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Search: id:A096811
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| A096811 |
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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-2). |
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5
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| 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 1, 1, 1, 2, 3, 3, 2, 1, 1, 1, 1, 1, 2, 4, 4, 3, 2, 1, 1, 1, 1, 1, 2, 4, 5, 4, 4, 2, 1, 1, 1, 1, 1, 2, 4, 6, 6, 5, 4, 2, 1, 1, 1, 1, 1, 2, 4, 7, 7, 8, 6, 4, 2, 1, 1, 1, 1, 1, 2, 4, 7, 9, 10, 9, 7, 4, 2, 1, 1
(list; table; graph; listen)
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OFFSET
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0,19
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COMMENT
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Two row convergents exist simultaneously. When the rows are read forwards, they converge to A096812. When the rows are read backwards, they converge to A096813. The row sums form A096814.
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FORMULA
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T(n, k) = Sum_{j=1..min(n-k, k-1)} T(n-k, j)*T(k-2, k-j-1), for n>=k>=1, with T(n, 0)=T(n+1, 1)=T(n, n)=1 for n>=0.
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EXAMPLE
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T(11,5) = 6 = 5-th term of convolution of row (11-5) with row (5-2) =
T(6,1)*T(3,3) + T(6,2)*T(3,2) + T(6,3)*T(3,1) + T(6,4)*T(3,0).
Rows begin with n=0:
[1],
[1,1],
[1,1,1],
[1,1,1,1],
[1,1,1,1,1],
[1,1,1,2,1,1],
[1,1,1,2,2,1,1],
[1,1,1,2,3,2,1,1],
[1,1,1,2,3,3,2,1,1],
[1,1,1,2,4,4,3,2,1,1],
[1,1,1,2,4,5,4,4,2,1,1],
[1,1,1,2,4,6,6,5,4,2,1,1],
[1,1,1,2,4,7,7,8,6,4,2,1,1],
[1,1,1,2,4,7,9,10,9,7,4,2,1,1],...
Forwards row convergent forms A096812:
[1,1,1,2,4,8,16,34,72,156,336,746,1652,3696,...].
Backwards row convergent forms A096813:
[0,1,1,2,4,8,18,40,92,210,490,1178,2834,6908,...].
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PROGRAM
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(PARI) T(n, k)=if(n<k|k<0, 0, if(k<=1|k==n, 1, sum(j=1, k-1, T(n-k, j)*T(k-2, k-j-1))))
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CROSSREFS
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Cf. A096813, A096814, A091499.
Sequence in context: A064531 A037829 A117546 this_sequence A082478 A083382 A025900
Adjacent sequences: A096808 A096809 A096810 this_sequence A096812 A096813 A096814
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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), Jul 20 2004
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