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Let v_p(n) denote the p-adic valuation of n.

The number of times that a prime p appears in all numbers <= n is given by:

$$ \nu_p(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor, $$ where [ x ] is the floor function of x

For example I have calculated this expression for some primes in a pull n = 10^6. $$ \nu_2(10^6!) = 999993 $$ $$ \nu_3(10^6!) = 499993 $$ $$ \nu_5(10^6!) = 249998 $$ $$ \nu_7(10^6!) = 166664 $$ and so on

My question

Is there any asymptotic solution for
$$ \nu_p(n!) $$

I would need a general expression for doing an stochastic model.

Thank you

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    $\begingroup$ $\nu_p(n!)=(n-s_p(n))/(p-1)$ where $s_p(n)$ denotes the sum of the $p$-ary digits of $n$. $\endgroup$ – user72870 Jun 15 '15 at 16:14
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Let $k=\lfloor\log_pn\rfloor$. The formula in the question gives an upper bound $$\nu_p(n!)\le\sum_{i=1}^\infty\frac n{p^i}=\frac n{p-1},$$ and a lower bound $$\nu_p(n!)\ge\sum_{i=1}^k\left(\frac n{p^i}-1\right)\ge\frac n{p-1}-1-k,$$ hence asymptotically, $$\nu_p(n!)=\frac n{p-1}+O(\log_pn).$$

With a bit more care, one can compute $$\frac n{p-1}-\log_p(n+1)\le\nu_p(n!)\le\frac{n-1}{p-1},$$ where both bounds are tight: the upper bound is attained when $n$ is a power of $p$, and the lower bound is attained when $n$ is one less than a power of $p$.

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