Legendre's formula

In mathematics, Legendre's formula gives an expression for the exponent of the largest power of a prime p that divides the factorial n ! . It is named after Adrien-Marie Legendre. It is also sometimes known as de Polignac's formula, after Alphonse de Polignac.

Statement
For any prime number p and any positive integer n, let $$\nu_p(n)$$ be the exponent of the largest power of p that divides n (that is, the p-adic valuation of n). Then
 * $$\nu_p(n!) = \sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor,$$

where $$\lfloor x \rfloor$$ is the floor function. While the sum on the right side is an infinite sum, for any particular values of n and p it has only finitely many nonzero terms: for every i large enough that $$p^i > n$$, one has $$\textstyle \left\lfloor \frac{n}{p^i} \right\rfloor = 0$$. This reduces the infinite sum above to
 * $$\nu_p(n!) = \sum_{i=1}^{L} \left\lfloor \frac{n}{p^i} \right\rfloor \, ,$$

where $$L = \lfloor \log_{p} n \rfloor$$.

Example
For n = 6, one has $$ 6! = 720 = 2^4 \cdot 3^2 \cdot 5^1$$. The exponents $$\nu_2(6!) = 4, \nu_3(6!) = 2$$ and $$\nu_5(6!) = 1$$ can be computed by Legendre's formula as follows:



\begin{align} \nu_2(6!) & = \sum_{i=1}^\infty \left\lfloor \frac 6 {2^i} \right\rfloor = \left\lfloor \frac 6 2 \right\rfloor + \left\lfloor \frac 6 4 \right\rfloor = 3 + 1 = 4, \\[3pt] \nu_3(6!) & = \sum_{i=1}^\infty \left\lfloor \frac 6 {3^i} \right\rfloor = \left\lfloor \frac 6 3 \right\rfloor = 2, \\[3pt] \nu_5(6!) & = \sum_{i=1}^\infty \left\lfloor \frac 6 {5^i} \right\rfloor = \left\lfloor \frac 6 5 \right\rfloor = 1. \end{align} $$

Proof
Since $$n!$$ is the product of the integers 1 through n, we obtain at least one factor of p in $$n!$$ for each multiple of p in $$\{1, 2, \dots, n\}$$, of which there are $$\textstyle \left\lfloor \frac{n}{p} \right\rfloor$$. Each multiple of $$p^2$$ contributes an additional factor of p, each multiple of $$p^3$$ contributes yet another factor of p, etc. Adding up the number of these factors gives the infinite sum for $$\nu_p(n!)$$.

Alternate form
One may also reformulate Legendre's formula in terms of the base-p expansion of n. Let $$s_p(n)$$ denote the sum of the digits in the base-p expansion of n; then
 * $$\nu_p(n!) = \frac{n - s_p(n)}{p - 1}.$$

For example, writing n = 6 in binary as 610 = 1102, we have that $$s_2(6) = 1 + 1 + 0 = 2$$ and so
 * $$\nu_2(6!) = \frac{6 - 2}{2 - 1} = 4.$$

Similarly, writing 6 in ternary as 610 = 203, we have that $$s_3(6) = 2 + 0 = 2$$ and so
 * $$\nu_3(6!) = \frac{6 - 2}{3 - 1} = 2.$$

Proof
Write $$n = n_\ell p^\ell + \cdots + n_1 p + n_0$$ in base p. Then $$\textstyle \left\lfloor \frac{n}{p^i} \right\rfloor = n_\ell p^{\ell-i} + \cdots + n_{i+1} p + n_i$$, and therefore

\begin{align} \nu_p(n!) &= \sum_{i=1}^{\ell} \left\lfloor \frac{n}{p^i} \right\rfloor \\ &= \sum_{i=1}^{\ell} \left(n_\ell p^{\ell-i} + \cdots + n_{i+1} p + n_i\right) \\ &= \sum_{i=1}^{\ell} \sum_{j=i}^{\ell} n_j p^{j-i} \\ &= \sum_{j=1}^{\ell} \sum_{i=1}^{j} n_j p^{j-i} \\ &= \sum_{j=1}^{\ell} n_j \cdot \frac{p^j - 1}{p - 1} \\ &= \sum_{j=0}^{\ell} n_j \cdot \frac{p^j - 1}{p - 1} \\ &= \frac{1}{p - 1} \sum_{j=0}^{\ell} \left(n_j p^j - n_j\right) \\ &= \frac{1}{p - 1} \left(n - s_p(n)\right). \end{align} $$

Applications
Legendre's formula can be used to prove Kummer's theorem. As one special case, it can be used to prove that if n is a positive integer then 4 divides $$\binom{2n}{n}$$ if and only if n is not a power of 2.

It follows from Legendre's formula that the p-adic exponential function has radius of convergence $$p^{-1/(p-1)}$$.