等冪求和

伯努利數是以雅各布 \(\cdot\) 伯努利的名字命名的，我們記

\[S_m(n) = \sum_{k = 0}^{n - 1} k^m \]

伯努利 暴力 算出了前幾項，並進行了觀察：

\[\begin{array}{ll} S_0(n) = n \\ S_1(n) = \dfrac{1}{2} n^2 - \dfrac{1}{2} n \\ S_2(n) = \dfrac{1}{3} n^3 - \dfrac{1}{2} n^2 + \dfrac{1}{6} n \\ S_3(n) = \dfrac{1}{4} n^4 - \dfrac{1}{2} n^3 + \dfrac{1}{4} n^2 + 0 n \\ S_4(n) = \dfrac{1}{5} n^5 - \dfrac{1}{2} n^4 + \dfrac{1}{3} n^3 + 0 n^2 - \dfrac{1}{30} n \\ S_5(n) = \dfrac{1}{6} n^6 - \dfrac{1}{2} n^5 + \dfrac{5}{12} n^4 + 0 n^3 - \dfrac{1}{12} n^2 + 0n \\ S_6(n) = \dfrac{1}{7} n^7 - \dfrac{1}{2} n^6 + \dfrac{1}{2} n^5 + 0 n^4 - \dfrac{1}{6} n^3 + 0 n^2 + \dfrac{1}{42} n \end{array} \]

我們發現，在 \(S_m(n)\)

中，\(n^{m + 1}\) 的係數總是 \(\dfrac{1}{m +1}\)，\(n^m\) 的係數是 \(-\dfrac{1}{2}\)，\(n^{m - 1}\) 的係數是 \(\dfrac{m}{12}\)，\(n^{m - 2}\) 的係數是 \(0\)，\(n^{m - 3}\) 的係數是 \(- \dfrac{m (m - 1) (m - 2)}{720}\)，\(n^{m - 4}\) 的係數是 \(0\cdots\cdots\) 而 \(n^{m - k}\) 的係數總是某個常數乘上 \(m^{\underline{k}} = \dfrac{m!}{(m - k)!}\)。

遞推公式

根據伯努利的發現（值得一提的是，他並沒有給出證明），我們有

\[S_m(n) = \dfrac{1}{m + 1} \sum_{k = 0}^m \dbinom{m + 1}{k} B_k n^{m + 1 - k} \]

其中 \(B_n\) 為伯努利數，其由一個遞迴關係定義：

\[\sum_{k = 0}^m \dbinom{m + 1}{k} B_k = [m = 0] \]

其實你也可以寫成

\[B_0 = 1 \\ \sum_{k = 0}^m \dbinom{m + 1}{k} B_k = 0, m \ge 1 \]

這樣方便你進行計算。

根據手算，不難得出前幾個值：

\(n\)	\(0\)	\(1\)	\(2\)	\(3\)	\(4\)	\(5\)	\(6\)	\(7\)	\(8\)	\(9\)	\(10\)	\(11\)	\(12\)
\(B_n\)	\(1\)	\(-\dfrac{1}{2}\)	\(\dfrac{1}{6}\)	\(0\)	\(-\dfrac{1}{30}\)	\(0\)	\(\dfrac{1}{42}\)	\(0\)	\(-\dfrac{1}{30}\)	\(0\)	\(\dfrac{5}{66}\)	\(0\)	\(-\dfrac{691}{2730}\)

（其中 \(B_{12}\) 這個奇怪分數的出現真是給那些有關 \(B_n\) 的簡單封閉形式的猜想潑冷水。）

證明：

首先令

\[\hat{S}_m(n) = \dfrac{1}{m + 1} \sum_{k = 0}^m \dbinom{m + 1}{k} B_k n^{m + 1 - k} \]

那麼現在我們就是要證明 \(S_m(n) = \hat{S}_m(n)\)，考慮使用數學歸納法。

首先

\[\begin{array}{ll} S_0(n) = \sum\limits_{i = 0}^{n - 1} n^0 = n \\ \hat{S}_0(n) = \dfrac{1}{1} \dbinom{1}{0} B_0 n^{1} = n \end{array} \]\[\begin{aligned} S_{m + 1}(n) + n^{m + 1} & = \sum_{k = 0}^n k^{m + 1} \\ & = \sum_{k = 0}^{n - 1} (k + 1)^{m + 1} \\ & = \sum_{k = 0}^{n - 1} \sum_{j = 0}^{m + 1} \dbinom{m + 1}{j} k^j \\ & = \sum_{j = 0}^{m + 1} \dbinom{m + 1}{j} \sum_{k = 0}^{n - 1} k^j \\ & = \sum_{j = 0}^{m + 1} \dbinom{m + 1}{j} S_j(n) \end{aligned} \]

當 \(m\ge 1\) 時，假設 \(\forall i \in [0, m)\) 均有 \(S_i(n) = \hat{S}_i(n)\)，將 \(S_{m +1}(n)\) 移項：

\[\begin{aligned} n^{m + 1} & = \sum_{j = 0}^{m + 1} \dbinom{m + 1}{j} S_j(n) - S_{m + 1}(n) \\ & = \sum_{j = 0}^m \dbinom{m + 1}{j} S_j(n) \\ & = \sum_{j = 0}^{m - 1} \dbinom{m + 1}{j} S_j(n) + \dbinom{m + 1}{m} S_m(n) \\ & = \sum_{j = 0}^{m - 1} \dbinom{m + 1}{j} \hat{S}_j(n) + (m + 1) S_m(n) \end{aligned} \]

設

\[\Delta = S_m(n) - \hat{S}_m(n) \]

那麼現在就是要證 \(\Delta = 0\)。

\[\begin{aligned} n^{m + 1} & = \left[\sum_{j = 0}^{m - 1} \dbinom{m + 1}{j} \hat{S}_j(n) + (m + 1) \hat{S}_m(n) \right] + (m + 1) S_m(n) - (m + 1) \hat{S}_m(n) \\ & = \sum_{j = 0}^m \dbinom{m + 1}{j} \hat{S}_j(n) + (m + 1) \Delta \end{aligned} \]

根據 \(\hat{S}_m(n)\) 的定義將其展開，有

\[n^{m + 1} = \sum_{j = 0}^m \dbinom{m +1}{j} \dfrac{1}{j + 1} \sum_{k = 0}^j \dbinom{j + 1}{k} B_k n^{j + 1 - k} + (m + 1) \Delta \]

將 \(k\) 改為倒序列舉 並 利用 對稱恆等式

\[\dbinom{n}{m} = \dbinom{n}{n - m}, n\in \mathbb{N}, k \in \mathbb{Z} \]\[\begin{aligned} n^{m +1} & = \sum_{j = 0}^m \dbinom{m + 1}{j} \dfrac{1}{j + 1} \sum_{k = 0}^j \dbinom{j + 1}{j - k} B_{j - k} n^{k + 1} + (m + 1) \Delta \\ & = \sum_{j = 0}^m \dbinom{m + 1}{j} \dfrac{1}{j + 1} \sum_{k = 0}^j \dbinom{j + 1}{k + 1} B_{j - k} n^{k + 1} + (m + 1) \Delta \end{aligned} \]

利用 吸收恆等式

\[\dbinom{r}{k} = \dfrac{r}{k} \dbinom{r - 1}{k - 1}, k \in \mathbb{Z}, k \ne 0 \]\[\begin{aligned} n^{m + 1} & = \sum_{j = 0}^m \dbinom{m + 1}{j} \dfrac{1}{j + 1} \sum_{k = 0}^j \dfrac{j + 1}{k + 1} \dbinom{j}{k} B_{j - k} n^{k + 1} + (m +1) \Delta \\ & = \sum_{j = 0}^m \dbinom{m + 1}{j} \sum_{k = 0}^j \dfrac{n^{k + 1}}{k + 1} \dbinom{j}{k} B_{j - k} + (m + 1) \Delta \\ & = \sum_{k = 0}^m \dfrac{n^{k + 1}}{k + 1} \sum_{j = k}^m \dbinom{m + 1}{j} \dbinom{j}{k} B_{j - k} + (m + 1) \Delta \end{aligned} \]

利用 三項式版恆等式

\[\dbinom{r}{m} \dbinom{m}{k} = \dbinom{r}{k} \dbinom{r - k}{m - k} \]\[\begin{aligned} n^{m + 1} & = \sum_{k = 0}^m \dfrac{n^{k + 1}}{k + 1} \sum_{j = k}^m \dbinom{m + 1}{k} \dbinom{m - k + 1}{j - k} B_{j - k} + (m + 1) \Delta \\ & = \sum_{k = 0}^m \dfrac{n^{k + 1}}{k + 1} \dbinom{m + 1}{k} \sum_{j = k}^m \dbinom{m - k + 1}{j - k} B_{j - k} + (m + 1) \Delta \end{aligned} \]

將 \(j - k\) 改為 \(j\)

\[n^{m + 1} = \sum_{k = 0}^m \dfrac{n^{k + 1}}{k + 1} \dbinom{m + 1}{k} \sum_{j = 0}^{m - k} \dbinom{m - k + 1}{j} B_j + (m + 1) \Delta \]

又根據伯努利數的遞迴定義

\[\sum_{k = 0}^m \dbinom{m + 1}{k} B_k = [m = 0] \]\[\begin{aligned} n^{m + 1} & = \sum_{k = 0}^m \dfrac{n^{k + 1}}{k + 1} \dbinom{m + 1}{k} [m - k = 0] + (m + 1) \Delta \\ & = \dfrac{n^{m + 1}}{m + 1} \dbinom{m + 1}{m} + (m + 1) \Delta \\ & = n^{m + 1} + (m + 1) \Delta \end{aligned} \]

啊哈！至此，我們推出了

\[(m + 1) \Delta = 0 \\ \because m + 1 \ne 0 \\ \therefore \Delta = 0 \]

證完嘍！

當然，還有一種用 指數生成函式 的更簡單的證明方法，因為筆者才疏學淺不會生成函式就先不證了，到時候在生成函式的筆記裡證（

程式碼實現

顯然這東西一般用在有模數的情況下。

//18 = 9 + 9 = 18.
#include <iostream>
#include <cstdio>
#define Debug(x) cout << #x << "=" << x << endl
typedef long long ll;
using namespace std;

const int MAXN = 1e4 + 5;
const int N = 1e4;
const int MOD = 1e9 + 7;

int add(int a, int b) {return (a + b) % MOD;}
int mul(int a, int b) {return (ll)a * b % MOD;}

int inv[MAXN], C[MAXN][MAXN], B[MAXN];

int main()
{
	freopen("mango.out", "w", stdout);
	
	for (int i = 0; i <= N; i++)
	{
		C[i][0] = C[i][i] = 1;
	}
	for (int i = 1; i <= N; i++)
	{
		for (int j = 1; j < i; j++)
		{
			C[i][j] = add(C[i - 1][j], C[i - 1][j - 1]);
		}
	}
	
	inv[1] = 1;
	for (int i = 2; i <= N; i++)
	{
		inv[i] = mul(MOD - MOD / i, inv[MOD % i]);
	}
	
	B[0] = 1;
	for (int m = 1; m <= N; m++)
	{
		int sum = 0;
		for (int k = 0; k < m; k++)
		{
			sum = add(sum, mul(C[m + 1][k], B[k]));
		}
		B[m] = mul(MOD - sum, inv[m + 1]);
	}
	
	for (int i = 1; i <= 10000; i++)
	{
		printf("%d\n", B[i]);
	}
	return 0;
}