# Coefficients in the sum $\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j=n^{2m+1}, \ m=1,2,…$

Consider a sum $$\sum_{k=0}^{n-1}\sum_{j=0}^{m}A_{j,m}(n-k)^jk^j$$ which returns an odd power $$n^{2m+1}$$ of $$n$$, for $$\ m=0,1,2,...$$ given fixed $$A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$$. The coefficients $$A_{0,m}, \ A_{1,m},....$$ are solutions of system of equations (refer to .txt-file with mathematica codes for $$m=1,2,...12$$). Coefficients $$A_{0,m}, \ A_{1,m},....$$ are arranged to the PDF-table. For example, $$\sum_{k=0}^{n-1}\sum_{j=0}^{2}A_{j,2}(n-k)^jk^j=\sum_{k=0}^{n-1}30k^2(n-k)^2+1=n^5$$ Our coefficients are closely related to to coefficients $$\beta_{mv}$$ from C. Jordan Calculus of Finite Differences, pp. 448 - 450.

Question: Could the coefficients $$A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$$ be reached in a recurrent way using $$\beta_{mv}$$? This question also can be stated as: does there exist such a function $$f(\beta_{mv})=A_{v,m}$$?

Additional Question How exactly are our coefficients $$A_{0,m}, \ A_{1,m}, \ ..., \ A_{m,m}$$ connected with $$\beta_{mv}$$?

UPDATE 10-Mar-2019

Let be a power function $$f_{r,n}(s)$$ defined as follows $$\begin{equation}\label{f1} f_{r,n}(s)= \begin{cases} s^r, \ &0\leq s\leq n,\\ 0, \ &\mathrm{otherwise}. \end{cases} \end{equation}$$

Now we want to connect our identity with the discrete convolution of $$f_{r,n}(s)$$. Since the $$f_{r,n}(s)$$ defined only on the interval $$[0,n]$$ of real set, we can conclude the following identity $$\begin{equation}\label{der1} (f_{r,n}*f_{r,n})[n]\equiv\sum_{m=0}^{n}m^r(n-m)^r \end{equation}$$ Definition of $$A_{m,j}$$ coefficients says that $$\begin{equation} (n+1)^{2m+1}=\sum_{r=0}^{m}A_{m,r}(f_{r,n}*f_{r,n})[n]\equiv\sum_{r=0}^{m}A_{m,r}\sum_{k=0}^{n}k^r(n-k)^r \end{equation}$$ Expanding $$\sum_{k=0}^{n}k^r(n-k)^r$$ and applying Faulhaber's formula, we get $$\begin{equation}\label{proof2} \begin{split} \sum_{k=0}^{n}k^r(n-k)^r &=\sum_{k=0}^{n} k^r \sum_{j} (-1)^j\binom{r}{j} n^{r-j}k^{j}=\sum_{j} (-1)^j\binom{r}{j} n^{r-j}\left(\sum_{k=0}^{n}k^{r+j}\right)\\ &=\sum_{j} \binom{r}{j} n^{r-j}\frac{(-1)^j}{r+j+1}\left[\sum_{s}\binom{r+j+1}{s}B_{s}n^{r+j+1-s}\right]\\ &=\sum_{j,s}\binom{r}{j}\frac{(-1)^j}{r+j+1}\binom{r+j+1}{s}B_{s}n^{2r+1-s}, \end{split} \end{equation}$$ where $$B_s$$ are Bernoulli numbers. Now, we notice that $$\begin{equation}\label{proof3} \sum_{j} \binom{r}{j}\frac{(-1)^j}{r+j+1}\binom{r+j+1}{s} =\begin{cases} \frac{1}{(2r+1)\binom{2r}r}, & \text{if } s=0;\\ \frac{(-1)^s}{s}\binom{r}{2r-s+1}, & \text{if } s>0. \end{cases} \end{equation}$$ In particular, the last sum is zero for $$0. Therefore, expression (2.3) takes the form $$\begin{equation} \sum_{k=0}^{n}k^r(n-k)^r=\frac{1}{(2r+1)\binom{2r}r}n^{2r+1}+\sum_{s>0}\frac{(-1)^s}{s}\binom{r}{2r-s+1}B_{s}n^{2r+1-s} \end{equation}$$ Using the definition of $$A_{m,j}$$, we obtain the following identity for polynomials in $$n$$ $$\begin{equation} (\star)\sum_{r}\frac{1}{(2r+1)\binom{2r}r}n^{2r+1}A_{m,r}+\sum_{r,s>0}\frac{(-1)^s}{s}\binom{r}{2r-s+1}B_{s}n^{2r+1-s}A_{m,r}\equiv (n+1)^{2m+1} \end{equation}$$

It seems that for inteval $$[0,n]$$ result of $$A_{m,r}$$ shpuld be different...

Question: How to evaluate the coefficients $$A_{m,r}$$ from last step $$(\star)$$ ?

EDIT 2018-04-16: Formulae are corrected.

I'm not sure about connection with $\beta_{mv}$, but we can obtain a recurrence formula for $A_{j,m}$ as follows.

First let us fix the unused values $A_{j,m}=0$ for $j<0$ or $j>m$, so we won't need to care about the summation range for $j$.

Expanding $(n-k)^j$ and using Faulhaber's formula, we get $$\sum_{k=0}^{n-1} (n-k)^j k^j = \sum_{k=0}^{n-1} \sum_i \binom{j}i n^{j-i} (-1)^i k^{i+j}$$ $$=\sum_{i} \binom{j}{i} n^{j-i} \frac{(-1)^i}{i+j+1} \left[ \sum_t \binom{i+j+1}t B_t n^{i+j+1-t} - B_{i+j+1}\right]$$ $$=\sum_{i,t} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t B_t n^{2j+1-t} - \sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} B_{i+j+1} n^{j-i}$$ where $B_t$ are Bernoulli numbers.

Now, we notice that $$\sum_{i} \binom{j}{i} \frac{(-1)^i}{i+j+1} \binom{i+j+1}t =\begin{cases} \frac{1}{(2j+1)\binom{2j}j}, & \text{if } t=0;\\ \frac{(-1)^j}{t}\binom{j}{2j-t+1}, & \text{if } t>0. \end{cases}$$ In particular, the last sum is zero for $0<t\leq j$.

Hence, introducing $\ell=2j+1-t$ and $\ell=j-i$, respectively, we get $$\sum_{k=0}^{n-1} (n-k)^j k^j = \frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + \sum_{\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell} - \sum_{\ell} \binom{j}{\ell} \frac{(-1)^{j-\ell}}{2j+1-\ell} B_{2j+1-\ell}n^{\ell}$$ $$=\frac{1}{(2j+1)\binom{2j}j} n^{2j+1} + 2\sum_{\text{odd }\ell} \frac{(-1)^j}{2j+1-\ell}\binom{j}{\ell}B_{2j+1-\ell}n^{\ell}.$$

Using the definition of $A_{j,m}$, we obtain the following identity for polynomials in $n$: $$(\star)\qquad\sum_{j} A_{j,m} \frac{1}{(2j+1)\binom{2j}j}n^{2j+1} + 2 \sum_{j,\text{ odd }\ell} A_{j,m} \binom{j}{\ell} \frac{(-1)^j}{2j+1-\ell} B_{2j+1-\ell}n^{\ell} \equiv n^{2m+1}.$$

Taking the coefficient of $n^{2m+1}$ in $(\star)$, we get $A_{m,m} = (2m+1) \binom{2m}{m},$ and taking the coefficient of $x^{2d+1}$ for an integer $d$ in the range $m/2 \leq d < m$, we get $A_{d,m} = 0$.

Taking the coefficient of $n^{2d+1}$ in $(\star)$ for $m/4 \leq d < m/2$, we get $$A_{d,m} \frac{1}{(2d+1)\binom{2d}{d}} + 2 (2m+1) \binom{2m}{m} \binom{m}{2d+1} \frac{(-1)^m}{2m-2d} B_{2m-2d} = 0,$$ i.e. $$A_{d,m} = (-1)^{m-1} \frac{(2m+1)!}{d!d!m!(m-2d-1)!}\frac{1}{m-d} B_{2m-2d}.$$

Continue similarly, we can express $A_{d,m}$ for each integer $d$ in the range $m/2^{s+1}\leq d< m/2^s$ (iterating consecutively $s=1,2,\dots$) via previously determined values of $A_{j,m}$ as follows: $$A_{d,m} = (2d+1)\binom{2d}{d} \sum_{j\geq 2d+1} A_{j,m} \binom{j}{2d+1} \frac{(-1)^{j-1}}{j-d} B_{2j-2d}.$$ The same formula holds also for $d=0$.

• thank you for your reply :) comment means to be very usefull – Petro Kolosov Apr 16 '18 at 8:13
• Dear Dr Alekseev, parhaps, do you see some connection between our $A_{j,m}$ and finite differences of power ? I asking because the zeros we have in this table seems to be like order of difference higher then power, i.e $D^m(x^n)=0, \ m>n$ – Petro Kolosov Apr 16 '18 at 9:18
• I dont understand your notation "Taking the coefficient of $n^{2d+1}$ in (⋆) for $m/4 \leq d<m/2$" i thought $d$ should be an integer? $x^{2d+1}$ is miss-typos? So are the $d$ and $m$ integers or ? In core of question they are meant to be integers. If its not difficult and you have free time, show, please, application of your result on existing coefficients, from above table, thank you – Petro Kolosov Apr 16 '18 at 12:32
• @KolosovPetro: $d$ is integer in the specified range. – Max Alekseyev Apr 16 '18 at 13:10
• oeis.org/A302971 OEIS sequence is finally accepted, thanks to everybody :) – Petro Kolosov May 1 '18 at 21:02