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A113340
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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.
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32
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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
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OFFSET
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0,5
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LINKS
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FORMULA
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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).
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EXAMPLE
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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,...].
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PROG
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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]
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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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