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A224228
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Matrix inverse of the triangle of Eulerian numbers T(n,k), 0<=k<=n, read by rows.
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0
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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
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OFFSET
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0,4
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COMMENTS
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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, ...
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LINKS
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Table of n, a(n) for n=0..54.
Eric W. Weisstein's World of Mathematics, Eulerian Number
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FORMULA
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Matrix inverse of Eulerian triangle as a matrix: Inverse(A123125)
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MATHEMATICA
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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]]
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CROSSREFS
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Cf. A008292, A123125.
Sequence in context: A108440 A102220 A279151 * A286680 A201680 A349409
Adjacent sequences: A224225 A224226 A224227 * A224229 A224230 A224231
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KEYWORD
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nonn,tabl
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AUTHOR
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Roger L. Bagula and Mats Granvik, Mar 10 2013
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STATUS
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approved
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