Transversal prime groups

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Proved uniform estimate

Let $P_1,\ldots,P_r$ be disjoint finite sets of primes, with $\sum_{p\in P_j}1/p\le c$ for a fixed $c$. Assign one residue $b_p$ to each prime. For every product $q$ choosing one prime from each group, give $q$ the residue induced by the $b_p$ via CRT. Let $B$ be the union of their delayed classes, activated only at $n\ge q$.

There is a function $g_c(r)=O_c(1/r)$ such that, for every positive integer $N$,

$$|B\cap(N,2N]|\le g_c(r)N. \tag{1}$$

The estimate is uniform in the prime groups and all their residue choices. In particular these coherent high-rank product families cannot have a fixed positive normalized harmonic deletion while $r\to\infty$, regardless of the sizes of the quotients $n/q$.

This controls the actual trace for the family underlying the divisor-face capacity obstruction. It shows why feasible abstract face weights need not be realizable as first-kill harmonic masses.

Proof

Put $\eta=e^{-2c}$, $C=2/\eta$, and $z=(2N)^{C/r}$. For each $j$, let $F_j(n)$ indicate avoidance of the prime classes in $P_j$ with $p\le z$, and set

$$p_j=\prod_{\substack{p\in P_j\\p\le z}}(1-1/p), \qquad M(n)=\sum_jF_j(n),\qquad m=\sum_jp_j.$$

Since $\log(1-1/p)\ge-2/p$, we have $p_j\ge\eta$ and $m\ge\eta r$.

Take $r$ sufficiently large in terms of $c$, and let $k$ be the largest even integer at most $r/(2C)=\eta r/4$. Thus $k\ge2$ and $z^k\le\sqrt{2N}$. For one group, or the union of any two groups, Bonferroni's inequalities give a uniform comparison between its avoidance probability on $(N,2N]$ and its Haar avoidance probability, with error at most

$$\epsilon_r(N)=\frac{(2c)^k}{k!}+\sqrt{\frac2N}. \tag{2}$$

To verify the error directly, compare the even inclusion--exclusion partial sum through order $k$ and the odd sum through order $k-1$. Each congruence intersection involved has a squarefree modulus at most $z^k\le\sqrt{2N}$. There is at most one term for each such modulus, because each prime has a single fixed residue. The total normalized counting error is therefore at most $\sqrt{2N}/N$. The Haar gap between the two partial sums is the degree-$k$ elementary symmetric sum of the prime reciprocals, at most $(2c)^k/k!$. This proves (2).

Haar independence between disjoint prime groups gives $\mathbb E_{\rm Haar}F_jF_\ell=p_jp_\ell$ for $j\ne\ell$. Using (2) for all one- and two-group expectations, with averages over $I=(N,2N]$, yields

$$\mathbb E_I(M-m)^2\le r/4+3r^2\epsilon_r(N).$$

Consequently

$$\Pr_I(M<\eta r/2) \le\eta^{-2}\bigl(1/r+12\epsilon_r(N)\bigr). \tag{3}$$

If $n\in B\cap I$, choose an active product $q\le n$ witnessing it. Every group counted by $M(n)$ must supply a prime greater than $z$ to this product. If $M(n)\ge\eta r/2$, then

$$q>z^{M(n)}\ge z^{\eta r/2}=2N\ge n,$$

a contradiction. Thus every actual deletion lies in the event in (3). If $B\cap I$ is nonempty, its witness satisfies $2N\ge q\ge2^r$, so $\sqrt{2/N}\le2^{1-r/2}$. We may therefore take

$$g_c(r)=\eta^{-2}\left( 1/r+12(2c)^k/k!+12\cdot2^{1-r/2}\right)=O_c(1/r).$$

For the finitely many smaller $r$, set $g_c(r)=1$. This proves (1). Summing (1) over dyadic intervals, and using $1/n\le1/N$ on each, gives the asserted vanishing normalized harmonic bound as $r\to\infty$.

Removing the bound on group reciprocal sums

For these same complete transversal families, a stronger result follows from the fundamental lemma of sieve theory. Define

$$u_j=\prod_{p\in P_j}(1-1/p),\qquad \mu=\sum_ju_j.$$

Uniformly over finite disjoint prime groups and their coherent residues,

$$\sup_{X\ge1}\frac{|B\cap[1,X]|}{X}\ll\frac1\mu \quad\text{as }\mu\to\infty. \tag{4}$$

External input, checked. Koukoulopoulos, The Distribution of Prime Numbers, Theorem 19.1, printed p. 195 (PDF page 206), supplies upper and lower sieve weights of absolute value at most one, supported on squarefree $d\le D$, with relative Euler-product error $O(s^{-s/2})$, where $s=\log D/\log z$ is sufficiently large. For any selected prime set and one forbidden residue per prime, its dimension hypothesis holds with absolute parameters: take $\nu(p)=1$, $\kappa=1$, and an absolute $C$ in Axiom 2, as explained on printed p. 187 (PDF page 198). Both passages were checked in the author-hosted full text. No supplied source file was altered.

Here is the application, including the interval error. On $(N,2N]$ take $z=(2N)^{2/\mu}$, $D=\sqrt N$, and define $F_j,p_j,m,M$ as in the elementary proof. Then $m\ge\mu$. For a single group or two groups together, each modulus in a sieve-weight sum prescribes one CRT class, with normalized counting error at most $1/N$. There are at most $D$ such moduli. The pointwise sieve inequalities apply to any vector of prime hits, by representing that vector by a squarefree integer. Thus the fundamental lemma gives

$$|\mathbb E_I F_j-p_j|\le\epsilon p_j+R, \qquad |\mathbb E_I F_jF_\ell-p_jp_\ell| \le\epsilon p_jp_\ell+R,$$

where $R=N^{-1/2}$ and $\epsilon\ll s^{-s/2}$, with absolute constants. For $N\ge2$, $s\ge\mu/8$. If the interval contains a deleted point, $N\ge2^{r-1}$ and $r\ge\mu$, so all size hypotheses hold when $\mu$ is sufficiently large.

Expansion of the centered second moment now gives

$$\mathbb E_I(M-m)^2\le m+4\epsilon m^2+3r^2R.$$

An active certificate requires $M<\mu/2$, since otherwise its prime product exceeds $z^{\mu/2}=2N$. This event is contained in $\{|M-m|>m/2\}$. Its probability is at most

$$\frac4\mu+16\epsilon+\frac{12r^2}{\mu^2\sqrt N} \ll\frac1\mu.$$

The last term is exponentially small in $\mu$, using $N\ge2^{r-1}$ and $r\ge\mu$. Dyadic summation proves (4), and partial summation gives the same bound for $H_B(X)/H_{\lfloor X\rfloor}$.

In particular a small periodic footprint always forces small deletion for one complete transversal family. Write

$$\beta=\prod_j(1-u_j),\qquad L=\log(1/\beta).$$

If some $u_j\ge1-1/L$, every deletion lies in that group's delayed prime union, whose count is at most $X\sum_{p\in P_j}1/p\le-X\log u_j\le2X/L$ for $L\ge2$. Otherwise the increasing function $-\log(1-u)/u$ is at most $2\log L$ on the possible range, so $L\le2\mu\log L$. Applying (4) yields

$$\sup_{X\ge1}\frac{H_B(X)}{H_{\lfloor X\rfloor}} \ll\frac{\log\log(1/\beta)}{\log(1/\beta)} \qquad(\beta\downarrow0).$$

The empty-family case $\beta=0$ is immediate.

Limitation

The proof uses one consistent prime residue $b_p$ throughout all product classes. Arbitrary singleton residues modulo the products need not have this consistency. Nor has a general small-footprint family been reduced to complete transversal families with controlled total error. These are specific missing steps, not consequences of this estimate.