OFFSET
4,3
COMMENTS
Row sums are: {2, 25, 234, 2058, 18444, 175005, 1790090, 19866022, 239148084, 3112158322, 43583945300,...}.
Noticing the Eulerian numbers and the binomial squared were the same for the first four rows, I subtracted them and extracted the zeros to get this sequence.
The resulting fractal can be obtained from the Mathematica code given in the Mathematica code section.
LINKS
G. C. Greubel, Table of n, a(n) for the first 50 rows
FORMULA
T(n,m) = ( Eulerian(n,m) - Binomial(n,m)^2 )/2 for n >= 4, and 2 <= m <= n-1.
EXAMPLE
{1, 1},
{5, 15, 5},
{16, 101, 101, 16},
{42, 483, 1008, 483, 42},
{99, 1926, 7197, 7197, 1926, 99},
{219, 6912, 42549, 75645, 42549, 6912, 219},
{466, 23272, 224068, 647239, 647239, 224068, 23272, 466},
{968, 75306, 1094544, 4847007, 7830372, 4847007, 1094544, 75306, 968},
{1981, 237623, 5080230, 33104787, 81149421, 81149421, 33104787, 5080230, 237623, 1981},
{4017, 737685, 22742525, 211518255, 752497122, 1137159114, 752497122, 211518255, 22742525, 737685, 4017},
{8100, 2265615, 99164495, 1285615475, 6420803247, 13984115718, 13984115718, 6420803247, 1285615475, 99164495, 2265615, 8100}
MATHEMATICA
p[x_, n_] = (x - 1)^(n + 1)*Sum[((-1)^(n + 1)*k^n)*x^k, {k, 0, Infinity}]/x; Table[Table[(CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m]] - Binomial[n - 1, m - 1]^2)/2, {m, 2, n - 1}], {n, 4, 14}]; Flatten[%]
(* fractal code *)
a = Table[Table[(CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x][[m]] - Binomial[n - 1, m - 1]^2)/2, {m, 2, n - 1}], {n, 4, 34}];
b = Table[If[m <= n + 1, Mod[a[[n]][[m]], 2], 0], {m, 1, Length[a]}, {n, 1, Length[a]}]; ListDensityPlot[b, Mesh -> False]
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Roger L. Bagula, Jan 07 2009
STATUS
approved