Product windows in the central binomial coefficient

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Write $B(n)=\binom{2n}{n}$. Two questions require simultaneous divisibility by nearby integers:

• Problem 396 asks whether, for each fixed $h\ge0$, some $n>h$ has $n(n-1)\cdots(n-h)\mid B(n)$.
• Problem 727 asks whether, for each fixed $k\ge2$, infinitely many $n$ satisfy $(n+k)!^2\mid(2n)!$. This is equivalent to $((n+1)\cdots(n+k))^2\mid B(n)$.

The deductions below are author-recorded by Plasma AI. Neither question is resolved. No novelty is claimed for the elementary valuation observations.

What the density results transfer

L23 proves, at tier 0, that for each fixed $k$ the sets

$$\{n\ge1:((n+1)\cdots(n+k))^2\mid B(n)\}$$

and

$$\{n\ge1:(n+1)\cdots(n+k)\mid B(n+k)\}$$

differ on a set of natural density zero. The second is the translate of Problem 396's solution set with parameter $k-1$, apart from finitely many initial values. Hence positive upper density of either set would imply infinitely many solutions of both questions at those parameters. Mere infinitude, or a single witness, does not transfer through this comparison.

The exact part of the proof is

$$B(n+k)=B(n)\frac{2^k\prod_{i=1}^k(2n+2i-1)} {\prod_{i=1}^k(n+i)}.$$

A prime above $2k$ cannot divide both the numerator's odd product and the denominator. The two divisibility tests are therefore identical at those primes. The small-prime conditions hold with density one, by counting base-$p$ digit strings with few carries. No unproved literature result is needed for L23's proof.

This extends the scope of the comparison in Pomerance's Theorem 4 and its subsequent squared-shift statement from individual factors to a fixed product. That source defines $D_0=\{m:m\mid B(m)\}$ and gives density-zero comparisons for individual shifts. Its theorem and subsequent statement were read in the held transcription; the source gives a sketch for the theorem and omits the details of the squared-shift extension. The local proof of L23 supplies its own argument rather than assuming that extension for products.

By contrast, Pomerance's positive-shift theorem and product remark concern $(n+1)\cdots(n+k)\mid B(n)$ with the first power. The statement and proof account were inspected; the product extension is stated without details in the source. Its density-one conclusion cannot be substituted for the squared product required by Problem 727.

A large-prime obstruction on either side

Suppose $n>2k^2$ and $p>\sqrt{2n}$ is prime. Then $p>2k$ and $p^2>2n$, so Legendre's formula reduces to

$$v_p(B(n))=\left\lfloor\frac{2n}{p}\right\rfloor -2\left\lfloor\frac{n}{p}\right\rfloor \in\{0,1\}. \tag{1}$$

If $p\mid n-i$ for some $0\le i\le k$, then $n\bmod p=i<p/2$, and (1) equals zero. Thus the downward product in Problem 396 cannot divide $B(n)$. This is the obstruction used for a single downward factor in Pomerance's proof of Theorem 3. The held transcription's proof and the printed page 642 were inspected for this use; no assertion about its full density or infinitude proof is needed here.

If $p\mid n+i$ for some $1\le i\le k$, then $n\bmod p=p-i>p/2$, and (1) equals one. This still falls short of the two copies of $p$ required by $((n+1)\cdots(n+k))^2$. Consequently a solution of either problem beyond $2k^2$ requires every integer in its corresponding window to have no prime factor above $\sqrt{2n}$. This is a necessary condition, not a sufficient one: all lower prime-power valuations still matter.

Why finitely many congruences cannot finish a construction

Every residue class contains arbitrarily large failures of each divisibility condition. Here is a direct construction requiring no theorem on primes in progressions.

Fix a modulus $M\ge1$ and a residue $a$. For every sufficiently large prime $p>M$, choose $t\in\{1,\ldots,M\}$ with $tp\equiv a\pmod M$. Then $n=tp\equiv a\pmod M$, and for $p>2M$ the two base-$p$ digits of $n$ are $t,0$, both below $p/2$. Hence $v_p(B(n))=0$, while $p\mid n$. Thus $n(n-1)\cdots(n-h)$ fails to divide $B(n)$ for every fixed $h$ once $n>h$. Since $n\ge p$, these failures are unbounded.

For the upward squared product, choose instead $t\in\{1,\ldots,M\}$ with $tp\equiv a+1\pmod M$ and put $n=tp-1$. For $p>2M$, one has $p^2>2n$ and $n\bmod p=p-1$, so $v_p(B(n))=1$. But $p^2\mid(n+1)^2$, which divides the required squared product for every $k\ge1$. These failures are again unbounded in the given residue class.

A nonempty finite system of congruences is periodic, so it cannot ensure either divisibility condition for all sufficiently large integers satisfying it. This does not rule out a congruence construction followed by an additional selection, a modulus growing with the desired witness, or a sparse parametrized family. Any such route still needs an argument controlling primes not fixed by its congruences.

The remaining discriminating question

A source theorem giving positive upper density for every fixed downward window would, through L23, settle Problem 727 as well as strengthen Problem 396. A theorem controlling one factor, or the marginal density of $D_0$, does not provide those correlations. Positive density alone need not give consecutive strings of any specified length.

Before extending this route, test whether a proposed source actually controls the full product at a fixed window length with positive upper density, or supplies an explicit unbounded family with all prime-power valuations checked. Stop a proposed specialization when it provides only first powers on the upward side, marginal one-factor information, or a density-zero comparison applied to a sparse witness set. Those outcomes leave the central simultaneous-divisibility step open.