|
%I #28 Apr 21 2015 09:32:50
%S 0,1,2,2,6,2,20,4,-10,334,-864,236,8356,22252,-450052,-347224,
%T 30131822,-54733650,-2300158256,10315132412,221235097892,
%U -1741162058468,-25836213977848,319542298035704,3570412755518780,-66526876971287412,-567916947391285280
%N G.f. A(x) satisfies A(A(x)) = (x+2*x^2)/(1-2*x-4*x^2).
%H Alois P. Heinz, <a href="/A199823/b199823.txt">Table of n, a(n) for n = 0..200</a>
%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 [math.CO], 2013.
%F a(n) = T(n,1), T(n,k) = 1 if n=k, else T(n,k) = 1/2*((sum(i=k..n, C(i-1,k-1) * C(i,n-i))) * 2^(n-k) - sum(i=k+1..n-1, T(n,i)*T(i,k))).
%p a:= n-> T(n, 1):
%p T:= proc(n, k) option remember;
%p `if`(n=k, 1, (add(binomial(i-1, k-1) *binomial(i, n-i), i=k..n)
%p *2^(n-k) -add(T(n, i)*T(i, k), i=k+1..n-1))/2)
%p end:
%p seq(a(n), n=0..30); # _Alois P. Heinz_, Nov 11 2011
%t a[n_] := T[n, 1]; T[n_, k_] := T[n, k] = If[n == k, 1, 1/2*(Sum[Binomial[i-1, k-1] * Binomial[i, n-i], {i, k, n}]*2^(n-k) - Sum[T[n, i]*T[i, k], {i, k+1, n-1}]) ]; Table[a[n], {n, 0, 30}] (* _Jean-François Alcover_, Apr 21 2015, from formula *)
%K sign
%O 0,3
%A _Vladimir Kruchinin_, Nov 11 2011
|
