A113370 Triangle P, read by rows, such that P^3 transforms column k of P into column k+1 of P, so that column k of P equals column 0 of P^(3*k+1), where P^3 denotes the matrix cube of P.

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

Offset: 0

Author

Paul D. Hanna, Nov 14 2005

Keywords

• nonn
• tabl

Comments

Triangle A114150 illustrates the identity: R^2*Q^-1 = Q^3*P^-2.

See also A114152 for the matrix product: R^3*P^-1.

Examples

			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,...].
		

Crossrefs

Cf. A113371 (column 1), A113372 (column 2), A113373 (column 3).

Cf. A113374 (P^2), A113378 (P^3), A113381 (Q), A113384 (Q^2), A113387 (Q^3), A113389 (R), A113392 (R^2), A113394 (R^3), A114156 (P^-1).

Cf. A114150 (R^2*Q^-1=Q^3*P^-2), A114152 (R^3*P^-1).

Cf. variants: A113340, A113350.

Programs

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]

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.