Coherent maximal matching

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A simultaneous finite trace bound

Let $D$ be a finite set of distinct moduli with common profinite center $\alpha$. Write $E_q$ for the actual delayed first-kill set and $e_q$ for its Haar first-kill mass. If $J\subseteq D$ and $\tau=\sum_{q\in J}e_q$, then, uniformly in all these data,

$$H_{\bigcup_{q\in J}E_q}(X) \ll \tau\log(2X)\log(2+\log(2X)). \tag{1}$$

The bound uses one common center simultaneously. It does not sum separate phase maxima. Its extra $\log\log X$ factor prevents it from establishing unrestricted coherent tail tightness. The positive kernel in (2) below already has order $\log X\log\log X$ for $D=J=\{1\}$. Thus bounding that kernel using only $G_J$ and its Haar mass cannot remove the loss; one would have to retain more of the numerical maximality condition.

Maximal matching multiples and a rough quotient

Choose an integer $b$ realizing $\alpha$ modulo $\operatorname{lcm}(1,\ldots,\lfloor X\rfloor)$. For each counted hit $n\le X$, with first modulus $q$, take the largest integer $m\le n$ satisfying

$$q\mid m,\qquad m\mid n-b.$$

Such an $m$ exists, since $q$ itself is eligible. In the zero-residue system on $D$, its first dividing generator is exactly $q$: an earlier generator dividing $m$ would cover $n$, and would be active there. Consequently

$$m\in G_J:=\{m\ge1:\operatorname{first}_D(m)\in J\}.$$

Maximality also implies that $(n-b)/m$ has no prime divisor at most $n/m$, since otherwise multiplying $m$ by that prime would give a larger admissible choice. If $m\le n/2$, this also excludes the quotient zero. Quotients may be negative; the upper sieve below applies on arbitrary translated intervals.

On a dyadic interval $(H,2H]$, fix $m$ and put $t=H/m$. For $t$ larger than a fixed absolute constant, the possible quotients lie in an interval of length $t$ and avoid every prime at most $t$. The uniform dimension-one upper sieve established in the power-cutoff notes, applied at $y=t^{1/B}$ for a sufficiently large fixed $B$, bounds their number by

$$O\bigl(t/\log(2+t)\bigr).$$

For bounded $t$, use $n\ge m$: the delayed progression $n\equiv b\pmod m$ has at most $2H/m=2t$ points below $2H$. Since $m\le2H$, here $t\ge1/2$, and the same bound holds after enlarging its absolute constant.

Weighting by $1/H$ and summing over possible $m$ therefore gives

$$H_{E_J\cap(H,2H]}(X) \ll\sum_{\substack{m\le\min(X,2H)\\m\in G_J}} \frac1{m\log(2+H/m)}.$$

For each fixed $m$, summation over dyadic $H$ is a harmonic sum in the dyadic index. Thus

$$H_{E_J}(X) \ll \tau+ \sum_{\substack{m\le X\\m\in G_J}} \frac{\log(2+\log(2X/m))}{m}. \tag{2}$$

The possible point $n=1$ costs at most $\tau$, since a hit there requires the modulus 1 and its full first cell has mass 1.

Haar control of the matching multiples

The geometric valuation law, as used in coherent transport, gives

$$T(Y):=\sum_{\substack{m\le Y\\m\in G_J}}\frac1m \le \tau Z_Y\ll\tau\log(2Y),\qquad Z_Y=\prod_{p\le Y}(1-1/p)^{-1}. \tag{3}$$

Only generators at most $Y$ matter in this calculation; their first-cell masses are unchanged by discarding later generators, and their selected masses sum to at most $\tau$.

For $K_X(t)=\log(2+\log(2X/t))$, partial summation yields

$$\sum_{\substack{m\le X\\m\in G_J}}\frac{K_X(m)}m =K_X(X)T(X)+ \int_1^X\frac{T(t)}{t(2+\log(2X/t))}\,dt.$$

Use (3) and $\log(2t)\le\log(2X)$ in the integral to obtain (1). In an infinite coherent system, restricting to moduli $q\le X$ leaves the active first kills and all their first-cell masses unchanged. Thus (1) applies to any selected tail of the infinite system as well.

Exact replacement exclusions

There is a stronger, checked formulation of numerical maximality, but no summation of it removing the logarithmic loss has been obtained. For the choice of $m$ used above, write $m=qa$, $t=n/m$, and $h=(n-b)/m$. Its maximality is equivalent to

$$k\nmid h\quad\text{whenever}\quad k\ge2,\quad \exists d\mid a:\ (d,k)=1, \quad d<k\le td. \tag{7}$$

Indeed, every competing multiple of $q$ is uniquely of the form $r=mk/d$, with $d\mid a$ and $(d,k)=1$, after reducing $r/m$. It divides $n-b=mh$ precisely when $k\mid h$; the two size conditions say $m<r\le n$.

For an ordered tail there is a still stronger choice. Let $D_0=D\cap[1,Y]$ and $G=M(D)\setminus M(D_0)$. At a tail hit choose the largest $m\le n$ dividing $n-b$ and belonging to $M(D)$, without requiring divisibility by the original first generator. Then $m\in G$: an old generator dividing $m$ would cover $n$. This retains the tail Haar mass $\mu(G)$, although the particular first label can change. Its exact quotient exclusion family is

$$\mathcal K(m,n)= \left\{\frac r{(r,m)}:m<r\le n,\ r\in M(D)\right\}. \tag{8}$$

The equivalence follows from $r\mid mh\iff r/(r,m)\mid h$. This includes replacements by incomparable generators.

On $H<n\le2H$, fix $m\le H$ and retain only $\mathcal K(m,H)$. It already contains every integer $2\le k\le H/m$, by taking $r=mk$. Thus every additional minimal generator of its multiples union exceeds $H/m$ and has no prime factor at most $H/m$. The quotient interval has length $H/m$. The new constraints therefore occur beyond that interval's length; a conventional per-quotient upper sieve using primes below $(H/m)^{1/B}$ sees no improvement. For $m>H$ the quotient interval has bounded length and there are no such retained competitors. These observations do not rule out a simultaneous divisor-switching argument. No such arithmetic estimate was proved.

A Haar-tail criterion for original activation

For an infinite coherent family suppose that

$$\tau(Y):=\sum_{q>Y}e_q =o\!\left(\frac1{\log\log Y}\right). \tag{4}$$

Then its delayed union $B$ has logarithmic density $\beta=\sum_qe_q$, with finite-periodic approximation in upper logarithmic density. Its survivor set has logarithmic density $1-\beta$. No natural-density conclusion is asserted.

Indeed the already-proved power-cutoff theorem gives natural density $\beta$ for

$$B_*:=\{n\in B:q(n)\le\sqrt n\}.$$

Fix $c>0$. If $X^c<n\le X$ lies in $B\setminus B_*$, then its first modulus exceeds $X^{c/2}$. By (1), these points have harmonic mass at most

$$O\bigl(\tau(X^{c/2})\log X\log\log X\bigr)=o(\log X).$$

The remaining points $n\le X^c$ have harmonic mass at most $c\log X+O(1)$. Divide by $\log X$, take the upper limit, and let $c\downarrow0$. This proves that $B\setminus B_*$ has logarithmic density zero. Subtracting the asymptotic for any finite prefix then proves the asserted finite-periodic approximation.

A convenient sufficient condition for (4) is

$$\sum_q e_q\log\log(3q)<\infty, \tag{5}$$

because $\tau(Y)\log\log(3Y)$ is bounded by the tail of this convergent series. Neither (4) nor (5) has been proved for an arbitrary coherent system. The criterion does not concern arbitrary incompatible residues, and does not resolve Problem 25.

Final status of the maximality approach

The continuation tried to retain the numerical maximality of $m$ instead of using only its rough-cofactor consequence. Numerical maximality also forbids larger matching multiples obtained by replacing factors, not just by adjoining a prime. No simultaneous estimate for these additional exclusions was obtained. Thus the logarithmic loss in (1) has not been removed. The example following (1) rules out removing it from the positive kernel alone; it does not rule out a different argument using maximality more fully.

One sufficient estimate considered, but not proved, was

$$H_B(X)\le\beta_D H_{\lfloor X\rfloor} +C\left(1+\sum_{\substack{q\in D\\q\le X}}e_q\log q\right) \tag{6}$$

uniformly over finite coherent families and their common centers, with an absolute $C$. Here $\beta_D$ is their full Haar union measure. This is only a possible coherent estimate, not an assertion for arbitrary residues; the general energy comparison has already been refuted.

For clarity, its sufficiency in the coherent case has no further analytic gap. For a fixed infinite family, $\sum_q e_q\le1$ implies

$$\frac1{\log X}\sum_{q\le X}e_q\log q\longrightarrow0.$$

For a fixed $M$, the terms $q\le M$ tend to zero after normalization, and the remaining normalized sum is at most $\sum_{q>M}e_q$, which tends to zero with $M$. Apply (6) to the finite truncation $q\le X$, whose active union agrees with the infinite union below $X$; its Haar mass is at most the limiting $\beta$. Finite prefixes give the reverse lower logarithmic bound. Hence (6), if established, would give logarithmic density $\beta$. No proof of (6), nor a coherent counterexample to the density statement, was obtained.