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I'm trying to prove this :

Let $f \in Z[X]$ then for sufficiently large $n$ we have $$s(f(n))<s(n!)$$ where $s$ is the sum of digits function.

What I have so far : I thought this must be true because for large $n$ , $n!$ has a lot more digits than $f(n)$ and so has a big chance of having a bigger digit sum .

We know that $f(n)$ has $\lfloor \lg(f(n)) \rfloor+1<\lg(f(n))+1$ digits so $s(f(n))<9\lg(f(n))+9<a\lg(f(n))$ for some super big constant $a$ .

Now if we could prove this we would be done : (let's denote with $x_n$ the number of digits of $n!$)

There exist a constant $c$ such that at least $cx_n$ of the digits of $n!$ are non-zero .

If this is true then :

$$s(n!)>cx_n>c\lg(n!)>a\lg(f(n))>s(f(n))$$ because $c\lg(n!)>a\lg(f(n))$ is true for large $n$ .

Even weaker such statements can work , for example :

At least $\frac{cx_n}{\ln n}$ of the digits of $n!$ are non-zero .

I would be interested in any thoughts regarding this or any other results regarding the digits of $n!$ . Thanks for help !

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Sum of digits is not easy to deal with. See OE S sequence A004152 and references there. Apparently the best lower bound so far (due to Sanna) is that $$ s(n!) >> \log n\; \log \log \log n $$ but this is much too weak for what you want.

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