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Search: id:A092905
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| A092905 |
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Triangle, read by rows, such that the partial sums of the n-th row form the n-th diagonal, for n>=0, where each row begins with 1. |
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4
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| 1, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 4, 4, 2, 1, 1, 5, 6, 4, 2, 1, 1, 6, 9, 7, 4, 2, 1, 1, 7, 12, 11, 7, 4, 2, 1, 1, 8, 16, 16, 12, 7, 4, 2, 1, 1, 9, 20, 23, 18, 12, 7, 4, 2, 1, 1, 10, 25, 31, 27, 19, 12, 7, 4, 2, 1, 1, 11, 30, 41, 38, 29, 19, 12, 7, 4, 2, 1, 1, 12, 36, 53, 53, 42, 30, 19, 12, 7, 4, 2, 1
(list; table; graph; listen)
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OFFSET
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0,5
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COMMENT
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Row sums form A000070, which is the partial sums of the partition numbers (A000041). Rows read backwards converge to the row sums (A000070).
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FORMULA
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T(n, k) = sum_{j=0..k} T(n-k, j), with T(0, n) = 1 for all n>=0. A000070(n) = sum_{k=0..n} T(n, k).
E.g.f.: (1/(1-y))*(1/Product(1-x*y^k, k=1..infinity)). - Vladeta Jovovic (vladeta(AT)eunet.rs), Jan 29 2005
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EXAMPLE
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The third row is {1,3,2,1} and the third diagonal is the partial sums of the third row: {1,4,6,7,7,7,7,7,...}.
Rows begin:
{1},
{1,1},
{1,2,1},
{1,3,2,1},
{1,4,4,2,1},
{1,5,6,4,2,1},
{1,6,9,7,4,2,1},
{1,7,12,11,7,4,2,1},
{1,8,16,16,12,7,4,2,1},
{1,9,20,23,18,12,7,4,2,1},
{1,10,25,31,27,19,12,7,4,2,1},
{1,11,30,41,38,29,19,12,7,4,2,1},
{1,12,36,53,53,42,30,19,12,7,4,2,1},...
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MAPLE
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T(n, k)=if(n<k|k<0, 0, if(n==k|k==0, 1, sum(j=0, min(k, n-k), T(n-k, j))))
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CROSSREFS
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Antidiagonal sums form the partition numbers (A000041).
Cf. A000070.
Sequence in context: A136575 A077592 A055794 this_sequence A052509 A093628 A114282
Adjacent sequences: A092902 A092903 A092904 this_sequence A092906 A092907 A092908
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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), Mar 12 2004
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