Basic reductions

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All residues are normalized as $0\le a_i<n_i$. The delayed excluded progression is $B_i=\{a_i+n_it:t\ge1\}$, including $n_i$ exactly when $a_i=0$. Write $A^{(k)}=\mathbb N\setminus\bigcup_{i\le k}B_i$ and $\delta_k=d(A^{(k)})$. This density exists because membership is periodic modulo $\operatorname{lcm}(n_1,\ldots,n_k)$ once $n\ge n_k$. Thus $\delta_k\downarrow\delta\in[0,1]$.

For a set $E$, write $H_E(X)=\sum_{n\le X,n\in E}1/n$. The lower and upper logarithmic densities are the liminf and limsup of $H_E(X)/\log X$; natural densities use $|E\cap[1,X]|/X$ instead. Always

$$\overline d(A)\le\delta,\qquad \overline{\operatorname{ld}}(A)\le\delta.$$

In particular $\delta=0$ already implies natural density zero. Neither bound supplies the missing lower logarithmic estimate.

Summable reciprocals: proved

For every $X$,

$$|B_i\cap[1,X]|=\max\{0,\lfloor(X-a_i)/n_i\rfloor\}\le X/n_i.$$

Only moduli $n_i\le X$ can contribute. Hence

$$|(A^{(k)}\setminus A)\cap[1,X]| \le X\sum_{i>k}1/n_i.$$

If $\sum_i1/n_i<\infty$, squeezing between the finite truncations proves $d(A)=\delta$, and partial summation gives logarithmic density $\delta$. This justifies the infinite union bound directly; countable subadditivity of upper natural density is not assumed. If the moduli are pairwise coprime, $\delta_k=\prod_{i\le k}(1-1/n_i)$, so either this case applies or $\delta=0$.

These results agree with Theorems 3.1–3.2 in the Chojecki source, whose relevant passages were checked.

Exact first-kill formulation: proved

Put $E_i=B_i\setminus\bigcup_{j<i}B_j$. The $E_i$ are disjoint and exhaust $\mathbb N\setminus A$. Their periodic densities are $e_i=\delta_{i-1}-\delta_i$, with $\delta_0=1$, and $\sum_i e_i=1-\delta$. Every element of $B_i$ is at least $n_i$, so all earlier congruences are already active there. Thus its quotient variable $t\ge1$ must avoid exactly the compatible quotient classes modulo $n_j/\gcd(n_i,n_j)$. No activation error remains in this quotient description.

The unresolved estimate is

$$\liminf_{X\to\infty}H_A(X)/\log X\ge\delta.$$

An equivalent formulation of this candidate-value conclusion is that $\lim_{k\to\infty}\limsup_{X\to\infty} H_{A^{(k)}\setminus A}(X)/\log X=0$. The ambient tail $\bigcup_{i>k}B_i$ need not have small logarithmic density (example: zero classes for all primes). Any general tail estimate must retain conditioning on the earlier survivors.

The conditioned excess does not depend on the prefix: proved

For every $k\ge0$, with $A^{(0)}=\mathbb N$ and $\delta_0=1$,

$$\limsup_{X\to\infty}\frac{H_{A^{(k)}\setminus A}(X)}{\log X}-(\delta_k-\delta) =\delta-\liminf_{X\to\infty}\frac{H_A(X)}{\log X}=:\Delta . \tag{1}$$

Indeed $A\subseteq A^{(k)}$ and $H_{A^{(k)}}(X)=\delta_k\log X+O_k(1)$, by eventual periodicity, so the limsup on the left equals $\delta_k-\liminf_X H_A(X)/\log X$. The conditioned tail therefore exceeds its Haar mass $\delta_k-\delta$ by the same amount $\Delta\ge0$ for every prefix, including the empty prefix, where it is the excess of the whole excluded union over $1-\delta$. Thus the numerical excess is independent of the prefix; proving that it vanishes is still the unresolved estimate. The identity does not replace a conditioned estimate by an ambient tail bound.

Consequently the unresolved estimate holds if and only if, for some constant $C$ that may depend on the family, and for infinitely many $k$,

$$\limsup_{X\to\infty}\frac{H_{A^{(k)}\setminus A}(X)}{\log X} \le C(\delta_k-\delta).$$

Necessity holds with $C=1$. For sufficiency, (1) gives $\Delta\le(C-1)(\delta_k-\delta)$ along those $k$, and $\delta_k\to\delta$. A bound $f(\delta_k-\delta)$ with $f(0)=0$ and $f(t)\to0$ as $t\downarrow0$ works equally well. More generally, a bound $C_k(\delta_k-\delta)+\eta_k$ suffices along any unbounded sequence of prefixes on which both $C_k(\delta_k-\delta)\to0$ and $\eta_k\to0$. The factors $C_k$ need not themselves stay bounded; their uncontrolled growth is the problem.

The zero class at every prime is the adversarial case for such factors. There $A=\{1\}$ and $\delta=0$, so $\Delta=0$ and the limsup in (1) equals $\delta_k$, while the relative tail $H_{A^{(k)}\setminus A}(X)/H_{A^{(k)}}(X)$ tends to one for every $k$. Hence a bound on the conditional probability of a later kill, given survival of the prefix, that tends to zero with the prefix is false in general. For zero classes the association argument for (2) applies to the increasing event of a later kill and the decreasing event of prefix survival; it bounds the conditioned tail by the density of the ambient tail times $H_{A^{(k)}}(X)$, and that density is one here. The fiber decomposition of the unresolved-estimates note loses a multiplicity factor up to $\tau(L_k)$ for the prefix period $L_k$; its harmonic change of variables costs only $O(\log L_k)$, which disappears in the limsup. For zero classes at the first $k$ primes, $\tau(L_k)=2^k$ and

$$2^k(\delta_k-\delta) =\prod_{i\le k}2(1-1/p_i)\ge(4/3)^{k-1}.$$

Thus the universal upper envelope $\tau(L_k)$ cannot simply be absorbed into the remaining mass. This is an obstruction to that bound, not an assertion that the prime family has actual quotient-modulus multiplicity $2^k$: for a new prime the actual multiplicity is one.

The same defect occurs with $\delta>0$, the unresolved case. Choose one prime $p_j$ in each interval $[2^j,2^{j+1})$, and again take zero residues. The reciprocals are summable and $\delta=\prod_j(1-1/p_j)>0$. With $\varepsilon_k=\delta_k-\delta$,

$$\varepsilon_k\ge\delta_k/p_{k+1}\ge\delta/p_{k+1},\qquad 2^k\varepsilon_k>\delta/4.$$

The first inequality is just the mass removed by the next prime, still conditioned on prefix survival. So even a hypothetical estimate with loss $\tau(L_k)$ does not force a vanishing bound on this already summable family. A successful general fiber estimate needs a controlled product $C_k\varepsilon_k$, for example a bounded factor, or a way to exploit actual multiplicity. Changing the activation shift alone does not provide it.

An exact finite harmonic inequality for multiples: proved

For a finite set $D$ of positive integers and $M(D)=\bigcup_{q\in D}q\mathbb N$,

$$H_{M(D)}(X)\le d(M(D))H_{\lfloor X\rfloor}. \tag{2}$$

It suffices to retain $q\le X$. Take independent geometric valuations $\Pr(V_p=e)=(1-1/p)p^{-e}$ for $p\le X$, and put $N=\prod_{p\le X}p^{V_p}$. The event $N\in M(D)$ is increasing in all valuations, whereas $N\le X$ is decreasing. The association inequality for product measures gives

$$\Pr(N\in M(D),N\le X)\le d(M(D))\Pr(N\le X).$$

The law of $N$, conditioned on $N\le X$, gives integer $n\le X$ mass $1/(nH_{\lfloor X\rfloor})$, proving (2).

For completeness, product association follows by induction on the number of independent coordinates. In one coordinate, $2\operatorname{Cov}(f,g)=\mathbb E[(f(V)-f(V'))(g(V)-g(V'))]\ge0$ for increasing $f,g$, where $V'$ is an independent copy. Condition on the last coordinate, apply induction to the conditional covariance, and then apply the one-coordinate inequality to the increasing conditional means. Negating the decreasing indicator gives the displayed reversed inequality. The same argument applies to the finite number of unbounded geometric coordinates, or by truncation and bounded convergence.

Inequality (2) also gives a direct proof of the logarithmic-density part of the Davenport–Erdős theorem. For arbitrary $D$, write $\beta=\lim_{Y\to\infty}d(M(D\cap[1,Y]))$. Apply (2) to $D\cap[1,X]$ to get $H_{M(D)}(X)\le\beta H_{\lfloor X\rfloor}$; each fixed finite union supplies the matching lower limit. Hence $\operatorname{ld}(M(D))=\beta$. This argument concerns multiples, not arbitrary translated congruences: the latter event is not increasing in the valuation coordinates.

A quantitative finite Dirichlet comparison for multiples: proved

The auxiliary estimate in Section 7 of the independent input is valid. For nonempty finite $D$, put $M=\max D$ and $\beta=d(M(D))$. For $s>1$,

$$M^{1-s}\beta \le\frac1{\zeta(s)}\sum_{n\in M(D)}n^{-s} \le\beta. \tag{3}$$

In particular its convergence to $\beta$ is uniform when $(s-1)\log M\to0$. The lower bound is an equality for singleton $D$.

Here is a self-contained proof. Under the probability law assigning integer $n$ mass $n^{-s}/\zeta(s)$, the prime valuations are independent and satisfy $\Pr(V_p\ge e)=p^{-se}$. For $s>1$ they are stochastically dominated by the independent Haar valuations, whose tails are $p^{-e}$. Thus every finite multiples union has Dirichlet probability at most its natural density, and its complement has probability at least its density. Only the finitely many primes dividing the generators matter for these comparisons. This proves the upper bound in (3).

Order the generators $m_i$ and partition $M(D)$ by its first dividing generator. Its $i$th piece is $m_iS_i$, where

$$S_i=\{t\ge1:m_j/(m_i,m_j)\nmid t\text{ for every }j<i\}.$$

If the piece has density $e_i$, then $d(S_i)=m_ie_i$. Applying the complement comparison to $S_i$ gives

$$\frac1{\zeta(s)}\sum_{n\in m_iS_i}n^{-s} \ge m_i^{-s}d(S_i) =e_im_i^{1-s}\ge e_iM^{1-s}.$$

Sum over $i$ to obtain the lower bound. This proof does not extend to moving residues, since the quotient sets are then not complements of multiples unions.

Why the finite-periodic candidate is the right universal target

The statement that every delayed singleton family has logarithmic density is equivalent to the stronger-looking statement that every such density is the finite-periodic value $\delta$. This is a subfamily extraction principle, not a proof that either statement holds.

Proof. Write $U=\mathbb N\setminus A$, let $\beta=1-\delta$, and suppose $\lambda=\overline{\operatorname{ld}}(U)>\beta$ for some family. Put $\varepsilon=(\lambda-\beta)/10$. Choose a finite original prefix $F_0$ whose excluded periodic density is at least $\beta-\varepsilon$. All subsequently selected families will contain $F_0$ and be subfamilies of the original.

Suppose the current selected family $F$ is finite. Choose a recovery cutoff $X$ beyond the previous cutoffs and all its moduli, large enough that $H_{U_F}(X)/\log X\le\beta+\varepsilon$. Let $m$ be the first original index with $n_m>X$. The selected family is contained in the original prefix with indices below $m$. There exist arbitrarily large $Y>X$ such that

$$\frac{H_U(Y)}{\log Y}\ge\beta+6\varepsilon, \quad \frac{H_{U^{(m-1)}}(Y)}{\log Y}\le\beta+\varepsilon, \quad \frac{H_{U_F}(Y)}{\log Y}\ge\beta-2\varepsilon.$$

The first inequality follows from the limsup, and the latter two hold at every sufficiently large $Y$ by finite periodicity. Thus the original tail removes harmonic mass at least $5\varepsilon\log Y$ outside the entire prefix $U^{(m-1)}$, hence outside $U_F$.

Add all original congruences with $X<n_i\le Y$. They capture that entire tail contribution up to $Y$, and the selected union at $Y$ has normalized harmonic mass at least $\beta+3\varepsilon$. No newly added class is active at or below $X$. Repeat, choosing the next recovery cutoff above $Y$.

The final subfamily still uses strictly increasing, unrepeated moduli. Activation protects both selected cutoff subsequences from every future addition. Its excluded union therefore has lower logarithmic density at most $\beta+\varepsilon$ and upper logarithmic density at least $\beta+3\varepsilon$, so neither it nor its survivor set has logarithmic density. This proves the extraction claim. Since every original $U$ has lower logarithmic density at least $\beta$ by finite truncation, the stated universal equivalence follows. $\square$

A different, structure-preserving version follows from the translation-dynamical dichotomy: any excess over $\beta$ produces a common profinite translate with logarithmic oscillation, keeping the entire modulus set and all cylinder compatibility relations unchanged. This does not establish the needed upper bound.