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The sequence https://oeis.org/A287326 - is Binomial distributed triangular array, that shows us necessary items to expand perfect cube $n^3$. Summation of $n$-th row of Triangle A287326 from $0$ to $n-1$ returns $n^3$. But is it exist simillar patterns in order to receive expansion of power $n>3$, where $n$ - positive integer?

$$ \begin{matrix} & & & & & 1\\ & & & & 1 & & 1\\ & & & 1 & & 7& & 1\\ & & 1 & & 13& & 13& & 1\\ & 1 & & 19& & 25& & 19& & 1\\ \end{matrix} $$ Figure 1. Triangle A287326.

It derived by means of identity $$ x^3=\sum\limits_{m=0}^{x-1}3!\cdot mx-3!\cdot m^2+1 $$

For detailed info on derivation, please, reffer to links below. Thank you !

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  • $\begingroup$ If you want your post to look less like spam and more like an inquiry, post only two links: one to your repository (and put your table of contents behind the link), and the other to some work done independently by a different researcher that you do not find on OEIS. Gerhard "We Must Keep Up Appearances" Paseman, 2018.02.19. $\endgroup$
    – Gerhard Paseman
    Commented Feb 19, 2018 at 22:21
  • $\begingroup$ Dear Gerhard, thank you for your reply, excessed links removed $\endgroup$
    – Petro Kolosov
    Commented Feb 20, 2018 at 12:29

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I'm not sure which generalization of this you might be interested in. But you might note that

$$ x^k = \sum_{m=0}^{x-1} ((m+1)^k - m^k) = \sum_{m=0}^{x-1} \sum_{j=0}^{k-1} {k \choose j} m^j $$

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  • $\begingroup$ Dear Robert, this fact is noticed and could be found at second link, on the page 2, lemma (1.4) $\endgroup$
    – Petro Kolosov
    Commented Feb 20, 2018 at 11:47
  • $\begingroup$ We looking not for generalization by means of finite differences (which for sure already known), we looking for generaliation of above triangle for powers $m>3$ $\endgroup$
    – Petro Kolosov
    Commented Feb 20, 2018 at 12:07