A111975 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.
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, 256, 32, 1, 13568, 13568, 13568, 11776, 5504, 1024, 64, 1, 345088, 345088, 345088, 317952, 178176, 43776, 4096, 128, 1, 15112192, 15112192, 15112192, 14422016
Offset: 0
Examples
Triangle P begins: 1; 1,1; 1,2,1; 4,4,4,1; 16,16,16,8,1; 96,96,96,64,16,1; 896,896,896,704,256,32,1; 13568,13568,13568,11776,5504,1024,64,1; 345088,345088,345088,317952,178176,43776,4096,128,1; ... where P^2 shifts columns left and up one place: 1; 2,1; 4,4,1; 16,16,8,1; 96,96,64,16,1; ... The matrix inverse, P^-1, equals signed P: 1; -1,1; 1,-2,1; -4,4,-4,1; 16,-16,16,-8,1; ...
Crossrefs
Programs
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PARI
P(n,k,q=2)=local(A=Mat(1),B);if(n
2,(A^q)[i-1,2],1), B[i,j]=(A^q)[i-1,j-1]));));A=B);return(A[n+1,k+1]))
Formula
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).
Comments