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Triangle T, read by rows, that satisfies the recurrence: T(n,k) = [T^3](n-1,k-1) + [T^3](n-1,k) for n>k>=0, with T(n,n)=1 for n>=0, where T^3 is the matrix third power of T.
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%I #5 Mar 30 2012 18:36:51

%S 1,1,1,3,4,1,21,33,13,1,331,586,294,40,1,11973,23299,13768,2562,121,1,

%T 1030091,2166800,1447573,333070,22569,364,1,218626341,490872957,

%U 361327779,97348117,8466793,200931,1093,1,118038692523,280082001078

%N Triangle T, read by rows, that satisfies the recurrence: T(n,k) = [T^3](n-1,k-1) + [T^3](n-1,k) for n>k>=0, with T(n,n)=1 for n>=0, where T^3 is the matrix third power of T.

%C Column 0 of the matrix power p, T^p, equals the number of 3-tournament sequences having initial term p.

%F Let GF[T] denote the g.f. of triangular matrix T. Then GF[T] = 1 + x*(1+y)*GF[T^3] and for all integer p>=1: GF[T^p] = 1 + x*Sum_{j=1..p} GF[T^(p+2*j)] + x*y*GF[T^(3*p)].

%e Triangle T begins:

%e 1;

%e 1,1;

%e 3,4,1;

%e 21,33,13,1;

%e 331,586,294,40,1;

%e 11973,23299,13768,2562,121,1;

%e 1030091,2166800,1447573,333070,22569,364,1; ...

%e Matrix square T^2 (A113088) begins:

%e 1;

%e 2,1;

%e 10,8,1;

%e 114,118,26,1;

%e 2970,3668,1108,80,1;

%e 182402,257122,96416,9964,242,1; ...

%e where column 0 equals A113089.

%e Matrix cube T^3 (A113090) begins:

%e 1;

%e 3,1;

%e 21,12,1;

%e 331,255,39,1;

%e 11973,11326,2442,120,1;

%e 1030091,1136709,310864,22206,363,1; ...

%e where adjacent sums in row n of T^3 forms row n+1 of T.

%o (PARI) {T(n,k)=local(M=matrix(n+1,n+1));for(r=1,n+1, for(c=1,r, M[r,c]=if(r==c,1,if(c>1,(M^3)[r-1,c-1])+(M^3)[r-1,c]))); return(M[n+1,k+1])}

%Y Cf. A113081; A097710, A113095, A113106; A113085 (column 0), A113088 (T^2), A113087 (row sums).

%K nonn,tabl

%O 0,4

%A _Paul D. Hanna_, Oct 14 2005