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%I #5 Mar 17 2013 05:01:01
%S 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,
%T 21,40,55,51,30,9,1,8,28,62,100,116,94,50,13,1,9,36,91,168,231,234,
%U 171,81,19,1,10,45,128,266,420,505,460,303,130,28,1,11,55,174,402,714,987,1065
%N Padovan ( A000931) version of A038137: expansion of polynomials as antidiagonal: p(x,n)=1/(1-x-x^3)^n.
%C Row sums are: 1, 2, 4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064, 11169, 24634, 54332 (cf. A008998).
%C These polynomials are sort of pseudo-combinations with the last element Padovan instead of one.
%C If you subtract the binomial triangle sequence you get:
%C {0},
%C {0, 0},
%C {0, 0, 0},
%C {0, 0, 0, 1},
%C {0, 0, 0, 2, 2},
%C {0, 0, 0, 3, 6, 3},
%C {0, 0, 0, 4, 12, 12, 5},
%C {0, 0, 0, 5, 20, 30, 23, 8},
%C {0, 0, 0, 6, 30, 60, 66, 42, 12}
%F p(x,n)=1/(1-x-x^3)^n; t(n,m)=anti_diagonal_expansion(p(x,n)).
%e {1},
%e {1, 1},
%e {1, 2, 1},
%e {1, 3, 3, 2},
%e {1, 4, 6, 6, 3},
%e {1, 5, 10, 13, 11, 4},
%e {1, 6, 15, 24, 27, 18, 6},
%e {1, 7, 21, 40, 55, 51, 30, 9},
%e {1, 8, 28, 62, 100, 116, 94, 50, 13},
%e {1, 9, 36, 91, 168, 231, 234, 171, 81, 19},
%e {1, 10, 45, 128, 266, 420, 505, 460, 303, 130, 28},
%e {1, 11, 55, 174, 402, 714, 987, 1065, 879, 527, 208, 41},
%e {1, 12, 66, 230, 585, 1152, 1792, 2220, 2175, 1640, 906, 330, 60},
%e {1, 13, 78, 297, 825, 1782, 3072, 4278, 4815, 4320, 3006, 1539, 520, 88},
%e {1, 14, 91, 376, 1133, 2662, 5028, 7752, 9807, 10122, 8391, 5424, 2586, 816, 129}
%t 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]
%Y Cf. A000931, A038137.
%K nonn,tabl,uned
%O 1,5
%A _Roger L. Bagula_ and _Gary W. Adamson_, Oct 03 2008
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