# Coefficients $U_m(n,k)$ in the identity $n^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k$

Review the main result of mathoverflow.net/questions/297900, that is the identity $$\label{f1} n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j,$$ where $A_{m,j}$ is from sequences A302971 and A304042. In this question we discuss the polynomial $$\label{f2} \sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k, \ m\geq0 \ \mathrm{integer}$$ That is generated by the identity $$\label{f3} (1.3)\quad\sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k,$$ where $T=1,2,3...$ and $m\geq0, \ m=\mathrm{const}$. The coefficient $A_{m,j}$ is generated by \begin{equation*}\label{gen_13} A_{m,j}:= \begin{cases} 0, & \mathrm{if } \ j<0 \ \mathrm{or } \ j>m \\ (2j+1)\binom{2j}{j} \sum_{d=2j+1}^{m} A_{m,d} \binom{d}{2j+1} \frac{(-1)^{d-1}}{d-j} B_{2d-2j}, & \mathrm{if } \ 0 \leq j < m \\ (2j+1)\binom{2j}{j}, & \mathrm{if } \ j=m \\ \end{cases} \end{equation*} Derivation of coefficients $A_{m,j}$ is discussed in mathoverflow.net/questions/297900. In particular, the right part of (1.3) returns odd power $2m+1$ of $T\in\mathbb{N}$ when $n=T$ \begin{equation*} T^{2m+1}=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(T,k)\cdot T^k \end{equation*}

Detailed derivation of the polynomials, consisting the coefficient $U_m(n,k)$.

Consider the identity discussed in mathoverflow.net/questions/297900, $$n^{2m+1}=\sum\limits_{1\leq k \leq n}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j,$$ Let show a few examples of polynomials $\sum\nolimits_{j\geq0}A_{m,j}k^j(n-k)^j$ for $m=1,2,3$. We denote the part $\sum\nolimits_{j\geq0}A_{m,j}k^j(n-k)^j$ of the left part of equation (1.3) as $$\label{f4} D_m(n,k)=\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j$$ Therefore, for $m=1,2,3$ we have corresponding $D_m(n,k)$ as $$(1.6)\quad\begin{cases} D_{1}(n,k)=1+6k(n-k), & \\ D_{2}(n,k)=1-0k(n-k)+30k^2(n-k)^2, & \\ D_{3}(n,k)=1-14k(n-k)+0k^2(n-k)^2+140k^3(n-k)^3, & \end{cases}$$ The coefficients in $D_{t=1,2,3}(n,k)$ are the terms of corresponding row of triangle https://oeis.org/A302971. Now, we show an example of generation of polynomials from the right part of (1.3) for $m=1$,

Example 1.

Let be $m=1$, then we rewrite the left hand side of (1.3) as $$(1.8)\quad\sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{1,j}k^j(n-k)^j$$ Next, let substitute the polynomial $D_1(n,k)$ from (1.6) into equation (1.8) and let be $T=1,...,10$, then $$\sum\limits_{1\leq k \leq T}1+6k(n-k)=\begin{cases} T=1 :& -5 + 6 n \\ T=2 :& -28 + 18 n \\ T=3 :& -81 + 36 n \\ T=4 :& -176 + 60 n \\ T=5 :& -325 + 90 n \\ T=6 :& -540 + 126 n \\ T=7 :& -833 + 168 n \\ T=8 :& -1216 + 216 n \\ T=9 :& -1701 + 270 n \\ T=10:& -2300 + 330 n \end{cases}$$ Coefficients of above polynomials are terms of sequences A028896 and A275709. Let show the case for $m=2$ and $T=1,...,10$, again we recall the corresponding polynomial $D_2(n,k)$ from (1.6) and substitute it into left part of (1.3), $$\sum\limits_{1\leq k \leq T}1-0k(n-k)+30k^2(n-k)^2=\begin{cases} T=1 :& 31 - 60 n + 30 n^2 \\ T=2 :& 512 - 540 n + 150 n^2 \\ T=3 :& 2943 - 2160 n + 420 n^2 \\ T=4 :& 10624 - 6000 n + 900 n^2 \\ T=5 :& 29375 - 13500 n + 1650 n^2 \\ T=6 :& 68256 - 26460 n + 2730 n^2 \\ T=7 :& 140287 - 47040 n + 4200 n^2 \\ T=8 :& 263168 - 77760 n + 6120 n^2 \\ T=9 :& 459999 - 121500 n + 8550 n^2 \\ T=10:& 760000 - 181500 n + 11550 n^2 \end{cases}$$ Similarly, let show an example for $m=3$ and $T=1,...,10$, $$\sum\limits_{1\leq k \leq T}1-14k(n-k)+0k^2(n-k)^2+140k^3(n-k)^3=\begin{cases} T=1 :& -125 + 406 n - 420 n^2 + 140 n^3\\ T=2 :& -9028 + 13818 n - 7140 n^2 + 1260 n^3\\ T=3 :& -110961 + 115836 n - 41160 n^2 + 5040 n^3\\ T=4 :& -684176 + 545860 n - 148680 n^2 + 14000 n^3\\ T=5 :& -2871325 + 1858290 n - 411180 n^2 + 31500 n^3\\ T=6 :& -9402660 + 5124126 n - 955500 n^2 + 61740 n^3\\ T=7 :& -25872833 + 12182968 n - 1963920 n^2 + 109760 n^3\\ T=8 :& -62572096 + 25945416 n - 3684240 n^2 + 181440 n^3\\ T=9 :& -136972701 + 50745870 n - 6439860 n^2 + 283500 n^3\\ T=10 :& -276971300 + 92745730 n - 10639860 n^2 + 423500 n^3 \end{cases}$$ We can observe, that in every upper example, the resulting polynomial for every $m, T$ has the following form $$\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k,$$ Therefore, the following question is stated

Question 1. Is there a recurrent that gives the coefficients $U_m(n,k)$ otherwise then by the identity $$\label{f3_1} \sum\limits_{1\leq k \leq T}\sum\limits_{j\geq0}A_{m,j}k^j(n-k)^j=\sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(n,k)\cdot n^k,$$ i.e is there any function $F(n,m)$ such that $F(m,n)=U_m(n,k)$ but different from relation (1.3) ?

Above examples could be generated using Mathematica code Um(n,k)_coefficients2.txt. The PDF-analog of this question with extended data of $U_m(n,k)$ coefficients up to $T=40$ is available at this link.

• Something is wrong with (1.3) - the l.h.s. depends on $T$, while the r.h.s. does not. – Max Alekseyev Jul 3 '18 at 15:09
• @max-alekseyev, Thank you for your reply, right hand side doesn't depend on $T$ as right side is namely 'closed form' of summation from left side, the $k$ runs in $0\leq k \leq T$ which results a polynomial in $n$, without $T$. – PKK Jul 3 '18 at 16:37
• One can verify results of (1.3) using Mathematica code kolosovpetro.github.io/mathematica_codes/… . – PKK Jul 3 '18 at 16:49
• I don't understand this. E.g., do you claim that the sum over $k$ with $1\leq k\leq 1$ (i.e., for $T=1$) is the same as the sum over $k$ with $1\leq k\leq 2$ (i.e., for $T=2$)? If so, then the summand for $k=2$ must be $0$. – Max Alekseyev Jul 3 '18 at 17:11
• In (1.3), shouldn't be $U_m(T,k)$ rather than $U_m(n,k)$ by any chance? – Max Alekseyev Jul 3 '18 at 17:24

First off, as I explained in the comments, the identity (1.3) should contain $U_m(T,k)$ rather than $U_m(n,k)$ (the latter does not make any sense), and so the correct identity (1.3) (for polynomials in $n$) states: $$(1.3)\quad\sum_{k=1}^T\sum_{j=0}^m A_{m,j}k^j(n-k)^j\equiv \sum\limits_{0\leq k \leq m}(-1)^{m-k}U_m(T,k)\cdot n^k.$$ Also, to avoid confusion, it is important to notice that while coefficients $A_{m,j}$ come from the previous question, there their indices are interchanged (i.e., $A_{j,m}$ instead of $A_{m,j}$).
Said that, it is not hard to express the polynomials $U_m(T,k)$ in terms of $A_{m,j}$ and Bernoulli numbers. To do so, let's expand the binomial $(n-k)^j$ in the l.h.s. of (1.3) and change of the order of summation: \begin{split} \sum_{k=1}^T\sum_{j=0}^m A_{m,j}k^j(n-k)^j &= \sum_{k=1}^T\sum_{j=0}^m A_{m,j}k^j\sum_{t=0}^j\binom{j}{t}n^t(-1)^{t-t}k^{j-t}\\ &=\sum_{t=0}^m n^t \sum_{k=1}^T\sum_{j=t}^m \binom{j}{t}A_{m,j}k^{2j-t}(-1)^{j-t}. \end{split} Now, taking the coefficient of $n^t$ in (1.3) gives: $$U_m(T,t) = (-1)^m \sum_{k=1}^T\sum_{j=t}^m \binom{j}{t}A_{m,j}k^{2j-t}(-1)^j.$$
From this formula it may be not immediately clear why $U_m(T,t)$ represent polynomials in $T$. However, this can be seen if we change the summation order again and use Faulhaber's formula to obtain: $$U_m(T,t) = (-1)^m \sum_{j=t}^m \binom{j}{t}A_{m,j} \frac{(-1)^j}{2j-t+1}\sum_{\ell=0}^{2j-t} \binom{2j-t+1}{\ell}B_{\ell}T^{2j-t+1-\ell}.$$ Introducing $k=2j-t+1-\ell$, we further get the formula: $$U_m(T,t) = (-1)^m \sum_{k=1}^{2m-t+1} T^k \sum_{j=t}^m \binom{j}{t}A_{m,j} \frac{(-1)^j}{2j-t+1}\binom{2j-t+1}{k}B_{2j-t+1-k},$$ which allows easily compute the coefficient of $T^k$ in $U_m(T,t)$ for each $k$.
N.B. In the above formulae, we assume that $B_1=+\frac{1}{2}$.
• Dear Dr. Alexeyev, your answer is totally correct and contains a details that I haven't noticed before, thank you for your time and power spent on this question, if you could be interested, you are welcome to revise (or add some info) to the sequences oeis.org/A316387 ($U_m(T,k), \ m=2$) oeis.org/A316387 ($U_m(T,k), \ m=3$) – PKK Sep 5 '18 at 9:57
• Typos issue: oeis.org/A316349 ($U_m(T,k), \ m=2$. – PKK Sep 5 '18 at 10:03