GPT 6 Astra Ultra attempt on Math-798

Erdős Problem 396: carry barriers and explicit certificates

Statement and outcome

For every $k\ge0$, must some $n>k$ satisfy \[ D_k(n):=\prod_{i=0}^k(n-i)\ \bigm|\ \binom{2n}{n}? \] We prove exact valuation criteria and structural barriers, and give fully checkable witnesses through $k=4$. The arbitrary-$k$ existence question remains unresolved. Source: https://www.erdosproblems.com/396. The older local draft left even $k=2$ without a witness; the candidates below are recorded in https://oeis.org/A375077 and are independently verified here. No minimality claim or new computational record is made.

Exact valuations and an exponential size barrier

Counting multiples of $p,p^2,\ldots$ in a factorial proves \[ v_p\binom{2n}{n} =\sum_{j\ge1}\left(\left\lfloor\frac{2n}{p^j}\right\rfloor -2\left\lfloor\frac n{p^j}\right\rfloor\right). \tag{1} \] Each summand is $0$ or $1$. Thus the desired divisibility is exactly the system \[ \sum_{i=0}^k v_p(n-i)\le v_p\binom{2n}{n} \quad\hbox{for every prime dividing }D_k(n). \tag{2} \] No other primes need checking.

For $p=2$, summing the binary expansion gives $v_2(n!)=n-s_2(n)$, where $s_2$ counts the one-bits. As $s_2(2n)=s_2(n)$, $1$ becomes $v_2\binom{2n}{n}=s_2(n)$. Since \[ D_k(n)=(k+1)!\binom n{k+1}, \] any witness must satisfy $s_2(n)\ge v_2((k+1)!)$. An integer with at least $r$ one-bits is at least $2^r-1$. Hence \[ n\ge2^{\,k+1-s_2(k+1)}-1. \tag{3} \] This is an exponential necessary lower bound up to a polynomial factor; it is not an existence construction.

If $n=p^a$ for an odd prime, $1$ is zero: for $j\le a$ the quotient $n/p^j$ is integral, and for $j>a$ one has $p^j>2n$. Thus even $n\mid\binom{2n}{n}$ fails. For $n=2^a$, its binary valuation is $1$, so only $a=1$ can work. Apart from $n=1$, the only prime-power witness for $k=0$ is therefore $n=2$.

A prime near the endpoint is impossible

Suppose $n>3k$ and one of $n,n-1,\ldots,n-k$ is a prime $p$. Then $2n/3<p\le n$, so $\lfloor2n/p\rfloor-2\lfloor n/p\rfloor=0$. For $n\ge5$, $p^2>2n$, and all higher terms in $1$ vanish. Therefore, for $n\ge5$ with $n>3k$, every entry of a successful block must be composite. Long composite blocks alone remain far from sufficient, because $2$ must hold at all their prime powers.

An explicit proof for $k=2$ and further witnesses

For $n=2480$, \[ D_2(n)=2^5\cdot3\cdot5\cdot7\cdot31\cdot37\cdot59\cdot67. \] Evaluating the finite sums $1$ gives respectively the exponents \[ \begin{array}{c|rrrrrrrr} p&2&3&5&7&31&37&59&67\\\hline v_p(D_2)&5&1&1&1&1&1&1&1\\ v_p\binom{4960}{2480}&5&4&3&1&1&1&1&1 \end{array} \] so $2$ proves this witness with no large-binomial rounding. For $k=3,n=8178$, the corresponding complete lists are \[ \begin{array}{c|rrrrrrrrrrr} p&2&3&5&7&13&17&29&37&47&73&109\\\hline v_p(D_3)&5&2&2&1&1&1&1&1&1&1&1\\ v_p\binom{2n}{n}&10&5&3&2&1&1&1&1&1&1&1 \end{array} \] and for $k=4,n=45153$ they are \[ \begin{array}{c|rrrrrrrrrrrrrr} p&2&3&5&7&13&17&23&29&43&83&151&163&173&277\\\hline v_p(D_4)&6&3&2&1&1&1&1&1&1&1&1&1&1&1\\ v_p\binom{2n}{n}&6&5&3&3&2&1&1&2&1&1&1&1&1&1 \end{array} \] Every displayed lower row dominates its upper requirement. Finally $k=0,n=1$ and $k=1,n=2$ are immediate, covering all five cases claimed.

Remaining gap

The valuation identities and all finite certificates are exact. They provide no uniform way to make the binomial valuations dominate the combined valuations of arbitrarily many consecutive factors. The known theorem treating a single shifted divisor, and the density-one theorem for factors $n+i$ with positive $i$, do not give this simultaneous negative-shift product. No such inference is used here. Completion rate estimate: 35%.