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A142724
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Irregular triangle read by rows: row n gives coefficients in expansion of Product_{k=1..n} (1 + x^(2*k + 1)) for n >= 0.
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14
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1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 0, 2, 1, 2, 1, 1, 2, 1, 2, 0, 2, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 1, 2
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OFFSET
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0,43
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COMMENTS
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For n >= 1, row n is the Poincaré polynomial for the Lie group A_n.
Row sums are powers of 2.
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REFERENCES
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Borel, A. and Chevalley, C., The Betti numbers of the exceptional groups, Mem. Amer. Math. Soc. 1955, no. 14, pp 1-9.
Samuel I. Goldberg, Curvature and Homology, Dover, New York, 1998, page 144
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LINKS
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FORMULA
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p(x,n) = Product[(1 + x^(2*k + 1)), {k, 1, n}]; t(n,m)=coefficients(p(x,n)).
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EXAMPLE
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Triangle begins:
{1} (the empty product)
{1, 0, 0, 1},
{1, 0, 0, 1, 0, 1, 0, 0, 1},
{1, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1},
{1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1},
{1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 0, 2, 1, 2, 1, 1, 2, 1, 2, 0, 2, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1},
...
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MAPLE
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A:=n->mul(1+x^(2*r+1), r=1..n);
for n from 1 to 12 do lprint(seriestolist(series(A(n), x, 10000))); od:
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MATHEMATICA
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Clear[p, x, n, m]; p[x_, n_] = Product[(1 + x^(2*k + 1)), {k, 1, n}]; Table[CoefficientList[p[x, n], x], {n, 1, 10}]; Flatten[%]
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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