A135880 Triangle P, read by rows, where column k of P^2 equals column 0 of P^(2k+2) such that column 0 of P^2 equals column 0 of P shift one place left, with P(0,0)=1.
1, 1, 1, 2, 2, 1, 6, 7, 3, 1, 25, 34, 15, 4, 1, 138, 215, 99, 26, 5, 1, 970, 1698, 814, 216, 40, 6, 1, 8390, 16220, 8057, 2171, 400, 57, 7, 1, 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1, 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1, 14563135
Offset: 0
Examples
Triangle P begins: 1; 1, 1; 2, 2, 1; 6, 7, 3, 1; 25, 34, 15, 4, 1; 138, 215, 99, 26, 5, 1; 970, 1698, 814, 216, 40, 6, 1; 8390, 16220, 8057, 2171, 400, 57, 7, 1; 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1; 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9, 1; 14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855, 1504, 126, 10, 1; where column k of P equals column 0 of R^(k+1) where R = A135894. Triangle Q = P^2 = A135885 begins: 1; 2, 1; 6, 4, 1; 25, 20, 6, 1; 138, 126, 42, 8, 1; 970, 980, 351, 72, 10, 1; 8390, 9186, 3470, 748, 110, 12, 1; 86796, 101492, 39968, 8936, 1365, 156, 14, 1; 1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ... where column k of Q equals column 0 of Q^(k+1) for k>=0; thus column k of P^2 equals column 0 of P^(2k+2). Triangle R = A135894 begins: 1; 1, 1; 2, 3, 1; 6, 12, 5, 1; 25, 63, 30, 7, 1; 138, 421, 220, 56, 9, 1; 970, 3472, 1945, 525, 90, 11, 1; 8390, 34380, 20340, 5733, 1026, 132, 13, 1; 86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ... where column k of R equals column 0 of P^(2k+1) for k>=0. Surprisingly, column 0 is also found in triangle A135879: 1; 1,1; 2,2,1,1; 6,6,4,4,2,2,1; 25,25,19,19,13,13,9,5,5,3,1,1; 138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ... and is generated by a process that seems completely unrelated.
Crossrefs
Programs
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PARI
{T(n,k)=local(P=Mat(1),R,PShR);if(n>0,for(n=0,n, PShR=matrix(#P,#P, r,c, if(r>=c,if(r==c,1,if(c==1,0,P[r-1,c-1]))));R=P*PShR; R=matrix(#P+1, #P+1, r,c, if(r>=c, if(r<#P+1,R[r,c], if(c==1,(P^2)[ #P,1],(P^(2*c-1))[r-c+1,1])))); P=matrix(#R, #R, r,c, if(r>=c, if(r<#R,P[r,c], (R^c)[r-c+1,1])))));P[n+1,k+1]}
Formula
Denote this triangle by P and define as follows.
Let [P^m]_k denote column k of matrix power P^m,
so that triangular matrix Q = A135885 may be defined by
[Q]_k = [P^(2k+2)]_0, for k>=0, such that
(1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
Define the dual triangular matrix R = A135894 by
[R]_k = [P^(2k+1)]_0, for k>=0.
Then columns of P may be formed from powers of R:
[P]_k = [R^(k+1)]_0, for k>=0.
Further, columns of powers of P, Q and R satisfy:
[R^(j+1)]_k = [P^(2k+1)]_j,
[Q^(j+1)]_k = [P^(2k+2)]_j,
[Q^(j+1)]_k = [Q^(k+1)]_j,
[P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
Also, we have the column transformations:
R * [P]k = [P]{k+1},
Q * [Q]k = [Q]{k+1},
Q * [R]k = [R]{k+1},
P^2 * [Q]k = [Q]{k+1},
P^2 * [R]k = [R]{k+1}, for all k>=0.
Other identities include the matrix products:
P^-1*R (A135898) = P shifted right one column;
P*R^-1*P (A135899) = Q shifted down one row;
R^-1*Q (A135900) = R shifted down one row.
Comments