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Search: id:A118931
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| A118931 |
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Triangle, read by rows, where T(n,k) = n!/[k!*(n-3*k)!*3^k)] for n>=3*k>=0. |
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+0
4
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| 1, 1, 1, 1, 2, 1, 8, 1, 20, 1, 40, 40, 1, 70, 280, 1, 112, 1120, 1, 168, 3360, 2240, 1, 240, 8400, 22400, 1, 330, 18480, 123200, 1, 440, 36960, 492800, 246400, 1, 572, 68640, 1601600, 3203200, 1, 728, 120120, 4484480, 22422400, 1, 910, 200200, 11211200
(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 n contains 1+floor(n/3) terms. Row sums yield A001470. Given column vector V = A118932, then V is invariant under matrix product T*V = V, or, A118932(n) = Sum_{k=0..n} T(n,k)*A118932(k). Given C = Pascal's triangle and T = this triangle, then matrix product M = C^-1*T yields M(3n,n) = (3*n)!/(n!*3^n), 0 otherwise (cf. A100861 formula due to Paul Barry).
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FORMULA
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E.g.f.: A(x,y) = exp(x + y*x^3/3).
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EXAMPLE
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Triangle T begins:
1;
1;
1;
1,2;
1,8;
1,20;
1,40,40;
1,70,280;
1,112,1120;
1,168,3360,2240;
1,240,8400,22400;
1,330,18480,123200;
1,440,36960,492800,246400; ...
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PROGRAM
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(PARI) T(n, k)=if(n<3*k, 0, n!/(k!*(n-3*k)!*3^k))
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CROSSREFS
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Cf. A001470 (row sums), A118932 (invariant vector); variants: A100861, A118933.
Sequence in context: A125911 A009385 A008308 this_sequence A101280 A008309 A131175
Adjacent sequences: A118928 A118929 A118930 this_sequence A118932 A118933 A118934
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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), May 06 2006
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