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%I #12 Feb 11 2013 02:11:21
%S 0,1,0,-1,0,-7,0,-23,0,-5947,0,-140759,0,-8265391,0,133286519,0,
%T 1088222737541,0,4970981405216383,0,7294918534710727,0,
%U -32299178524632916333,0,-944164720798082858723567,0,2252653730296347607326319,0,1968938229271096381309083587
%N Numerators of A(x) where A(x) satisfies A(A(x)) = (2-2*cos(x))/x.
%H Dmitry Kruchinin, Vladimir Kruchinin, <a href="http://arxiv.org/abs/1302.1986">Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$</a>, arXiv:1302.1986
%F a(n) = numerator(T(n,1)); T(n,m) = if n=m then 1 else 1/2*(((-1)^m*2^m*((-1)^(n+m)+1)*sum(j=1..m, ((sum(i=0..(j-1)/2,(j-2*i)^(n+m) *C(j,i))) *C(m,j) *(-1)^((n+m)/2+m-j))/2^j))/(n+m)! -sum(i=m+1..n-1, T(n,i) *T(i,m))).
%e x - 1/24*x^3 - 7/5760*x^5 - 23/193536*x^7 - 5947/464486400*x^9 + ...
%K sign
%O 0,6
%A _Vladimir Kruchinin_, Nov 12 2011