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%I #3 Mar 30 2012 18:37:07
%S 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,
%T 40,6,1,8390,16220,8057,2171,400,57,7,1,86796,182714,93627,25628,4740,
%U 666,77,8,1,1049546,2378780,1252752,348050,64805,9080,1029,100,9,1,14563135
%N 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.
%C Amazingly, column 0 (A135881) also equals column 0 of tables A135878 and A135879, both of which have unusual recurrences seemingly unrelated to this triangle.
%F Denote this triangle by P and define as follows.
%F Let [P^m]_k denote column k of matrix power P^m,
%F so that triangular matrix Q = A135885 may be defined by
%F [Q]_k = [P^(2k+2)]_0, for k>=0, such that
%F (1) Q = P^2 and (2) [Q]_0 = [P]_0 shifted left.
%F Define the dual triangular matrix R = A135894 by
%F [R]_k = [P^(2k+1)]_0, for k>=0.
%F Then columns of P may be formed from powers of R:
%F [P]_k = [R^(k+1)]_0, for k>=0.
%F Further, columns of powers of P, Q and R satisfy:
%F [R^(j+1)]_k = [P^(2k+1)]_j,
%F [Q^(j+1)]_k = [P^(2k+2)]_j,
%F [Q^(j+1)]_k = [Q^(k+1)]_j,
%F [P^(2j+2)]_k = [P^(2k+2)]_j, for all j>=0, k>=0.
%F Also, we have the column transformations:
%F R * [P]_k = [P]_{k+1},
%F Q * [Q]_k = [Q]_{k+1},
%F Q * [R]_k = [R]_{k+1},
%F P^2 * [Q]_k = [Q]_{k+1},
%F P^2 * [R]_k = [R]_{k+1}, for all k>=0.
%F Other identities include the matrix products:
%F P^-1*R (A135898) = P shifted right one column;
%F P*R^-1*P (A135899) = Q shifted down one row;
%F R^-1*Q (A135900) = R shifted down one row.
%e Triangle P begins:
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 6, 7, 3, 1;
%e 25, 34, 15, 4, 1;
%e 138, 215, 99, 26, 5, 1;
%e 970, 1698, 814, 216, 40, 6, 1;
%e 8390, 16220, 8057, 2171, 400, 57, 7, 1;
%e 86796, 182714, 93627, 25628, 4740, 666, 77, 8, 1;
%e 1049546, 2378780, 1252752, 348050, 64805, 9080, 1029, 100, 9,
%e 1;
%e 14563135, 35219202, 19003467, 5352788, 1004176, 140908, 15855,
%e 1504, 126, 10, 1;
%e where column k of P equals column 0 of R^(k+1) where R =
%e A135894.
%e Triangle Q = P^2 = A135885 begins:
%e 1;
%e 2, 1;
%e 6, 4, 1;
%e 25, 20, 6, 1;
%e 138, 126, 42, 8, 1;
%e 970, 980, 351, 72, 10, 1;
%e 8390, 9186, 3470, 748, 110, 12, 1;
%e 86796, 101492, 39968, 8936, 1365, 156, 14, 1;
%e 1049546, 1296934, 528306, 121532, 19090, 2250, 210, 16, 1; ...
%e where column k of Q equals column 0 of Q^(k+1) for k>=0;
%e thus column k of P^2 equals column 0 of P^(2k+2).
%e Triangle R = A135894 begins:
%e 1;
%e 1, 1;
%e 2, 3, 1;
%e 6, 12, 5, 1;
%e 25, 63, 30, 7, 1;
%e 138, 421, 220, 56, 9, 1;
%e 970, 3472, 1945, 525, 90, 11, 1;
%e 8390, 34380, 20340, 5733, 1026, 132, 13, 1;
%e 86796, 399463, 247066, 72030, 13305, 1771, 182, 15, 1; ...
%e where column k of R equals column 0 of P^(2k+1) for k>=0.
%e Surprisingly, column 0 is also found in triangle A135879:
%e 1;
%e 1,1;
%e 2,2,1,1;
%e 6,6,4,4,2,2,1;
%e 25,25,19,19,13,13,9,5,5,3,1,1;
%e 138,138,113,113,88,88,69,50,50,37,24,24,15,10,5,5,2,1; ...
%e and is generated by a process that seems completely unrelated.
%o (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]}
%Y Cf. columns: A135881, A135882, A135883, A135884.
%Y Cf. related tables: A135885 (Q=P^2), A135894 (R).
%Y Cf. A135888 (P^3), A135891 (P^4), A135892 (P^5), A135893 (P^6).
%Y Cf. A135898 (P^-1*R), A135899 (P*R^-1*P), A135900 (R^-1*Q).
%Y Cf. A135878, A135879.
%K nonn,tabl
%O 0,4
%A _Paul D. Hanna_, Dec 15 2007
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