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Triangle read by rows: T(n,k) = (n+1,k)-th element of (M^5-M)/4, where M is the infinite lower Pascal's triangle matrix, 1<=k<=n.
1

%I #13 May 24 2016 03:11:25

%S 1,6,2,31,18,3,156,124,36,4,781,780,310,60,5,3906,4686,2340,620,90,6,

%T 19531,27342,16401,5460,1085,126,7,97656,156248,109368,43736,10920,

%U 1736,168,8,488281,878904,703116,328104,98406,19656,2604,216,9,2441406

%N Triangle read by rows: T(n,k) = (n+1,k)-th element of (M^5-M)/4, where M is the infinite lower Pascal's triangle matrix, 1<=k<=n.

%e Triangle begins:

%e 1;

%e 6, 2;

%e 31, 18, 3;

%e 156, 124, 36, 4;

%e 781, 780, 310, 60, 5;

%e 3906, 4686, 2340, 620, 90, 6;

%p P:= proc(n) option remember; local M; M:= Matrix(n, (i, j)-> binomial(i-1, j-1)); (M^5-M)/4 end: T:= (n, k)-> P(n+1)[n+1, k]: seq(seq(T(n, k), k=1..n), n=1..11); # _Alois P. Heinz_, Oct 07 2009

%t max = 11; M = Table[If[k > n, 0, Binomial[n, k]], {n, 0, max}, {k, 0, max} ];

%t T = (MatrixPower[M, 5] - M)/4;

%t Table[T[[n + 1]][[1 ;; n]] , {n, 1, max}] // Flatten (* _Jean-François Alcover_, May 24 2016 *)

%Y Cf. A007318. First column gives A003463. Row sums give A016129.

%K nonn,tabl

%O 1,2

%A _Gary W. Adamson_, Jun 17 2004

%E Edited with more terms by _Alois P. Heinz_, Oct 07 2009