gpt pro 5.4 attempt on Math-786

Problem 374. Growth of $D_k$ for products of factorials that are squares --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction via a genuine strengthening of the Round-1 lower bounds.

Round 1 already established \[ D_3(X)\gg \sqrt{X},\qquad D_4(X)\gg X,\qquad D_5(X)\gg \frac{X}{\log X},\qquad D_6(X)\gg \frac{X}{\log X}, \] but left the original question "is $D_6(X)\gg X$?" unresolved. In this round I will not restart from scratch. Instead, I will use the Round-1 semiprime strategy and strengthen it in two ways:

• obtain an explicit leading constant for the $D_5(X)$ lower bound, and
• prove the genuinely stronger statement \[ D_6(X)=\omega\!\left(\frac{X}{\log X}\right). \]

This does not settle the original linear-growth question for $D_6$, but it is a strict advance over Round 1.

2) Round-1 FOUNDATION USED

I use the following Round-1 results without reproving them.

• The factorial identities \[ x!(x-1)!=\frac{(x!)^2}{x}, \qquad x!(x-1)!y!(y-1)!=\frac{(x!y!)^2}{xy}. \]
• The Round-1 consequence that if $n=ab$ with $a,b>1$ and $a\neq b$, then $n\in F_6$.
• The Round-1 use of Erdős--Graham's statements that:
  • if a proper divisor of $n$ belongs to $\{2,3,5,7,11\}$, then $n\in F_5$;
  • for any fixed prime $q$, almost all numbers of the form $pq$ (with $p$ prime) do not belong to $F_4$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredient is that Erdős--Graham's Theorem 3 is not really special to $13$: their proof is explicitly said to work for $13$ and all larger primes. Therefore, for every fixed prime $q\ge 13$, almost all numbers of the form $pq$ fail to lie in $F_5$.

Once this is combined with the Round-1 observation $pq\in F_6$, one can union over arbitrarily many fixed primes $q\ge 13$. Because the reciprocal sum of the primes diverges, this forces \[ D_6(X)=\omega\!\left(\frac{X}{\log X}\right). \]

4) ATTACK PLAN (ROUND-2)

Round 1 left two concrete gaps:

• the $D_5(X)\gg X/\log X$ lower bound used only one semiprime family;
• the $D_6(X)\gg X/\log X$ lower bound used only the family $13p$.

To improve these, I will:

1. union the five $D_5$-families $qp$ with $q\in\{2,3,5,7,11\}$ and show the overlaps are only $O(1)$;
2. union any fixed finite collection of $D_6$-families $qp$ with $q\ge 13$ prime, again with only $O(1)$ overlap;
3. let the finite collection grow after $X\to\infty$ to force an arbitrarily large constant in front of $X/\log X$.

5) WORK (ROUND-2)

Lemma 5.1 (semiprimes with a small fixed factor give many elements of $D_5$)

Let \[ S_5:=\{2,3,5,7,11\}. \] Then \[ D_5(X)\ge \left(\sum_{q\in S_5}\frac1q+o(1)\right)\frac{X}{\log X}. \]

Proof.

Fix $q\in S_5$, and define \[ A_q(X):=\{qp\le X:\ p \text{ prime},\ p\neq q,\ qp\notin F_4\}. \] Because $q$ is a proper divisor of $qp$ and $q\in\{2,3,5,7,11\}$, Erdős--Graham's Fact 7 gives \[ qp\in F_5. \] Since by definition $qp\notin F_4$, every element of $A_q(X)$ lies in \[ F_5\setminus F_4=D_5. \] Hence \[ A_q(X)\subseteq D_5. \]

Now Erdős--Graham also prove that for any fixed prime $q$, almost all numbers of the form $pq$ do not belong to $F_4$. Therefore \[ |A_q(X)|=\pi(X/q)-o(\pi(X/q)) =\left(\frac1q+o(1)\right)\frac{X}{\log X} \] by the prime number theorem.

Next, if $q_1\neq q_2$ belong to $S_5$ and \[ q_1p_1=q_2p_2 \] with $p_1,p_2$ prime, then primality forces \[ p_1=q_2,\qquad p_2=q_1. \] So for each ordered pair $(q_1,q_2)$ there is at most one overlap, and therefore \[ \sum_{\substack{q_1,q_2\in S_5\\ q_1\neq q_2}} |A_{q_1}(X)\cap A_{q_2}(X)| = O(1). \] Hence \[ D_5(X)\ge \left|\bigcup_{q\in S_5} A_q(X)\right| = \sum_{q\in S_5}|A_q(X)| - O(1) = \left(\sum_{q\in S_5}\frac1q+o(1)\right)\frac{X}{\log X}. \] This proves the lemma. $\square$

Lemma 5.2 (fixed semiprime families in $D_6$)

Let $q\ge 13$ be a fixed prime. Then, for almost all primes $p\neq q$, \[ qp\in D_6. \]

Proof.

Fix such a prime $q$.

First, $qp\in F_6$ for every prime $p\neq q$. Indeed, taking the six-term set \[ \{qp,\ qp-1,\ q,\ q-1,\ p,\ p-1\}, \] we have \[ (qp)!(qp-1)!q!(q-1)!p!(p-1)! = \frac{\big((qp)!q!p!\big)^2}{(qp)\,q\,p} = \frac{\big((qp)!q!p!\big)^2}{q^2p^2}, \] which is a perfect square. Thus $qp\in F_6$.

Second, Erdős--Graham's Theorem 3 proves that for almost all primes $p$, one has $13p\notin F_5$; and immediately after the proof they explain that the same proof works for $13$ and all larger primes. Therefore, for our fixed $q\ge 13$, almost all primes $p$ satisfy \[ qp\notin F_5. \]

Combining the two statements, for almost all primes $p\neq q$, \[ qp\in F_6\setminus F_5=D_6. \] This proves the lemma. $\square$

Lemma 5.3 (finite unions of $D_6$-families)

Let $S$ be any fixed finite set of primes, all at least $13$. Then \[ D_6(X)\ge \left(\sum_{q\in S}\frac1q+o(1)\right)\frac{X}{\log X}. \]

Proof.

For each $q\in S$, define \[ B_q(X):=\{qp\le X:\ p \text{ prime},\ p\neq q,\ qp\notin F_5\}. \] By Lemma 5.2, for each fixed $q\in S$, \[ |B_q(X)|=\pi(X/q)-o(\pi(X/q)) = \left(\frac1q+o(1)\right)\frac{X}{\log X}, \] and every element of $B_q(X)$ belongs to $D_6$.

As in Lemma 5.1, if $q_1\neq q_2$ and \[ q_1p_1=q_2p_2 \] with $p_1,p_2$ prime, then necessarily \[ p_1=q_2,\qquad p_2=q_1. \] Hence the total overlap among the finitely many families $\{B_q(X)\}_{q\in S}$ is only $O_S(1)$. Therefore \[ D_6(X)\ge \left|\bigcup_{q\in S} B_q(X)\right| = \sum_{q\in S}|B_q(X)| - O_S(1) = \left(\sum_{q\in S}\frac1q+o(1)\right)\frac{X}{\log X}. \] This proves the lemma. $\square$

Corollary 5.4 (strict Round-2 advance)

One has \[ D_6(X)=\omega\!\left(\frac{X}{\log X}\right). \]

Proof.

Let $A>0$ be arbitrary. Because the reciprocal sum of the primes diverges, there exists a finite set $S$ of primes $q\ge 13$ such that \[ \sum_{q\in S}\frac1q > 2A. \] Now $S$ is fixed, so Lemma 5.3 applies: \[ D_6(X)\ge \left(\sum_{q\in S}\frac1q+o(1)\right)\frac{X}{\log X}. \] For all sufficiently large $X$ the parenthesis is at least $A$, and thus \[ D_6(X)\ge A\frac{X}{\log X}. \] Since $A>0$ was arbitrary, this is exactly \[ D_6(X)=\omega\!\left(\frac{X}{\log X}\right). \] $\square$

6) ADVERSARIAL VERIFICATION

Potential weak points and checks:

• Overlap problem. Two different semiprime families $q_1p$ and $q_2p'$ can coincide only in the trivial swapped-factor case; for a fixed finite set of $q$ this contributes only $O(1)$ points. So no density is lost here.

• Uniformity in $q$. I never need an error term uniform in $q$ over all large primes. In Lemma 5.3 the set $S$ is fixed before letting $X\to\infty$, so the individual $o(1)$ terms may depend on $q$.

• Could some $qp$ lie in $F_5$ and still belong to $D_6$? Yes, and I do not use them. The lower bound counts only the subfamily explicitly known to be outside $F_5$.

• Does this prove $D_6(X)\gg X$? No. The argument only yields arbitrarily large multiples of $X/\log X$, not a positive proportion of all integers. So the original question from Round 1 remains open.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Relative to Round 1, the new unconditional statements are: \[ D_5(X)\ge \left(\frac12+\frac13+\frac15+\frac17+\frac1{11}+o(1)\right)\frac{X}{\log X}, \] and \[ D_6(X)=\omega\!\left(\frac{X}{\log X}\right). \] The original question "is $D_6(X)\gg X$?" remains open.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 68%

9) REFERENCES

[EG76] P. Erdős and R. L. Graham, On products of factorials, Bull. Inst. Math. Acad. Sinica 4 $1976$, 337--355.

[DNSS14] A. Dujella, F. Najman, N. Saradha, and T. N. Shorey, Products of three factorials, Publ. Math. Debrecen 85 $2014$, 3--13.