%I #9 Feb 16 2025 08:33:19

%S 1,1,1,5,4,1,65,55,11,1,1993,1668,352,26,1,131665,110421,23084,1784,

%T 57,1,18010589,15099584,3161545,242630,8031,120,1,4981731137,

%U 4176740483,874335823,67166609,2212739,33933,247,1,2752004906353,2307299944904,483009645152,37100594596,1222926298,18699092,138602,502,1,3017738824824961,2530094418968969,529648104311800,40683406518208,1340952746858,20509601522,151765114,556366,1013,1

%N Matrix inverse of the triangle of Eulerian numbers T(n,k), 0<=k<=n, read by rows.

%C This sequence arose from discussion of a binomial inverse matrix made by Mats Granik and posted to Active Mathematica Yahoo group.

%C Row sums are:1, 2, 10, 132, 4040, 267012, 36522500, 10102220972, 5580656855500, 6119526369294812, ...

%H Eric W. Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/EulerianNumber.html">Eulerian Number</a>

%F Matrix inverse of Eulerian triangle as a matrix: Inverse(A123125)

%t As posted in discussion on Active Mathematica yahoo group:

%t It gives a Sierpinski triangle...<< Combinatorica`

%t << Notation`

%t Table[If[k <= n, Eulerian[n + 1, k], 0], {n, 0, 5}, {k, 0, 5}]

%t f[n_, k_] = If[k <= n, Eulerian[n + 1, k], 0]

%t Clear[x, n, k, A, B, nn]

%t A[m_] := Table[

%t Table[f[n - 1, k - 1]*If[n > k, (-1)^Floor[(n - k + 1)/2], 1], {k, 1,

%t m}], {n, 1, m}]

%t B[m_] := Inverse[A[m]]

%t Table[TableForm[B[m]], {m, 1, 10}]

%t Table[Apply[Plus, B[10][[m]]], {m, 1, Length[B[10]]}]

%t a = Table[Table[B[10][[m, n]], {n, 1, m}], {m, 1, Length[B[10]]}]

%t Flatten[a]

%t ListDensityPlot[Mod[B[128], 4]]

%t MatrixPlot[Mod[B[128], 4]]

%Y Cf. A008292, A123125.

%K nonn,tabl

%O 0,4

%A _Roger L. Bagula_ and _Mats Granvik_, Mar 10 2013