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A113381
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Triangle Q, read by rows, such that Q^3 transforms column k of Q^2 into column k+1 of Q^2, so that column k of Q^2 equals column 0 of Q^(3*k+2), where Q^3 denotes the matrix cube of Q.
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24
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1, 2, 1, 6, 5, 1, 37, 45, 8, 1, 429, 635, 120, 11, 1, 7629, 12815, 2556, 231, 14, 1, 185776, 343815, 71548, 6556, 378, 17, 1, 5817106, 11651427, 2508528, 233706, 13391, 561, 20, 1, 224558216, 480718723, 106427700, 10069521, 579047, 23817, 780, 23, 1
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OFFSET
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0,2
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COMMENTS
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LINKS
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FORMULA
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Let [Q^m]_k denote column k of matrix power Q^m,
so that triangular matrix Q may be defined by
[Q]_k = [P^(3*k+2)]_0, k>=0,
where the triangular matrix P = A113370 satisfies:
[P]_k = [P^(3*k+1)]_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.
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EXAMPLE
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Triangle Q begins:
1;
2,1;
6,5,1;
37,45,8,1;
429,635,120,11,1;
7629,12815,2556,231,14,1;
185776,343815,71548,6556,378,17,1;
5817106,11651427,2508528,233706,13391,561,20,1;
224558216,480718723,106427700,10069521,579047,23817,780,23,1;
Matrix square Q^2 (A113384) starts:
1;
4,1;
22,10,1;
212,130,16,1;
3255,2365,328,22,1;
70777,57695,8640,616,28,1; ...
1;
6,1;
48,15,1;
605,255,24,1;
11196,5630,624,33,1;
280440,159210,19484,1155,42,1; ...
where Q^3 transforms column k of Q^2 into column k+1:
at k=0, [Q^3]*[1,4,22,212,3255,...] = [1,10,130,2365,...];
at k=1, [Q^3]*[1,10,130,2365,...] = [1,16,328,8640,...].
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PROG
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(PARI) Q(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^(3*k+2))[n-k+1, 1]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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