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Let $C_n$ denote the number of (monic) degree $n$ polynomials $P(x) = \prod_{t=1}^{n}(x-t) \pmod {n!}$ such that $n! | P(x)$ for all integers $x$. Here are the first four examples:

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Polynomials of degree 2 (mod 2) that are divisible by 2 for all x
x^2+x
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Polynomials of degree 3 (mod 6) that are divisible by 6 for all x
x^3 + 5*x
x^3 + 3*x^2 + 2*x
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Polynomials of degree 4 (mod 24) that are divisible by 24 for all x
x^4 + 2*x^3 + 11*x^2 + 10*x
x^4 + 2*x^3 + 23*x^2 + 22*x
x^4 + 6*x^3 + 11*x^2 + 6*x
x^4 + 6*x^3 + 23*x^2 + 18*x
x^4 + 10*x^3 + 11*x^2 + 2*x
x^4 + 10*x^3 + 23*x^2 + 14*x
x^4 + 14*x^3 + 11*x^2 + 22*x
x^4 + 14*x^3 + 23*x^2 + 10*x
x^4 + 18*x^3 + 11*x^2 + 18*x
x^4 + 18*x^3 + 23*x^2 + 6*x
x^4 + 22*x^3 + 11*x^2 + 14*x
x^4 + 22*x^3 + 23*x^2 + 2*x
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Trivially, we have

$C_1 = 1$ (since $x$ is the only such polynomial)

From the tables we have,

$C_2 = 1, C_3 = 2, C_4 = 12$.

For $C_5$, manual computations showed there are $288$ such polynomials.

For general $n$, a counting technique can be used where all possible polynomials $P(x)$ $\mod {}$ prime powers $p^k | n!$ are considered and using the Chinese Remainder Theorem.

The answer seems to be that $C_n = \prod_{t=1}^{n-1} t!$, but how would this be proven?

J. Linne
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  • How can you count the polynomials of degree $n$ $\bmod p^k$ that are always $0\bmod p^k$? – Asinomás Apr 03 '21 at 20:58
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    You see that the cubics vanishing modulo six are in the span of the descending factorial $x(x-1)(x-2)$ and $3x(x-1)$ (also $6x$, but that vanishes when coefficients are reduced modulo six). Similarly the polynomials vanishing modulo $4!$ are in the span of $x(x-1)(x-2)(x-3)$, $4x(x-1)(x-2)$ and $12x(x-1)$ et cetera. I think this leads to a proof of your observed formula. The top degree ($\deg =n$) generator can only appear with multiplier $1$, the one with degree one less has $(n-1)!$ possible multipliers leading to polynomials with distinct coefficients etc. – Jyrki Lahtonen Apr 03 '21 at 21:04
  • See this old thread for more explanations and a few references. – Jyrki Lahtonen Apr 03 '21 at 21:10

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