伯努利数 – 源码巴士

参考文献资料：
小蒟蒻yyb的博客
 百度百科-伯努利数
 伯努利数(Bernoulli)——学习笔记
 康复计划#3 简单常用的几种计算自然数幂和的方法
Bernoulli number has been found for about 200 years.Its name is named after Bernoulli, its discoverer.

Recursive definition：

$\sum_{k=0}^n{n+1\choose k}B_k=0$

Take out item n,we can get:
$\sum_{k=0}^{n-1}{n+1\choose k}B_k+{n+1\choose n}B_n=0$

$\sum_{k=0}^{n-1}{n+1\choose k}B_k+(n+1)B_n=0$

$B_n=-\frac{1}{n+1}\sum_{k=0}^{n-1}{n+1\choose k}B_k$

Generating Function Definition:

$\hat B(z)=\frac{z}{e^z-1}$

where $\hat B(z)$ is the generating function of ${B_n\}$ :
$\hat B(z)=\sum_{k=0}^{\infty}B_k\frac{z^k}{k!}$

The Sum of the k-th Powers:

We define :
$S_t(n)=\sum_{k=0}^{n-1}k^t$

Be care about this:the generating function of ${k^t\}$ is
$\sum_{t=0}^{\infty}k^t\frac{z^t}{t!}=\sum_{t=0}^{\infty}\frac{(kz)^t}{t!}=e^{kz}$

The result can be got by Taylor expansion.
Considering the Generation Function of $S_t(z)$ :
$\hat S_t(z)=\sum_{p=0}^{\infty}S_p(z)\frac{z^p}{p!}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$

$=\sum_{p=0}^{\infty}\bigg(\sum_{k=0}^{n-1}k^p\bigg)\frac{z^p}{p!}$

$=\sum_{k=0}^{n-1}\sum_{p=0}^{\infty}\frac{(kz)^p}{p!}\ \ \ \ \ \$

$=\sum_{k=0}^{n-1}e^{kz}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$

According to the summation formula of equal ratio series, we can get:
$\hat S_t(z)=\frac{e^{nz}-1}{e^z-1}$

Considering the definition of the generating function of Bernoulli number:
$\hat S_t(z)=\hat B(z)\frac{e^{nz}-1}{z}$

So we have:
$S_t(n)=\frac{1}{t+1}\sum_{k=0}^t{t+1\choose k}B_kn^{t+1-k}$

Where $S_t(n)$ is a t-th polynomial of $n$

原文链接：https://blog.csdn.net/alpha202/article/details/87199799