# Difference between revisions of "Legendre's Formula"

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=== Part 2 === | === Part 2 === | ||

− | Let the base <math>p</math> representation of <math>n</math> be <cmath>e_xe_{x-1}e_{x-2}\dots e_0</cmath> where the <math>e_i</math> are digits in base <math>p.</math> Then, the base <math>p</math> representation of <math>\lfloor \frac{n}{p^i}\rfloor</math> is <cmath>e_xe_{x-1}\dots e_{x-i}.</cmath> 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 <cmath>e_xe_{x-1}e_{x-2}\dots e_0</cmath> where the <math>e_i</math> are digits in base <math>p.</math> Then, the base <math>p</math> representation of <math>\left\lfloor \frac{n}{p^i}\right\rfloor</math> is <cmath>e_xe_{x-1}\dots e_{x-i}.</cmath> Note that the infinite sum of these numbers (which is <math>e_p(n!)</math>) is |

<cmath>\begin{align*} \sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1) &= \sum_{j=1}^{x} e_j \left( \frac{p^j-1}{p-1} \right) \\ | <cmath>\begin{align*} \sum_{j=1}^{x} e_j(p^{j-1}+p^{j-2}+\cdots +1) &= \sum_{j=1}^{x} e_j \left( \frac{p^j-1}{p-1} \right) \\ |

## Revision as of 21:19, 13 December 2017

**Legendre's Formula** states that

where is a prime and is the exponent of in the prime factorization of and is the sum of the digits of when written in base .

## Proofs

### Part 1

We use a counting argument.

We could say that is equal to the number of multiples of less than , or . But the multiples of are only counted once, when they should be counted twice. So we need to add on. But this only counts the multiples of twice, when we need to count them thrice. Therefore we must add a on. We continue like this to get . This makes sense, because the terms of this series tend to 0.

### Part 2

Let the base representation of be where the are digits in base Then, the base representation of is Note that the infinite sum of these numbers (which is ) is

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