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Exponent of $p$ in the prime factorization of $n!$ is given by $\large \sum \limits_{i=1}^{\lfloor\log_p n \rfloor } \left\lfloor \dfrac{n}{p^i}\right\rfloor $. Can this sum be simplified further to some direct expression so that the number of calculations are reduced?

Martin Sleziak
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AgentS
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    This is an incredibly small calculation... –  Aug 09 '14 at 17:18
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    I am looking at @robjohn 's reply in below thread which shows that the sum simplifies to $\dfrac{n-\lfloor \log_p n\rfloor }{p-1}$ , but I don't really understand it and it is not giving correct answer for n=20 and p=2 http://math.stackexchange.com/questions/590885/proving-p-nmid-dbinomprmpr-where-p-nmid-m/590901#590901 – AgentS Aug 09 '14 at 17:19
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    robjohn's answer says that the sum simplifies to $\dfrac{n-\sigma_p(n)}{p-1}$, where $\sigma_p(n)$ is the sum of the digits of $n$ in the base-$p$ representation, just like David Holden's answer here. Where did you get the $\lfloor \log_p n\rfloor$ from? – Daniel Fischer Aug 09 '14 at 17:31
  • @DanielFischer, he seems to have figured out about the digit sum... if $i > \log_p n$ then $p^i > n$ and the floor of the fraction is zero. – Will Jagy Aug 09 '14 at 17:39
  • @WillJagy I was referring to the formula in the comment, not the bound on the sum. (Didn't even notice that there appeared a $\lfloor \log_p n\rfloor$ too.) – Daniel Fischer Aug 09 '14 at 17:41
  • @DanielFischer, you are absolutely right. I did not notice the incorrect log in his comment... – Will Jagy Aug 09 '14 at 17:44
  • FYI: There is an approximation for this found in Chapter 4 of Concrete Mathematics. Only really useful for very large numbers, though. (My book is packed up right now, and I don't know the exact formula off-hand.) – apnorton Aug 15 '14 at 21:01

3 Answers3

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yes:

$$ \frac{N-\sigma_p(N)}{p-1} $$ where $\sigma_p(N)$ is the sum of digits in the $p$-ary expression of $N$

David Holden
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  • I think $\sigma_p (N) $ is same as $\lfloor \log_p N\rfloor$ ?

    i am trying to work below example : exponent of 2 in prime factorization of $20!$ is $18$ : http://www.wolframalpha.com/input/?i=prime+factorization+of+20%21

    but your expression gives a different number : $\dfrac{20-\lfloor \log_2 20\rfloor }{2-1} = 16$

     – AgentS Aug 09 '14 at 17:33
  • @GaneshTadi No. In binary, $20$ is $10100$, so the digit sum of $20$ in base-$2$ is $2$. $\frac{20-2}{2-1} = 18$, as it should be. $\lfloor \log_p n\rfloor$ is one less than the number of digits in base $p$. – Daniel Fischer Aug 09 '14 at 17:35
  • if in binary $N=1101101$ then $\sigma_2(N) = 1+1+0+1+1+0+1=5$ – David Holden Aug 09 '14 at 17:35
  • Gotcha :) I see my error in interpreting the formula, thanks a lot xD – AgentS Aug 09 '14 at 17:37
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note what Daniel says. $\lfloor \log_p N\rfloor$ is the exponent of $p$ in $N$ not in $N!$. let $P_N$ be the exponent of $p$ in $N!$ and consider $P_{N+1}$.

if $N+1$ is not divisible by $p$ then the least significant $p$-ary digit of $N$ increases by $1$ and so $$ N+1 - \sigma_p(N+1) = N - \sigma_p(N) $$ and the exponent is unchanged.

suppose $N+1$ is divisible by $p^r$ for $r \gt 0$ but not by $p^{r+1}$ then each of the $r$ least significant binary digits must take the value $p-1$ but $1$ is added to the $r^{\text{th}} $ digit, which is not equal to $p-1$ hence: $$ \sigma_p(N+1) = \sigma_p(N)+1 - r(p-1) $$ and $$ N+1 - \sigma_p(N+1) = N - \sigma_p(n) + r(p-1) $$ hence $$ P_{N+1} = \frac{N+1 - \sigma_p(N+1) }{p-1} \\ = P_N + r $$

David Holden
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The idea of this theorm is to reduce manual calculation , try to find exp of 2 for (23263662!) :P so i guess its fair to follow the theorm . Origion theorm was $\large \sum \limits_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor$ , the floor log n function come from condition that i<=n thus any i after n give term of zero.

Bswan
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