Common shift blocks

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Checking Section 5 of the independent input gives the following new obstruction. The proof below makes its local estimate finite and also controls all quotient ranges in a single block. It does not settle the unrestricted coherent-residue problem.

The later weighted prime-residue estimate allows growing numbers of local residues and gives a uniform harmonic bound from their weighted support sizes. Its single-residue specialization also verifies the quantitative rate $T(\log Q)^{-0.086071\ldots+o(1)}$ on intervals of length $T\ge Q^{1+\varepsilon}$. The qualitative interval range in (1) below is wider, so neither formulation is discarded.

Uniform interval bound: proved

For integers $Q\ge2$ and $b$, put

$$U_Q(b)=\bigcup_{Q\le q\le2Q}(b+q\mathbb Z).$$

Uniformly in $b$ and in intervals $I$ of length $H$,

$$\frac{|U_Q(b)\cap I|}{H}\longrightarrow0 \quad\text{as }Q\to\infty\text{ and }H/Q\to\infty. \tag{1}$$

There is no restriction on the location of the interval or on the size of the common shift.

Finite estimate. For squarefree $P\le Q$, define

$$A(P)=\sqrt3\prod_{p\mid P} \left(\frac{\sqrt{2-1/p}}p+(1-1/p)^{3/2}\right).$$

Then

$$\frac{|U_Q(b)\cap I|}{H}\le A(P)+\frac{2PQ}{H}. \tag{2}$$

Proof. On a residue $r\pmod P$, the incidence sum is

$$S_Q(r)=\sum_{Q\le q\le2Q}\frac{(q,P)}q \mathbf1_{(q,P)\mid r-b}.$$

For fixed $t\pmod P$, the nonnegative periodic function $f_t(q)=(q,P)\mathbf1_{(q,P)\mid t}$ has mean

$$W_P(t)= \prod_{\substack{p\mid P\\p\mid t}}(2-1/p) \prod_{\substack{p\mid P\\p\nmid t}}(1-1/p). \tag{3}$$

Indeed the residues with $(q,P)=d$ have proportion $\varphi(P/d)/P$; summing $d\varphi(P/d)/P$ over $d\mid(P,t)$ factors as (3). Splitting $[Q,2Q]$ into periods of length $P$ and using $1/q\le1/Q$ gives $S_Q(r)\le3W_P(r-b)$ when $P\le Q$.

CRT and the elementary count in one progression give

$$|U_Q(b)\cap I\cap(r+P\mathbb Z)| \le \frac HP S_Q(r)+2Q.$$

The trivial bound in the same residue class is $H/P+1$. Taking the smaller bound and summing over $r$ yields

$$\frac{|U_Q(b)\cap I|}{H} \le\frac1P\sum_{r\bmod P}\min(1,S_Q(r))+\frac{2PQ}{H}.$$

Since $\min(1,x)\le\sqrt x$ for $x\ge0$, the first term is at most $\sqrt3\,\mathbb E_{t\bmod P}\sqrt{W_P(t)}=A(P)$. The product formula for this expectation follows from CRT independence. This proves (2).

For $P$ the product of successively more primes,

$$A(P)=\exp\left[-(3/2-\sqrt2)\sum_{p\mid P}\frac1p+O(1)\right] \longrightarrow0.$$

Here $3/2-\sqrt2>0$ and the prime reciprocal sum diverges. The latter also follows directly because the Euler product over $p\le z$ dominates $H_{\lfloor z\rfloor}$, while its logarithm is $\sum_{p\le z}1/p+O(1)$. In (2), first fix such a $P$, then let $Q,H/Q\to\infty$, and finally increase $P$. This proves (1). $\square$

The same bounds apply to any coherent subfamily with moduli in $[Q,2Q]$: finite CRT supplies a common integer shift, even when the residues originally come from an arbitrary profinite center. Both the center and the subfamily may vary with $Q$. Delayed activation only reduces the counted set.

Uniform harmonic bound for one block: proved

Let $B_Q$ be any delayed coherent singleton union with moduli in $[Q,2Q]$. Then

$$\sup_{B_Q}\sup_{X\ge2} \frac{H_{B_Q}(X)}{H_{\lfloor X\rfloor}}\longrightarrow0 \qquad(Q\to\infty). \tag{4}$$

Thus a single coherently shifted dyadic block cannot retain positive normalized harmonic deletion by moving its hits to arbitrarily large quotients.

For a proof, write $N_Q(t)=|B_Q\cap[1,t]|$. It vanishes for $t<Q$, and (2), applied to $(0,t]$, gives $N_Q(t)\le A(P)t+2PQ$. Partial summation, for $X\ge Q$, gives

$$H_{B_Q}(X) \le A(P)(1+\log(X/Q))+2P.$$

Consequently, for $Q\ge\max(P,3)$,

$$\sup_{B_Q,X\ge2}\frac{H_{B_Q}(X)}{H_{\lfloor X\rfloor}} \le A(P)+\frac{2P}{\log Q}. \tag{5}$$

First let $Q\to\infty$ and then increase the primorial $P$. This proves (4).

Scope and limitations

The condition $H/Q\to\infty$ in (1) cannot simply be removed. For $b=Q$, every integer in $[2Q,3Q]$ is an active hit, witnessed by $q=n-Q\in[Q,2Q]$. Such natural-density spikes therefore remain possible, although (4) shows their uniformly vanishing normalized harmonic effect.

An unbounded number of dyadic blocks need not obey a summable version of (4). Neither this result nor the earlier bounded-quotient compression bound controls a general conditioned tail. The unrestricted coherent case is still unresolved.

Coherence is essential for the long-interval conclusion

For arbitrary singleton residues the interval assertion (1) is false, even when the Haar footprint tends to zero and $H/Q\to\infty$. This is a natural-counting obstruction, not a logarithmic-density counterexample.

Let $L\ge12$ be divisible by $12$, and choose $Q>L$, $Q\ge12$. Independently color each integer $q\in[Q,2Q]$ uniformly by $j_q\in\{1,\ldots,L\}$ and use the class $j_q\pmod q$. Write $b_Q=\mu(\bigcup_{Q\le q\le2Q}q\widehat{\mathbb Z})$. For every coloring its Haar union has measure at most $Lb_Q$, since it is contained in the union of $L$ common translates of this multiples window. The proved bound (2) implies $b_Q\to0$.

Fix an integer $n\in I=[3LQ/4,LQ]$. For each integer $L/2\le k\le2L/3$, choose $j\in[1,L]$ with $j\equiv n\pmod k$. Such a $j$ exists because $k\le L$. Then

$$q=(n-j)/k\in[Q,2Q]:\qquad q\ge9Q/8-3/2\ge Q,\quad q\le2Q.$$

Different chosen pairs give different moduli: if $n=kq+j=k'q+j'$, then $q|j-j'$, and $|j-j'|<L<Q\le q$ forces $j=j'$ and $k=k'$. Thus at least $L/6$ independent colors can delete $n$, each with probability $1/L$. Consequently

$$\Pr(n\text{ is deleted})\ge1-e^{-1/6}.$$

Averaging over $n$ supplies a coloring deleting at least this fraction of the integer points of $I$. All these hits are active, since $n>q$.

For each increasing $L$, choose $Q$ so large that $Lb_Q\le1/L$. Then the Haar footprint tends to zero, whereas the interval length is $H=LQ/4$ and its deleted proportion stays bounded away from zero. The moduli are distinct and all lie in a single dyadic window.

Why the consecutive-shift container does not amplify harmonically

Let

$$V=\bigcup_{1\le j\le K}\bigcup_{Q\le q\le2Q}[j]_q, \qquad 1\le K<Q.$$

Writing $s=\sum_{Q\le q\le2Q}1/q\ge1/2$, its full incidence count $F$ has mean $Ks$. For $d=(q,r)$, at most $K^2/d+K$ pairs $(j,\ell)\in[1,K]^2$ satisfy $j\equiv\ell\pmod d$. Therefore

$$\mathbb EF^2\le K^2s^2+K G_Q, \qquad G_Q=\sum_{Q\le q,r\le2Q}\frac{(q,r)}{qr} \le4(1+\log(2Q)).$$

The last bound follows from $(q,r)=\sum_{d\mid q,r}\varphi(d)$. Cauchy--Schwarz gives

$$\mu(V)\ge\frac{Ks^2}{Ks^2+G_Q} \ge\frac{K}{K+16(1+\log(2Q))}.$$

Thus making this entire container's footprint tend to zero requires $K=o(\log Q)$. Even the full range $[Q,KQ]$ contributes only $O(1+\log K)=o(\log Q)$ harmonic mass, not a positive normalized harmonic anomaly. This does not give a lower bound for every colored subunion, whose measure can be much smaller than that of its container.