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Search: id:A098568
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| A098568 |
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Triangle of triangular binomial coefficients, read by rows, where column k has the g.f.: 1/(1-x)^((k+1)*(k+2)/2) for k>=0. |
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+0
11
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| 1, 1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 10, 21, 10, 1, 1, 15, 56, 55, 15, 1, 1, 21, 126, 220, 120, 21, 1, 1, 28, 252, 715, 680, 231, 28, 1, 1, 36, 462, 2002, 3060, 1771, 406, 36, 1, 1, 45, 792, 5005, 11628, 10626, 4060, 666, 45, 1, 1, 55, 1287, 11440, 38760, 53130, 31465, 8436
(list; table; graph; listen)
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OFFSET
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1,5
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COMMENT
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The row sums form A098569: {1,2,5,14,43,143,510,1936,7775,32869,...}. How do the terms of row k tend to be distributed as k grows?
Remarkably, column k of the matrix inverse (A121434) equals signed column k of the triangular matrix power: A107876^(k*(k+1)/2) for k>=0. - Paul D. Hanna (pauldhanna(AT)juno.com), Aug 25 2006
Surprisingly, the row sums (A098569) equal the row sums of triangle A131338. - Paul D. Hanna (pauldhanna(AT)juno.com), Aug 30 2007
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REFERENCES
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Nate Kube and Frank Ruskey, Sequences That Satisfy a(n-a(n))=0, Journal of Integer Sequences, Vol. 8 (2005), Article 05.5.5.
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FORMULA
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T(n, k) = C( (k+1)*(k+2)/2 + n-k-1, n-k).
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EXAMPLE
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G.f. of columns are: 1/(1-x), 1/(1-x)^3, 1/(1-x)^10, 1/(1-x)^15, ...
Rows begin:
[1],
[1,1],
[1,3,1],
[1,6,6,1],
[1,10,21,10,1],
[1,15,56,55,15,1],
[1,21,126,220,120,21,1],
[1,28,252,715,680,231,28,1],
[1,36,462,2002,3060,1771,406,36,1],
[1,45,792,5005,11628,10626,4060,666,45,1],
[1,55,1287,11440,38760,53130,31465,8436,1035,55,1],
[1,66,2002,24310,116280,230230,201376,82251,16215,1540,66,1],...
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PROGRAM
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(PARI) T(n, k)=binomial((k+1)*(k+2)/2+n-k-1, n-k)
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CROSSREFS
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Cf. A098569.
Cf. A121434 (inverse); variants: A122175, A122176, A122177; A107876.
Cf. A131338.
Sequence in context: A107105 A088925 A100862 this_sequence A131235 A157243 A146769
Adjacent sequences: A098565 A098566 A098567 this_sequence A098569 A098570 A098571
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KEYWORD
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nonn,tabl
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Sep 15 2004
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