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Search: id:A118354
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| A118354 |
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Convolution triangle, read by rows, where diagonals are successive self-convolutions of A118351. |
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6
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| 1, 1, 0, 1, 1, 0, 1, 2, 6, 0, 1, 3, 13, 42, 0, 1, 4, 21, 96, 325, 0, 1, 5, 30, 153, 770, 2688, 0, 1, 6, 40, 244, 1353, 6530, 23286, 0, 1, 7, 51, 340, 2093, 11760, 57612, 208659, 0, 1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314, 0, 1, 9, 76, 581, 4125, 27441, 170580
(list; table; graph; listen)
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OFFSET
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0,8
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COMMENT
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A118351 equals the central terms of pendular triangle A118350 and the lower diagonals of this triangle form the semi-diagonals of the triangle A118350.
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FORMULA
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Since g.f. G=G(x) of A118351 satisfies: G = 1 - 3*x*G + 3*x*G^2 + x*G^3 then T(n,k) = T(n-1,k) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1). Recurrence involving antidiagonals: T(n,k) = T(n-1,k) + Sum_{j=1..k} [4*T(n-1+j,k-j) - 3*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) - 3*T(n-1,k-1) + 3*T(n,k-1) + T(n+1,k-1)
at n=8,k=4: T(8,4) = T(7,4) - 3*T(7,3) + 3*T(8,3) + T(9,3)
or: 2093 = 1353 - 3*244 + 3*340 + 452.
Triangle begins:
1;
1, 0;
1, 1, 0;
1, 2, 6, 0;
1, 3, 13, 42, 0;
1, 4, 21, 96, 325, 0;
1, 5, 30, 163, 770, 2688, 0;
1, 6, 40, 244, 1353, 6530, 23286, 0;
1, 7, 51, 340, 2093, 11760, 57612, 208659, 0;
1, 8, 63, 452, 3010, 18636, 105681, 523446, 1918314, 0;
1, 9, 76, 581, 4125, 27441, 170580, 973953, 4864795, 17994264, 0; ...
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PROGRAM
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(PARI) {T(n, k)=polcoeff((serreverse(x*(1-3*x+sqrt((1-3*x)*(1-7*x)+x*O(x^k)))/2/(1-3*x))/x)^(\ n-k), k)}
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CROSSREFS
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Cf. A118350, A118351, A118352, A118353.
Row sums: A151616.
Sequence in context: A165733 A039907 A072340 this_sequence A080730 A016590 A079461
Adjacent sequences: A118351 A118352 A118353 this_sequence A118355 A118356 A118357
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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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