Capped prime fractions

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This continues prime filtration, using the exact forbidden fraction rather than its possibly very large union-bound potential. The result controls two prime ranges uniformly for arbitrary singleton residues. It does not establish logarithmic density: an intermediate range and the conditioned arithmetic digit transfer both remain unresolved.

Definitions and theorem

Let $\mathcal F$ be a finite family of distinct moduli $q>1$, with one residue $a_q$ for each. Write a modulus whose largest prime factor is $p$ uniquely as $q=p^e m$, where $e\ge1$ and $P^+(m)<p$; take $P^+(1)=1$. For an integer $n$, define

$$\alpha_p(n)=\mu_p\!\left( \bigcup_{\substack{q=p^em\in\mathcal F,\ q\le n\\ n\equiv a_q\ (m)}}[a_q]_{p^e} \right), \tag{1}$$

where $\mu_p$ is Haar probability on $\mathbb Z_p$. Thus $0\le\alpha_p(n)\le1$, and $\alpha_p(n)=0$ for $p>n$. The condition $q\le n$ is the original activation, not a power delay.

Put

$$U_p=\frac1{(p-1)^2} \prod_{\ell<p}\frac{1+1/\ell}{(1-1/\ell)^2} \ll\frac{\log^3(2p)}{p^2}, \qquad T_K=\sum_{p>K}U_p. \tag{2}$$

Products and sums indexed by $p,\ell$ are over primes. In particular $T_K\to0$.

For all sufficiently large $n$, set

$$y(n)=\exp\!\left(\frac{\log n}{8\log\log n}\right),$$

and set $y(n)=1$ on the remaining finite initial segment. Then

$$\limsup_{X\to\infty}\ \sup_{\mathcal F} \frac1{H_{\lfloor X\rfloor}} \sum_{n\le X}\frac1n \sum_{\substack{p>K\\p\le y(n)\ \text{or}\ p>n^{1/3}}} \alpha_p(n)^2 \le \frac{T_K}{\log2} \qquad(K\ge60). \tag{3}$$

The supremum can depend on $X$. Consequently the left side tends to zero as $K\to\infty$. The same assertion applies to infinite singleton systems, since only finitely many moduli are active below any fixed cutoff.

The optimized refinement below replaces $y(n)$ by $\exp(c\log n\,\log\log\log n/\log\log n)$ for every fixed $0<c<1/2$, and improves the right side of (3) to $T_K$.

Truncating the smaller part

Average over $I_N=\{N+1,\ldots,2N\}$. For $M\ge1$, let

$$A_{p,M}(n)= \sum_{\substack{q=p^em\in\mathcal F\\m\le M}} p^{-e}\mathbf1_{n\equiv a_q\ (m)}, \qquad R_{p,M}(n)= \sum_{\substack{q=p^em\in\mathcal F\\m>M,\ q\le n}} p^{-e}\mathbf1_{n\equiv a_q\ (m)}.$$

The small-part potential deliberately omits activation. The large-part potential retains it. Since $\alpha_p\le\min(1,A_{p,M}+R_{p,M})$, the elementary inequality

$$\min(1,A+R)^2\le A^2+2R\qquad(A,R\ge0) \tag{4}$$

gives a useful capped estimate. To check (4), it is immediate for $A\ge1$; for $A<1$, the capped square increases with $R$ at rate at most two.

Two compatible smaller-part congruences form one class modulo $[m,r]$, and hence have at most $N/[m,r]+1$ simultaneous solutions in $I_N$. There is at most one original congruence at each $p^em$. Therefore

$$\mathbb E_{I_N}A_{p,M}^2 \le U_p+\frac{M^2}{N(p-1)^2}. \tag{5}$$

For the main term, enlarge to all smaller parts supported on primes below $p$ and use

$$\sum_{P^+(m),P^+(r)<p}\frac1{[m,r]} =\prod_{\ell<p}\sum_{i,j\ge0}\ell^{-\max(i,j)} =\prod_{\ell<p}\frac{1+1/\ell}{(1-1/\ell)^2}.$$

For the error, the sum of the coefficients of all small-part terms is at most $M\sum_{e\ge1}p^{-e}=M/(p-1)$.

For a term of $R_{p,M}$, normalize $a_q\bmod m$ as $c\in[0,m)$. Writing $n=c+km$, activation is exactly $k\ge p^e$. Its active count up to $2N$ is therefore at most $2N/m$, with no extra root error. It follows that

$$\mathbb E_{I_N}R_{p,M} \le \frac2{p-1} \sum_{\substack{m>M\\P^+(m)<p}}\frac1m.$$

Combining this with (4)--(5) proves

$$\mathbb E_{I_N}\alpha_p^2 \le U_p+\frac{M^2}{N(p-1)^2} +\frac4{p-1} \sum_{\substack{m>M\\P^+(m)<p}}\frac1m. \tag{6}$$

The small-prime range

For $p\ge60$, Rankin's inequality with $\theta=1/\log p\le1/4$ gives

$$\sum_{\substack{m>M\\P^+(m)<p}}\frac1m \le M^{-\theta} \prod_{\ell<p}(1-\ell^{-1+\theta})^{-1} \ll (\log p)\exp\!\left(-\frac{\log M}{\log p}\right). \tag{7}$$

Here the product is $O(\log p)$ uniformly in $p$. Indeed, its logarithmic difference from $\prod_{\ell<p}(1-1/\ell)^{-1}$ has first-power contribution

$$\sum_{\ell<p}\frac{\ell^\theta-1}{\ell} \le e\theta\sum_{\ell<p}\frac{\log\ell}{\ell}=O(1),$$

and higher-power contribution bounded by $O(\sum_{\ell}\ell^{-3/2})=O(1)$. The elementary prime-product and prime-log-sum estimates used here are proved in the coherent power-cutoff notes.

Summing (6)--(7), and using $\sum_{p\le Y}(\log p)/(p-1)\ll\log Y$, gives

$$\mathbb E_{I_N}\sum_{K<p\le Y}\alpha_p^2 \le T_K+\frac{CM^2}{NK} +C\log Y\exp\!\left(-\frac{\log M}{\log Y}\right) \qquad(60\le K<Y). \tag{8}$$

Take

$$M=N^{1/3},\qquad Y(N)=\exp\!\left(\frac{\log N}{6\log\log N}\right).$$

For sufficiently large $N$, (8) becomes

$$\mathbb E_{I_N}\sum_{K<p\le Y(N)}\alpha_p^2 \le T_K+\frac{C N^{-1/3}}K +\frac{C}{\log N\log\log N}. \tag{9}$$

If $Y(N)\le K$, the sum is empty and the same upper bound is harmless.

The large-prime range

Let $A_p(n)$ be the full union-bound potential, including activation in every term. Then $\alpha_p^2\le A_p^2$. For a point of $I_N$, any active modulus $p^em$ has $m\le2N/p^e$. Counting compatible pairs on $[1,2N]$, rather than $I_N$, gives

$$\mathbb E_{I_N} A_p^2 \le 2U_p+\frac1N \left(\sum_{e\ge1}p^{-e}\lfloor2N/p^e\rfloor\right)^2 \le 2U_p+\frac{4N}{(p^2-1)^2}. \tag{10}$$

Chebyshev's prime-counting upper bound and partial summation imply

$$\sum_{p>Z}(p^2-1)^{-2}\ll\frac1{Z^3\log(2Z)}.$$

Consequently

$$\mathbb E_{I_N}\sum_{p>N^{1/3}}\alpha_p^2 \ll \frac{(\log N)^3}{N^{1/3}}+\frac1{\log N}=o(1), \tag{11}$$

uniformly in the family.

Harmonic summation and the remaining gaps

For $N<n\le2N$, eventually $y(n)\le Y(N)$. Also $p>n^{1/3}$ implies $p>N^{1/3}$. Thus (9)--(11) bound the ordinary average on $I_N$ of the inner sum in (3) by $T_K+o_N(1)$, with a uniform error. Multiplication by $1/n\le1/N$ shows that its harmonic mass on that interval has the same upper bound. Sum over $N=2^j$; a final partial interval can be enlarged to the whole interval. The number of intervals is $(\log X)/\log2+O(1)$, and their uniform error has vanishing Cesàro average. Finitely many small integers contribute a family-independent bounded amount, since $\alpha_p(n)=0$ for $p>n$. Division by $H_X\sim\log X$ proves (3).

The preceding proof leaves the prime interval

$$n^{1/(8\log\log n)}<p\le n^{1/3}$$

untreated. The optimized refinement below shortens this interval but does not eliminate it. More fundamentally, $\alpha_p(n)$ averages over a $p$-adic fiber, whereas the integer $n$ uses one possibly correlated digit in that fiber. Even a bound for the omitted squared fractions would not by itself identify their sum with actual deletions. An arithmetic transfer retaining the old survivor condition is still required. No such transfer, general logarithmic-density theorem, or counterexample is asserted here.

The bounded-height center theorem now controls this middle range when each prime's smaller-part residues admit moderately many integer centers of bounded height. It gives full arithmetic square tails for that subclass, with arbitrary new-prime digits. The unrestricted middle range remains open. Furthermore, bounded full-survivor conditional weights do not turn even pointwise square tails into signed arithmetic cancellation; that specific infinite-marginal transfer is false.

The cap, rather than the distribution of digits, is the extra input

In the squarefree-largest-prime subfamily $q=pm$, a uniform bound for the exact squared fractions is equivalent up to an absolute constant to a bound for the squared capped union-bound potential. This concerns unconditioned nonnegative averages, not the missing survivor-conditioned digit transfer.

Fix arbitrary smaller-part residues $c_{p,m}\pmod m$. Independently choose a uniform residue $b_{p,m}\pmod p$ and use CRT to install one class modulo $pm$ with these two projections. Moduli are still globally distinct, since $P^+(m)<p$. At a fixed $n$, let

$$C_p(n)=\#\{m:pm\text{ is selected},\ pm\le n, \ n\equiv c_{p,m}\pmod m\}, \qquad A_p(n)=C_p(n)/p.$$

The exact forbidden fraction is the fraction of the $p$ digits occupied by $C_p(n)$ independent uniform choices. Hence

$$\mathbb E\alpha_p(n) =1-(1-1/p)^{C_p(n)} \ge(1-e^{-1})\min(1,A_p(n)).$$

Jensen gives

$$\mathbb E\alpha_p(n)^2 \ge(1-e^{-1})^2\min(1,A_p(n))^2. \tag{12}$$

Conversely $\alpha_p(n)\le\min(1,A_p(n))$ for every assignment. After any finite nonnegative weighted summation in $n,p$, some deterministic digit assignment attains at least the expectation in (12). Thus one cannot resolve the uniform intermediate-prime square estimate merely by seeking additional cancellation between its forbidden digits. A bound for the high multiplicities of the smaller-part congruences, with their activation retained, is needed.

An optimized small-prime cutoff

Theorem. Fix $0<c<1/2$ and $K\ge60$, and for sufficiently large $n$ put

$$y_c(n)=\exp\!\left( c\,\frac{\log n\,\log\log\log n}{\log\log n}\right);$$

set $y_c(n)=1$ on the finite initial segment. Then

$$\limsup_{X\to\infty}\sup_{\mathcal F} \frac1{H_{\lfloor X\rfloor}} \sum_{n\le X}\frac1n \sum_{\substack{p>K\\p\le y_c(n)\ \mathrm{or}\ p>n^{1/3}}} \alpha_p(n)^2 \le T_K. \tag{13}$$

All exponents $e\ge1$, arbitrary singleton residues, and original activation are retained. In particular the remaining middle range is $y_c(n)<p\le n^{1/3}$, not the larger range left by (3).

A uniform tilted Euler product

For $0<\theta\le1/4$, $p\le Y$, and $s=\theta\log Y$,

$$\prod_{\ell<p}(1-\ell^{-1+\theta})^{-1} \ll \log(2p)\exp(Cse^s). \tag{14}$$

Indeed, relative to the product at $\theta=0$, the first-power logarithmic contribution is at most

$$\sum_{\ell<p}\frac{\ell^\theta-1}{\ell} \le \theta Y^\theta \sum_{\ell<p}\frac{\log\ell}{\ell} \ll se^s.$$

All higher powers contribute $O(1)$, uniformly, since their sum is bounded by $C\sum_\ell\ell^{-3/2}$. The elementary product and prime-log-sum estimates already used in (7) complete the proof. Thus, with $S_p(M)=\sum_{m>M,\ P^+(m)<p}1/m$, Rankin's inequality gives

$$S_p(M)\ll\log(2p) \exp(-\theta\log M+Cse^s). \tag{15}$$

A useful specialization, also used in the very-smooth modulus reduction, is $M=p^u$, $s=\log u-2\log\log u$. Whenever $u$ is sufficiently large and $0<s/\log p\le1/4$, it gives

$$S_p(p^u)\ll\log(2p) \exp\!\left(-u\log u+2u\log\log u +O\!\left(\frac{u}{\log u}\right)\right). \tag{16}$$

No asymptotic theorem for smooth numbers is required.

Harmonic truncation with its exact main coefficient

For any residue class modulo $d$, discrete partial summation gives

$$\sum_{\substack{N<n\le2N\\n\equiv b\ (d)}}\frac1n \le \frac{H_{2N}-H_N}{d}+\frac1N. \tag{17}$$

The counting discrepancy on each initial subinterval of $\{N+1,\ldots,2N\}$ has absolute value at most one; its weighted discrepancy is therefore at most $1/(N+1)$.

Use the same $A_{p,M},R_{p,M}$ as above. Pairwise CRT in (17) and the coefficient sum $M/(p-1)$ imply

$$\sum_{N<n\le2N}\frac{A_{p,M}(n)^2}{n} \le (H_{2N}-H_N)U_p+\frac{M^2}{N(p-1)^2}.$$

Activation gives the unchanged remainder estimate

$$\sum_{N<n\le2N}\frac{R_{p,M}(n)}n \le\frac2{p-1}S_p(M).$$

Consequently the capped inequality (4) yields

$$\sum_{N<n\le2N}\frac1n\sum_{K<p\le Y}\alpha_p(n)^2 \le (H_{2N}-H_N)T_K+\frac{CM^2}{NK} +4\sum_{p\le Y}\frac{S_p(M)}{p-1}. \tag{18}$$

Take

$$M=\frac{\sqrt N}{\log N},\qquad Y=y_c(2N),\qquad u=\frac{\log M}{\log Y}.$$

Then

$$u\log u=\left(\frac1{2c}+o(1)\right)\log\log N.$$

Choose a fixed $a$ with $2c<a<1$, set $s=a\log u$, and take $\theta=s/\log Y$. Formula (15), uniformly over $p\le Y$, gives

$$S_p(M)\ll\log(2p) \exp\!\left(-a u\log u+Cau^a\log u\right) =\log(2p)(\log N)^{-a/(2c)+o(1)}.$$

Using $\sum_{p\le Y}\log(2p)/(p-1)\ll\log Y\ll\log N$, the final term in (18) is $(\log N)^{1-a/(2c)+o(1)}=o(1)$. The other error is $O(1/(K(\log N)^2))$. The function $y_c$ is eventually increasing, so $p\le y_c(n)$ implies $p\le Y$ on this interval.

The already proved large-prime estimate (11) contributes $o(1)$ in harmonic mass on the same interval. For $c=1/4$, one may take $a=3/4$, and the combined error is $O((\log N)^{-1/4})$. For every fixed $c<1/2$, it is still a uniform vanishing error.

Finally sum (18) over $N=2^j$, enlarging the last partial interval. The main terms telescope; the final dyadic endpoint changes $H_X$ by only $O(1)$. The errors have vanishing Cesàro average, and the finite initial segment is uniformly bounded since $\alpha_p(n)=0$ for $p>n$. This proves (13).

The sharper cutoff does not address the survivor-conditioned digit transfer. Even a proof for all omitted primes would leave that separate arithmetic step unresolved.