The compatibility condition in the sum is exactly $C_q\cap[x]_L\ne\varnothing$. The complement of $D$ is open: a clopen fiber witnessing convergence does so for all its points. On any such fiber, only a reciprocal-summable collection of classes can contribute. The ordinary activated union bound therefore gives finite-periodic approximation on that fiber. In (1), the defect is consequently supported on $[0,1]\times(K\cap D)$ as well.

Theorem. If $\mu(K\cap D)=0$, the delayed survivor set has natural density $\delta$. Convergence is uniform when all residues are changed simultaneously by

$$a_q\longmapsto ua_q+z\pmod q, \qquad u\in\widehat{\mathbb Z}^{\times},\quad z\in\widehat{\mathbb Z}. \tag{3}$$

Proof. Given $\varepsilon>0$, choose a clopen $O\supset K\cap D$ with $\mu(O)<\varepsilon$. The compact set $K\setminus O$ has a finite cover by clopen fibers whose intersecting cylinder families have convergent reciprocal sums. Their union $V$ satisfies

$$\sum_{q:C_q\cap V\ne\varnothing}\frac1q<\infty.$$

The compact set $W=G\setminus(O\cup V)$ is contained in $U$, so a finite prefix, with moduli at most $Q_0$, covers $W$. Choose $Q\ge Q_0$ such that

$$\sum_{\substack{q>Q\,: C_q\cap V\ne\varnothing}}\frac1q <\varepsilon,\qquad \beta-\mu(U_{\le Q})<\varepsilon.$$

Every actual deletion outside the prefix has a witnessing modulus $q>Q$. It cannot lie in $W$, since $n\ge q>Q_0$ makes the old covering class there active. If $P$ is a period of $O$, then

$$|(B\setminus B_{\le Q})\cap[1,N]| \le \mu(O)N+P+ N\sum_{\substack{q>Q\,: C_q\cap V\ne\varnothing}}\frac1q \le2\varepsilon N+P. \tag{4}$$

The bound $N/q$ per class is exact for delayed classes; there is no accumulated inactive-root error. Finite-prefix counting and (4) prove natural density $\beta$ for $B$.

Under (3), transport $O,V,W$ by $x\mapsto ux+z$. Their measures, periods, cylinder intersections, and reciprocal sums are unchanged. The finite-prefix counting error is uniform over these transformations, as is (4). Thus the conclusion is uniform in $(u,z)$. $\square$

A noncoherent example with unbounded prime rank

Let $R_j$ be the product of the first $j$ odd primes and set $M_j=2^{2j}R_j$. Choose $c_j\pmod{M_j}$ with

$$c_j\equiv2^{2j-1}\pmod{2^{2j}},\qquad c_j\equiv0\pmod{R_j}.$$

The cylinders $W_j=[c_j]_{M_j}$ are pairwise disjoint. For every $j$ and every prime $p\nmid M_j$, use modulus $M_jp$, with residue $c_j$ modulo $M_j$ and any chosen residue modulo $p$. All moduli are distinct: their $2$-adic valuations distinguish different $j$, and within a fixed $j$ the prime $p$ distinguishes them. They can therefore be enumerated strictly increasingly.

For a fixed $j$, independent prime coordinates and $\sum_{p\nmid M_j}1/p=\infty$ show that these classes cover $W_j$ almost everywhere. Hence

$$\beta=\sum_{j\ge1}\frac1{M_j}<\frac13.$$

Moreover

$$D\subseteq\bigcup_j W_j\ \cup\ \{x:x_2=0\}.$$

To see this, a point of finite $2$-adic valuation can have a neighborhood meeting only its matching $W_j$, if there is one. If it is outside that $W_j$, refine the neighborhood to avoid it too. The last displayed $2$-adic zero set is Haar-null, and $K\cap W_j$ is null for each $j$. Thus the theorem gives natural density $1-\sum_j1/M_j$, uniformly under (3). The reciprocal sum diverges, the survivor density is positive, and both modulus prime rank and pairwise-gcd prime rank are unbounded. The residues need not be coherent.

The geometric argument permits multi-residue families as well, with reciprocal sums counting congruence pairs with multiplicity. The next example shows that its local reciprocal hypothesis is not automatic, even for primitive zero-residue singleton families whose natural density exists.

Positive natural density with $D=G$: proved obstruction

For every $0<\varepsilon<1$, there is a primitive family of distinct zero-residue moduli such that

$$\mu(K)>1-\varepsilon,\qquad D=G,\qquad\partial U=K, \tag{5}$$

and the original delayed survivor set nevertheless has natural density $\mu(K)$. Thus positive survivor measure, primitivity, and even actual natural-density existence do not force local reciprocal summability on any positive-measure part of the survivor set.

Construction and Haar measure

Choose sufficiently large integers $k_j\to\infty$ such that

$$\sum_j\frac1{k_j}<\infty, \qquad \sum_j\exp\bigl(-(2\log2-1)k_j+k_j^{-3}\bigr)<\varepsilon. \tag{6}$$

For example, a sufficiently large multiple of $2^j$ works. Choose pairwise disjoint finite prime blocks $P_j$, all primes in $P_j$ larger than $k_j^3$, with

$$k_j\le\lambda_j:=\sum_{p\in P_j}\frac1p <k_j+k_j^{-3}. \tag{7}$$

They can be chosen successively, beyond all primes already used, by stopping when the prime reciprocal sum first reaches $k_j$.

Let $\mathcal Q_j$ consist of all products of $2k_j$ distinct primes from $P_j$, and install the zero class at every $q\in\bigcup_j\mathcal Q_j$. All moduli are distinct and primitive: within a block they are squarefree of the same rank, and different blocks use disjoint prime sets. They therefore admit the required strictly increasing enumeration.

Let $U_j=\bigcup_{q\in\mathcal Q_j}qG$ and $\beta_j=\mu(U_j)$. For Haar-random $z$, let $Y_j(z)$ count the primes of $P_j$ dividing $z$. Then $U_j=\{Y_j\ge2k_j\}$, and independent prime coordinates give

$$\beta_j \le 2^{-2k_j}\mathbb E2^{Y_j} =2^{-2k_j}\prod_{p\in P_j}(1+1/p) \le\exp\bigl(-(2\log2-1)k_j+k_j^{-3}\bigr). \tag{8}$$

Consequently $\mu(U)\le\sum_j\beta_j<\varepsilon$.

Reciprocal divergence in every neighborhood

The reciprocal sum of one block is

$$\sum_{q\in\mathcal Q_j}\frac1q =e_{2k_j}\bigl((1/p)_{p\in P_j}\bigr) =\bigl(1+O(k_j^{-2})\bigr) \frac{\lambda_j^{2k_j}}{(2k_j)!}. \tag{9}$$

To check the approximation, draw $2k_j$ primes independently with probabilities $(1/p)/\lambda_j$. The probability of no collision is $(2k_j)!e_{2k_j}/\lambda_j^{2k_j}$. Each draw has maximal point mass at most $k_j^{-4}$, so the union bound over pairs gives collision probability $O(k_j^{-2})$. Stirling's formula shows that the right side of (9) tends to infinity. Increase the initial choice of the $k_j$ if necessary so that every block reciprocal sum is at least one.

Fix a modulus $L$. All sufficiently late blocks consist entirely of primes coprime to $L$. For every $q$ in these blocks, the cylinder $qG$ meets every residue fiber modulo $L$, by CRT. Summing (9) over these blocks proves divergence in every fiber. This is exactly $D=G$. The same observation shows that $U$ is dense in $G$. Since $U$ is open, $\partial U=G\setminus U=K$, establishing (5).

Uniform counting bound for one block

Fix one block, abbreviating $k=k_j$, $P=P_j$, and $\lambda=\lambda_j$. For an integer $X\ge1$, let

$$Z_X(n)=\sum_{\substack{p\in P\\p\le\sqrt X}}\mathbf1_{p\mid n}, \qquad \lambda_X=\sum_{\substack{p\in P\\p\le\sqrt X}}\frac1p,$$

and let $s$ be the number of primes in this sum. Average over $1\le n\le X$. Exact floor counts imply

$$\lambda_X-s/X\le\mathbb E Z_X\le\lambda_X, \qquad \mathbb E Z_X^2\le\lambda_X+\lambda_X^2.$$

Therefore

$$\mathbb E(Z_X-\lambda_X)^2 \le\lambda_X+2\lambda_Xs/X \le\lambda+1, \tag{10}$$

using $\lambda_X\le s/2$ and $s\le\sqrt X$.

If a positive integer $n\le X$ is deleted by a modulus in this block, that modulus divides $n$ and hence is at most $X$. At most one of its $2k$ prime factors can exceed $\sqrt X$. Thus $Z_X(n)\ge2k-1$. By (7), for sufficiently large $k$, $2k-1-\lambda_X\ge k/2$. Chebyshev's inequality and (10) give the uniform bound

$$\frac1X\#\{n\le X:\exists q\in\mathcal Q_j,\ q\mid n\} \le\frac{C}{k_j}, \tag{11}$$

with an absolute constant $C$.

Natural-density conclusion and activation

For zero residues, a positive full hit $q\mid n$ automatically has $n\ge q$, so original activation introduces no exception at all. By (11), the union of blocks after $J$ has at most

$$CX\sum_{j>J}\frac1{k_j}$$

hits below $X$, uniformly in $X$. For each $X$ only finitely many distinct moduli can be active, so this countable union bound follows directly from the finite one. Its normalized upper bound tends to zero as $J\to\infty$ by (6).

Every finite union of blocks is periodic, and its Haar measure tends to $\mu(U)$ as the blocks exhaust the family. The last tail estimate therefore proves natural density $\mu(U)$ for the excluded set and $\mu(K)>1-\varepsilon$ for its complement. This construction is an obstruction to making the local reciprocal criterion automatic, not a counterexample to the target density statement.