Smooth largest prime reduction

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These are support reductions and a construction obstruction, not a logarithmic-density theorem for general residues. They use the optimized Rankin estimate from capped prime fractions.

A reciprocal-summability threshold

For sufficiently large primes $p$, define

$$u_c(p)=c\,\frac{\log\log p}{\log\log\log p}.$$

Let $\mathcal S_c$ consist of the integers

$$q=p^e m,\qquad e\ge1,\quad P^+(m)<p,\quad m>p^{u_c(p)}. \tag{1}$$

Any fixed finite set of smaller primes may be omitted from this definition. Including all integers supported on those primes would still add only a reciprocal-summable set.

Theorem. For every fixed $c>1$,

$$\sum_{q\in\mathcal S_c}\frac1q<\infty. \tag{2}$$

For every fixed $0<c<1$, that sum diverges, already with $e=1$. No assertion at $c=1$ is needed here. The sharp threshold concerns reciprocal summability only, not existence of density.

Convergence. The optimized reciprocal smooth-tail estimate is

$$\sum_{\substack{m>p^u\\P^+(m)<p}}\frac1m \ll\log(2p)\exp\!\left( -u\log u+2u\log\log u+O(u/\log u)\right). \tag{3}$$

Its hypotheses hold for $u=u_c(p)$ and all sufficiently large $p$. Since

$$u_c(p)\log u_c(p)=c\log\log p+o(\log\log p),\qquad u_c(p)\log\log u_c(p)=o(\log\log p),$$

the left side of (3) is at most $(\log p)^{1-c+o(1)}$. For $\eta=(c-1)/2>0$, it is therefore $O_c((\log p)^{-\eta})$. Summing the $p$-power exponent gives

$$\sum_{q\in\mathcal S_c}\frac1q =\sum_p\frac1{p-1} \sum_{\substack{m>p^{u_c(p)}\\P^+(m)<p}}\frac1m<\infty.$$

The last prime sum converges by Chebyshev's upper bound and partial summation. Its part with $p>K$ is $O_c((\log K)^{-\eta})$.

Divergence. Fix $a$ with $c<a<1$, put $y=p^a$, and take $k=\lceil u_c(p)/a\rceil$. Consider products

$$m=b\,r_1\cdots r_k, \qquad P^+(b)\le y,\qquad y<r_1<\cdots<r_k<p.$$

Their factorization into $b$ and the displayed primes is unique, and $m>p^{ak}\ge p^{u_c(p)}$. Chebyshev's upper and lower prime-counting bounds, followed by partial summation, give

$$\lambda_p:=\sum_{p^a<r<p}\frac1r\asymp_a1.$$

For example the integral of the lower bound $\pi(t)\gg t/\log t$ on $[p^a,p]$ supplies a positive constant, while the endpoint error is $O_a(1/\log p)$.

Choose $k$ primes independently with probabilities $1/(r\lambda_p)$. Their collision probability is at most $O_a(k^2/p^a)=o(1)$, so the elementary symmetric sum obeys

$$e_k((1/r)_{p^a<r<p}) \ge (1-o(1))\frac{\lambda_p^k}{k!}.$$

Also $\prod_{\ell\le y}(1-1/\ell)^{-1}\ge H_{\lfloor y\rfloor} \gg_a\log p$. Consequently

$$\begin{aligned} \sum_{\substack{m>p^{u_c(p)}\\P^+(m)<p}}\frac1m &\ge H_{\lfloor y\rfloor} (1-o(1))\frac{\lambda_p^k}{k!}\\ &\ge(\log p)^{1-c/a+o(1)}. \end{aligned} \tag{4}$$

Here $k\log k=(c/a+o(1))\log\log p$. Since $c/a<1$, the last bound eventually exceeds one. The divergence of $\sum_p1/p$ now proves the assertion for $q=pm$. $\square$

Reductions retaining original activation

For one normalized residue at each selected $q\in\mathcal S_c$,

$$|B_q\cap[1,X]|\le X/q,\qquad H_{B_q}(X)\le H_{\lfloor X\rfloor}/q.$$

Thus the numerical tails of this family are negligible in upper natural and logarithmic density. Its fixed initial classes may have positive density and must still be retained when approximating a full system. Combining (2) with the repeated-largest-prime reduction shows that, for any fixed $c>1$, the remaining general tail problem can be restricted to

$$q=pm,\qquad P^+(m)<p,\qquad m\le p^{u_c(p)}. \tag{5}$$

This is a reduction to a large-prime support, not an estimate for its remaining congruences.

There is also a uniform reduction for the capped-square target. Let $\widetilde\alpha_p$ retain only the nonexceptional moduli. The fraction removed by exceptional moduli is at most their incidence potential $R_p$, and

$$0\le\alpha_p(n)^2-\widetilde\alpha_p(n)^2\le2R_p(n).$$

For each term $p^{-e}\mathbf1_{n\equiv a_q\ (m)}$ of $R_p$, activation $n\ge p^em$ implies

$$\sum_{\substack{n\le X\\n\ge p^em\\n\equiv a_q\ (m)}}\frac1n \le \frac{H_{\lfloor X\rfloor}}m.$$

Hence the normalized harmonic contribution of the squared-fraction difference at $p>K$ is at most

$$2\sum_{\substack{q\in\mathcal S_c\\P^+(q)>K}}\frac1q=o_K(1), \tag{6}$$

uniformly over the finite family and $X$. The same calculation handles the reciprocal-summable repeated-largest-prime exception.

Bounded polynomial CRT templates have vanishing square tails

The quadratic CRT construction in Green--Ruzsa, Section 5 suggests assigning a starting point $m^2\bmod Q$ to a progression of direction $m$. The following obstruction to its direct use here is independent of any theorem in that paper.

Fix $d\ge1$, $C\ge1$, and $\gamma>0$. For each sufficiently large prime $p$, let $Q_p$ be an integer with $\omega(Q_p)\ge\gamma\log p$, and let $f_p\in\mathbb Z[t]$ satisfy

$$f_p(0)=0,\qquad 0\le f_p(m)\le C m^d\quad(m\ge1). \tag{7}$$

For selected $m$ with $P^+(m)<p$, prescribe the smaller-coordinate residue at modulus $pm$ by

$$c_{p,m}\equiv (f_p(m)\bmod Q_p)\pmod m, \tag{8}$$

where the inner residue is the least nonnegative representative. The $p$-digits may be arbitrary, independently at every selected modulus. All these original moduli are globally distinct.

Proposition. The exact fractions for such systems obey

$$\lim_{K\to\infty}\sup_{\mathcal F,X\ge2} \frac1{H_{\lfloor X\rfloor}} \sum_{n\le X}\frac1n\sum_{p>K}\alpha_p(n)^2=0, \tag{9}$$

uniformly over $Q_p,f_p$ subject to the fixed parameters in (7).

Proof. Put $M_p=(Q_p/(2C))^{1/d}$. If $m<M_p$, then $0\le f_p(m)<Q_p$, and $m\mid f_p(m)$ by $f_p(0)=0$. Thus the smaller-coordinate residue is zero.

Let $r=\omega(Q_p)$. Distinct prime divisors give $Q_p\ge(r+1)!$, without any squarefreeness assumption. Therefore

$$\frac{\log M_p}{\log p} \ge\frac{\gamma}{d}\log\log p-O_{C,d,\gamma}(1).$$

For any fixed $c>1$, all sufficiently large $p$ consequently satisfy $M_p>p^{u_c(p)}$, uniformly in the choices.

Split the incidence potential as $A_{0,p}+R_p$, with

$$A_{0,p}(n)=\frac1p\sum_{\substack{m<M_p\\pm\ {\rm selected}}} \mathbf1_{m\mid n},$$

and with $R_p$ containing the remaining, activated incidences. The zero part deliberately omits activation. Then $\alpha_p^2\le A_{0,p}^2+2R_p$. Every pair in the zero part intersects in a multiples set, so the exact harmonic multiples bound gives

$$\frac1{H_{\lfloor X\rfloor}}\sum_{n\le X}\frac{A_{0,p}(n)^2}{n} \le \frac1{p^2} \prod_{\ell<p}\frac{1+1/\ell}{(1-1/\ell)^2} \le U_p.$$

All remaining moduli belong to $\mathcal S_c$. Their contribution is bounded by (6). Summing $U_p$ and letting $K\to\infty$ proves (9). $\square$

For the literal quadratic template, a proposed image compression $\prod_{\ell\mid Q_p}\frac{1+1/\ell}{2}\le p^{-2}$ forces $\omega(Q_p)\ge2\log p/\log2$, so the proposition applies. It controls the entire installed system, not only a selected finite progression trace.

Large varying coefficients or constant terms are not covered: $f_p(m)=B_pm^2+A_p$ need not satisfy (7), and its small smaller-part residues need not be zero. No obstruction to all arithmetic-Kakeya constructions, no general capped-square theorem, and no conditioned arithmetic digit transfer is asserted.