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Search: id:A118349
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| A118349 |
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Convolution triangle, read by rows, where diagonals are successive self-convolutions of A118346. |
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5
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| 1, 1, 0, 1, 1, 0, 1, 2, 5, 0, 1, 3, 11, 30, 0, 1, 4, 18, 70, 201, 0, 1, 5, 26, 121, 487, 1445, 0, 1, 6, 35, 184, 873, 3592, 10900, 0, 1, 7, 45, 260, 1375, 6606, 27600, 85128, 0, 1, 8, 56, 350, 2010, 10672, 51728, 218566, 682505, 0, 1, 9, 68, 455, 2796, 15996, 85182
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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A118346 equals the central terms of pendular triangle A118345, and the diagonals of this triangle form the semi-diagonals of the triangle A118345.
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FORMULA
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Since g.f. G=G(x) of A118346 satisfies: G = 1 - 2*x*G + 2*x*G^2 + x*G^3 then T(n,k) = T(n-1,k) - 2*T(n-1,k-1) + 2*T(n,k-1) + T(n+1,k-1). Also, a recurrence involving antidiagonals is: T(n,k) = T(n-1,k) + Sum_{j=1..k} [3*T(n-1+j,k-j) - 2*T(n-2+j,k-j)] for n>k>=0.
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EXAMPLE
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Show: T(n,k) = T(n-1,k) - 2*T(n-1,k-1) + 2*T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - 2*T(7,3) + 2*T(8,3) + T(9,3)
or: 1375 = 873 - 2*184 + 2*260 + 350.
Triangle begins:
1;
1, 0;
1, 1, 0;
1, 2, 5, 0;
1, 3, 11, 30, 0;
1, 4, 18, 70, 201, 0;
1, 5, 26, 121, 487, 1445, 0;
1, 6, 35, 184, 873, 3592, 10900, 0;
1, 7, 45, 260, 1375, 6606, 27600, 85128, 0;
1, 8, 56, 350, 2010, 10672, 51728, 218566, 682505, 0;
1, 9, 68, 455, 2796, 15996, 85182, 415629, 1771367, 5585115, 0; ...
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PROGRAM
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(PARI) {T(n, k)=polcoeff((serreverse(x*(1-2*x+sqrt((1-2*x)*(1-6*x)+x*O(x^k)))/2/(1-2*x))/x)^(n-k), k)}
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CROSSREFS
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Cf. A118346, A118345.
Adjacent sequences: A118346 A118347 A118348 this_sequence A118350 A118351 A118352
Sequence in context: A065452 A004598 A091086 this_sequence A011183 A005671 A127863
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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), Apr 26 2006
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