# If a function is a product of factorials of polynomials, how can I prove that this representation is unique?

If a function is a product of factorials of polynomials, how can I prove that this representation is unique?

Specifically, let $F(x) = \prod_i{P_i(x)!^{a_i}}$, $F:\mathbb{N} \rightarrow \mathbb{Q}$, where each $P_i$ is a distinct non-negative, non-constant polynomial with integer coefficients (ordered lexically to exclude permutations as solutions), and each $a_i$ is a non-zero integer.

For example, $\binom{2 \cdot x}{x}$ and $\frac{(x^2+1)!^3}{(x^5+1)!}$ are in this class of functions. So is $x^7+9 = \frac{(x^7+9)!}{(x^7+8)!}$.

I'm requiring that the polynomials be non-negative only to avoid the issue of factorials of negative numbers (which I don't really want to discuss, although I am interested in any ways to simplify the problem statement). I'm requiring that they be non-constant because I realized that $(n!)! = (n!-1)! \cdot n!$ and this gives a class of solutions; for example, $F(x) = 6! = 5! \cdot 3!$.

So I am wondering: is such a representation unique? I can prove it for various cases (e.g. if the maximum degrees of the polynomials in the products differ, then the functions differ) but I don't know what to do in general.

Edit: After discussion with Dan in chat, I’m adding this alternative wording of the definition and question, in an effort to make the question a bit clearer. $-$Brian M. Scott.

Let $P$ be the set of non-constant polynomials in $\Bbb Z[x]$ with positive coefficients, viewed as functions from $\Bbb N$ to $\Bbb Q$. For $p\in P$ denote by $p!$ the function $n\mapsto p(n)!$. Let $$F=\left\{\prod_{k=1}^mp_k!^{a_k}:m\in\Bbb Z^+\;\land\;p_1,\dots,p_m\in P\;\land\;a_1,\dots,a_m\in\Bbb Z\setminus\{0\}\right\}\;.$$ $F$ is a set of functions from $\Bbb N$ to $\Bbb Q$.

Question: Suppose that $f(n)=\prod_{k=1}^mp_k!^{a_k}\in F$. Is this representation of $f$ as a member of $F$ unique up to a permutation of the $p_k$?

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## 1 Answer

I found a counterexample:

$\frac{(x^2+4x+4)!}{(x^2+4x+3)!} = x^2+4x+4 = (x+2)^2 = \frac{(x+2)!^2}{(x+1)!^2}$.

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This construction works for any $f(n) = f(n)!/(f(n)-1)!$ that resolves to a rational function, e.g. $(x+4)!^2/x!/(x+6)! = (x^3+9x^2+26x+24)/(x^2+11x+30)$ $= (x^3+9x^2+26x+24)!(x^2+11x+29)!/(x^3+9x^2+26x+23)!(x^2+11x+30)!$. – Zander May 22 '12 at 12:12
Sorry, example in previous comment should be $(x+4)!^2/(x+1)!/(x+6)!$. – Zander May 22 '12 at 12:20