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Consider the following matrix where the rows are coefficients of sums of powers of $n$: $$\left( \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{1}{6} & \frac{1}{2} & \frac{1}{3} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 \\ -\frac{1}{30} & 0 & \frac{1}{3} & \frac{1}{2} & \frac{1}{5} & 0 & 0 & 0 & 0 & 0 \\ 0 & -\frac{1}{12} & 0 & \frac{5}{12} & \frac{1}{2} & \frac{1}{6} & 0 & 0 & 0 & 0 \\ \frac{1}{42} & 0 & -\frac{1}{6} & 0 & \frac{1}{2} & \frac{1}{2} & \frac{1}{7} & 0 & 0 & 0 \\ 0 & \frac{1}{12} & 0 & -\frac{7}{24} & 0 & \frac{7}{12} & \frac{1}{2} & \frac{1}{8} & 0 & 0 \\ -\frac{1}{30} & 0 & \frac{2}{9} & 0 & -\frac{7}{15} & 0 & \frac{2}{3} & \frac{1}{2} & \frac{1}{9} & 0 \\ 0 & -\frac{3}{20} & 0 & \frac{1}{2} & 0 & -\frac{7}{10} & 0 & \frac{3}{4} & \frac{1}{2} & \frac{1}{10} \\ \end{array} \right)$$ Now consider that the diagonals of the above matrix are easily computed as: $$B_n\cdot\frac{\binom{k+n}{k}}{k+1}$$

where $n$ is the row and $k$ is the $k_{th}$ element of that row's diagonal.

Now take the partial sums of the diagonals (without $B_n$) and taking the coefficients once again to generate:

$$\left( \begin{array}{cccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{3}{4} & \frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{11}{18} & \frac{1}{3} & \frac{1}{18} & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{25}{48} & \frac{35}{96} & \frac{5}{48} & \frac{1}{96} & 0 & 0 & 0 & 0 & 0 & 0 \\ \frac{137}{300} & \frac{3}{8} & \frac{17}{120} & \frac{1}{40} & \frac{1}{600} & 0 & 0 & 0 & 0 & 0 \\ \frac{49}{120} & \frac{203}{540} & \frac{49}{288} & \frac{35}{864} & \frac{7}{1440} & \frac{1}{4320} & 0 & 0 & 0 & 0 \\ \frac{363}{980} & \frac{67}{180} & \frac{967}{5040} & \frac{1}{18} & \frac{23}{2520} & \frac{1}{1260} & \frac{1}{35280} & 0 & 0 & 0 \\ \frac{761}{2240} & \frac{29531}{80640} & \frac{267}{1280} & \frac{1069}{15360} & \frac{9}{640} & \frac{13}{7680} & \frac{1}{8960} & \frac{1}{322560} & 0 & 0 \\ \frac{7129}{22680} & \frac{6515}{18144} & \frac{4523}{20412} & \frac{95}{1152} & \frac{3013}{155520} & \frac{5}{1728} & \frac{29}{108864} & \frac{1}{72576} & \frac{1}{3265920} & 0 \\ \frac{7381}{25200} & \frac{177133}{504000} & \frac{16819}{72576} & \frac{341693}{3628800} & \frac{8591}{345600} & \frac{7513}{1728000} & \frac{121}{241920} & \frac{11}{302400} & \frac{11}{7257600} & \frac{1}{36288000} \\ \end{array} \right)$$

Considering each column, the portion of the total for each column seems to be Stirling numbers. For example, consider column 2:

$1,6,35,225..$ -> Unsigned Stirling numbers of first kind $s(n,3)$.

In addition, there are some interesting properties of the inverses of this matrix (and the former, but that is known to be pascal's triangle):

$$\left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ -3 & 4 & 0 & 0 & 0 & 0 \\ 7 & -24 & 18 & 0 & 0 & 0 \\ -15 & 100 & -180 & 96 & 0 & 0 \\ 31 & -360 & 1170 & -1440 & 600 & 0 \\ -63 & 1204 & -6300 & 13440 & -12600 & 4320 \\ \end{array} \right)$$

The first column is $2^n-1$, the second column is $2\cdot 3^{n-1}-2^{n+1}+2$ (this is related to pan digital numbers for some reason OEIS) ... and so on.

What is the connection between these concepts? I don't see it clearly.

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A couple of years ago I've investigated this question in terms of a study of combinatorical matrixes and have found a truly nice relation.

Let's call your matrix as Gp (as in my old texts).
The top left looks like

      1      .    .     .    .    .
    1/2    1/2    .     .    .    .
    1/6    1/2  1/3     .    .    .  = top left of "Gp"
      0    1/4  1/2   1/4    .    .
  -1/30      0  1/3   1/2  1/5    .
      0  -1/12    0  5/12  1/2  1/6

Let's define the matrix St1 as matrix of the Stirling Numbers of first kind, and its top-left part is

     1    .     .    .    .  .
    -1    1     .    .    .  .
     2   -3     1    .    .  .
    -6   11    -6    1    .  .  = top-left of St1
    24  -50    35  -10    1  .
  -120  274  -225   85  -15  1

Let's define the matrix St2 as matrix of the Stirling Numbers of second kind, and its top-left part is

  1   .   .   .   .  .
  1   1   .   .   .  .
  1   3   1   .   .  . = top-left of "St2"
  1   7   6   1   .  .
  1  15  25  10   1  .
  1  31  90  65  15  1

Note that they are mutual inverses $St1 = St2^{-1}$
Now let's define the diagonal vector of first reciprocal numbers , $\;^dZ(1)$

  1    .    .    .    .    .
  .  1/2    .    .    .    .
  .    .  1/3    .    .    . = top-left of "dZ(1)" 
  .    .    .  1/4    .    .
  .    .    .    .  1/5    .
  .    .    .    .    .  1/6

Then that three matrices are also the eigensystem of Gp

$$ St2 \cdot \;^dZ(1) \cdot St1 = Gp \tag 1$$


P.s.: note, that using the coefficients along rows gives polynomials to determine the sums-of-like-powers of consecutive integers. But is not widely knwon, that the coefficients along columns gives powerseries to determine the sums-of-like-powers of reciprocal consecutive integers - only we need techniques for divergent summation. This gives in principle an implementation for Hurwitz-zeta at positive and negative integer parameter (see my extension of Gp which I called for obvious reasons ZETA-matrix).

If you like, there is much interesting explorative material related to this in my old, in many parts amateurish written, investigation on related matrices. See for instance
- binomialmatrix see pg.14
- SummingOfLikePowers
- pmatrix
- stirling pg. 13

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