OFFSET
0,5
COMMENTS
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^(3*k+1)]_0, k>=0.
Define the triangular matrix Q = A113381 by
[Q]_k = [P^(3*k+2)]_0, k>=0.
Define the triangular matrix R = A113389 by
[R]_k = [P^(3*k+3)]_0, k>=0.
Then P, Q and R are related by:
Q^2 = R*P = R*Q*(R^-2)*Q*R = P*Q*(P^-2)*Q*P,
P^2 = Q*(R^-2)*Q^3, R^2 = Q^3*(P^-2)*Q.
Amazingly, columns in powers of P, Q, R, obey:
[P^(3*j+1)]_k = [P^(3*k+1)]_j,
[Q^(3*j+1)]_k = [P^(3*k+2)]_j,
[R^(3*j+1)]_k = [P^(3*k+3)]_j,
[Q^(3*j+2)]_k = [Q^(3*k+2)]_j,
[R^(3*j+2)]_k = [Q^(3*k+3)]_j,
[R^(3*j+3)]_k = [R^(3*k+3)]_j,
for all j>=0, k>=0.
Also, we have the column transformations:
P^3 * [P]_k = [P]_{k+1},
P^3 * [Q]_k = [Q]_{k+1},
P^3 * [R]_k = [R]_{k+1},
Q^3 * [P^2]_k = [P^2]_{k+1},
Q^3 * [Q^2]_k = [Q^2]_{k+1},
Q^3 * [R^2]_k = [R^2]_{k+1},
R^3 * [P^3]_k = [P^3]_{k+1},
R^3 * [Q^3]_k = [Q^3]_{k+1},
R^3 * [R^3]_k = [R^3]_{k+1},
for all k>=0.
EXAMPLE
Triangle P begins:
1;
1,1;
1,4,1;
1,28,7,1;
1,326,91,10,1;
1,5702,1722,190,13,1;
1,136724,43764,4945,325,16,1;
1,4226334,1415799,163705,10751,496,19,1;
1,161385532,56096733,6617605,437723,19896,703,22,1;
1,7378504140,2644883675,317416204,21179483,960696,33136,946,25,1;
Matrix cube P^3 (A113378) starts:
1;
3,1;
15,12,1;
136,168,21,1;
1998,3190,483,30,1;
41973,80136,13615,960,39,1; ...
where P^3 transforms column k of P into column k+1 of P:
at k=0, [P^3]*[1,1,1,1,1,...] = [1,4,28,326,5702,...];
at k=1, [P^3]*[1,4,28,326,5702,...] = [1,7,91,1722,43764,...].
PROG
(PARI) P(n, k)=local(A, B); A=Mat(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^(3*j-2))[i-j+1, 1])); )); A=B); A[n+1, k+1]
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Paul D. Hanna, Nov 14 2005
STATUS
approved