gpt pro 5.2 attempt on Math-6160

FORMAL RESTATEMENT. Let $p(n)$ denote the partition function of $n$ (number of integer partitions of $n$), with $p(0)=1$ and $p(n)\in\mathbb N$ for $n\ge 1$. For $n\ge 1$ define \[P(n):=\prod_{k=1}^n p(k),\qquad F(n):=\omega(P(n)), \] where $\omega(N)$ is the number of distinct prime divisors of a positive integer $N$. Questions:

• Does $F(n)\to\infty$ as $n\to\infty$?
• Is $F(n)>n$ for all sufficiently large $n$?

QUICK LITERATURE/CONTEXT CHECK (from the problem file only). The text reports: (i) Schinzel observed (Oberwolfach problem book) that $F(n)\to\infty$ follows from the asymptotic for $p(n)$ and a theorem of Tijdeman; details in Erdős--Ivić $1990$. (ii) Schinzel--Wirsing proved $F(n)\gg \log n$. (iii) Ono proved that every prime divides $p(n)$ for some $n\ge 1$ (indeed for a positive density of $n$ for each fixed prime). I do not reprove any of these external statements.

ATTACK PLAN.

• Prove basic structural facts about $F(n)$ (monotonicity) and give a clean conditional implication from (iii) to $F(n)\to\infty$.
• Provide computational sanity checks for $F(n)$ for moderate $n$.

WORK.

Lemma 1 (monotonicity). The function $F(n)$ is nondecreasing in $n$.

Proof. We have $P(n+1)=P(n)\,p(n+1)$. Thus the set of prime divisors of $P(n+1)$ is the union of the set of prime divisors of $P(n)$ with those of $p(n+1)$. In particular it contains the prime divisors of $P(n)$, so $\omega(P(n+1))\ge \omega(P(n))$, i.e. $F(n+1)\ge F(n)$. $\square$

Lemma 2 (conditional: Ono-type statement $\Rightarrow F(n)\to\infty$). Assume the following statement:

For every prime $q$ there exists an integer $m(q)\ge 1$ such that $q\mid p(m(q))$.

Then $F(n)\to\infty$.

Proof. Fix $y\ge 2$ and let $\mathcal P(y)$ be the set of primes $q\le y$. For each $q\in\mathcal P(y)$ choose $m(q)$ with $q\mid p(m(q))$. Let \[N(y):=\max\{m(q): q\le y,\ q\text{ prime}\}. \] Then for every $n\ge N(y)$ and every prime $q\le y$ we have $q\mid P(n)$ because $q\mid p(m(q))$ and $p(m(q))$ is a factor of $P(n)$. Therefore $F(n)=\omega(P(n))\ge |\mathcal P(y)|=\pi(y)$ for all $n\ge N(y)$. Since $\pi(y)\to\infty$ as $y\to\infty$, it follows that $F(n)\to\infty$ as $n\to\infty$. $\square$

FAST REALITY CHECK (computation of $F(n)$). I computed $p(k)$ for $1\le k\le n$, factored each $p(k)$, and tracked the union of prime factors. The exact values obtained include: \[ F(10)=5,\qquad F(50)=34,\qquad F(100)=96,\qquad F(200)=242,\qquad F(300)=412. \] In this computation, the first $n$ for which $F(n)>n$ was \[n=116\quad\text{with}\quad F(116)=117. \] These data strongly suggest $F(n)\to\infty$ and support the possibility that $F(n)>n$ eventually, but they are not a proof.

VERIFICATION.

• Lemma 1 is a direct set-inclusion argument on prime divisors.
• Lemma 2 is purely logical and does not use any properties of $p(n)$ beyond the assumption.
• Computation: $F(n)$ was computed as the size of the union of $\mathrm{primefactors}(p(k))$ for $k\le n$.

FINAL: UNRESOLVED. (i) Strongest proved partial result here. $F(n)$ is nondecreasing (Lemma 1). Moreover, if one assumes the statement "every prime divides some partition number" (which the problem text attributes to Ono), then $F(n)\to\infty$ follows immediately (Lemma 2).

(ii) First gap (crisp). I do not have a self-contained proof (without importing the cited deep results) that infinitely many distinct primes divide the set $\{p(1),p(2),\dots\}$, equivalently that $F(n)\to\infty$ unconditionally.

(iii) Top 3 next moves. 1. Use (and explicitly write out) the asymptotic formula for $p(n)$ together with a suitable "many small prime factors" theorem (as suggested in the problem text) to force new primes to appear in $\prod_{k\le n}p(k)$. 2. Attempt to show that for each fixed prime $q$, the congruence $p(n)\equiv 0\pmod q$ holds for infinitely many $n$ by elementary modular-form arguments (this is essentially what the text attributes to Ono). 3. Extend the computation to larger $n$ (with certified integer factorizations) to test the conjecture $F(n)>n$ and estimate growth rate.

(iv) Minimal counterexample structure. A counterexample to $F(n)\to\infty$ would require that all $p(1),p(2),\dots$ are composed of primes from a fixed finite set $S$; equivalently, $p(n)$ would be an $S$-smooth integer for all $n$, which seems incompatible with the rapid growth and varied congruence behavior of $p(n)$, but proving this incompatibility is the main obstacle.