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%I #7 Jun 13 2017 22:38:35
%S 1,1,1,1,2,1,4,4,4,1,16,16,16,8,1,96,96,96,64,16,1,896,896,896,704,
%T 256,32,1,13568,13568,13568,11776,5504,1024,64,1,345088,345088,345088,
%U 317952,178176,43776,4096,128,1,15112192,15112192,15112192,14422016
%N Triangle P, read by rows, that satisfies [P^2](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(2*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(k,k)=1 and P(k+2,2)=P(k+2,0) for k>=0.
%C Terms of column 0, column 1 and column 2 in row n are equal for n>2.
%F The g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*2^k)^n/n! * Product_{j=0..n-1} L(2^j*x) where L(x) is the g.f. of column 0 of the matrix log of P (A111979) and satisfies: x-x^2 = Sum_{j>=1}(1-2^j*x)*Prod_{i=0..j-1}L(2^i*x).
%e Triangle P begins:
%e 1;
%e 1,1;
%e 1,2,1;
%e 4,4,4,1;
%e 16,16,16,8,1;
%e 96,96,96,64,16,1;
%e 896,896,896,704,256,32,1;
%e 13568,13568,13568,11776,5504,1024,64,1;
%e 345088,345088,345088,317952,178176,43776,4096,128,1; ...
%e where P^2 shifts columns left and up one place:
%e 1;
%e 2,1;
%e 4,4,1;
%e 16,16,8,1;
%e 96,96,64,16,1; ...
%e The matrix inverse, P^-1, equals signed P:
%e 1;
%e -1,1;
%e 1,-2,1;
%e -4,4,-4,1;
%e 16,-16,16,-8,1; ...
%o (PARI) P(n,k,q=2)=local(A=Mat(1),B);if(n<k || k<0,0, for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1,if(j==1,B[i,j]=if(i>2,(A^q)[i-1,2],1), B[i,j]=(A^q)[i-1,j-1]));));A=B);return(A[n+1,k+1]))
%Y Cf. A111976 (column 0), A111977 (row sums), A111978 (matrix log), A098539 (variant), A078536 (variant).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Aug 24 2005
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