Coherent counting obstruction

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Counting domination fails: proved

For a finite set $D\subseteq\mathbb N$ and a common center $\alpha\in\widehat{\mathbb Z}$, define

$$B_{D,\alpha} =\{n\ge1:\exists q\in D,\ q\le n,\ n\equiv\alpha\pmod q\}, \qquad M(D)=\{n\ge1:\exists q\in D,\ q\mid n\}.$$

Theorem. For every pair of constants $A,c\ge1$, there are a finite primitive set $D$, an integer center $\alpha$, and an integer $X$ such that

$$|B_{D,\alpha}\cap[1,X]| >A\,|M(D)\cap[1,cX]|. \tag{1}$$

In particular, no absolute constant $C\ge1$ gives the proposed bound

$$|B_{D,\alpha}\cap[1,X]| \le C\,|M(D)\cap[1,CX]|$$

uniformly over finite coherent systems.

The proof uses a smooth-number band and an elementary slow-growth argument and is self-contained.

Choosing a smooth band

Let $R=2^r$, where $r\ge1$ is an integer to be chosen sufficiently large, and put

$$\kappa=\lfloor R/c\rfloor.$$

We require $\kappa\ge3$. Let $\mathcal P$ be the finite set of primes at most $R$, and write

$$H(t)=\#\{d\le t:d\text{ has only odd prime factors in }\mathcal P\}, \qquad w=\prod_{\substack{p\in\mathcal P\\p>2}}(1-1/p).$$

The integer $1$ is included in $H(t)$.

If $k$ is the number of odd primes in $\mathcal P$, then

$$H(R^j)\le(rj+1)^k\qquad(j\ge0), \tag{2}$$

because each prime exponent is at most $\log_2(R^j)=rj$. Consequently there are arbitrarily large integers $j$ such that

$$H(R^{j+1})\le2H(R^j). \tag{3}$$

Indeed, if (3) failed for every $j\ge j_0$, repeated doubling would give $H(R^{j_0+m})>2^mH(R^{j_0})\ge2^m$, contradicting the polynomial bound (2) for large $m$.

Choose such a $j$, set $T=R^j$, and define

$$D=\{q:T<q\le2T,\ q\text{ has all its prime factors in }\mathcal P\}, \qquad X=\kappa T. \tag{4}$$

The set $D$ is nonempty, since $2T$ is a power of two. It is primitive: if distinct $q,q'\in(T,2T]$ satisfied $q\mid q'$, then $q'\ge2q>2T$, a contradiction.

Bounding the zero-residue count

Suppose $n\in M(D)$ and $n\le cX\le RT$. Write $n=qm$ with $q\in D$. Since $q>T$, the positive integer $m$ satisfies $m<R$. Every prime factor of $m$ therefore lies in $\mathcal P$. Thus $n$ is $\mathcal P$-smooth and $T<n\le RT$, proving

$$|M(D)\cap[1,cX]| \le\#\{n\in(T,RT]:n\text{ is }\mathcal P\text{-smooth}\}. \tag{5}$$

Write every integer counted on the right uniquely as $2^ed$, where $d$ is odd and $\mathcal P$-smooth. Necessarily $d\le RT$. For a fixed $d$, the condition $T<2^ed\le RT$ puts $e$ in a half-open interval of length $\log_2R=r$. It therefore permits at most $r$ nonnegative integer exponents. By (3),

$$|M(D)\cap[1,cX]|\le rH(RT)\le2rH(T). \tag{6}$$

Lower-bounding the full Haar footprint

Let

$$\beta_D=\mu\left(\bigcup_{q\in D}q\widehat{\mathbb Z}\right).$$

For each odd $\mathcal P$-smooth $d\le T$, let $e(d)\ge1$ be minimal such that $2^{e(d)}d>T$. Then

$$a_d:=2^{e(d)}d\in(T,2T],\qquad a_d\in D.$$

Consider the Haar event

$$E_d=\{z\in\widehat{\mathbb Z}: v_p(z)=v_p(d)\text{ for every odd }p\in\mathcal P, \quad v_2(z)\ge e(d)\}.$$

Haar prime coordinates are independent, with $\mu(v_p(z)=a)=(1-1/p)p^{-a}$ and $\mu(v_p(z)\ge a)=p^{-a}$. It follows that

$$\mu(E_d)=\frac{w}{d\,2^{e(d)}}=\frac{w}{a_d} \ge\frac{w}{2T}.$$

The events $E_d$ are pairwise disjoint, because their prescribed odd prime valuations distinguish the integers $d$. Every $E_d$ is contained in $a_d\widehat{\mathbb Z}$, and hence in the full union. Therefore

$$\beta_D\ge\frac{wH(T)}{2T}. \tag{7}$$

Averaging the coherent center

Let $L=\operatorname{lcm}(D)$, and choose an integer $\alpha$ uniformly from $\{0,\ldots,L-1\}$. For every fixed integer $n\ge2T$, all moduli in $D$ are active, and translation invariance in $\mathbb Z/L\mathbb Z$ gives

$$\Pr(n\in B_{D,\alpha})=\beta_D.$$

There are $(\kappa-2)T+1$ integers in $[2T,X]$. By averaging, some integer center $\alpha$ satisfies

$$|B_{D,\alpha}\cap[1,X]| \ge(\kappa-2)T\beta_D \ge\frac{(\kappa-2)wH(T)}2. \tag{8}$$

The neglected extra $+1$ only weakens the estimate.

An elementary bound for the Euler product

Since the odd primes at most $R$ are a subset of the odd integers from $3$ to $R-1$,

$$w\ge\prod_{i=1}^{R/2-1}\frac{2i}{2i+1} \ge\frac1{\sqrt{R-1}}>\frac1{\sqrt R}. \tag{9}$$

For completeness, if $P_m=\prod_{i=1}^m2i/(2i+1)$, induction gives $P_m^2\ge1/(2m+1)$. The induction step follows from

$$\frac1{2m-1}\left(\frac{2m}{2m+1}\right)^2 \ge\frac1{2m+1},$$

which is equivalent to $4m^2\ge(2m-1)(2m+1)$.

Combining (6), (8), and (9), and noting that the denominator is positive, gives

$$\frac{|B_{D,\alpha}\cap[1,X]|}{|M(D)\cap[1,cX]|} \ge\frac{\lfloor R/c\rfloor-2}{4\sqrt R\log_2R} \longrightarrow\infty \qquad(R\to\infty\text{ through powers of two}). \tag{10}$$

For the prescribed $A,c$, choose $R$ so large that the last lower bound exceeds $A$. This proves (1). After choosing $R$, the parameter $T$ can still be taken arbitrarily large. $\square$

Scope

This disproves a uniform natural-counting comparison that might otherwise have been integrated to obtain the finite harmonic small-footprint estimate discussed in coherent power cutoffs. It does not disprove that harmonic estimate, nor construct a logarithmic-density anomaly, nor resolve Erdős Problem 25. Each system in the construction is finite and therefore eventually periodic.

Separate exact-cover barriers

The following elementary facts delimit proposed amplifications to a finite covering system. They are not additional obstructions to the harmonic small-footprint estimate and do not rule out every possible amplification.

Coherence-preserving operations cannot give a cover. Every coherent family with moduli greater than one misses the profinite point $\alpha+1$. For a finite family, choose an integer $a$ representing the center modulo its common period; then $a+1$ is an integer survivor. Thus an exact-cover reduction intended also to handle a hypothetical coherent anomaly would have to break coherence.

Integer-orbit coverage does not imply profinite coverage. Given $N\ge1$ and $0<\varepsilon<1$, enumerate $\mathbb Z=\{z_1,z_2,\ldots\}$ and choose distinct primes $p_j>\max(N,|z_j|)$ with $\sum_j1/p_j<\varepsilon$. CRT supplies $\alpha\in\widehat{\mathbb Z}$ with $\alpha\equiv z_j\pmod{p_j}$ for every $j$. The full cylinder union contains every ordinary integer but has Haar measure less than $\varepsilon$. Every finite subfamily consequently leaves positive Haar measure. Compactness cannot extract a finite cover from coverage of the dense integer orbit: it requires coverage of the whole profinite space. This is not a delayed counterexample; the designed hits have $p_j>|z_j|$ and are inactive, and the reciprocal sum is finite.

Fresh-prime affine tagging leaves a surviving cylinder. Suppose finitely many copies are embedded as $p_jB_j+b_j$, where the primes $p_j$ are distinct. Every class in the $j$-th copy is contained in $b_j\pmod{p_j}$. Choose one residue different from $b_j\pmod{p_j}$ for each $j$, and apply CRT. The resulting integer residue cylinder avoids all copies. Thus this standard way of distinguishing moduli cannot produce an exact finite cover, even if every base set $B_j$ were all of $\mathbb Z$.

These facts explain why orbit-coverage amplification and ordinary fresh-prime tagging do not supply the exact finite cover needed for an application of the minimum-modulus theorem. No reduction from a positive logarithmic tail anomaly to such a cover has been established.