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A144384
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Anti-diagonal expansion of: f(t,n)=If[n == 1, 1/(1 - t), (1 - t)/(1 - t^n)].
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0
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1, 1, 1, 1, -1, 1, 1, -1, 1, 1, 1, -1, 0, -1, 1, 1, -1, 0, 1, 1, 1, 1, -1, 0, 0, -1, -1, 1, 1, -1, 0, 0, 1, 0, 1, 1, 1, -1, 0, 0, 0, -1, 1, -1, 1, 1, -1, 0, 0, 0, 1, 0, -1, 1, 1, 1, -1, 0, 0, 0, 0, -1, 0, 0, -1, 1, 1, -1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, -1, 0, 0, 0, 0, 0, -1, 0, -1, -1, -1, 1, 1, -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, -1, 0, 0, 0, 0, 0, 0, -1, 0, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Row sums are:
{1, 2, 1, 2, 0, 3, -1, 3, 0, 2, -1, 5, -3, 3, 1}.
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FORMULA
| f(t,n)=If[n == 1, 1/(1 - t), (1 - t)/(1 - t^n)]; T(n,m)=anti_diagonal(f(t,n)).
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EXAMPLE
| {1},
{1, 1},
{1, -1, 1},
{1, -1, 1, 1},
{1, -1, 0, -1, 1},
{1, -1, 0, 1, 1, 1},
{1, -1, 0, 0, -1, -1, 1},
{1, -1, 0, 0, 1, 0, 1, 1},
{1, -1, 0, 0, 0, -1, 1, -1, 1},
{1, -1, 0, 0, 0, 1, 0, -1, 1,1},
{1, -1, 0, 0, 0, 0, -1, 0, 0, -1, 1},
{1, -1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1},
{1, -1, 0, 0, 0, 0, 0, -1, 0, -1, -1, -1, 1},
{1, -1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1},
{1, -1, 0, 0, 0, 0, 0, 0, -1, 0, 1, 0, 1, -1, 1}
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MATHEMATICA
| Clear[f, b, a, g, h, n, t]; f[t_, n_] =If[n == 1, 1/(1 - t), (1 - t)/(1 - 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]
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CROSSREFS
| Sequence in context: A105567 A114213 A108358 * A144475 A011758 A015088
Adjacent sequences: A144381 A144382 A144383 * A144385 A144386 A144387
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KEYWORD
| sign,uned
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AUTHOR
| Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Oct 01 2008
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