Representing sum of power in different way.

Question

Definition . For $m ≥ 0$, the $m$th power-sum denominator is the smallest positive integer $d_m$ such that $d_m · (1^m + 2^m + · · · + n^m) $is a polynomial in $n$ with integer coefficients. The first few values of $d_m$ (see , Sequence A064538) are $d_m = 1, 2, 6, 4, 30, 12, 42, 24, 90, 20, 66, 24, 2730, 420, 90, 48, 510, . . . .$

Define $$\begin{split} S_m(n)&= 1^m+2^m+...+n^m \\ &= \frac{1}{d_m}\sum_{l=1}^{m+1} b_l n^{l} \end{split}$$

Can it be shown that

Let $n=d_mt+r$ where $t\ge0,r\ge 0$ then $S_m(n)$ may represent as

$$\begin{split} S_m(n)&= \sum_{l=1}^{m+1}(x_lt+y_l)n^{l-1}\end{split}$$

With $x_l$ and $ y_l$ are integers.

Example

Let $m=2$ gives $S_2(n)=\frac{n(n+1)(2n+1)}{6}\to d_2=6$

let $r=2\to n=6t+2$

And $S_2(n)=(2t+1)n^2+(t)n^1+(3t+1)n^0$

Here $x_1,x_2,x_3,y_1,y_2,y_3$ are $2,1,3,1,0,1$ respectively.

Answer 1

Theorem 1: Given $b_1,b_2,...,b_m,b_{m+1}$ finite integers. Denominator $d$ is smallest positive integer for $b_l$ integer coefficient. consider $n=dt+r$ then

$$\sum_{0<l\le m+1}\binom{n}l a_l= \frac{1}d\sum_{0<l\le m+1}b_ln^{l}=\sum_{0<l\le m+1}(x_lt+y_l)n^{l-1}$$

With $x_u$ and $y_u$ integers.

Proof: Denote $$\sum_{0<l\le m+1} b_l(h+r)^{l}=\sum_{0<l\le m+1} c_l h^{l}$$

The numbers $c_l$ are still integers, and we have, denoting $dt+r=n$, $$ \frac1d\sum_{0<l\le m+1} b_l(dt+r)^{l}\\ =\frac1d\sum_{0<l\le m+1} c_l (dt)^l\\ =\sum_{0<l\le m+1} c_l t\cdot (dt)^{l-1}\\ =\sum_{0<l\le m+1} c_l t\cdot (n-r)^{l-1}\\ = \sum_{0<l\le m+1\\ 0\leqslant j\leqslant l-1} c_l {l-1\choose j}(-r)^{l-1-j}n^jt, \\ =\sum_{0<l\le m+1} (x_lt+y_l)n^{l-1}$$

Source link (MO post)

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Formula

Let $n$ and $m$ are the integers with $n\geq 1$ and $m\geq 0$

$$\sum_{k=1}^{n} k^{m}=\sum_{l=1}^{m+1} \binom{n}l\sum_{i=0}^{l-1} (-1)^{i}(l-i)^{m}\binom{l-1}i$$

Proof link

Consider $$a_{l,m}=\sum_{i=0}^{l-1} (-1)^{i}(l-i)^{m}\binom{l-1}i$$

By theorem 1 $$S_m(n)=\sum_{l=1}^{m+1} \binom{n}la_{l,m}=\sum_{l=1}^{m+1}(x_lt+y_l)n^{l-1}$$