A224228 Matrix inverse of the triangle of Eulerian numbers T(n,k), 0<=k<=n, read by rows.

1, 1, 1, 5, 4, 1, 65, 55, 11, 1, 1993, 1668, 352, 26, 1, 131665, 110421, 23084, 1784, 57, 1, 18010589, 15099584, 3161545, 242630, 8031, 120, 1, 4981731137, 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

Offset

0,4

Comments

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

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

Links

Table of n, a(n) for n=0..54.

Eric W. Weisstein's World of Mathematics, Eulerian Number

Formula

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

Mathematica

As posted in discussion on Active Mathematica yahoo group:
It gives a Sierpinski triangle...<< Combinatorica`
<< Notation`
Table[If[k <= n, Eulerian[n + 1, k], 0], {n, 0, 5}, {k, 0, 5}]
f[n_, k_] = If[k <= n, Eulerian[n + 1, k], 0]
Clear[x, n, k, A, B, nn]
A[m_] := Table[
  Table[f[n - 1, k - 1]*If[n > k, (-1)^Floor[(n - k + 1)/2], 1], {k, 1,
    m}], {n, 1, m}]
B[m_] := Inverse[A[m]]
Table[TableForm[B[m]], {m, 1, 10}]
Table[Apply[Plus, B[10][[m]]], {m, 1, Length[B[10]]}]
a = Table[Table[B[10][[m, n]], {n, 1, m}], {m, 1, Length[B[10]]}]
Flatten[a]
ListDensityPlot[Mod[B[128], 4]]
MatrixPlot[Mod[B[128], 4]]

Crossrefs

Cf. A008292, A123125.

Sequence in context: A108440 A102220 A279151 * A286680 A201680 A349409

Adjacent sequences: A224225 A224226 A224227 * A224229 A224230 A224231

Keyword

nonn,tabl

Author

Roger L. Bagula and Mats Granvik, Mar 10 2013

Status

approved