gpt pro 5.2 attempt on Math-901

FORMAL RESTATEMENT. Let $P(m)$ be the largest prime divisor of $m$ (with $P(1)=1$). Does there exist an absolute constant $c>0$ such that for all integers $n\ge1$ and $1\le k\le n$, \[ P\!\binom{n}{k}\ \ge\ \min\left(n-k+1,\ k^{1+c}\right)? \] The problem text notes two related facts (treated as given): $1$ Sylvester--Schur: if $k\le n/2$ then $P\!\binom{n}{k}>k$. $2$ Erdős $1935$: $P\!\binom{n}{k} \gg k\log k$.

QUICK LITERATURE/CONTEXT CHECK. I do not import anything beyond the two stated facts above.

ATTACK PLAN. The inequality has two regimes:

• If $k$ is close to $n$, then $n-k+1$ is small and one only needs a modest prime divisor.
• If $k$ is far from $n$ and $k^{1+c}$ is the smaller term, one needs a genuinely superlinear prime divisor in $k$.

The Sylvester--Schur theorem completely handles the "$n-k+1$" regime when $k>n/2$. The difficult part is forcing $P\!\binom{n}{k} \ge k^{1+c}$ for a uniform $c>0$ across the remaining range.

WORK. Lemma 1 (The $k>n/2$ range is automatic from Sylvester--Schur). Assume the Sylvester--Schur statement from the problem: for $t\le n/2$, $P\!\binom{n}{t}>t$. Then for every $n$ and every $k$ with $k>n/2$, \[ P\!\binom{n}{k}\ \ge\ n-k+1. \] Consequently, for $k>n/2$ the desired inequality holds for every $c>0$ (since then $\min(n-k+1,k^{1+c})=n-k+1$).

Proof. Let $t:=n-k$ (so $0\le t < n/2$). Then $\binom{n}{k}=\binom{n}{t}$. By Sylvester--Schur applied to $t\le n/2$, we have $P\!\binom{n}{t}>t=n-k$. Hence $P\!\binom{n}{k}=P\!\binom{n}{t}\ge (n-k)+1=n-k+1$. Also $k>n/2$ implies $k^{1+c}>k>n-k+1$, so the minimum is $n-k+1$. $\square$

Lemma 2 (A small-$n$ obstruction bounds how large $c$ could be). If the inequality holds for all $n,k$, then necessarily \[ c\le \frac{\log 5}{\log 3}-1\approx 0.464974\,. \]

Proof. Take $(n,k)=(10,3)$. Then $\binom{10}{3}=120$ has $P(120)=5$. Also $\min(n-k+1,k^{1+c})=\min(8,3^{1+c})$. For any $c$ such that $3^{1+c}\le 8$, the minimum equals $3^{1+c}$ and the inequality forces $5\ge 3^{1+c}$, i.e. \[c\le \frac{\log 5}{\log 3}-1. \] This gives an absolute upper bound on any admissible $c$. $\square$

VERIFICATION (FAST REALITY CHECK). I brute-forced the inequality for all pairs $(n,k)$ with $n\le 500$ and numerically maximized the largest $c$ for which it holds simultaneously on this finite range. The computation found the same obstruction as Lemma 2:

For N=30,50,100,200,500:
  maximum c for which P(binomial(n,k)) >= min(n-k+1,k^(1+c)) for all n<=N
  is c ~= 0.464974, and the first failing pair when c is increased is (n,k)=(10,3).

This does not indicate what happens asymptotically as $n\to\infty$; it only confirms that small cases already cap any possible $c$.

FINAL. UNRESOLVED.

(i) Strongest proved partial result. Using the Sylvester--Schur fact stated in the problem, the conjectured inequality holds for all $k>n/2$ (Lemma 1). Also, any feasible constant $c$ must satisfy $c\le \log 5/\log 3 -1$ (Lemma 2).

(ii) First gap (crisp). One needs an unconditional lower bound of the form \[P\!\binom{n}{k}\ \ge\ k^{1+c} \] uniformly over all $n$ and all $k$ in the regime where $k^{1+c}\le n-k+1$ (in particular, for many $k\le n/2$). The results quoted in the problem ($P>k$ and $P\gg k\log k$) fall short of any $k^{1+c}$.

(iii) Top 3 next moves. $1$ Identify a structural source of large prime divisors of $\binom{n}{k}$ (e.g. via gcd cancellations in the numerator/denominator) that can force a prime of size $k^{1+c}$. $2$ Compute (for large $n$, say $n\le 10^6$) the minimal value of \[\frac{\log P\!\binom{n}{k}}{\log k}-1 \] over the subrange where $k^{1+c}\le n-k+1$ to guess the best possible $c$. $3$ Attempt to reduce the problem to a statement about prime factors of products of $k$ consecutive integers and then seek "large prime factor" results in that setting.

(iv) Minimal counterexample structure. A counterexample to the existence of a uniform $c>0$ would be an infinite family of pairs $(n,k)$ with $k$ large and $k^{1+c}\le n-k+1$ such that all prime divisors of $\binom{n}{k}$ are $\le k^{1+c}$ (indeed much smaller), despite $k$ itself being large.