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A144401 Padovan ( A000931) version of A038137: expansion of polynomials as antidiagonal: p(x,n)=1/(1-x-x^3)^n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 2, 1, 4, 6, 6, 3, 1, 5, 10, 13, 11, 4, 1, 6, 15, 24, 27, 18, 6, 1, 7, 21, 40, 55, 51, 30, 9, 1, 8, 28, 62, 100, 116, 94, 50, 13, 1, 9, 36, 91, 168, 231, 234, 171, 81, 19, 1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28, 1, 11, 55, 174, 402, 714, 987, 1065
Offset: 1

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Author

Roger L. Bagula and Gary W. Adamson, Oct 03 2008

Keywords

Comments

Row sums are: 1, 2, 4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332 (cf. A008998).
These polynomials are sort of pseudo-combinations with the last element Padovan instead of one.
If you subtract the binomial triangle sequence you get:
{0},
{0, 0},
{0, 0, 0},
{0, 0, 0, 1},
{0, 0, 0, 2, 2},
{0, 0, 0, 3, 6, 3},
{0, 0, 0, 4, 12, 12, 5},
{0, 0, 0, 5, 20, 30, 23, 8},
{0, 0, 0, 6, 30, 60, 66, 42, 12}

Examples

			{1},
{1, 1},
{1, 2, 1},
{1, 3, 3, 2},
{1, 4, 6, 6, 3},
{1, 5, 10, 13, 11, 4},
{1, 6, 15, 24, 27, 18, 6},
{1, 7, 21, 40, 55, 51, 30, 9},
{1, 8, 28, 62, 100, 116, 94, 50, 13},
{1, 9, 36, 91, 168, 231, 234, 171, 81, 19},
{1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28},
{1, 11, 55, 174, 402, 714, 987, 1065, 879, 527, 208, 41},
{1, 12, 66, 230, 585, 1152, 1792, 2220, 2175, 1640, 906, 330, 60},
{1, 13, 78, 297, 825, 1782, 3072, 4278, 4815, 4320, 3006, 1539, 520, 88},
{1, 14, 91, 376, 1133, 2662, 5028, 7752, 9807, 10122, 8391, 5424, 2586, 816, 129}
		

Crossrefs

Programs

  • Mathematica
    Clear[f, b, a, g, h, n, t]; f[t_, n_] = 1/(1 - t - t^3)^n; a = Table[Table[SeriesCoefficient[Series[f[t, m], {t, 0, 30}], n], {n, 0, 30}], {m, 1, 31}]; b = Table[Table[a[[n - m + 1]][[m]], {m, 1, n }], {n, 1, 15}]; Flatten[b]

Formula

p(x,n)=1/(1-x-x^3)^n; t(n,m)=anti_diagonal_expansion(p(x,n)).