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Difference between revisions of "Legendre's Formula"

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Let the base <math>p</math> representation of <math>n</math> be <math>e_xe_{x-1}e_{x-2}\dots e_0</math>, where <math>e_i</math> is a digit for all <math>0\leq i\leq x</math>. Therefore the base <math>p</math> representation of <math>\lfloor \frac{n}{p^i}\rfloor</math> is <math>e_xe_{x-1}\dots e_{x-i}</math>. Note that the infinite sum of these numbers (which is <math>e_p(n!)</math>) is
 
Let the base <math>p</math> representation of <math>n</math> be <math>e_xe_{x-1}e_{x-2}\dots e_0</math>, where <math>e_i</math> is a digit for all <math>0\leq i\leq x</math>. Therefore the base <math>p</math> representation of <math>\lfloor \frac{n}{p^i}\rfloor</math> is <math>e_xe_{x-1}\dots e_{x-i}</math>. Note that the infinite sum of these numbers (which is <math>e_p(n!)</math>) is
  
−
<math>\sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1)=\sum_{j=1}^{x} e_j(\frac{p^j-1}{p-1})=\frac{\sum_{j=1}^{x} e_jp^j -\sum{j=1}^{x} e_j}{p-1}</math>
+
<math>\sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1)=\sum_{j=1}^{x} e_j(\frac{p^j-1}{p-1})=\frac{\sum_{j=1}^{x} e_jp^j -\sum_{j=1}^{x} e_j}{p-1}</math>
  
 
<math>=\frac{(n-e_0)-(S_p(n)-e_0)}{p-1}=\frac{n-S_p(n)}{p-1}</math>.
 
<math>=\frac{(n-e_0)-(S_p(n)-e_0)}{p-1}=\frac{n-S_p(n)}{p-1}</math>.

Revision as of 09:03, 31 August 2009

Legendre's Formula states that

\[e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}\]

where $p$ is a prime and $e_p(n)$ is the exponent of $p$ in the prime factorization of $n$ and $S_p(n)$ is the sum of the digits of $n$ when written in base $p$.

Proof

Part 1

We could say that $e_p(n!)$ is equal to the number of multiples of $p$ less than $n$, or $\lfloor \frac{n}{p}\rfloor$. But the multiples of $p^2$ are only counted once, when they should be counted twice. So we need to add $\lfloor \frac{n}{p^2}\rfloor$ on. But this only counts the multiples of $p^3$ twice, when we need to count them thrice. Therefore we must add a $\lfloor \frac{n}{p^3}\rfloor$ on. We continue like this to get $e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor$. This makes sense, because the terms of this series tend to 0.

Part 2

Let the base $p$ representation of $n$ be $e_xe_{x-1}e_{x-2}\dots e_0$, where $e_i$ is a digit for all $0\leq i\leq x$. Therefore the base $p$ representation of $\lfloor \frac{n}{p^i}\rfloor$ is $e_xe_{x-1}\dots e_{x-i}$. Note that the infinite sum of these numbers (which is $e_p(n!)$) is

$\sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1)=\sum_{j=1}^{x} e_j(\frac{p^j-1}{p-1})=\frac{\sum_{j=1}^{x} e_jp^j -\sum_{j=1}^{x} e_j}{p-1}$

$=\frac{(n-e_0)-(S_p(n)-e_0)}{p-1}=\frac{n-S_p(n)}{p-1}$.

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