

A118931


Triangle, read by rows, where T(n,k) = n!/[k!*(n3*k)!*3^k)] for n>=3*k>=0.


3



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
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OFFSET

0,5


COMMENTS

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).


LINKS

Table of n, a(n) for n=0..48.


FORMULA

E.g.f.: A(x,y) = exp(x + y*x^3/3).


EXAMPLE

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; ...


PROG

(PARI) T(n, k)=if(n<3*k, 0, n!/(k!*(n3*k)!*3^k))


CROSSREFS

Cf. A001470 (row sums), A118932 (invariant vector); variants: A100861, A118933.
Sequence in context: A008308 A176889 A208753 * A101280 A008309 A131175
Adjacent sequences: A118928 A118929 A118930 * A118932 A118933 A118934


KEYWORD

nonn,tabl


AUTHOR

Paul D. Hanna, May 06 2006


STATUS

approved



