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G.f. A(x) satisfies: A(A(x)) = x + 2*A(x)^2.
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%I #21 Dec 16 2017 01:09:40

%S 1,1,1,0,-2,0,13,0,-145,0,2328,0,-49784,0,1358965,0,-46076544,0,

%T 1902202515,0,-94104681660,0,5503867176832,0,-376096374571125,0,

%U 29714871818774044,0,-2689473418781240320,0,276562260699626541509,0,-32073434441440654231749,0

%N G.f. A(x) satisfies: A(A(x)) = x + 2*A(x)^2.

%C Inspired by peculiar functional equations suggested by Michael Somos. Unexpectedly, the even-indexed terms are all zeros after index 2; see A107699 for odd-indexed terms. The self-COMPOSE equals A107701.

%H P. Majer, <a href="http://mathoverflow.net/questions/49724/">The functional equation f(f(x))=x+f(x)^2</a>, Dec. 2010

%F G.f. satisfies: A(-A(-x)) = x.

%F G.f. satisfies: A( A(x) - 2*x^2 ) = x. [_Paul D. Hanna_, Aug 20 2008]

%F a(n)=T(n,1), T(n,m)=sum(j=max(2*m-n,0)..m-1,binomial(m,j)*2^(m-j-1) *T(n-j,2*(m-j)))-1/2*sum(i=m+1..n-1, T(n,i)*T(i,m)), n>m, T(n,n)=1. [_Vladimir Kruchinin_, Mar 12 2012]

%t T[n_, n_] = 1; T[n_, m_] := T[n, m] = Sum[Binomial[m, j]*2^(m-j-1)*T[n-j, 2*(m-j)], {j, Max[2*m-n, 0], m-1}] - 1/2*Sum[T[n, i]*T[i, m], {i, m+1, n-1}]; Table[T[n, 1], {n, 1, 34}] (* _Jean-François Alcover_, Mar 03 2014, after _Vladimir Kruchinin_ *)

%o (PARI) {a(n) = local(A,B,F); A=x+x^2+x*O(x^n); if(n<1, 0, for(i=0, n, F=x+2*A^2; B=serreverse(A); A=(A+subst(B,x,F))/2); polcoeff(A,n,x))}

%o (PARI) {a(n) = my(A); if( n<0, 0, A = x + O(x^2); for(k=2, n, A = truncate(A) + x * O(x^k); A += (x + A^2 - subst(A, x, A))/2); polcoeff(A, n) * 2^(n-1))}; /* _Michael Somos_, Dec 15 2017 */

%o (Maxima)

%o T(n,m):=if n=m then 1 else sum(binomial(m,j)*2^(m-j-1)*T(n-j,2*(m-j)),j,max(2*m-n,0),m-1)-1/2*sum(T(n,i)*T(i,m),i,m+1,n-1);

%o makelist(T(n,1),n,1,9); /* Vladimir Kruchinin, Mar 12 2012 */

%Y Cf. A107699, A107701.

%K sign

%O 1,5

%A _Paul D. Hanna_, May 21 2005