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A144409 Antidiagonal expansion of: f(t,n) = If[n == 1, 1/(1 - t), 1/(1 - t^floor(n/2) - t^n)]. 0

%I #4 Dec 07 2016 23:17:31

%S 1,1,1,1,1,1,1,1,2,1,1,0,1,3,1,1,0,1,2,5,1,1,0,1,0,3,8,1,1,0,0,0,2,4,

%T 13,1,1,0,0,1,1,0,6,21,1,1,0,0,1,0,1,3,9,34,1,1,0,0,0,0,0,1,0,13,55,1,

%U 1,0,0,0,1,0,2,2,5,19,89,1,1,0,0,0,1,0,1,0,1,0,28,144,1,1,0,0,0,0,0,0,1,0,3

%N Antidiagonal expansion of: f(t,n) = If[n == 1, 1/(1 - t), 1/(1 - t^floor(n/2) - t^n)].

%C Row sums are {1, 2, 3, 5, 6, 10, 14, 21, 31, 50, 71, 120, 177, 288, 445}.

%F f(t,n) = If[n == 1, 1/(1 - t), 1/(1 - t^floor(n/2) - t^n)); t(n,m) = antidiagonal_expansion(f(t,n)).

%e {1},

%e {1, 1},

%e {1, 1, 1},

%e {1, 1, 2, 1},

%e {1, 0, 1, 3, 1},

%e {1, 0, 1, 2, 5, 1},

%e {1, 0, 1, 0, 3, 8, 1},

%e {1, 0, 0, 0, 2, 4, 13, 1},

%e {1, 0, 0, 1, 1, 0, 6, 21, 1},

%e {1, 0, 0, 1, 0, 1, 3, 9, 34, 1},

%e {1, 0, 0, 0, 0, 0, 1, 0, 13, 55, 1},

%e {1, 0, 0, 0, 1, 0, 2, 2, 5, 19, 89, 1},

%e {1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 28, 144, 1},

%e {1, 0, 0, 0, 0, 0, 0, 1, 0, 3, 8, 41, 233, 1},

%e {1, 0, 0, 0, 0, 1, 0, 0, 0, 3, 2, 0, 60, 377, 1}

%t f[t_, n_] = If[n == 1, 1/(1 - t), 1/(1 - t^Floor[n/2] - t^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. A099238.

%K nonn,uned

%O 1,9

%A _Roger L. Bagula_ and _Gary W. Adamson_, Sep 30 2008

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