A113084 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.
1, 1, 1, 3, 4, 1, 21, 33, 13, 1, 331, 586, 294, 40, 1, 11973, 23299, 13768, 2562, 121, 1, 1030091, 2166800, 1447573, 333070, 22569, 364, 1, 218626341, 490872957, 361327779, 97348117, 8466793, 200931, 1093, 1, 118038692523, 280082001078
Offset: 0
Examples
Triangle T begins: 1; 1,1; 3,4,1; 21,33,13,1; 331,586,294,40,1; 11973,23299,13768,2562,121,1; 1030091,2166800,1447573,333070,22569,364,1; ... Matrix square T^2 (A113088) begins: 1; 2,1; 10,8,1; 114,118,26,1; 2970,3668,1108,80,1; 182402,257122,96416,9964,242,1; ... where column 0 equals A113089. Matrix cube T^3 (A113090) begins: 1; 3,1; 21,12,1; 331,255,39,1; 11973,11326,2442,120,1; 1030091,1136709,310864,22206,363,1; ... where adjacent sums in row n of T^3 forms row n+1 of T.
Crossrefs
Programs
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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])}
Formula
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)].
Comments