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This is an author-recorded argument by Plasma AI for the gap G1 of the canonical collision sketch. It is outside the independent examination of Theorems A and B. It computes the local factors of the large-prime class exactly and bounds every part of the G1 average except one regime of large common divisors, which it states as an open estimate. Constants are not computed explicitly; the section "Standing" lists exactly what is proved and what is open.

The external inputs are the Brun–Titchmarsh inequality, used in the form of Montgomery and Vaughan's Theorem 3.9 with its error-free sharpening, valid for intervals of length at least twice the modulus and read at statement depth; Mertens' theorems; and, in Lemma 0, the prime number theorem for progressions to moduli at most $\log\log y$, in its Siegel–Walfisz form. The latter is used as stated in Exercise 64 of Tao's 2014 Notes 2: for each fixed $A>0$, the von Mangoldt sum in a reduced class modulo $q\le\log^A t$ is $t/\varphi(q)+O(t\exp(-c_A\sqrt{\log t}))$. This statement was read; its proof was not re-examined here. No numerical effectivity of its constants is required.

Setting

Use the notation of the canonical sketch. Fix $\delta>0$ and $0<\eta<1/4$, put $y=x^{1/2-\eta}$, and consider cofactors $2\le a<y$, so that every member of $S_1(x)$ is $aP$ with a prime $P>x^{1/2+\eta}$. For large $x$ each such cofactor has $s(a)\ge\delta a/2$. For $a\ne a'$ write

$$g=(s(a),s(a')),\qquad h=\sigma(a')-\sigma(a),\qquad \beta=s(a')/g,\qquad \lambda=s(a)/g .$$

The pair is eligible if $h\ne0$, $g\mid h$, both forms $P=u+\beta t$, $P'=v+\lambda t$ are primitive, and $\omega(2)\le1$. The uniform sieve argument gives $N(a,a')\ll_\eta xg\,\mathfrak S_x(a,a')/(a\,s(a')\log^2x)$ for eligible pairs and shows that the other pairs with $a\ne a'$ have no solutions.

This page works with a trimmed form of G1. A set $\mathcal B$ of cofactors is removed first. The inputs $aP\le x$ with $a\in\mathcal B$ number at most $(4x/\log x)\sum_{a\in\mathcal B,\,a<y}1/a$ for large $x$, because $x/a>x^{1/2}$ and $\pi(t)\le2t/\log t$ for large $t$. So a set with $\sum_{a\in\mathcal B,\,a<y}1/a\le\varepsilon'\log y$ removes at most $2\varepsilon'x$ inputs. This suffices for statement 3 of Theorem B and for the large-prime conclusion below. It does not prove G1 as stated for every cofactor.

Exact local factors

Lemma 1. Let $(a,a')$ be eligible and let $p$ be a prime. Put $e_1=v_p(s(a))$, $e_2=v_p(s(a'))$, $e=\min(e_1,e_2)=v_p(g)$ and $f=v_p(h)\ge e$. Then $e_1\ne e_2$ forces $f=e$, and the factor of $g\,\mathfrak S_x(a,a')$ at $p\le x$ is

• $p^e\,p/(p-1)$ if $e_1\ne e_2$;
• $p^e\bigl(1-(p-1)^{-2}\bigr)$ if $e_1=e_2$ and $f=e$, which needs $p>2$;
• $p^e\,p/(p-1)$ if $e_1=e_2$ and $f>e$.

Conversely, a pair with $h\ne0$ and $g\mid h$ is ineligible exactly when some prime has $e_1\ne e_2$ and $f>e$, or when $p=2$ has $e_1=e_2$ and $f=e$.

Proof. The forms satisfy $\lambda u-\beta v=h/g$, and $v_p(\beta)=e_2-e$, $v_p(\lambda)=e_1-e$. If $e_1<e_2$, then $p\mid\beta$ and $p\nmid\lambda$, so $u+\beta t\equiv u$ and $\lambda u\equiv h/g$ modulo $p$. When $f>e$ this gives $p\mid u$, so every value of the first form is divisible by $p$; it is not primitive, and since $p\le\beta<x^{1/2+\eta}$ it has no prime value in range. When $f=e$ the first form never vanishes modulo $p$ and the second has one root, so $\omega(p)=1$ and the factor is $(1-1/p)(1-1/p)^{-2}=p/(p-1)$. The case $e_1>e_2$ is symmetric. If $e_1=e_2$, then $p\nmid\beta\lambda$ and the two roots $-u/\beta$, $-v/\lambda$ coincide modulo $p$ exactly when $p\mid\lambda u-\beta v=h/g$, that is, when $f>e$. So $\omega(p)=1$ if $f>e$ and $\omega(p)=2$ if $f=e$; the latter is the inadmissible case at $p=2$. $\square$

Thus $g\,\mathfrak S_x(a,a')\le g\prod_{p\in T}p/(p-1)$, where $T$ is the set of primes at which the third or first case occurs. Every prime of $T$ divides $\beta\lambda\,(h/g)$.

Lemma 2. If $v_p(\sigma(a))>v_p(a)$ and $v_p(\sigma(a'))>v_p(a')$, then $e_1=v_p(a)$, $e_2=v_p(a')$ and $f>e$. Hence an eligible pair has $v_p(a)=v_p(a')$, and its factor at $p$ is exactly $p^{v_p(a)}p/(p-1)$.

Proof. $v_p(s(a))=v_p(\sigma(a)-a)=v_p(a)$, and $f\ge\min(v_p(\sigma(a)),v_p(\sigma(a')))\ge e+1$. Lemma 1 excludes $e_1\ne e_2$ and gives the third case. $\square$

Good cofactors

Let $L=\log\log y$ and $W=L/(4\log L)$, and fix $0<\nu<1/10$. For a cofactor $a$ put $Q=P(a)$, $c=a/Q$, $Q_2=P(c)$ and $c_2=c/Q_2$, and write $W(a)=\prod_{p\le W}p^{v_p(a)}$. Call $a$ good if

1. $Q>y^\nu$, $Q_2>y^\nu$, $Q\nmid c$ and $Q_2\nmid c_2$;
2. $v_p(\sigma(c))>v_p(c)$ for every prime $p\le W$;
3. $(a,\sigma(a))$ has no prime factor larger than $W$.

Then $a=c_2Q_2Q$ with pairwise coprime factors, and condition 2 implies $v_p(\sigma(a))>v_p(a)$ for $p\le W$, since $\sigma(a)=\sigma(c)(Q+1)$ and $v_p(a)=v_p(c)$. Condition 3 has two consequences used below. If $p>W$ divides $s(a)$, then $p\nmid a$ and $p\nmid\sigma(a)$; and $p\nmid s(c)$, since otherwise $p\mid\sigma(c)$ by $s(a)=s(c)Q+\sigma(c)$, hence $p\mid c$, and $p$ would divide both $a$ and $\sigma(a)$.

Lemma 0 (density of good cofactors). As $y\to\infty$, the cofactors $2\le a<y$ that are not good satisfy $\sum1/a\le\bigl(e^\gamma\nu(1+\log(1/\nu))+o(1)\bigr)\log y$.

Proof. Write $\mu(S)=\sum_{a\in S,\,a<y}1/a$, and use Euler products throughout: for a nonnegative multiplicative $f$, $\sum_{n<y}f(n)/n\le\prod_{q<y}\sum_{j\ge0}f(q^j)q^{-j}$.

Condition 1. The $a$ with $P(a)\le y^\nu$ have $\mu\le\prod_{q\le y^\nu}(1-1/q)^{-1}=e^\gamma\nu\log y\,(1+o(1))$ by Mertens' product theorem. Those with exactly one prime factor $q>y^\nu$ are $q^jm$ with $P(m)\le y^\nu$, so $\mu\le\sum_{y^\nu<q\le y,,j\ge1}q^{-j}\cdot e^\gamma\nu\log y,(1+o(1)) =e^\gamma\nu(\log(1/\nu)+o(1))\log y$. Those divisible by $q^2$ for some prime $q>y^\nu$ have $\mu\le(1+\log y)\sum_{q>y^\nu}q^{-2}=o(1)$. Every remaining $a$ satisfies condition 1.

Condition 3. The $a$ divisible by a prime $p\in(W,L]$ have $\mu\le(1+\log y)\sum_{W<p\le L}1/p=o(\log y)$, because $\log W=\log L-\log(4\log L)$. For a prime $p>L$ with $p\mid a$ and $p\mid\sigma(a)$, either $p^2\mid a$, costing $(1+\log y)/p^2$, or $p\,\|\,a$ and some prime power $q^j\,\|\,a/p$ has $p\mid\sigma(q^j)$, costing at most $(1+\log y)p^{-1}\sum_{q^j<y,\,p\mid\sigma(q^j)}q^{-j}$. The terms with $j\ge2$ have $q^j\ge p/2$ and sum to $O(p^{-1/2})$. The terms with $j=1$ have $q\equiv-1\pmod p$ and $q\ge2p-1$; the cited Brun–Titchmarsh inequality and partial summation give $\sum_{q<y,\,q\equiv-1\ (p)}1/q\le(2L+3)/(p-1)$. Summing over $p>L$ gives $(1+\log y)\cdot O(1/\log L+L^{-1/2})=o(\log y)$.

Condition 2. For $a$ satisfying condition 1, the $a$ with a given $c$ have weight $c^{-1}\sum_{P(c)<Q\le y/c}1/Q\le(\log(1/\nu)+o(1))/c$, since $P(c)>y^\nu$. So it suffices to show that the $c<y$ violating condition 2 have $\sum1/c=o(\log y)$. Fix a prime $p\le W$, let $\mathcal P_p$ be the primes $q\equiv-1\pmod p$ in $(W,y^\nu]$, and let $\omega_1(n)$ count the primes of $\mathcal P_p$ dividing $n$ exactly once. Each of them contributes at least $1$ to $v_p(\sigma(n))$. So if $c=p^kc_0$ with $p\nmid c_0$ violates the condition at $p$, then $\omega_1(c_0)\le k$. With $z=1/2$, the Euler product bound and Mertens' product theorem give $\sum_{n<y}z^{\omega_1(n)}/n\le e^\gamma\log y, \exp(-E_p/2+O(1/W))(1+o(1))$, where $E_p=\sum_{q\in\mathcal P_p}1/q$, because each $q\in\mathcal P_p$ changes its Euler factor by the factor $1-1/(2q)+O(q^{-2})$. Hence the violations with $v_p(c)=k$ have $\sum1/c\le p^{-k}2^ke^\gamma\log y\,e^{-E_p/2}(1+o(1))$, and those with $v_p(c)>k_0$ have $\sum1/c\le2(1+\log y)p^{-k_0-1}$. By the prime number theorem for progressions to moduli at most $\log\log y$, $E_p\ge(1-o(1))L/(p-1)\ge(4-o(1))\log L$ for $p\le W$. Take the lower part of this estimate only from primes above $T_0=\exp(L^2)$: partial summation of Siegel–Walfisz on $[T_0,y^\nu]$ gives $(L-2\log L+\log\nu)/(p-1)+O(\exp(-cL))$, uniformly for $p\le W$, which justifies the stated lower bound. Take $k_0=\lceil2\log L\rceil$. For $p=2$ the first bound sums over $k\le k_0$ to $(k_0+1)e^{-E_2/2}\log y\,O(1)$, with $E_2\ge(1-o(1))L$; for $p\ge3$ it sums to at most $3e^\gamma e^{-E_p/2}\log y\,(1+o(1))$. Summing over $p\le W$ gives $\bigl(k_0e^{-L/3}+\pi(W)L^{-2+o(1)}+2^{-k_0}\bigr)O(\log y)=o(\log y)$. $\square$

Compatibility

For an eligible pair of good cofactors, Lemma 2 forces $W(a)=W(a')$, and then

$$g\,\mathfrak S_x(a,a')\le W(a)\prod_{p\le W}\frac p{p-1}\cdot g'\cdot F, \qquad F=\prod_{\substack{p>W\\ p\mid\beta\lambda(h/g)}}\frac p{p-1},$$

where $g'$ is the part of $g$ composed of primes above $W$.

Lemma 3. Let $(a,a')$ be an eligible pair of good cofactors and let $d\ge1$ divide $g'$. Then:

1. $(s(c),d)=(c,d)=1$, and $Q\equiv q_c(d)\pmod d$, where $q_c(d)\equiv-\sigma(c)\,s(c)^{-1}\pmod d$; likewise for $a'$.
2. $d\mid a'-a$.
3. $d$ divides $\Delta'(c;a')=a'\,s(c)+c\,\sigma(c)$.
4. The polynomial $F_{c_2,a'}(t)=c_2\sigma(c_2)t^2+(c_2\sigma(c_2)+a's(c_2))t +a'\sigma(c_2)$ satisfies $F_{c_2,a'}(Q_2)=\Delta'(c;a')$, and it is nonzero modulo every prime $p\mid d$. Hence $Q_2$ lies in at most $2^{\omega(d)}$ residue classes modulo the radical of $d$.

Proof. Every prime of $d$ exceeds $W$ and divides $s(a)$, so the consequences of condition 3 give the coprimality, and $d\mid s(a)=s(c)Q+\sigma(c)$ gives the class of $Q$. Item 2 follows from $s(a)-s(a')+h=a'-a$ and $d\mid g\mid h$. Then $a'\equiv a=cQ\equiv-c\,\sigma(c)s(c)^{-1}\pmod d$, which is item 3. For item 4, substitute $c=c_2t$, $\sigma(c)=\sigma(c_2)(t+1)$ and $s(c)=s(c_2)t+\sigma(c_2)$. If all three coefficients vanished modulo $p$, then $p\mid a's(c_2)$. If $p\mid a'$, then $p\mid\sigma(a')$ because $p\mid s(a')$, against condition 3 for $a'$. Otherwise $p\mid s(c_2)$ and $p\mid\sigma(c_2)$, so $p$ divides $c_2\mid a$ and $\sigma(c_2)\mid\sigma(a)$, against condition 3 for $a$. A nonzero polynomial of degree at most two has at most two roots modulo $p$, and $Q_2$ is a root modulo each $p\mid d$ by item 3. $\square$

The point of item 4 is that the congruence $a\equiv a'\pmod d$ costs a second prime of $a$, not only its largest one. The largest prime supplies $d\mid s(a)$ and the second supplies the compatibility with $a'$.

Lemma 6 (simple roots after a negligible removal). Among good cofactors, the ones for which some prime $p>W$ divides both $s(a)$ and $Q-Q_2$ have total reciprocal weight

$$\ll_\nu \frac{\log y}{W} +y^{-\nu/2}(1+\log y)^3=o(\log y).$$

Call a good cofactor separated if it survives this removal. For an eligible pair of separated good cofactors and every $d\mid g'$, the admissible values of $Q_2$ lie in at most $2^{\omega(d)}$ classes modulo $d$, not merely modulo its radical.

Proof of the removal bound. Put $c=a/Q$, so $c<y^{1-\nu}$, and let $Q_2=P(c)$. A prime counted in the removal satisfies

$$p\mid T(c):=s(c)Q_2+\sigma(c),\qquad Q\equiv Q_2\pmod p.$$

The positive integer $T(c)$ is at most $2y^2(1+\log y)$ and has $O(1+\log y)$ distinct prime factors. For $p>y^{\nu/2}$, a single residue class modulo $p$, restricted to integers above $y^\nu$ and at most $y$, has reciprocal sum at most $y^{-\nu}+(1+\log y)/p$. Summing this bound over $p\mid T(c)$ and then over $c$ with weight $1/c$ gives the second term of the assertion. Primality is not needed for this part.

For $W<p\le y^{\nu/2}$ write $c=c_2Q_2$. Multiplicativity gives

$$T(c)=s(c_2)Q_2^2+2\sigma(c_2)Q_2+\sigma(c_2).$$

Since $p\mid s(a)$ and $a$ is good, $p\nmid a\sigma(a)$, and therefore $p\nmid c_2\sigma(c_2)$. The displayed polynomial is nonzero modulo $p$ and has at most two roots. For fixed $c_2,p$, Brun–Titchmarsh and partial summation bound the reciprocal sum over the allowed $Q_2>y^\nu$ by $O_\nu(1/(p-1))$. For each such $Q_2$ the same bound applies to $Q>y^\nu$ in its class modulo $p$. Both uses have $p\le y^{\nu/2}$; the prime parameters are larger than $p^2$. Dropping ordering and product restrictions only increases these positive sums. The resulting bound is

$$O_\nu(1+\log y)\sum_{p>W}\frac1{(p-1)^2} \ll_\nu \frac{\log y}{W}.$$

Proof of the prime-power root bound. Write $A=c_2$ and $F(t)=F_{A,a'}(t)$ as in Lemma 3. The identities $a'\equiv A Q_2Q\pmod d$ and

$$s(a)=s(A)Q_2Q+\sigma(A)(Q_2+Q+1)\equiv0\pmod d$$

give

$$F'(Q_2)\equiv A\sigma(A)(Q_2-Q)\pmod d.$$

Every factor on the right is a unit modulo $d$: the first two by goodness, and the last by separation. Thus every root that can represent a retained cofactor is simple modulo each prime dividing $d$. A simple root modulo $p^j$ has exactly one lift modulo $p^{j+1}$, since $F(t+p^j u)\equiv F(t)+p^j u F'(t)\pmod{p^{j+1}}$. There are at most two simple roots modulo each $p$, and the Chinese remainder theorem gives at most $2^{\omega(d)}$ classes modulo $d$. $\square$

The extra removal costs $o(x)$ inputs in $S_1(x)$ by the reciprocal-weight estimate in "Setting". It does not assume that the large-prime part of $s(a)$ is squarefree or cubefree. Proposition 4 below holds for all good cofactors; the subsequent reductions of its remaining core use the separated subset.

The accessible divisors

Fix $0<\kappa<\nu/3$. Since $g'=\sum_{d\mid g'}\varphi(d)$, the G1 sum over eligible pairs of good cofactors is at most

$$\sum_{(a,a')}\frac{W(a)\prod_{p\le W}\frac p{p-1}\,F}{aa'} \sum_{d\mid g'}\varphi(d).$$

Call a divisor $d$ of $g'$ accessible for the pair if $\max(P(a),P(a'))>d^{1/(1-\kappa)}$, and write $\Lambda_{\rm acc}$ and $\Lambda_{\rm core}$ for the parts of this sum over accessible and inaccessible $d$.

Proposition 4. $\Lambda_{\rm acc}\ll_{\nu,\kappa}\log^2y$.

Main estimate. Consider the divisors with $P(a)>d^{1/(1-\kappa)}$; the case $P(a')>d^{1/(1-\kappa)}$ is symmetric. Fix a good $a'$ and $d\mid s(a')$ composed of primes above $W$. By Lemma 3, the good $a=c_2Q_2Q$ with $d\mid s(a)$, $a\equiv a'\pmod d$ and $Q>d^{1/(1-\kappa)}$ satisfy

$$\sum\frac1a\le\sum_{c_2\le y}\frac1{c_2} \sum_{\substack{Q_2>y^\nu\\ F_{c_2,a'}(Q_2)\equiv0\ (\mathrm{rad}\,d)}} \frac1{Q_2} \sum_{\substack{d^{1/(1-\kappa)}<Q\le y\\ Q\equiv q_c(d)\ (d)}}\frac1Q .$$

If $(q_c(d),d)>1$, a prime $p\mid d$ divides $\sigma(c)$, and $d\mid s(a)$ forces $Q=p<d^{1/(1-\kappa)}$; so the inner class is reduced whenever it is nonempty. For $t\ge\max(y^\nu,d^{1/(1-\kappa)})$ we have $t\ge2d$ for large $y$ and $\log(t/d)\ge\kappa\log t$, and the cited Brun–Titchmarsh inequality gives $\pi(t;d,b)\le2t/(\kappa\varphi(d)\log t)$. Partial summation from $z\ge y^\nu$ to $Z\le y$ bounds the inner sum by $K_1/\varphi(d)$, with $K_1=2(1+\log(1/\nu))/\kappa$.

Put $r=\mathrm{rad}\,d$. The $Q_2>r^{1/(1-\kappa)}$ in the at most $2^{\omega(d)}$ classes contribute at most $2^{\omega(d)}K_1/\varphi(r)$ in the same way; a class containing a multiple of a prime $p\mid r$ contains no such prime. Each class contains at most $2r^{2\kappa}$ integers up to $r^{1/(1-\kappa)}$, each above $y^\nu$. Therefore the $a$-sum is at most

$$(1+\log y)\,2^{\omega(d)} \Bigl(\frac{K_1}{\varphi(r)}+\frac{2r^{2\kappa}}{y^\nu}\Bigr) \frac{K_1}{\varphi(d)} .$$

Multiply by $\varphi(d)$, sum over $d\mid s(a')$ and then over $a'$ with weight $1/a'$. The second term contributes at most $y^{2\kappa-\nu+o(1)}$, because $d\le y\log y$ and $\sum_{a'<y}\tau(s(a'))^2/a'\le y^{o(1)}$. The first term contributes $K_1^2(1+\log y)\sum_{a'}\Phi(s(a'))/a'$, where

$$\Phi(N)=\sum_{\substack{d\mid N\\ p\mid d\Rightarrow p>W}} \frac{2^{\omega(d)}}{\varphi(\mathrm{rad}\,d)} =\prod_{\substack{p^k\|N\\ p>W}}\Bigl(1+\frac{2k}{p-1}\Bigr) =\sum_{e\mid N}\psi(e),$$

with $\psi$ multiplicative, supported on integers whose primes exceed $W$, and $\psi(p^j)=2/(p-1)$ for $j\ge1$. Swap the sums. For $e\le y^{\nu/2}$, the good $a'=c'Q'$ with $e\mid s(a')$ have $Q'$ in one class modulo $e$, reduced because otherwise $Q'$ would be a prime factor of $e$, and $Q'>y^\nu\ge e^2$, so their weight is at most $(1+\log y)K_1'/\varphi(e)$ with $K_1'$ depending only on $\nu$; and $\sum_e\psi(e)/\varphi(e)\le\prod_{p>W}(1+O(p^{-2}))$. For $e>y^{\nu/2}$, the primes $Q'>y^\nu$ of one residue class modulo $e$ have reciprocal sum at most $y^{-\nu}+(1+\log y)/e$, because after the first of them the class has one element in each interval of length $e$; so the weight is at most $(1+\log y)(y^{-\nu}+(1+\log y)/e)$. Since $\sum_{e\le y\log y}\psi(e)=y^{o(1)}$ and $\sum_e\psi(e)e^{-1/2}\ll1$, these $e$ contribute $o(1)$. Hence $\sum_{a'}\Phi(s(a'))/a'\ll\log y$, and the divisors with $P(a)>d^{1/(1-\kappa)}$ contribute $O(\log^2y)$ before the weights $W(a)\prod_{p\le W}p/(p-1)$ and $F$ are restored.

Restoring the small-prime weight. Restrict $c_2$ and $a'$ to $W(c_2)=W(a')=w$; this is forced, because $W(a)=W(c_2)$ when $Q,Q_2>W$. Writing $c_2=wm$ with $m$ free of primes up to $W$, the Euler product and Mertens' product theorem give $\sum_{c_2<y,,W(c_2)=w}1/c_2\le w^{-1}\prod_{W<q<y}(1-1/q)^{-1} =w^{-1}e^\gamma\log y\prod_{p\le W}(1-1/p),(1+o(1))$. The bounds for $Q_2$ and $Q$ do not depend on $w$, and the same factor appears for $a'$ in every step of the $\Phi$ estimate. So the weight $w\prod_{p\le W}p/(p-1)$ times the two factors $w^{-1}\prod_{p\le W}(1-1/p)$, summed over $W$-smooth $w$, gives $\prod_{p\le W}(1-1/p)\sum_w1/w=1$ times the unweighted bound.

The singular-series weight. Pointwise, $F\ll\log\log y$: the integer $\beta\lambda h/g<y^4$ has $O(\log y)$ prime factors, and the product of $p/(p-1)$ over the first $O(\log y)$ primes above $W$ is $O(\log\log y/\log W)$. This suffices wherever a power saving is available, in particular for the reductions of the core below. In the main term $F$ must be averaged, as follows.

Put $Z_0=(\log y)^2$. Primes of $T$ above $Z_0$ contribute a factor $1+o(1)$, since there are $O(\log y)$ of them. For $W<p\le Z_0$ let $h_3(p)=(p/(p-1))^3-1\le7/p$. Every such prime of $T$ divides $s(a)$, or $s(a')$, or $h$ while dividing neither $s(a)$ nor $s(a')$. By the arithmetic–geometric mean inequality,

$$F\le(1+o(1))\,\tfrac13\bigl(A(a)^3+A(a')^3+H^3\bigr),$$

where $A(n)=\prod_{p\mid s(n),\,W<p\le Z_0}p/(p-1)$ and $H=\prod_{p\mid h,\ p\nmid s(a)s(a'),\ W<p\le Z_0}p/(p-1)$. Expand each cube as $\sum_eh_3(e)$ over squarefree $e$ with the indicated prime factors.

For $A(a')^3$: $\Phi(N)A(N)^3=\sum_{e\mid N}\psi_3(e)$ with $\psi_3$ multiplicative and $\psi_3(p^j)\ll1/p$, and the $\Phi$ estimate applies unchanged.

For $A(a)^3$: a prime $p\mid e$ with $p\nmid d$ puts $Q$ in one class modulo $p$, because $p\nmid s(c)$, and $p\mid\sigma(c)$ would force $Q=p\le Z_0$. For $e\le E_0=y^{\kappa\nu/16}$, $Q$ then lies in one class modulo $de$, and $\log E_0\le(\kappa/4)\log Q$ lets Brun–Titchmarsh give $K_2/(\varphi(d)\varphi(e))$ in place of $K_1/\varphi(d)$, with $K_2$ depending on $\kappa,\nu$; and $\sum_eh_3(e)/\varphi(e)\ll1$. Primes $p\mid e$ that divide $d$ impose nothing new and contribute $\prod_{p\mid d}(1+7/p)$, which joins $\Phi$. For $e>E_0$, Rankin's trick gives $\sum_{e\mid s(a),,e>E_0}h_3(e)\le E_0^{-1/2}\prod_{p\le Z_0}(1+7p^{-1/2})=y^{-\kappa\nu/32+o(1)}$.

For $H^3$: a prime $p\mid e$ divides $\sigma(a')-\sigma(a)$ but neither $s(a)$ nor $s(a')$, so $p\nmid d$. If $p\nmid\sigma(c)$, then $Q\equiv\sigma(a')\sigma(c)^{-1}-1\pmod p$, one class, handled as for $A(a)^3$. If $p\mid\sigma(c)$, then $p\mid\sigma(a)$, so $p\mid\sigma(a')=\sigma(c')(Q'+1)$. If moreover $p\nmid\sigma(c')$, then $Q'\equiv-1\pmod p$, one class, absorbed by the Brun–Titchmarsh step for $Q'$ in the $\Phi$ estimate. The remaining primes divide both $\sigma(c)$ and $\sigma(c')$, and their factors multiply to at most $G(c)G(c')$, where $G(n)=\prod_{p\mid\sigma(n),\,W<p\le Z_0}(1+7/p)^{1/2}$.

The weight $G$ has bounded mean. Since $\sigma(c)=\sigma(c_2)(Q_2+1)$, $G(c)\le G(c_2)\prod_{p\mid Q_2+1,\,W<p\le Z_0}(1+7/p)^{1/2}$, and the second factor expands into conditions $Q_2\equiv-1\pmod f$, absorbed by the Brun–Titchmarsh step for $Q_2$.

Here the simultaneous congruences can be kept throughout the sums. A prime dividing $d$ divides neither $\sigma(a)$ nor $\sigma(a')$ on the good set. Thus the new moduli from $H$ and from $Q_2+1$ are coprime to $d$. On the long $Q_2$ range in the main estimate, a modulus $f\le E_0$ therefore changes its root-class bound to $O_{\nu,\kappa}(2^{\omega(d)}/(\varphi(r)\varphi(f)))$. Indeed $rf\le Q_2^{1-\kappa}y^{\kappa\nu/16} \le Q_2^{1-15\kappa/16}$, and the Chinese remainder theorem combines the polynomial roots modulo $r$ with the class $-1$ modulo $f$. The short $Q_2$ range is already a power-saving error and can use the pointwise polylogarithmic bound for $F$.

The primes assigned to a congruence on $Q$ multiply its original modulus $d$ by at most $E_0$, with the same margin. The primes assigned to $Q'\equiv-1$ multiply the small modulus $e\le y^{\nu/2}$ in the $\Phi$ calculation by at most $E_0$. These two moduli on $Q'$ are coprime: a common prime would divide both $s(a')$ and $\sigma(a')$, contradicting goodness. Their product is at most $Q'^{1/2+\kappa/16}$. All reciprocal prime sums consequently keep their uniform Brun–Titchmarsh bounds, with the additional factors $1/\varphi(f)$ for the added moduli. Summing their nonnegative expansion coefficients converges, since each new prime contributes $O(p^{-2})$. For any added modulus above $E_0$, the same Rankin bound as for $A(a)^3$ is a negative power of $y$ times $y^{o(1)}$; pointwise polylogarithmic bounds handle the other weights. This also covers the large-$e$ remainder of the $\Phi$ calculation.

For $c_2=wm$ with $m$ free of primes up to $W$, $G(c_2)\le G(w)G(m)$, and the Euler product gives $\sum_{m<y}G(m)/m\le\prod_{W<q<y}(1-1/q)^{-1} \exp\bigl(\sum_q(g(q)-1)/q+O(1/W)\bigr)$ with $g(q)=\prod_{p\mid q+1,\,W<p\le Z_0}(1+7/p)^{1/2}=\sum_{f\mid q+1}\xi(f)$, where $\xi$ is multiplicative on squarefree $f$ with primes in $(W,Z_0]$ and $\xi(p)\le4/p$. The partial summation of Lemma 0 gives $\sum_{q<y,\,q\equiv-1\ (f)}1/q\le(2L+5)/\varphi(f)$ for $f\le y^{1/2}$, and the trivial bound $2\log y/f$ holds for larger $f$. Hence $\sum_q(g(q)-1)/q\le(2L+5)\sum_{f>1}\xi(f)/\varphi(f)+o(1) =O(L/(W\log W))=O(1)$. The same applies to $c'$ on the $a'$-side. Finally $\sum_wG(w)^2/w\le(1+o(1))\prod_{p\le W}(1-1/p)^{-1}$ over $W$-smooth $w$, because $\sigma(p^k)<p^{k+1}$ has at most $(k+1)\log p/\log W$ prime factors above $W$. So the factors $G(w)^2$ cost $1+o(1)$ against $\sum_w1/w$. Large moduli $e$ and $f$ are removed by Rankin's trick as above. Hence the main estimate holds with the weight $F$, up to a constant factor depending on $\nu$ and $\kappa$. $\square$

The open core

The remaining part is

$$\Lambda_{\rm core} =\sum_{(a,a')}\frac{W(a)\prod_{p\le W}\frac p{p-1}\,F}{aa'} \sum_{\substack{d\mid g'\\ d^{1/(1-\kappa)}\ge\max(P(a),P(a'))}} \varphi(d),$$

Here and throughout this section the sum is restricted to eligible pairs of separated good cofactors. The discarded reciprocal weight is still at most the bound in Lemma 0 plus $o(\log y)$. We have $d>y^{\nu(1-\kappa)}$, and $d\le\min(|a'-a|,s(a),s(a'))<y$. When $d>P(a)$, item 1 of Lemma 3 pins the largest prime of $a$:

$$P(a)\ \text{is the least positive residue of}\ -\sigma(c)\,s(c)^{-1} \pmod d ,$$

and likewise for $a'$. So the core counts pairs of cofactors whose values of $s$ share a divisor that also divides their difference and is nearly as large as every prime factor of either cofactor; when it exceeds them, each largest prime is a least residue of a modular fraction formed from the rest of its cofactor.

A narrower core. Fix $\kappa''>0$ with $2\kappa+\kappa''<\nu$. Fix $a'$ and $d\mid s(a')$. Lemma 6 puts $Q_2$ in at most $2^{\omega(d)}$ classes modulo $d$, and the primes above $y^\nu$ in one class have reciprocal sum at most $y^{-\nu}+(1+\log y)/d$.

If $Q=P(a)\ge d^{1-\kappa''}$, then $Q\le d^{1/(1-\kappa)}$ lies in one class modulo $d$, so at most $2d^{2\kappa}$ values occur, each with $1/Q\le d^{\kappa''-1}$. After the factor $\varphi(d)\le d$ and the sum over $c_2$, the $a$-sum is at most $(1+\log y)2^{\omega(d)+1}d^{2\kappa+\kappa''} (y^{-\nu}+(1+\log y)d^{-1})$. Since $y^{\nu(1-\kappa)}\le d\le y\log y$, summing over $d\mid s(a')$ and over $a'$ gives $y^{-c+o(1)}$ for some $c=c(\nu,\kappa,\kappa'')>0$.

Lemma 5. Let $c,c'\ge2$ satisfy $v_p(\sigma(c))>v_p(c)$ and $v_p(\sigma(c'))>v_p(c')$ for every prime $p\le W$, let $(c,\sigma(c))$ and $(c',\sigma(c'))$ have no prime factor above $W$, and let $W(c)=W(c')$. If $c\ne c'$, then $\Delta(c,c')=c\,\sigma(c)s(c')-c'\sigma(c')s(c)\ne0$. The hypotheses hold for $c=a/P(a)$ and $c'=a'/P(a')$ when $a,a'$ are good and $W(a)=W(a')$.

Proof. $\Delta(c,c')=0$ means $c+c^2/s(c)=c'+c'^2/s(c')$. The first hypothesis gives $v_p(s(c))=v_p(c)$ for $p\le W$, and by the second no prime above $W$ divides both $c$ and $s(c)$. So $(c^2,s(c))=W(c)$, and $(c^2/W(c))/(s(c)/W(c))$ is $c^2/s(c)$ in lowest terms. Two rationals that differ by an integer have the same reduced denominator, so $s(c)/W(c)=s(c')/W(c')$, and $W(c)=W(a)=W(a')=W(c')$ gives $s(c)=s(c')$. Since $t\mapsto t+t^2/s$ is increasing for $s>0$, $c=c'$. For good cofactors, condition 2 is the first hypothesis, condition 3 gives the second because $c\mid a$ and $\sigma(c)\mid\sigma(a)$, and $W(c)=W(a)$ because $P(a)>W$. $\square$

Both largest primes below $d$. If $P(a),P(a')<d$, then each is the least positive residue of its class, so $c\ne c'$ (otherwise $a=a'$), the pair is determined by $(c,c',d)$, and $d\mid\Delta(c,c')$, which follows from item 3 of Lemma 3 after multiplying by $s(c')$. Lemma 5 makes $\Delta(c,c')$ a nonzero integer of size $y^{O(1)}$, so it has $y^{o(1)}$ divisors. Hence the pairs with $P(a)P(a')\ge d^{1+\epsilon}$ contribute at most $\sum_{c,c'}(cc')^{-1}\sum_{d\mid\Delta(c,c')}d^{-\epsilon}$ over $d\ge y^{\nu(1-\kappa)}$, which is $y^{-c+o(1)}$. Since $P(a)P(a')>y^{2\nu}$, this covers every such pair with $d\le y^{2\nu/(1+\epsilon)}$.

Two large primes on one side. If $P(a)<d$ and $P(a)P_2(a)\ge d^{1+\epsilon}$, fix $a'$ and $d\mid s(a')$ and restrict $Q$ and $Q_2$ to dyadic ranges $[R,2R)$ and $[R_2,2R_2)$ with $RR_2\ge d^{1+\epsilon}/4$. Then $Q$ is determined by $(c_2,Q_2)$, and $Q_2$ takes at most $2^{\omega(d)}(R_2/d+1)$ values, so $\sum1/(QQ_2)\le2^{\omega(d)}(1/(dR)+1/(RR_2))\le2^{\omega(d)+3} d^{-1-\epsilon/2}$, using $R\ge(RR_2/2)^{1/2}$. After the factor $\varphi(d)$, the sums over $c_2$, the dyadic ranges, $d\mid s(a')$ and $a'$ give $y^{-c+o(1)}$. The same holds with $a$ and $a'$ exchanged.

These parts contribute $o(1)$, also with the weights restored: the small-prime weight as in Proposition 4, and $F$ through its pointwise bound $F\ll\log\log y$ against the power savings. What remains consists of divisors exceeding $P(a)^{1/(1-\kappa'')}$ and $P(a')^{1/(1-\kappa'')}$, for which the two largest of $P(a),P_2(a),P(a'),P_2(a')$ have product below $d^{1+\epsilon}$. Necessarily $d>y^{2\nu/(1+\epsilon)}$. Both largest primes are least residues of their classes, determined by the rest of their cofactors. There is no restriction on the radical of $d$: Lemma 6 supplies the prime-power root count for all retained pairs.

Gap G1-core. Prove $\Lambda_{\rm core}\ll\log^2y$ for fixed small $\nu$ and $\kappa$. By the reductions above it suffices to bound the remaining regime, with its weights. In a model where $s(a)$ and $s(a')$ are independent of $a$ and $a'$ modulo large $d$, the inner condition has probability about $d^{-3}$, and the core is $O(y^{-\nu/2}\log^2y)$. No such independence is proved. The estimate is therefore expected but unknown: it is supported only by that model, and no source read here supplies it.

The tools above stop exactly here. Brun–Titchmarsh needs a free prime whose range exceeds the modulus $d$, and in the core no prime of either cofactor has that range. Enumerating $d$ as a divisor of one value of $s$ and bounding the progression $a'\equiv a\pmod d$ trivially loses the factor $d$ that the condition $d\mid s(a')$ should supply; the result is too large by about the average number of divisors of $s(a)$ above $y^{\nu/2}$. The two-dimensional large sieve bounds the total variance of $(a,\sigma(a))$ over all residue pairs modulo $d$, but that variance may sit on the diagonal residues $a\equiv\sigma(a)$ counted here, and it reaches only $d$ up to about $y^{1/2}$. A proof must show that these least residues fall in short initial intervals no more often than their length predicts, on average over the pairs.

Lemma 7 (separation of bounded reduced ratios). Put $\Pi_W=\prod_{p\le W}p$ and $C_\delta=1+2/\delta$. For an eligible pair of good cofactors in $\mathcal A_\delta(y)$, let $u=s(a)/g$, $v=s(a')/g$ and $H=h/g$. Then

$$\Pi_W\mid H,\qquad 0<|H|<C_\delta\max(u,v),\qquad (uv,\Pi_W)=1.$$

In particular $\max(u,v)>\Pi_W/C_\delta$. For every fixed $K$, no such pair has $\max(u,v)\le K$ for sufficiently large $y$. This includes $s(a)=s(a')$, for which $u=v=1$.

Proof. At every $p\le W$, Lemma 2 gives $v_p(s(a))=v_p(s(a'))=v_p(g)$ and $v_p(h)>v_p(g)$. This proves the divisibility and coprimality. Also $\sigma(a)\le C_\delta s(a)$ and $\sigma(a')\le C_\delta s(a')$, while both divisor sums are positive. Consequently $0<|h|<\max(\sigma(a),\sigma(a'))\le C_\delta g\max(u,v)$. Finally $W\to\infty$, so $\Pi_W\to\infty$. $\square$

The core is a scaled collision problem at the cofactor scale. For a core divisor $d$ put $\lambda=s(a)/d$ and $\lambda'=s(a')/d$. Then

$$\lambda'\,s(a)=\lambda\,s(a'),\qquad a'\equiv a\pmod{s(a)/\lambda},\qquad \lambda\le\frac{s(a)}{P(a)^{1-\kappa}}\le\frac{(1+\log y)\,a}{P(a)^{1-\kappa}},$$

and likewise for $\lambda'$. These coefficients have size at most $y^{1-\nu/2}$ for large $y$. Lemma 7 excludes bounded coefficients after their common factor is cancelled; in particular ordinary cofactor collisions contribute no eligible pair once $y$ is large. An induction using only ordinary fiber sizes therefore has no supplied bridge to this core. It needs a bound for the displayed weighted scaled collisions with unbounded reduced coefficients and with the extra congruence retained.

Removing the largest prime does not close that induction by itself: substitution of $a=cQ$ and $a'=c'Q'$ changes the scaled equality to

$$\lambda'(Q+1)s(c)-\lambda(Q'+1)s(c') =\lambda c'-\lambda'c.$$

The right side need not vanish, and the new coefficients depend on the removed primes. Thus the zero-right-side scaled equality alone is not an invariant induction hypothesis. A successful descent must control these affine equations, their congruence constraints and their summed weights; no contracting recurrence is established here.

Luca and Pomerance avoid this regime in the proof of their positive-density range theorem (15-page preprint, Section 3.1). Their cofactors are $m=q\ell$ with $\ell\le x^{1/10}$, and a common divisor $h$ of $s(m)$ and $s(m')$ divides $\ell\sigma(\ell)s(\ell')-\ell'\sigma(\ell')s(\ell)$, the analogue of $\Delta'$ with both cofactors decomposed. That integer is smaller than $2x^{3/10}$ in absolute value, so a divisor $h>x^{1/3}$ forces it to vanish, and the gcd properties of their Lemma 1 then force $\ell=\ell'$. For general cofactors the analogous integer has size up to about $y^3$, and no such forcing occurs.

Consequence for the large-prime class

G0, Proposition 4 and the core estimate together give, for each $\varepsilon>0$ and the corresponding removal of at most $\varepsilon x$ inputs, $C_{S_1}(x)\ll_{\delta,\eta,\varepsilon}x$ for the remaining inputs. With the truncation argument of Theorem B this would show that for every set $A$ of density zero, the $n\le x$ with $s(n)\in A$ and $P(n)>x^{1/2+\eta}$ number $o(x)$, and hence that $s^{-1}(A)$ has upper density at most $1-\log2$. This conclusion is conditional on Gap G1-core; it is not proved here.

Standing

Proved here, author-recorded: Lemmas 0, 1, 2, 3, 5, 6 and 7; Proposition 4, including both weights; and the stated reductions of the core on separated good cofactors. Constants were not computed explicitly; each step records the inequality it uses.

Open: Gap G1-core in the remaining large-modulus regime. Nothing on this page is independently reviewed, and the trimmed estimate it targets is not G1 as stated for every cofactor.