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%I #4 Mar 30 2012 18:36:50
%S 0,1,0,16,0,1536,0,-319488,0,36007575552,0,-53682434054553600,0,
%T 1790644857560674043166720,0,-1280831660558056667387645027942400,0,
%U 18961467116136182692294341450867551502336000,0
%N Column 0 of the matrix logarithm (A111978) of triangle A111975, which shifts columns left and up under matrix square; these terms are the result of multiplying the element in row n by n!.
%C Let q=2; the g.f. of column k of A111975^m (matrix power m) is: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} A(q^j*x).
%F E.g.f. A(x): x-x^2 = Sum_{j>=1}(1-2^j*x)/j!*Prod_{i=0..j-1}A(2^i*x). E.g.f. A(x): x+x^2 = Sum_{j>=1}(1-4^j*x^2)/j!*Prod_{i=0..j-1}A(2^i*x).
%e E.g.f. A(x) = x + 16/3!*x^3 + 1536/5!*x^5 - 319488/7!*x^7
%e + 36007575552/9!*x^9 - 53682434054553600/11!*x^11 +...
%e where A(x) satisfies:
%e x*(1-x) = (1-2*x)*A(x) + (1-2^2*x)*A(x)*A(2*x)/2!
%e + (1-2^3*x)*A(x)*A(2*x)*A(2^2*x)/3! +...
%o (PARI) {a(n,q=2)=local(A=Mat(1),B);if(n<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); B=sum(i=1,#A,-(A^0-A)^i/i);return(n!*B[n+1,1]))}
%Y Cf. A111978 (matrix log), A111975 (triangle), A111976, A111811 (variant), A111814 (variant).
%K sign
%O 0,4
%A _Paul D. Hanna_, Aug 25 2005
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