Consider a recent arXiv preprint 1805.11445. The author of 1805.11445 has done an overview of classical problem of simplifying of power sum

$$\sum_{1\leq k\leq n}k^m, \ (n,m)\geq 0, \ m=\mathrm{const}$$

and proposed "A Fast Algorithm to Calculate Power Sum of Natural Numbers". The follow identites are found in 1805.11445 for $\Sigma_{1\leq k\leq n} k^4$ and $\Sigma_{1\leq k\leq n} k^{12}$,

$$\sum_{1\leq k\leq n} k^4=\binom{n+4}{n-1}+11\binom{n+3}{n-2}+11\binom{n+2}{n-3}+\binom{n+1}{n-4}$$

And, respectively

\begin{eqnarray*} % \nonumber to remove numbering (before each equation) \sum_{1\leq k\leq n} k^{12} &=& \binom{n+12}{n-1}+4083\binom{n+11}{n-2}+478271\binom{n+10}{n-3} \\ &+&10187685\binom{n+9}{n-4}+66318474\binom{n+8}{n-5}+162512286\binom{n+7}{n-6}\\ &+&162512286\binom{n+6}{n-7}+66318474\binom{n+5}{n-8}+10187685\binom{n+4}{n-9}\\ &+&478271\binom{n+3}{n-10}+4083\binom{n+2}{n-11}+\binom{n+1}{n-12} \end{eqnarray*}

Now we have to mention that the coefficients in corresponding sums $\Sigma_{1\leq k\leq n} k^4$ and $\Sigma_{1\leq k\leq n} k^{12}$ are terms of forth and twelfth rows of Eulerian Triangle. Therefore, we have right to generalize above identities as follows

$$(\star)\quad\sum_{1\leq k\leq n} k^{m}=\sum_{k\geq0}E_{m,k}\binom{n+m-k}{n-1-k}$$

that is analog of Faulhaber's formula for every positive integers $(n,m)$.

Question: Is there any reference to equation $(\star)$ ?


1 Answer 1


Result in this question is direct consequence of Worpitzky Identity and Symmetry of Binomial coeficients. Recall the power sum, see Wikipedia

$$(\ast)\quad\sum_{k=1}^{n} k^{m}=\sum_{k=0}^{m-1}E_{m,k}\binom{n+1+k}{m+1}$$

Now, let compare the binomial coefficients in $(\star)$ and $(\ast)$, we start to check from the values of corresponding binomial coefficients for $k=0$ and $k=m-1$ respectively, we have

$$\binom{n+m-(m-1)}{n-1-(m-1)}\equiv \binom{n+1}{m+1}\rightarrow \binom{n+1}{n-m}=\binom{n+1}{m+1}$$ Therefore, denote $j=n+1$ and $r=m+1$, symmetry of binomial coefficients holds $$\binom{j}{r}=\binom{j}{r-j}$$ Now if we substitute the step $a$ instead $1$, the identity holds again

$$\binom{n+a}{n-m}=\binom{n+a}{m+a}$$ Therefore, Worpitzky Identity implies $$\sum_{k=1}^{n} k^{m}=\sum_{k=0}^{m-1}E_{m,k}\binom{n+1+k}{m+1}=\sum_{k=0}^{m-1}E_{m,k}\binom{n+m-k}{n-1-k}$$


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