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A142724
Irregular triangle read by rows: row n gives coefficients in expansion of Product_{k=1..n} (1 + x^(2*k + 1)) for n >= 0.
14
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
OFFSET
0,43
COMMENTS
For n >= 1, row n is the Poincaré polynomial for the Lie group A_n.
Row sums are powers of 2.
REFERENCES
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
LINKS
FORMULA
p(x,n) = Product[(1 + x^(2*k + 1)), {k, 1, n}]; t(n,m)=coefficients(p(x,n)).
EXAMPLE
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},
...
MAPLE
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:
MATHEMATICA
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[%]
KEYWORD
nonn,tabf
AUTHOR
EXTENSIONS
Edited by N. J. A. Sloane, Dec 25 2010
STATUS
approved