Prime filtration

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This approach exposes entire prime-adic coordinates rather than installing congruences in increasing modulus order. The following estimates are proved, but their transfer to the full activated integer orbit remains incomplete. All moduli are distinct unless indicated. The case $q=1$ is trivial and is omitted.

Write $G=\prod_p\mathbb Z_p$, let $S$ be the full Haar survivor set, and suppose $\mu(S)=\delta>0$. Put $\nu=\delta^{-1}\mathbf1_S\mu$. For $q=p^e m$ with $P^+(m)<p$, and an earlier-coordinate configuration $x$, define the forbidden new-coordinate fraction

$$\alpha_p(x)=\mu_p\left( \bigcup_{\substack{q=p^e m\,: x\equiv a_q\pmod m}}[a_q]_{p^e}\right). \tag{1}$$

Use $P^+(1)=1$. This includes every chosen modulus with largest prime factor $p$, regardless of the size of its smaller-prime part.

A finite conditional entropy budget

With earlier-coordinate expectation under $\nu$,

$$\sum_p\mathbb E_\nu[-\log(1-\alpha_p)]\le\log(1/\delta). \tag{2}$$

In particular $\sum_p\mathbb E_\nu\alpha_p<\infty$.

For a proof, the conditional law $K_p(x,\cdot)$ of the $p$-coordinate under $\nu$ is supported on a set of $\mu_p$-measure at most $1-\alpha_p(x)$. Relative entropy therefore satisfies

$$D(K_p(x,\cdot)\Vert\mu_p)\ge-\log(1-\alpha_p(x)).$$

This follows by conditioning Haar on the allowed set and using nonnegativity of relative entropy. Fibers with $\alpha_p=1$ have zero earlier marginal under $\nu$. The chain rule, over finitely many prime coordinates, gives

$$\sum_{p\le P}\mathbb E_\nu D(K_p\Vert\mu_p) =D(\nu_{\le P}\Vert\mu_{\le P})\le\log(1/\delta).$$

For completeness, if $h_P=\mathbb E_\mu(\mathbf1_S\mid\mathcal F_{\le P})$, the right side equals

$$\log(1/\delta)+\delta^{-1}\mathbb E_\mu[h_P\log h_P].$$

The martingale $h_P$ converges to $\mathbf1_S$; bounded convergence of $h_P\log h_P$ shows that this expression tends to $\log(1/\delta)$. Summing the preceding conditional lower bounds proves (2). This entropy statement itself permits multiple residues.

A summable Haar second moment using singleton distinctness

The union bound in the new coordinate gives

$$\alpha_p(x)\le A_p(x):= \sum_{q=p^em}p^{-e}\mathbf1_{[a_q]_m}(x).$$

Since each $p^em$ occurs at most once, pairwise CRT and enlargement to all $p$-smooth smaller parts give

$$\begin{aligned} \mathbb E_\mu\alpha_p^2 &\le\mathbb E_\mu A_p^2\\ &\le\frac1{(p-1)^2} \sum_{P^+(m),P^+(r)<p}\frac{(m,r)}{mr}\\ &=U_p:=\frac1{(p-1)^2} \prod_{\ell<p}\frac{1+1/\ell}{(1-1/\ell)^2} \ll\frac{\log^3(2p)}{p^2}. \end{aligned} \tag{3}$$

Indeed the local series is $\sum_{i,j\ge0}\ell^{-\max(i,j)} =(1+1/\ell)/(1-1/\ell)^2$. The product in (3) is at most $\prod_{\ell<p}(1-1/\ell)^{-3}$, so the already checked elementary Euler-product bound applies. Thus $\sum_p\mathbb E_\mu\alpha_p^2<\infty$ universally. Repeated residues would add multiplicities in this calculation.

An activated arithmetic second moment

For an integer $n$, truncate using the actual activation condition:

$$A_p(n)=\sum_{\substack{q=p^em\le n\,: P^+(m)<p}} p^{-e}\mathbf1_{n\equiv a_q\pmod m}.$$

Then, for $X\ge2$,

$$\frac1{H_{\lfloor X\rfloor}}\sum_{n\le X}\frac{A_p(n)^2}{n} \le U_p+\frac{2X}{H_{\lfloor X\rfloor}(p^2-1)^2}. \tag{4}$$

For each compatible pair $q=p^em$, $q'=p^fr$, the smaller-coordinate conditions form one class modulo $[m,r]$, counted only for $n\ge T=\max(q,q')$. Its harmonic sum is at most

$$H_{\lfloor X\rfloor}/[m,r]+1/T.$$

To check the error, normalize that class modulo $[m,r]$: its only possible positive root below the modulus contributes at most $1/T$, and the later terms are bounded by the harmonic progression of zero residue. The main terms sum to at most $U_p$ as in (3).

For fixed $e,f$, discarding smoothness only increases the error. The number of pairs with $\max(p^em,p^fr)\le t$ is at most $t^2/p^{e+f}$. Partial summation therefore bounds the sum of $1/\max(p^em,p^fr)$ below $X$ by $2X/p^{e+f}$. Multiplication by $p^{-e-f}$ and summation over $e,f\ge1$ gives $2X/(p^2-1)^2$, proving (4).

For any fixed threshold $a>0$, Markov's inequality and a union bound now show that the normalized harmonic mass of integers $n\le X$ with $A_p(n)>a$ for some $p>Y$ is at most

$$a^{-2}\left[ O\left(\frac{\log^3(2Y)}Y\right) +O\left(\frac{X}{H_{\lfloor X\rfloor}Y^3}\right)\right]. \tag{5}$$

This is useful, for example, at $Y=X^{1/3}$. It bounds large-potential fibers, not the total number of deletions by those primes. It gives no vanishing bound for the remaining moving prime range. In fact, a uniform original-activation second-moment bound for the uncapped $A_p$ is false. The counterexample below does not refute such a bound for the exact forbidden fraction $\alpha_p$.

Positive survivor measure gives no pointwise distortion cap

For a prime $p$, use moduli $q_j=pj$, $1\le j<p$, and residues

$$a_j\equiv0\pmod j,\qquad a_j\equiv j\pmod p.$$

These moduli are distinct and all have largest prime factor $p$. The full survivor density is at least $1-H_{p-1}/p\to1$. Nevertheless, on the earlier-coordinate fiber $x\equiv0\pmod L$, where $L=\operatorname{lcm}(1,\ldots,p-1)$, the stage removes $p-1$ of the $p$ digits. Conditional survival forces the zero digit, so its density relative to Haar in that coordinate is $p$. The earlier fiber has positive $\nu$-mass $1/(\delta pL)$. Thus (2) cannot be replaced by a uniform pointwise bound based on $\delta$, even when $\delta$ tends to one.

The remaining difficulty is specific: (2) is averaged under the full survivor law $\nu$, while (4)--(5) sample actual integers. Nonlinear events determined by earlier prime coordinates can have periods far larger than $X$, despite individual active moduli being at most $X$. No transfer estimate connecting these two controls, and no proof of the required lower logarithmic density, was obtained.

The later full-survivor trace obstruction shows that directly sampling the canonical earlier-coordinate densities of this infinite law $\nu$ can in fact give a nonzero signed trace, even at a fixed prime. Uniform high-prime cancellation also fails for these weights despite $\delta>1/4$, weights bounded by four, and the pointwise bound $\alpha_p(n)\le1/p$. The example's actual delayed sieve has natural density. Thus a viable transfer must retain finite activation or impose additional arithmetic information; the infinite conditional marginals themselves are not available arithmetic weights.

Exact arithmetic moments with stronger power activation

There is a useful exact transfer under a different activation rule. It does not establish density even for that stronger rule; the remaining conditional-trace obstruction is stated below.

Fix an integer $k\ge2$ and replace activation by $n\ge q^k$. Put

$$A_{p,k}(n)=\sum_{\substack{q=p^em,\ P^+(m)<p\\q^k\le n}} p^{-e}\mathbf1_{n\equiv a_q\pmod m}.$$

For every integer $X\ge1$, both natural and harmonic moments obey

$$\frac1X\sum_{n\le X}A_{p,k}(n)^k\le U_{p,k},\qquad \frac1{H_X}\sum_{n\le X}\frac{A_{p,k}(n)^k}{n}\le U_{p,k}, \tag{6}$$

where

$$U_{p,k}=\frac1{(p-1)^k} \prod_{\ell<p}\left(\sum_{a\ge0} \frac{(a+1)^k-a^k}{\ell^a}\right) \ll_k\frac{\log^{2^k-1}(2p)}{p^k}. \tag{7}$$

In particular $U_{p,2}=U_p$: quadratic activation removes the entire $X$-dependent error from (4).

To prove this, expand the $k$th power. A compatible tuple $q_j=p^{e_j}m_j$ specifies one residual class modulo $d=[m_1,\ldots,m_k]$. Every active point in that tuple satisfies

$$n\ge\max_jq_j^k\ge\prod_jq_j\ge d.$$

The exact delayed progression bounds are therefore at most $X/d$ for counting and $H_X/d$ for harmonic sums. Discarding compatibility and enlarging the modulus collection increases the resulting positive sum. Distinctness ensures that each $p^em$ occurs at most once. The sums over $e_1,\ldots,e_k\ge1$ give $(p-1)^{-k}$; at a smaller prime $\ell$, the number of exponent tuples with maximum exactly $a$ is $(a+1)^k-a^k$. This gives the product (7). Its local factor is $1+(2^k-1)/\ell+O_k(\ell^{-2})$, so the elementary Euler-product bound proves the last estimate in (7).

Consequently, for $Y\ge2$ and $t>0$, the natural or normalized harmonic mass below $X$ of

$$\{n:\exists p>Y,\ A_{p,k}(n)>t\}$$

is at most

$$C_k t^{-k}\frac{\log^{2^k-1}(2Y)}{Y^{k-1}}, \tag{8}$$

uniformly in $X$. This follows from Markov's inequality, a union bound, and summing (7), even over all integers greater than $Y$.

The missing transfer is still not a moment estimate of this kind. Haar prime-stage normalization uses cancellation while all earlier coordinates are fixed. An arithmetic distortion argument would require control of signed expressions of the form

$$\frac1{H_X}\sum_{n\le X} \frac{w_{p-}(n)(\mathbf1_{B_p}(n)-\alpha_p(n))}{n},$$

where $w_{p-}$ is a nonlinear accumulated weight depending on arbitrarily many earlier congruences. Bounding products of $k$ active congruences does not bound this trace. Their accumulated CRT conductor can exceed $n$ despite $q^k\le n$ for every individual active modulus. No arithmetic conditional cancellation has been proved, so neither (6) nor (8) is a density theorem for arbitrary residues.

Removing repeated largest prime factors from the tail

There is a simple exact reduction, valid for arbitrary residues. Put

$$\mathcal R=\{q\ge2:P^+(q)^2\mid q\}.$$

Unique factorization and the established elementary Euler-product bound give

$$\begin{aligned} \sum_{q\in\mathcal R}\frac1q &=\sum_p\left(\sum_{e\ge2}p^{-e}\right) \sum_{P^+(m)<p}\frac1m\\ &=\sum_p\frac1{p(p-1)} \prod_{\ell<p}(1-1/\ell)^{-1} \ll\sum_p\frac{\log(2p)}{p^2}<\infty. \end{aligned} \tag{9}$$

Here $P^+(1)=1$. Moreover, the portion with $P^+(q)>Y$ is $O(\log(2Y)/Y)$, by comparison with the corresponding sum over all integers greater than $Y$.

For each normalized residue, activation gives at most $X/q$ hits below $X$. Thus moduli from $\mathcal R$ beyond a numerical cutoff $Q$ have, uniformly in $X$, at most

$$X\sum_{\substack{q\in\mathcal R\\q>Q}}\frac1q$$

hits. Their tails are negligible in upper natural density and hence in upper logarithmic density. This does not say that deleting the whole family $\mathcal R$ leaves the density unchanged: its fixed initial classes can have positive density and must be retained in finite approximations.

Consequently, to prove finite-congruence logarithmic approximation in general, it suffices to treat the core

$$q=pm,\qquad P^+(m)<p,$$

and then add back the reciprocal-summable family by the preceding tail estimate. In this core the prime-stage forbidden fraction is exactly the number of distinct forbidden $p$-digits represented by matching smaller-coordinate classes, divided by $p$. This reduction does not bound that digit count on the activated integer orbit and does not supply the missing conditional trace estimate.

A uniform original-activation second-moment bound is false

Despite the summable Haar bound (3), the original-activation arithmetic second moments of $A_p$ need not be uniformly bounded, even for systems whose moduli all have total prime-factor rank five and whose largest prime factor occurs only once.

Let $p$ tend to infinity through primes, put $X=p^5$, and let $\mathcal D_p$ consist of the products of four distinct primes at most $p/2$. Unique factorization and Chebyshev's prime-counting bounds give

$$M_p:=|\mathcal D_p| =\binom{\pi(p/2)}4\asymp \frac{p^4}{(\log p)^4}.$$

For each $m\in\mathcal D_p$, install the single modulus $q=pm$ with normalized residue

$$a_{pm}=X-pm\left\lfloor\frac{X}{pm}\right\rfloor.$$

These moduli are distinct, satisfy $q\le p(p/2)^4<X$, and have $P^+(q)=p$, $v_p(q)=1$, and $\Omega(q)=5$. Thus they already belong to the core retained in the preceding reduction.

At the integer $n=X$, every smaller-coordinate condition matches: $X\equiv a_{pm}\pmod m$, and every modulus is active. Hence

$$A_p(X)=\frac{M_p}{p},\qquad \frac1{H_X}\sum_{n\le X}\frac{A_p(n)^2}{n} \ge\frac{M_p^2}{p^2XH_X} \gg\frac{p}{(\log p)^9}\longrightarrow\infty.$$

In particular, for every fixed $Y$, the supremum over finite systems and cutoffs of

$$\sum_{p>Y}\frac1{H_X}\sum_{n\le X}\frac{A_p(n)^2}{n}$$

is infinite. The natural-average second moment also diverges, since its corresponding one-point lower bound is $M_p^2/(p^2X)\gg p/(\log p)^8$.

This construction does not produce a large exact forbidden fraction. Every assigned residue has the same $p$-digit, namely $a_{pm}\equiv X\equiv0\pmod p$. Consequently $\alpha_p\le1/p$ on every smaller-coordinate fiber, with equality at $n=X$. The obstruction is multiplicity in the union-bound potential $A_p$, not a failure of a uniform estimate for $\alpha_p$. No conclusion about the remaining conditioned arithmetic trace follows from this example.

Information as finite congruences activate

The obstruction to infinite-survivor marginals does not invalidate the following finite information identity. It still supplies no arithmetic trace bound.

For a finite increasing family, let

$$S_j=\bigcap_{i\le j}C_i^c,\quad \delta_j=\mu(S_j)>0,\quad \nu_j=\delta_j^{-1}\mathbf1_{S_j}\mu, \quad S_0=G,\quad\delta_0=1.$$

Work modulo a common finite period, exposing its prime-power coordinates in increasing prime order. Write $K_p^{(j)}$ for the conditional law of the coordinate at $p$ under $\nu_j$, given earlier coordinates. Then

$$\sum_{j=1}^J\sum_p\mathbb E_{\nu_j} D(K_p^{(j)}\Vert K_p^{(j-1)}) =\log\frac1{\delta_J}. \tag{10}$$

The expectation is over the earlier-coordinate marginal of $\nu_j$. Conditional versions on null fibers do not matter, since $\nu_j\ll\nu_{j-1}$.

Indeed, on $S_j$ the density ratio $d\nu_j/d\nu_{j-1}$ equals $\delta_{j-1}/\delta_j$, so $D(\nu_j\Vert\nu_{j-1})=\log(\delta_{j-1}/\delta_j)$. The finite relative-entropy chain rule, followed by telescoping in $j$, proves (10). In particular, if the infinite survivor measure is $\delta>0$, every finite budget is at most $\log(1/\delta)$. The identity does not use singleton distinctness.

There are also exact projection recursions. Put

$$h_{p,j}=\mathbb E_\mu(\mathbf1_{S_j}\mid\mathcal F_{\le p}), \qquad w_{p,j}=h_{p,j}/\delta_j, \qquad F_j=S_{j-1}\setminus S_j,\quad e_j=\mu(F_j).$$

Linearity and $\delta_j=\delta_{j-1}-e_j$ give

$$h_{p,j}-h_{p,j-1} =-\mathbb E_\mu(\mathbf1_{F_j}\mid\mathcal F_{\le p}),$$ $$w_{p,j}-w_{p,j-1} =\frac{e_jw_{p,j-1} -\mathbb E_\mu(\mathbf1_{F_j}\mid\mathcal F_{\le p})} {\delta_j}, \qquad 0\le w_{p,j}\le1/\delta_J. \tag{11}$$

To retain actual activation, let $S_n$ use exactly the moduli $q\le n$, with $\delta_n=\mu(S_n)$, and define $h_{p,n}$ accordingly. For each finite cutoff $X$, both telescopes are finite and yield

$$\begin{aligned} \sum_{n\le X}\frac{\mathbf1_A(n)-\delta_n}{n} &=\sum_p\sum_{n\le X} \frac{h_{p,n}(n)-h_{p-,n}(n)}n\\ &=-\sum_{q\le X}\sum_{q\le n\le X} \frac{\mathbf1_{F_q}(n)-e_q}{n}. \end{aligned} \tag{12}$$

Here $h_{p-,n}$ fixes primes below $p$, and the initial prime projection is the constant $\delta_n$. At an active point all earlier congruences are active, so the full first-kill cells $F_q$ give exactly the second equality.

Thus exchanging the prime and activation sums reproduces the original first-kill discrepancy. The bounded information in (10) is integrated under changing Haar-conditioned laws, whereas (12) samples their changing functions at $x=n$. No estimate transferring the former to this diagonal trace was obtained. Since $H_X^{-1}\sum_{n\le X}\delta_n/n\to\delta$, an $o(\log X)$ bound for (12) is still precisely the required candidate-density conclusion, not a consequence of the telescoping.