Difference between revisions of "Legendre's Formula"
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<cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | <cmath>e_p(n!)=\sum_{i=1}^{\infty} \left\lfloor \dfrac{n}{p^i}\right\rfloor =\frac{n-S_{p}(n)}{p-1}</cmath> | ||
− | where <math>p</math> is a prime and <math>e_p(n)</math> is the [[exponent]] of <math>p</math> in the [[prime factorization]] of <math>n</math> and <math>S_p(n)</math> is the [[sum]] of the [[digit]]s of <math>n</math> when written in [[base]] <math>p</math>. | + | where <math>p</math> is a prime and <math>e_p(n)</math> is the [[exponent]] of <math>p</math> in the [[prime factorization]] of <math>n!</math> and <math>S_p(n)</math> is the [[sum]] of the [[digit]]s of <math>n</math> when written in [[base]] <math>p</math>. |
==Proofs== | ==Proofs== |
Revision as of 15:33, 15 February 2018
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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