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%I #5 Jun 13 2017 22:50:03
%S 1,1,1,1,3,1,1,12,5,1,1,69,35,7,1,1,560,325,70,9,1,1,6059,3880,889,
%T 117,11,1,1,83215,57560,13853,1881,176,13,1,1,1399161,1030751,258146,
%U 36051,3421,247,15,1,1,28020221,21763632,5633264,805875,77726,5629,330,17,1
%N Triangle P, read by rows, such that P^2 transforms column k of P into column k+1 of P, so that column k of P equals column 0 of P^(2*k+1), where P^2 denotes the matrix square of P.
%F Let [P^m]_k denote column k of matrix power P^m,
%F so that triangular matrix P may be defined by
%F [P]_k = [P^(2*k+1)]_0, for k>=0.
%F Define the dual triangular matrix Q = A113350 by
%F [Q]_k = [P^(2*k+2)]_0, for k>=0.
%F Then, amazingly, powers of P and Q satisfy:
%F [P^(2*j+1)]_k = [P^(2*k+1)]_j,
%F [P^(2*j+2)]_k = [Q^(2*k+1)]_j,
%F [Q^(2*j+2)]_k = [Q^(2*k+2)]_j,
%F for all j>=0, k>=0.
%F Also, we have the column transformations:
%F P^2 * [P]_k = [P]_{k+1},
%F P^2 * [Q]_k = [Q]_{k+1},
%F Q^2 * [P^2]_k = [P^2]_{k+1},
%F Q^2 * [Q^2]_k = [Q^2]_{k+1},
%F for all k>=0.
%F Further, g.f.s of P and Q satisfy:
%F GF(P) = 1/(1-x) + x*y*GF(Q^2*P^-1),
%F GF(Q^-1*P^2) = 1 + x*y*GF(Q).
%e Triangle P begins:
%e 1;
%e 1,1;
%e 1,3,1;
%e 1,12,5,1;
%e 1,69,35,7,1;
%e 1,560,325,70,9,1;
%e 1,6059,3880,889,117,11,1;
%e 1,83215,57560,13853,1881,176,13,1;
%e 1,1399161,1030751,258146,36051,3421,247,15,1;
%e 1,28020221,21763632,5633264,805875,77726,5629,330,17,1;
%e 1,654110586,531604250,141487178,20661609,2023461,147810,8625,425,19,1;
%e Matrix square P^2 (A113345) starts:
%e 1;
%e 2,1;
%e 5,6,1;
%e 19,39,10,1;
%e 113,327,105,14,1;
%e 966,3556,1315,203,18,1; ...
%e where P^2 transforms column k of P into column k+1 of P:
%e at k=0, [P^2]*[1,1,1,1,1,...] = [1,3,12,69,560,...];
%e at k=1, [P^2]*[1,3,12,69,560,...] = [1,5,35,325,3880,...].
%o (PARI) P(n,k)=local(A,B);A=matrix(1,1);A[1,1]=1;for(m=2,n+1,B=matrix(m,m); for(i=1,m, for(j=1,i,if(i<3 || j==i || j>m-1,B[i,j]=1,if(j==1, B[i,1]=1,B[i,j]=(A^(2*j-1))[i-j+1,1]));));A=B);A[n+1,k+1]
%Y Cf. A113341 (column 1), A113342 (column 2), A113343 (column 3), A113344 (column 4); A113345 (P^2), A113360 (P^3), A113350 (Q).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 08 2005
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