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Search: id:A091499
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| A091499 |
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Triangle, read by rows, such that T(n,k) equals the k-th term of the convolution of the two prior rows indexed by (n-k) and (k-1). |
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4
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| 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 3, 3, 1, 1, 1, 2, 4, 4, 3, 1, 1, 1, 2, 4, 6, 5, 4, 1, 1, 1, 2, 4, 7, 8, 7, 4, 1, 1, 1, 2, 4, 8, 10, 12, 8, 5, 1, 1, 1, 2, 4, 9, 13, 15, 16, 10, 5, 1, 1, 1, 2, 4, 9, 15, 20, 22, 21, 12, 6, 1, 1, 1, 2, 4, 9, 17, 25, 31, 31, 27, 14, 6, 1, 1, 1, 2, 4, 9
(list; table; graph; listen)
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OFFSET
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0,9
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COMMENT
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Row sums are A091501. Convergent of rows is A091500: {1,1,2,4,9,20,47,113,275,676,1685,4271,10843,27801,71611,185795,...}.
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FORMULA
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T(n, k) = Sum T(n-k, j)*T(k-1, k-j-1) {j=0..min(n-k, k)}, with T(n, 0)=1, T(n, n)=1 for n>=0.
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EXAMPLE
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T(7,3) = 4 = third term of convolution[{1,1,2,2,1},{1,1,1}] = 1*1 + 1*1 + 2*1 = T(4,0)*T(2,2) + T(4,1)*T(2,1) + T(4,2)*T(2,0).
T(14,5) = 19 = 5-th term of convolution[{1,1,2,4,8,10,12,8,5,1},{1,1,2,2,1}] = 1*1 + 1*2 + 2*2 + 4*1 + 8*1 = T(9,0)*T(4,4) + T(9,1)*T(4,3) + T(9,2)*T(4,2) + T(9,3)*T(4,1) + T(9,4)*T(4,0).
Rows begin:
{1},
{1,1},
{1,1,1},
{1,1,2,1},
{1,1,2,2,1},
{1,1,2,3,3,1},
{1,1,2,4,4,3,1},
{1,1,2,4,6,5,4,1},
{1,1,2,4,7,8,7,4,1},
{1,1,2,4,8,10,12,8,5,1},
{1,1,2,4,9,13,15,16,10,5,1},
{1,1,2,4,9,15,20,22,21,12,6,1},
{1,1,2,4,9,17,25,31,31,27,14,6,1},...
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PROGRAM
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(PARI) {T(n, k)=if(k>n|n<0|k<0, 0, if(k<=1|k==n, 1, sum(j=0, min(n-k, k), T(n-k, j)*T(k-1, k-j-1))))}
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CROSSREFS
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Cf. A091500, A091501.
Sequence in context: A089723 A055215 A058398 this_sequence A137350 A166240 A114087
Adjacent sequences: A091496 A091497 A091498 this_sequence A091500 A091501 A091502
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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), Jan 16 2004
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