伯努利数 伯努利数 B_n
B_0=1,B_1=-\frac{1}{2},B_2=\frac{1}{6},B_3=0,B_4=-\frac{1}{30},\dots
等幂求和 伯努利数是由雅各布·伯努利的名字命名的，他在研究 m 次幂和的公式时发现了奇妙的关系。我们记
S_{m}(n)=\sum_{k=0}^{n-1}k^m=0^m+1^m+\dots+(n-1)^m 伯努利观察了如下一列公式，勾画出一种模式：
\begin{aligned} S_0(n)&=n\\ S_1(n)&=\frac{1}{2}n^2-\frac{1}{2}n\\ S_2(n)&=\frac{1}{3}n^3-\frac{1}{2}n^2+\frac{1}{6}n\\ S_3(n)&=\frac{1}{4}n^4-\frac{1}{2}n^3+\frac{1}{4}n^2\\ S_4(n)&=\frac{1}{5}n^5-\frac{1}{2}n^4+\frac{1}{3}-\frac{1}{30}n \end{aligned} 可以发现，在 S_m(n) n^{m+1} \frac{1}{m+1} n^m -\frac{1}{2} n^{m-1} \frac{m}{12} n^{m-3} -\frac{m(m-1)(m-2)}{720} n_{m-4} n^{m-k} m^{\underline{k}} m^{\underline{k}} \frac{m!}{(m-k)!}
递推公式 \begin{aligned} S_m{(n)}&=\frac{1}{m+1}(B_0n^{m+1}+\binom{m+1}{1}B_1 n^m+\dots+\binom{m+1}{m}B_m n) \\ &=\frac{1}{m+1}\sum_{k=0}^{m}\binom{m+1}{k}B_kn^{m+1-k} \end{aligned} 伯努利数由隐含的递推关系定义：
\begin{aligned} \sum_{j=0}^{m}\binom{m+1}{j}B_j&=0,(m>0)\\ B_0&=1 \end{aligned} 例如， \binom{2}{0}B_0+\binom{2}{1}B_1=0
n 0 1 2 3 4 5 6 7 8 \dots B_n 1 -\frac{1}{2} \frac{1}{6} 0 -\frac{1}{30} 0 \frac{1}{42} 0 -\frac{1}{30} \dots
证明 利用归纳法证明 这个证明方法来自 Concrete Mathematics 6.5 BERNOULLI NUMBER。
运用二项式系数的恒等变换和归纳法进行证明：
\begin{aligned} S_{m+1}(n)+n^{m+1}&= \sum_{k=0}^{n-1}(k+1)^{m+1}\\ &=\sum_{k=0}^{n-1}\sum_{j=0}^{m+1}\binom{m+1}{j}k^j\\ &=\sum_{j=0}^{m+1}\binom{m+1}{j}S_j(n) \end{aligned} 令 \hat{S}_{m}(n)=\frac{1}{m+1} \sum_{k=0}^{m} \binom{m+1}{k}B_kn^{m+1-k} S_m(n)=\hat{S}_m(n) j\in[0,m) S_j(n)=\hat{S}_j(n)
将原式中两边都减去 S_{m+1}(n)
\begin{aligned} S_{m+1}(n)+n^{m+1}&=\sum_{j=0}^{m+1}\binom{m+1}{j}S_j(n)\\ n^{m+1}&=\sum_{j=0}^{m}\binom{m+1}{j}S_j(n)\\ &=\sum_{j=0}^{m-1}\binom{m+1}{j}\hat{S}_j(n)+\binom{m+1}{m}S_m(n) \end{aligned} 尝试在式子的右边加上 \binom{m+1}{m}\hat{S}_m(n)-\binom{m+1}{m}\hat{S}_m(n)
n^{m+1}=\sum_{j=0}^{m}\binom{m+1}{j}\hat{S}_j(n)+(m+1)(S_m(n)-\hat{S}_m(n)) 不妨设 \Delta = S_m(n)-\hat{S}_m(n) \hat{S}_j(n)
\begin{aligned} n^{m+1}&=\sum_{j=0}^{m}\binom{m+1}{j}\hat{S}_j(n)+(m+1)\Delta\\ &=\sum_{j=0}^{m}\binom{m+1}{j}\frac{1}{j+1}\sum_{k=0}^{j}\binom{j+1}{k}B_kn^{j+1-k}+(m+1)\Delta\\ \end{aligned} 将第二个 \sum
\begin{aligned} n^{m+1}&=\sum_{j=0}^{m}\binom{m+1}{j}\frac{1}{j+1}\sum_{k=0}^{j}\binom{j+1}{j-k}B_{j-k}n^{k+1}+(m+1)\Delta\\ &=\sum_{j=0}^{m}\binom{m+1}{j}\frac{1}{j+1}\sum_{k=0}^{j}\binom{j+1}{k+1}B_{j-k}n^{k+1}+(m+1)\Delta\\ &=\sum_{j=0}^{m}\binom{m+1}{j}\frac{1}{j+1}\sum_{k=0}^{j}\frac{j+1}{k+1}\binom{j}{k}B_{j-k}n^{k+1}+(m+1)\Delta\\ &=\sum_{j=0}^{m}\binom{m+1}{j}\sum_{k=0}^{j}\binom{j}{k}\frac{B_{j-k}}{k+1}n^{k+1}+(m+1)\Delta \end{aligned} 对两个求和符号进行交换，可以得到：
n^{m+1}=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\sum_{j=k}^{m}\binom{m+1}{j}\binom{j}{k}B_{j-k}+(m+1)\Delta 对 \binom{m+1}{j}\binom{j}{k}
\binom{m+1}{j}\binom{j}{k}＝\binom{m+1}{k}\binom{m-k+1}{j-k} ＝ 那么式子就变成了：
\begin{aligned} n^{m+1}&=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\sum_{j=k}^{m}\binom{m+1}{k}\binom{m-k+1}{j-k}B_{j-k}+(m+1)\Delta\\ &=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\binom{m+1}{k}\sum_{j=k}^{m}\binom{m-k+1}{j-k}B_{j-k}+(m+1)\Delta\\ \end{aligned} 将所有的 j-k j
n^{m+1}=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\binom{m+1}{k}\sum_{j=0}^{m-k}\binom{m-k+1}{j}B_{j}+(m+1)\Delta 考虑我们前面提到过的递归关系
\begin{aligned} \sum_{j=0}^{m}\binom{m+1}{j}B_j&=0,(m>0)\\ B_0&=1\\ \sum_{j=0}^{m}\binom{m+1}{j}B_j&=[m = 0] \end{aligned} 代入后可以得到：
\begin{aligned} n^{m+1}&=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\binom{m+1}{k}[m - k = 0]+(m+1)\Delta\\ &=\sum_{k=0}^{m}\frac{n^{k+1}}{k+1}\binom{m+1}{k}+(m+1)\Delta\\ &=\frac{n^{m+1}}{m+1}\binom{m+1}{m}+(m+1)\Delta\\ &=n^{m+1}+(m+1)\Delta \end{aligned} 于是 \Delta=0 S_m(n)=\hat{S}_m(n)
利用指数生成函数证明 对递推式 \sum_{j=0}^{m}\binom{m+1}{j}B_j=[m=0]
两边都加上 B_{m + 1}
\begin{aligned} \sum_{j=0}^{m+1}\binom{m+1}{j}B_j&=[m=0]+B_{m+1}\\ \sum_{j=0}^{m}\binom{m}{j}B_j&=[m=1]+B_{m}\\ \sum_{j=0}^{m}\dfrac{B_j}{j!}\cdot\dfrac{1}{(m-j)!}&=[m=1]+\dfrac{B_{m}}{m!} \end{aligned} 设 B(z) = \sum\limits_{i\ge 0}\dfrac{B_i}{i!}z^i
\begin{aligned} B(z)e^z &= z+B(z)\\ B(z)&=\dfrac{z}{e^z - 1} \end{aligned} 设 F_n(z) = \sum_{m\ge 0}\dfrac{S_m(n)}{m!}z^m
\begin{aligned} F_n(z) &= \sum_{m\ge 0}\dfrac{S_m(n)}{m!}z^m\\ &= \sum_{m\ge 0}\sum_{i=0}^{n-1}\dfrac{i^mz^m}{m!}\\ \end{aligned} 调换求和顺序：
\begin{aligned} F_n(z) &=\sum_{i=0}^{n-1}\sum_{m\ge 0}\dfrac{i^mz^m}{m!}\\ &=\sum_{i=0}^{n-1}e^{iz}\\ &=\dfrac{e^{nz} - 1}{e^z - 1}\\ &=\dfrac{z}{e^z - 1}\cdot\dfrac{e^{nz} - 1}{z} \end{aligned} 代入 B(z)=\dfrac{z}{e^z - 1}
\begin{aligned} F_n(z) &= B(z)\cdot\dfrac{e^{nz} - 1}{z}\\ &= \left(\sum_{i\ge 0}\dfrac{B_i}{i!} \right)\left(\sum_{i\ge 1}\dfrac{n^i z^{i - 1}}{i!}\right)\\ &= \left(\sum_{i\ge 0}\dfrac{B_i}{i!} \right)\left(\sum_{i\ge 0}\dfrac{n^{i+1} z^{i}}{(i+1)!}\right) \end{aligned} 由于 F_n(z) = \sum_{m\ge 0}\dfrac{S_m(n)}{m!}z^m S_m(n)=m![z^m]F_n(z)
\begin{aligned} S \times m(n)&=m![z^m]F_n(z)\\ &= m!\sum_{i=0}^{m}\dfrac{B \times i}{i!}\cdot\dfrac{n^{m-i+1}}{(m-i+1)!}\\ &=\dfrac{1}{m+1}\sum_{i=0}^{m}\binom{m+1}{i}B_in^{m-i+1} \end{aligned} 故得证。
参考实现 typedef long long ll;
const int maxn = 10000;
const int mod = 1e9 + 7;
ll B[maxn]; // 伯努利数
ll C[maxn][maxn]; // 组合数
ll inv[maxn]; // 逆元（计算伯努利数）
void init() {
// 预处理组合数
for (int i = 0; i < maxn; i++) {
C[i][0] = C[i][i] = 1;
for (int k = 1; k < i; k++) {
C[i][k] = (C[i - 1][k] % mod + C[i - 1][k - 1] % mod) % mod;
}
}
// 预处理逆元
inv[1] = 1;
for (int i = 2; i < maxn; i++) {
inv[i] = (mod - mod / i) * inv[mod % i] % mod;
}
// 预处理伯努利数
B[0] = 1;
for (int i = 1; i < maxn; i++) {
ll ans = 0;
if (i == maxn - 1) break;
for (int k = 0; k < i; k++) {
ans += C[i + 1][k] * B[k];
ans %= mod;
}
ans = (ans * (-inv[i + 1]) % mod + mod) % mod;
B[i] = ans;
}
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