Probabilistic smoothing

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This approach randomizes the sieve rather than estimating its original first-kill cells. Smoothing gives a strong convergence theorem, but the passage back to the unsmoothed union remains unproved. Two explicit obstructions explain why basic thinning and random reordering do not remove the difficulty.

Summable clocks: proved convergence

For the full cylinders $C_q=[a_q]_q$, choose positive rates such that

$$S=\sum_q\lambda_q/q<\infty.$$

Distinctness permits the universal choice $\lambda_q=(1+\log q)^{-2}$. For a common translation $z\in G= \widehat{\mathbb Z}$, define

$$W_n(z)=\sum_{q\le n}\lambda_q\mathbf1_{C_q}(n-z),\qquad W(\omega)=\sum_q\lambda_q\mathbf1_{C_q}(\omega).$$

The Haar random variable $W$ is integrable, with mean $S$.

Theorem. The law of $W_n(z)$ under harmonic sampling $\Pr(n)=1/(nH_N)$, $1\le n\le N$, converges to the Haar law of $W$ in $1$-Wasserstein distance, uniformly in $z$. The same assertion holds for uniform sampling on $[1,N]$.

Proof. Truncate both potentials to $q\le K$. On the integer side, the harmonic $L^1$ error is at most

$$\frac1{H_N}\sum_{q>K}\lambda_q H_{B_{q,z}}(N) \le\sum_{q>K}\lambda_q/q,$$

because $H_{B_{q,z}}(N)\le H_{\lfloor N/q\rfloor}/q$. The Haar $L^1$ error is exactly the same tail sum. These bounds are uniform in $z$. The truncated integer potential is eventually periodic. Its distribution converges to its Haar distribution uniformly over the finitely many possible translations modulo the fixed common period. The triangle inequality for $1$-Wasserstein distance proves the result. Equivalently, apply the same estimates to every $1$-Lipschitz test function. The uniform-sampling version uses $|B_{q,z}\cap[1,N]|\le N/q$. $\square$

In particular, for every fixed $t\ge0$,

$$\frac1{H_N}\sum_{n\le N}\frac{1-e^{-tW_n(z)}}n \longrightarrow\int_G(1-e^{-tW})\,d\mu, \tag{1}$$

uniformly in the common translation.

This has an exact clock interpretation: independently retain class $q$ when an exponential clock of rate $\lambda_q$ rings before $t$. An integer with active potential $W_n(z)$ is deleted with probability $1-e^{-tW_n(z)}$. For fixed $t$, the expected reciprocal sum of retained moduli is at most $tS$, hence that sum is finite almost surely. Every fixed-time random sieve therefore has natural density by the summable case.

The unresolved limit is an atom, not a convergence theorem for the clocks

Let $\beta=\mu(\bigcup_qC_q)$. Positivity of all rates gives

$$\mu(W=0)=1-\beta,\qquad \mathbf1_{B_z}(n)=\mathbf1_{\{W_n(z)>0\}}.$$

Thus taking $t\to\infty$ in (1) asks for convergence at the possible atom $0$ of the limiting distribution. Wasserstein convergence does not control this: the constants $1/N$ converge to $0$ in that distance, although their probabilities of being positive are all one.

For fixed $t$ the exact omitted mass is

$$\frac1{H_N}\sum_{\substack{n\le N\\W_n(z)>0}} \frac{e^{-tW_n(z)}}n.$$

The required estimate, for each fixed $z$, is

$$\lim_{t\to\infty}\limsup_{N\to\infty} \frac1{H_N}\sum_{\substack{n\le N\\W_n(z)>0}} \frac{e^{-tW_n(z)}}n=0.$$

It has not been established. This is the remaining small-positive-potential problem, not a consequence of dominated convergence.

The proof even permits repeated moduli if summable rates are assigned to the individual congruence pairs. It cannot, by itself, distinguish Problem 25 from the multi-residue setting. Summable rates regularize each finite time precisely by suppressing remote moduli.

Fixed-probability thinning preserves hypothetical anomalous blocks

Let a finite singleton family have all $q\ge Q>1$, full footprint $\beta$, and normalized harmonic deletion $\alpha$ at $N$. Retain each class independently with fixed probability $p>0$, obtaining $\alpha_p,\beta_p$. Then

$$\mathbb E\alpha_p\ge p\alpha,\qquad \beta_p\le\beta. \tag{2}$$

Each originally deleted integer survives the thinning as a deletion with probability at least $p$, which proves the first inequality.

Changing the retention of class $q$ changes $\alpha_p$ by at most $1/q$. The elementary product variance bound therefore gives

$$\operatorname{Var}(\alpha_p) \le p(1-p)\sum_q q^{-2}\le\frac{p(1-p)}{Q-1}.$$

For completeness, reveal the independent retention bits in order. The corresponding martingale differences have squared means at most $p(1-p)/q^2$, because the two conditional expectations differ by at most $1/q$. Orthogonality and summation give the variance bound. Chebyshev then yields

$$\Pr\{\alpha_p<p\alpha-\eta\} \le\frac{p(1-p)}{\eta^2(Q-1)}. \tag{3}$$

Hence hypothetical late blocks with fixed positive deletion and vanishing footprint retain positive deletion with probability tending to one after every fixed-probability thinning. Such thinning is not a mechanism for destroying the counterexample obstruction.

Random-order charges can escape to wholly redundant classes

For a finite family, let $M(\omega)$ count full hits and $m(n)$ count active hits. Averaging a uniform random order gives symmetric allocations

$$e_q=\int_{C_q}\frac1{M(\omega)}\,d\mu, \qquad c_q(N)=\frac1{H_N}\sum_{\substack{n\le N\\n\in B_q}} \frac1{n\,m(n)}.$$

Their totals are $\beta$ and the actual normalized harmonic deletion. They are not consistent as the finite family grows.

Take zero residues at $2$ and at $2p$ for every odd prime $p$. All later classes lie inside $C_2$, so every genuine increasing-order tail increment is zero. On $C_2$, however, infinitely many odd primes divide a Haar integer almost surely. Indeed the probability of avoiding all sufficiently large primes is zero by CRT and divergence of their reciprocal sum. Thus $M\to\infty$ on $C_2$, and dominated convergence gives $e_q\to0$ for every fixed $q$, while $\sum_qe_q=1/2$ throughout. For every fixed cutoff $K$,

$$\sum_{q>K}e_q\longrightarrow\frac12.$$

All allocation mass escapes to late classes that add no footprint at all. Keeping an old prefix first avoids this defect but restores the original conditioned-survivor difficulty.

A weaker entropy bound does hold for the symmetric allocation

If all moduli are at most $N$, then

$$\sum_q c_q(N)\log_+\frac{c_q(N)}{e_q}\ll\log\log(3N). \tag{4}$$

To check this, conditional Jensen gives

$$e_q\ge\frac1{qD_q},\qquad D_q:=\mathbb E(M\mid C_q) =\sum_{\substack{r\\C_q\cap C_r\ne\varnothing}}\frac{(q,r)}r \le H_N\sum_{d\mid q}\frac{\varphi(d)}d \le H_N2^{\Omega(q)}.$$

Since $qc_q(N)\le1$, $\log_+(c_q/e_q)\le\log H_N+\Omega(q)\log2$. Distinctness and the established prime-rank generating bound give

$$\sum_{\Omega(q)>r}c_q(N) \le\min\{1,C(3/2)^{-r}(\log(3N))^{3/2}\}.$$

Summing this tail bound over $r\ge0$ proves $\sum_q\Omega(q)c_q(N)\ll\log\log(3N)$ and hence (4).

This growing bound and the allocation escape above prevent it from replacing the unresolved constant entropy bound for the original disjoint first-kill potential. No conclusion about the full logarithmic density has been obtained from these probabilistic regularizations.