Proposition 1.1 For every integers $m,n\geq 0$ the following identity holds \begin{equation} n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^m A_{m,j}k\strut^j(n-k)\strut^j=\sum_{k=0}^{m}(-1)^{m-k}U_m(n, k)\cdot n\strut^k, \end{equation} where $U_m(n, k), \ A_{m,j}$ are coefficients defined as follows

Definition 1.2 (Definition of $A_{m,j}$ coefficients.) \begin{equation*} 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)\strut^{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*}

Definition 1.3 (Definition of $U_m(\ell, k)$ coefficients.) \begin{equation*} U_m(\ell, k) := (-1)\strut^m \sum_{k=1}^{\ell}\sum_{j=t}^m (-1)\strut^j\binom{j}{t}A_{m,j}k\strut^{2j-t}. \end{equation*}

The sense of problem: Entire problem is to prove that Proposition (1.1) directly follows from Faulhaber's formula.

Below we provide our try to solve this problem. Consider the Faulhaber's formula Knuth, page 9: \begin{equation}\label{knuth1} (2.1) \quad \quad n^{2m-1}=\begin{cases} n^1 = \binom{n}{1}, \ &\mathrm{if} \ m=1;\\ n^3 = 6\binom{n+1}{3}+\binom{n}{1}, \ &\mathrm{if} \ m=2;\\ n^5 = 120\binom{n+2}{5}+30\binom{n+1}{3}+\binom{n}{1}, \ &\mathrm{if} \ m=3;\\ \vdots\\ n^{2m-1} = \sum\limits_{1\leq k\leq m}(2k-1)!T(2m,2k)\binom{n+k-1}{2k-1}, \ &\mathrm{if} \ m\in\mathbb{N}; \end{cases} \end{equation} The coefficients $(2k-1)!T(2m,2k)$ in the Faulhaber's identities (2.1) can be calculated using following formula \begin{equation} (2.2) \quad \quad (2k-1)!T(2n,2k)=\frac{1}{r}\sum_{j=0}^{r}(-1)^j\binom{2r}{j}(r-j)^{2n}, \end{equation} where $r=n-k+1$ and $T(2n,2k)$ is central factorial number, see A303675. Consider the partial case of (2.1) for $m=2$, for instance \begin{equation} (2.3) \quad \quad n^3=\sum_{k=1}^{n} 6\binom{k}{2}+1=\sum_{k=1}^{n}6k(n-k)+1 \end{equation} The l.h.s of identity (2.3) is reached by means of Hockey Stick pattern in terms of Binomial coefficients \begin{equation} \sum_{k=s}^{n}\binom{k}{s}=\binom{n+1}{s+1} \end{equation} It follows from the identity (2.3) that \begin{equation} \sum_{k=1}^{n}\binom{k}{2}=\sum_{k=1}^{n}k(n-k) \end{equation} Now, lets keep our attention to the identity (2.3) again. As we said, this is the partial case of Faulhaber's formula for $m=2$, and we can notice that the following structure of (2.3) can be assumed \begin{equation*} n^{3}=\sum_{k=1}^{n}\sum_{j=0}^{m=1} A_{m,j}k\strut^j(n-k)\strut^j \end{equation*} and, consequently, for each non-negative integer $m$ \begin{equation*} (2.4)\quad \quad n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j \end{equation*} The coefficients $A_{m,j}$ in (2.4) are terms of sequences: A302971 - numerators; A304042 - denominators. More info concerning the identity $n^{2m+1}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j$ can be found at this link.

Now we get a structure, which seems to be direct consequence of the Faulhaber's formula, with only one difference, the corresponding binomial coefficients are replaced by the polynomials $k^j(n-k)^j$. So, for the moment, the task is to represent the binomial coefficients in terms of polynomial $k^j(n-k)^j$. Let's introduce the function \begin{equation} (2.5)\quad \quad F_s(n,k)=k\binom{n-k}{s-1}=k\sum_{j=0}^{n-k-1}\binom{j}{s-2} \end{equation} For instance, \begin{equation} \begin{split} s=1: \ F_1(n,k) &=0\\ s=2: \ F_2(n,k) &=k(n-k) \\ s=3: \ F_3(n,k) &=k(n-k)(n-k-1)/2\\ s=4: \ F_4(n,k) &=k(n-k)(n-k-1)(n-k-2)/6 \end{split} \end{equation} And most importantly \begin{equation} (\star)\quad \quad \sum_{k=0}^{n}\binom{k}{s}=\sum_{k=0}^{n}F_s(n,k) \end{equation} As we already have the run-algorithm for $m=2$ in Faulhaber's formula that is $n^3$, lets take another example, for $m=3$ and corresponding result $n^5$, we will apply recently received results (1.6) and (1.7). By Faulhaber's formula, the fifth power is \begin{equation} (2.6)\quad \quad n^5 = 120\tbinom{n+2}{5}+30\tbinom{n+1}{3}+\tbinom{n}{1} \end{equation} Our main aim is to compile expression (2.6) to the form $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{2} A_{2,j}k\strut^j(n-k)\strut^j$. By the identity $(\star)$ and definition (2.5) we have \begin{equation} \begin{split} &\sum_{k=1}^{n}\tbinom{k+1}{4}=\sum_{r=1}^{n}F_4(n,k+1)=\sum_{r=1}^{n}-(1/6) k (-1 + k - n) (k - n) (1 + k - n)\\ &=1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2) \\ &\sum_{k=1}^{n}\tbinom{k}{2}=\sum_{k=1}^{n}F_2(n,k)=\sum_{k=1}^{n}k(n-k)\\ &\sum_{k=1}^{n}\tbinom{k-1}{0}=\tbinom{n}{1}=n \end{split} \end{equation} Let be $a=n-k$ in $F_4(n,k)=1/6k(n-k)(n-k+1)(n-k-1)$, thus \begin{equation} \begin{split} F_4(n,k) &=1/6k(a^3-a)\\ &=1/6k((n-k)^3-(n-k))\\ &=1/6k[n^3-3kn^2+3k^2n-k^3-n+k]\\ &=1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2) \end{split} \end{equation} Let's substitute $F_2(n,k)$, and $F_4(n,k)$ to the equation (2.6), respectively \begin{equation} \begin{split} n^5 &= 120\sum_{k=1}^{n}1/6(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2)+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}(kn^3-3k^2n^2+3k^3n-k^4-kn+k^2)+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}(kn^3-3k^2n^2+3k^3n-k^4)-20\sum_{k=1}^{n}kn+k^2+30\sum_{k=1}^{n}kn-k^2+\sum_{k=1}^{n}1\\ &= 20\sum_{k=1}^{n}k(n-k)^3+10\sum_{k=1}^{n}k(n-k)+\sum_{k=1}^{n}1\\ &= \sum_{k=1}^{n}20k(n-k)^3+10k(n-k)+1 \end{split} \end{equation} As we can see, the form of the our transformation is different from $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{2} A_{2,j}k\strut^j(n-k)\strut^j$, but still is near to the result $n^{5}=\sum_{k=1}^{n}30k^2(n-k)^2+1$. We can see that if the variable of polynomial at $n^5=\sum_{k=1}^{n}20k(n-k)^3+10k(n-k)+1$ would be $k^2(n-k)^2$ then we get desired result $n^{5}=\sum_{k=1}^{n}30k^2(n-k)^2+1$.

The problem: Revise the function $F_s(n,k)$ such way, that the construction $n^{5}=\sum_{k=1}^{n}\sum_{j=0}^{m} A_{m,j}k\strut^j(n-k)\strut^j$ directly follows when corresponding binomial coefficient sums in Faulhaber's formula (2.1) is replaced by sums of $F_s(n,k)$.

PS For those who interested, we collected a few questions concenring, everybody are invited to participate, visit https://kolosovpetro.github.io/math_project/

  • $\begingroup$ Is it possible to precisely formulate the open question, i.e. what the statement to prove is? (At the moment it seems somehow crowded with additional information; it is good in general to have it, but makes it hard to extract the actual question.) $\endgroup$ – Daniel Krenn Feb 4 at 13:31
  • $\begingroup$ Thank you for your comment, see the PDF with more clear stated question, now i will revise this one as well $\endgroup$ – Petro Kolosov Feb 4 at 16:21

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