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Let $\displaystyle\sigma_m=\sum_{r=1}^n r^m$.

Refer to the tabulation of the power sum of integers here.

It is interesting to note that

$$\begin{align} \color{green}{\sigma_1}\ &=\frac 12 n(n+1)\\ \color{blue}{\sigma_2}\ &=\frac 16 n(n+1)(2n+1)\\ \color{red}{\sigma_3}\ &=\frac 14 n^2(n+1)^2&&=\color{green}{\sigma_1}^2\\ \sigma_4\ &=\frac 1{30}n(n+1)(2n+1)(3n^2+3n-1)&&=\frac 15\; \color{blue}{\sigma_2} \ (3n^2+3n-1)\\ \sigma_5\ &=\frac 1{12}n^2(n+1)^2(2n^2+2n-1)&&=\frac 13\; \color{red}{\sigma_3}\ (2n^2+2n-1)\\ \sigma_6\ &=\frac 1{42}n(n+1)(2n+1)(3n^4+6n^3-3n+1)&&=\frac 17\;\color{blue}{\sigma_2}\ (3n^4+6n^3-3n+1)\\ \sigma_7\ &=\frac 1{24}n^2(n+1)^2 (\cdots)&&=\frac 16\; \color{red}{\sigma_3}\ (\cdots)\\ \sigma_8\ &=\frac 1{90}n(n+1)(2n+1)(\cdots)&&=\frac 1{15}\color{blue}{\sigma_2}\ (\cdots)\\ \sigma_9\ &=\frac 1{20}n^2(n+1)^2(n^2+n-1)(\cdots)&&=\frac 15\; \color{red}{\sigma_3}\ (n^2+n-1)(\cdots)\\ \sigma_{10}&=\frac 1{66}n(n+1)(2n+1)(n^2+n-1)(\cdots)&&=\frac 1{11}\color{blue}{\sigma_2}\ (n^2+n-1)(\cdots) \end{align}$$ i.e.

  • the sum of squares, $\sigma_2$, is a factor of sum of even powers greater than $2$, and
  • the sum of cubes, $\sigma_3$, is a factor of sum of odd powers greater than $3$.

Is there a simple explanation for this, if possible without using Faulhaber's formula and Bernoulli numbers, etc?

and also,

Why does this occur only for $\sigma_2, \sigma_3$ but not for $\sigma_4, \sigma_5$, etc?

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Disclaimer

Its kinda hand wavy if I can't use anything really advanced, but here's an intuitive look on the situation:


Let $\sigma_m(x)$ be a polynomial of $x$ such that on $x\in\mathbb N$, it agrees with your $\sigma_m$. Note this polynomial satisfies the recursive relation

$$\sigma_m(x)=\sigma_m(x-1)+x^m$$

which extends it to negative values.

The phenomenon of $\sigma_2$ and $\sigma_3$ appearing in $\sigma_{m>3}$ is not too surprising, since it is easy to note that

$$\sigma_m(-1)=\sigma_m(0)=0$$

for any $m\in\mathbb N_{>0}$.

One can see from the recursive relation that $\sigma_m(x)$ is symmetric along $x=-\frac12$.

For even $m$, the symmetry is odd, so there is a root at $x=-\frac12$.

For odd $m$, the symmetry is even, so every other root reflects over. This makes $x=0$ and $x=-1$ roots with a multiplicity of $2$.

Combine these two and you can see that

$$\sigma_{2m}(x)=x(x+1)(2x+1)P_m(x)\\\sigma_{2m+1}(x)=x^2(x+1)^2Q_m(x)$$

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  • $\begingroup$ Very elegant solution! (+1) $\endgroup$ – hypergeometric Aug 10 '17 at 16:26
  • $\begingroup$ @Simply Beautiful Art: Can you exaplain how do you get symmetry about $x=-\frac 12$, please? We have $\sigma_m(-1/2+x)=\sigma_m(-3/2+x)+(-1/2+x)^m$. $\endgroup$ – Fabio Lucchini Jul 9 '18 at 9:01
  • $\begingroup$ @FabioLucchini notice that $$\sigma_m(-x)=\sigma_m(-x+1)-(-x)^m$$ which is almost the same as the original recursive relation but for negative integers. $\endgroup$ – Simply Beautiful Art Jul 12 '18 at 12:01
  • $\begingroup$ Sorry, but I don't get it ... it would to be $\sigma_m(-x)=\sigma_m(-x+1)-(1-x)^m$. But how this proves $\sigma_m(-1-x)=-(-1)^m\sigma_m(x)$? $\endgroup$ – Fabio Lucchini Jul 12 '18 at 15:59
  • $\begingroup$ @FabioLucchini yes my bad. Use also the fact that $\sigma_m(-1)=\sigma_m(0)=0$ and you may prove it by induction. $\endgroup$ – Simply Beautiful Art Jul 12 '18 at 16:24

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