Weighted prime residue windows

---

The weighted-coherence sketch in Section 5 of the third independent record can be made rigorous and gives the following one-block estimate. It assumes only control of residue supports modulo primes, not common prime-power centers or a bounded number of global profiles. It does not control an arbitrary conditioned tail or an unbounded number of blocks.

Uniform harmonic theorem

Let $Q\ge2$ be an integer, let $D\subseteq[Q,2Q]\cap\mathbb N$, and choose one normalized residue $a_q$ for each distinct $q\in D$. Put

$$B=\bigcup_{q\in D}\{a_q+tq:t\ge1\}.$$

For every prime $p$ dividing some selected modulus, define

$$R_p=\{a_q\bmod p:q\in D,\ p\mid q\},\qquad r_p=|R_p|, \qquad \mathcal H=\sum_{p:r_p>0}\frac1{p r_p}.$$

If $D$ is nonempty, there is an absolute constant $C$ such that

$$\sup_{X\ge2}\frac{H_B(X)}{H_{\lfloor X\rfloor}} \le \min\{1,C/\mathcal H\}. \tag{1}$$

In particular, any sequence of such blocks with $\mathcal H\to\infty$ has uniformly vanishing normalized harmonic mass, even at unbounded quotients. The empty family is trivial.

More generally, (1) holds for any chosen set of primes at most $2Q$ and nonempty sets $R_p$ containing all required residues at those primes. One may therefore adjoin missing primes with arbitrary singleton supports. If every actual support size is at most $R$, this gives the corollary

$$\sup_{X\ge2}\frac{H_B(X)}{H_{\lfloor X\rfloor}} \ll\min\{1,R/\log\log(3Q)\}. \tag{1a}$$

Here $R$ may grow with $Q$. This uses only prime-level support bounds.

A finite interval estimate

First use any finite set $\mathcal P$ of primes with nonempty supports $R_p$ satisfying the same containment condition; it is permissible to enlarge a support. Write

$$m=|\mathcal P|,\qquad h=\sum_{p\in\mathcal P}\frac1{p r_p}>0, \qquad \ell=\sum_{p\in\mathcal P}\frac1p.$$

For an interval $I$ of $T\ge Q$ consecutive integers, the full union $U=\bigcup_{q\in D}[a_q]_q$ obeys

$$|U\cap I|\ll \frac{T}{h} +\left(\frac{T}{Q}+Q\right)\frac{m^2}{h^2}. \tag{2}$$

All constants are absolute, including with arbitrarily large $T$.

Define the prime score and the extra score forced by a modulus by

$$K(n)=\sum_{p\in\mathcal P}\frac1{r_p} \mathbf1_{n\bmod p\in R_p},\qquad c_p=\frac1{r_p}-\frac1p\ge0,\qquad G(q)=\sum_{p\mid q,\ p\in\mathcal P}c_p.$$

Under uniform residue sampling their means are respectively $\ell$ and

$$v=\sum_{p\in\mathcal P}\frac{c_p}{p} =h-\sum_{p\in\mathcal P}\frac1{p^2}, \qquad h-1<v\le h.$$

The variance of $K$ is exactly $v$; that of $G$ is at most $h$.

For any interval $J$ of $S$ consecutive integers, counting single and double prime congruences by CRT gives

$$\sum_{n\in J}(K(n)-\ell)^2\ll Sh+m^2, \qquad \sum_{q\in J}(G(q)-v)^2\ll Sh+m^2. \tag{3}$$

Here the counting error for one residue is at most one. In the first identity, a prime pair has $r_pr_{p'}$ residue choices but coefficient $1/(r_pr_{p'})$, so its error is $O(1)$; the total is $O(m^2)$. The mean-count error is at most $m$, and multiplication by $\ell\le m/2$ preserves this bound. The second identity is the same calculation with the single zero residue at each prime and coefficients $0\le c_p\le1$. These observations also cover the diagonal terms.

It suffices to prove (2) for $h\ge8$, since otherwise $|U\cap I|\le T$ proves it after adjusting the constant. By (3), at most

$$O(Q/h+m^2/h^2) \tag{4}$$

integers $q\in[Q,2Q]$ have $G(q)<h/2$. Each selected exceptional modulus contributes at most $T/Q+1$ points of $I$.

For any remaining selected $q$, parametrize its points in $I$ as $n=a_q+qt$, with $t$ in an interval of $T_q\le T/q+1$ integers. Primes dividing $q$ force membership in $R_p$. At primes not dividing $q$, the allowed residues of $t$ again form sets of size $r_p$. The mean of $K(a_q+qt)$ in this residue model is therefore exactly

$$\sum_{p\mid q}\frac1{r_p} +\sum_{p\nmid q}\frac1p=\ell+G(q)\ge\ell+h/2.$$

The same CRT variance bound shows that at most $O(T_q/h+m^2/h^2)$ of these points have $K(n)<\ell+h/4$. Summing over the at most $Q+1$ distinct selected moduli costs $O(T/h+Qm^2/h^2)$, since $\sum_qT_q=O(T+Q)=O(T)$. The remaining covered points belong to $K(n)\ge\ell+h/4$, whose count in $I$ is $O(T/h+m^2/h^2)$ by (3). Combining these estimates with (4) proves (2).

The term $Tm^2/(Qh^2)$ must be retained before imposing a size condition on $m$. Discarding a modulus-window error before multiplying by $T/Q+1$ would not be uniform in the length of the interval. Also, using $c_p=1/r_p-1/p$ avoids subtracting a possibly larger $O(\log\log\log Q)$ loss from a slowly growing $h$.

Passing to harmonic measure

Assume $m^2\le Qh$, and put $Y=Q\max(1,m^2/h)$. Equation (2) implies $|B\cap[1,t]|\ll t/h$ for $t\ge Y$ (integer rounding only changes the absolute constants). There are no active points below $Q$. Below $Y$, use the trivial harmonic bound $1+\log(Y/Q)$; above $Y$, use partial summation with this counting estimate. Uniformly for $X\ge2$, this gives

$$\frac{H_B(X)}{H_{\lfloor X\rfloor}} \ll \frac1h+ \frac{1+\log^+(m^2/h)}{\log Q}. \tag{5}$$

For $X<Y$, apply the trivial estimate directly up to $X$; for $X\ge Y$ the remaining contribution is $O(h^{-1}(1+\log(X/Y)))$. Thus (5) also covers every cutoff within the initially uncontrolled range.

To deduce (1), use the full support parameter $\mathcal H$. The elementary Euler-product estimates imply, uniformly for $2\le y\le2Q$,

$$\sum_{y<p\le2Q}\frac1p \le\log\frac{\log(2Q)}{\log y}+C_0, \qquad \mathcal H\le\log\log(2Q)+C_0. \tag{6}$$

Indeed the logarithm of the finite Euler product differs from the prime reciprocal sum by $O(1)$, while that product is bounded above and below by absolute multiples of $\log y$.

For $\mathcal H$ larger than a fixed absolute constant choose

$$y=Q^{\exp(-\mathcal H/2)},\qquad \mathcal P=\{p\le y:r_p>0\}.$$

The second bound in (6) ensures $y\ge2$ in this range. The first ensures

$$h\ge\mathcal H/2-O(1)\ge\mathcal H/4.$$

Also $m^2\le y^2\le Q\le Qh$. Applying (5) gives

$$\frac{H_B(X)}{H_{\lfloor X\rfloor}} \ll \frac1{\mathcal H}+\frac{1+2\log y}{\log Q} \ll \frac1{\mathcal H},$$

since $\log y/\log Q=e^{-\mathcal H/2}$ and $\mathcal H\ll\log Q$. Bounded $\mathcal H$ is handled by the trivial bound one. This proves (1).

Comparison and limits

For the full coherent dyadic family all $r_p=1$, so $\mathcal H\asymp\log\log Q$ and (1) supplies an explicit harmonic rate. The common-shift interval theorem has the stronger qualitative length range $T/Q\to\infty$; (2) instead gives a quantitative estimate with finitely many prime coordinates. Under $\mathcal P\subseteq[2,z]$ and $z^2\le Qh$, it implies the input's bound $O(T/h+Qz^2/h^2)$.

The parameter can tend to infinity even with unbounded $r_p$; prime-power coherence is unnecessary. Conversely, small periodic footprint alone has not been related to this parameter in a way that controls the general problem. No summation over arbitrarily many windows or localization to a first-kill tail is justified by (1). The power-law assertion mentioned separately in the input also admits a complete proof, given next; none of its unavailable companion files is needed.

Power-law interval bound with one residue at each prime

Suppose that, for each prime $p$, all selected moduli divisible by $p$ have the same residue $\alpha_p\pmod p$. There need not be coherence at higher prime powers. For each fixed $\varepsilon>0$, uniformly over these families and intervals $I$ of $T\ge Q^{1+\varepsilon}$ consecutive integers,

$$|U\cap I|\le T(\log Q)^{-\delta_*+o_\varepsilon(1)},\qquad \delta_*=1-\frac{1+\log\log2}{\log2}=0.086071\ldots. \tag{7}$$

Unrepresented primes may receive arbitrary $\alpha_p$.

We first record the finite truncation estimate used in the proof. For any prime subset of $[2,z]$, coefficients $0\le d_p\le1$, one residue $b_p$ at each prime, and an even integer $k$, Bonferroni and CRT imply on any interval $J$ of $S$ integers

$$\sum_{t\in J}\prod_p(1-d_p\mathbf1_{t\equiv b_p\ (p)}) \le S\left(\prod_p(1-d_p/p)+E\right)+z^k, \qquad E=\frac{L^{k+1}}{(k+1)!},\quad L=\sum_{p\le z}\frac1p. \tag{8}$$

Indeed the even inclusion--exclusion truncation is an upper bound. Its Haar excess above the full product is at most the next elementary symmetric sum, bounded by $E$. There are at most $z^k$ squarefree products of at most $k$ primes at most $z$, each giving a counting error at most one. This proves (8), including its uniformity in $S$.

Fix $1<b<2$, set $s=b/2$ and $a=1-s$, and put $K(n)=\sum_{p\le z}\mathbf1_{n\equiv\alpha_p\ (p)}$. Choose $k$ to be the least even integer at least $20L$; then $E\le e^{-30L}$ for large $L$. On the progression for a selected $q$, primes dividing $q$ force $A(q):=\omega_z(q)$ matches. Applying (8) to the other primes gives

$$\sum_{n\in I\cap[a_q]_q}s^{K(n)} \le T_q F(q)+T_qE+z^k,\qquad F(q)=s^{A(q)}\prod_{\substack{p\le z\\p\nmid q}}(1-a/p), \tag{9}$$

where $T_q\le T/q+1$. To average over $q$, write

$$F(q)=P_s\prod_{p\le z}(1-d_p\mathbf1_{p\mid q}),\quad P_s=\prod_{p\le z}(1-a/p),\quad d_p=1-\frac{s}{1-a/p}\in[0,1].$$

A second application of (8), now on $[Q,2Q]$, yields

$$\sum_{Q\le q\le2Q}F(q) \le(Q+1)(D_s+E)+z^k,\quad D_s=\prod_{p\le z}(1-2a/p+a/p^2) =\exp(-2aL+O_b(1)). \tag{10}$$

Using $T_q\le T/Q+1$ and $\sum_qT_q=O(T)$ in (9)--(10), and allowing incidence overcounting, bounds the covered points with $K(n)\le bL$ by

$$O_b\!\left(T e^{-I_2(b)L}+T s^{-bL}E +s^{-bL}(T/Q+Q)z^k\right),\quad I_2(b)=2-b+b\log(b/2). \tag{11}$$

The factor $s^{-bL}$ is valid since $s^{K(n)}\ge s^{bL}$ on these points. In particular the $Tz^k/Q$ error from averaging the moduli has not been dropped.

For the remaining points, put $B_0=\lceil bL\rceil$ and $j=\lfloor(b-1)L\rfloor$. CRT counting of the factorial moment gives

$$\#\{n\in I:K(n)\ge bL\} \le\frac{T L^j/j!+z^j}{\binom{B_0}{j}} \le T\exp(-I_1(b)L+O_b(\log L))+z^k, \quad I_1(b)=b\log b-b+1. \tag{12}$$

The last exponent follows by Stirling's formula from $B_0-j=L+O(1)$. We work with large $Q$, so $j\ge1$; smaller values can be absorbed in the constants.

Finally choose

$$z=Q^{\min(\varepsilon,1)/(1000\log\log Q)}.$$

The elementary prime estimates give $L=\log\log Q-\log\log\log Q+O_\varepsilon(1)$, and $z^k\le Q^{\min(\varepsilon,1)/40}$ for sufficiently large $Q$. All CRT errors in (11)--(12), divided by $T$, are at most

$$(\log Q)^{O_b(1)} \left(Q^{-1+\min(\varepsilon,1)/40} +Q^{-\varepsilon+\min(\varepsilon,1)/40}\right).$$

They decay as a fixed negative power of $Q$, uniformly for every $T\ge Q^{1+\varepsilon}$. The term $s^{-bL}E$ is also negligible. Taking $b=1/\log2$ makes $I_1(b)=I_2(b)=\delta_*$ and proves (7).

This verifies the independent note's quantitative exponent and slightly weakens its coherence hypothesis. The estimate remains a one-window bound; it provides neither the missing inverse structural theorem nor a summable estimate for all windows of a general system.