对于质数\(p\)，函数\(v_p(n)\)为\(n\)标准分解后\(p\)的次数
显然有
\[v_p(n!) = \sum\limits_{i = 1}^{\infty} \lfloor \frac{n}{p^i} \rfloor\]
令函数\(s_p(n)\)为\(n\)在\(p\)进制下的数位和
有:
\[v_p(n!) = \frac{n - s_p(n)}{p - 1}\]
证明:
设\(n = \sum\limits_{i = 0}^{\infty} c_i p^i\)，
有\(v_p(n!) = \sum\limits_{i = 1}^{\infty} \lfloor \frac{n}{p^i} \rfloor\)
\(= \sum\limits_{i = 1}^{\infty} \sum\limits_{j = i}^{\infty} c_j p^{j - i}\)
\(= \sum\limits_{j = 1}^{\infty} c_j \sum\limits_{i = 0}^{j - 1} p^i\)
\(= \sum\limits_{j = 1}^{\infty} \frac{c_j(p^j - 1)}{p - 1}\)
\(= \frac{1}{p - 1} (\sum\limits_{i = 0}^{\infty} c_i p^i - \sum\limits_{i = 0}^{\infty} c_i)\)
$= \frac{n - s_p{n}}{p - 1} $