So I have heard countless times of Bernoulli Numbers and its relation to the sums of powers. Inspired by this (as well as Chebyshev polynomials), I decided to look at the sums of powers myself and find patterns in the coefficients. Let $$S_{n,k}=1^k+2^k+\cdots + n^k$$ then, thanks to wikipedia, these are the first five formulae. However, I will highlight the following. $$\begin{align}S_{n,1}&=\frac 12n^2 + \frac 12n\\S_{n,2}&=\frac 13n^3+\frac 12n^2+\color{red}{\frac 16}n\\ S_{n,3}&=\frac 14n^4+\frac 12n^3+\color{red}{\frac 14}n^2+0n\\ S_{n,4}&= \frac 15n^5+\frac 12n^4+\color{red}{\frac 13}n^3+0n^2-\frac 1{30}n\\S_{n,5}&=\frac 16n^6+\frac 12n^5+\color{red}{\frac 5{12}}n^4+0n^3-\frac{1}{12}n^2+0n\\&{\;}\vdots\end{align}$$ Let $R_k=$ the red coefficient of $S_{n,k}$. Then, it appears that, for all $m\in\{0,1,2,3,4\}$, we have $$R_{6+m}+R_{6-m}=1.$$
e.g. $R_6=\frac 12$, thus $m=0\Rightarrow \frac 12 + \frac 12 = 1$; and $R_7=\frac 7{12}$, thus $m=1\Rightarrow \frac 7{12}+\frac 5{12}=1$.
Is there some kind of "reason" why this is happening? I mean, upon speculation, I think $R_k=\frac k{12}$ so if that is true, then my conjecture is true. How can this be (dis)proven?
Any thoughts?
Thanks.
Incidentally, I did not include bernoulli-numbers because, although they may be related, I do not specifically focus on them in this question.
Yes, according to Faulhaber's formula, the coefficient on $n^{k-1}$ for $S_{n,k}$ is $\dfrac1{k+1}{\binom{k+1}{2}}B_2,$ for $k>1$.
The second Bernoulli number $B_2=\dfrac16$, so $R_k=\dfrac k{12}$, and that's why your conjecture is true.