Recently, I figured out that for all $n,k\in\mathbb{N}$: $$\sum_{i=1}^{n}i^k=\sum_{i=1}^{k+1}\frac{c_{k+1-i}k!}{i!}n^i$$ where: $$c_0=1$$ $$c_n=\frac1{n!}-\sum_{i=0}^{n-1}\frac{c_i}{(m-i+1)!}$$ (proof down below). The first few values of $c_n$ are: $$c_0=1$$ $$c_1=\frac12$$ $$c_2=\frac1{12}$$ $$c_3=0$$ $$c_4=-\frac1{720}$$ $$c_5=0$$ $$c_6=\frac1{30240}$$ $$c_7=0$$ $$c_8=-\frac1{1209600}$$ $$c_9=0$$ $$c_{10}=\frac1{47900160}$$ $$c_{11}=0$$ I would like to know what is known about $\{c_k\}_{k=0}^{\infty}$. Some specific questions would be:
- Does the pattern of $c_n=0$ for odd $n\ge 3$ continue?
- Does the pattern of non-zero $c$ alternating between positive and negative values continue?
- For non-zero $c$, do we always have a numerator that is equal to $1$?
- Is there a closed formula?
Proof of the formula
We have for all $n,k\in\mathbb{N}$: \begin{align*} n^k &=\sum_{i=1}^{n}i^k-\sum_{i=0}^{n-1}i^k\\ &=\sum_{i=0}^{n-1}(i+1)^k-i^k\\ &=\sum_{i=0}^{n-1}\sum_{j=0}^{k-1}{k\choose j}i^j \end{align*} Switching the $\Sigma$'s gives: \begin{align*} n^k&=\sum_{j=0}^{k-1}\sum_{i=1}^{n-1}{k\choose j}i^j\\ &=\sum_{j=0}^{k-1}{k\choose j}\sum_{i=0}^{n-1}i^j \end{align*} At this point, we define for every $n,k\in \mathbb{N}$: $$f(n,k)=\sum_{i=1}^{n}i^k$$ So: $$n^k=\sum_{j=0}^{k-1}{k\choose j}f(n-1,j)$$ Now, we perform a change of variable. $n$ becomes $n+1$ and $k$ becomes $k+1$: \begin{align*} (n+1)^{k+1} &=\sum_{j=0}^{k}{k+1\choose j}f(n,j)\\ &={k+1\choose}f(n,k)+\sum_{j=0}^{k-1}{k+1\choose j}f(n,j)\\ f(n,k) &= \frac1{k+1}\left((n+1)^{k+1}-\sum_{j=0}^{k-1}{k+1\choose j}f(n,j)\right) \end{align*} The RHS is a polynomial of degree $k+1$ and $f(n,k-1)$ is the first time a term of degree $k$ appears. Let $a_k(m)$ be the coefficient of $n^k$ in $f(n,k-1+m)$. Since: $$\deg\left(\sum_{j=0}^{k-1}{k+1\choose j}f(n,j)\right)=k$$ we have $a_k(0)=\frac1k{k\choose k}=\frac1k$ for all $k\in\mathbb{N}$. From the origial equation it now follows that for all $m\in \Bbb{N}$. $$a_k(m)=\frac1{m+1}\left({m+k\choose k}-\sum_{j=k-1}^{m+k-2}{m+k\choose j}a_k(j-k+1)\right)$$ A change of variable in $j$ gives: $$a_k(m)=\frac{1}{m+k}\left({m+k\choose k}-\sum_{j=0}^{m-1}{m+k\choose j+k-1}a_k(j)\right)$$ Now, we have for all $n,k\in\mathbb{N}$: $$\sum_{i=1}^{n}i^k=\sum_{i=1}^{k+1}a_i(k+1-i)n^i$$ Finally, we'll prove using strong induction that: $$a_k(m)=c_m\cdot\frac{(m+k-1)!}{k!}$$ The base case ($m=0$) is pretty clear, since: $$a_k(0)=\frac1k = \frac{(k-1)!}{k!}\cdot c_0$$ Now, assume that for all $j<m$, we have: $$a_k(j)=\frac{(j+k-1)!}{k!}\cdot c_j$$ the induction step becomes: \begin{align*} a_k(m)&=\frac1{m+1}\left({m+k\choose k}-\sum_{j=0}^{m-1}{m+k\choose j+k-1}a_k(j)\right)\\ &=\frac1{m+k}\left({m+k\choose k}-\sum_{j=0}^{m-1}\frac{(m+k)!}{(j+k-1)!(m-j+1)!}\cdot \frac{(j+k-1)!}{k!}\cdot c_j\right)\\ &= \frac1{m+k}\left(\frac{(m+k)!}{m!k!}-\sum_{j=0}^{m-1}\frac{(m+k)!}{k!}\cdot\frac{c_j}{(m-j+1)!}\right)\\ &= \frac1{m+k}\frac{(m+k)!}{k!}\left(\frac1{m!}-\sum_{j=0}^{m-1}\frac{c_j}{(m-j+1)!}\right)\\ &=\frac{(m+k-1)!}{k!}\cdot c_m \end{align*} So now we proven using induction that $a_k(m)=\frac{(m+k-1)!}{k!}c_m$. It follows that for all $n,k\in\mathbb{N}$: $$f(n,k)=\sum_{i=1}^{n}i^k=\sum_{i=1}^{k+1}a_i(k+1-i)x^i=\sum_{i=1}^{k+1}\frac{c_{k+1-i}k!}{i!}n^i$$ And that is exactly what we wanted to prove. Q.E.D.