A113340 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.
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, 117, 11, 1, 1, 83215, 57560, 13853, 1881, 176, 13, 1, 1, 1399161, 1030751, 258146, 36051, 3421, 247, 15, 1, 1, 28020221, 21763632, 5633264, 805875, 77726, 5629, 330, 17, 1
Offset: 0
Examples
Triangle P begins: 1; 1,1; 1,3,1; 1,12,5,1; 1,69,35,7,1; 1,560,325,70,9,1; 1,6059,3880,889,117,11,1; 1,83215,57560,13853,1881,176,13,1; 1,1399161,1030751,258146,36051,3421,247,15,1; 1,28020221,21763632,5633264,805875,77726,5629,330,17,1; 1,654110586,531604250,141487178,20661609,2023461,147810,8625,425,19,1; Matrix square P^2 (A113345) starts: 1; 2,1; 5,6,1; 19,39,10,1; 113,327,105,14,1; 966,3556,1315,203,18,1; ... where P^2 transforms column k of P into column k+1 of P: at k=0, [P^2]*[1,1,1,1,1,...] = [1,3,12,69,560,...]; at k=1, [P^2]*[1,3,12,69,560,...] = [1,5,35,325,3880,...].
Crossrefs
Programs
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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]
Formula
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix P may be defined by
[P]_k = [P^(2*k+1)]_0, for k>=0.
Define the dual triangular matrix Q = A113350 by
[Q]_k = [P^(2*k+2)]_0, for k>=0.
Then, amazingly, powers of P and Q satisfy:
[P^(2*j+1)]_k = [P^(2*k+1)]_j,
[P^(2*j+2)]_k = [Q^(2*k+1)]_j,
[Q^(2*j+2)]_k = [Q^(2*k+2)]_j,
for all j>=0, k>=0.
Also, we have the column transformations:
P^2 * [P]k = [P]{k+1},
P^2 * [Q]k = [Q]{k+1},
Q^2 * [P^2]k = [P^2]{k+1},
Q^2 * [Q^2]k = [Q^2]{k+1},
for all k>=0.
Further, g.f.s of P and Q satisfy:
GF(P) = 1/(1-x) + x*y*GF(Q^2*P^-1),
GF(Q^-1*P^2) = 1 + x*y*GF(Q).