A220110 Expansion of A(x) satisfying A(A(x)) = x+2x^2+4x^3.

1, 1, 1, -3, 5, 1, -39, 117, 13, -1311, 3441, 9525, -78603, 16961, 1520521, -3649323, -28760163, 144787265, 601582689, -5374096875, -15170850555, 225456060897, 461284881657, -11141961064971, -15963771799251, 647040052660257, 569313149887057

Offset: 1

Author

Dmitry Kruchinin, Dec 05 2012

Keywords

• sign

Examples

			First column of
1;
1,1;
1,2,1;
-3,3,3,1;
5,-4,6,4,1;
1,5,-2,10,5,1;
-39,6,3,4,15,6,1;
117,-57,9,3,15,21,7,1;
13,128,-56,8,10,32,28,8,1;
-1311,201,84,-44,6,30,56,36,9,1;
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..552
• Dmitry Kruchinin, Vladimir Kruchinin, Method for solving an iterative functional equation $A^{2^n}(x)=F(x)$, arXiv:1302.1986

Programs

Mathematica

t[n_, m_] := t[n, m] = 1/2*(2^(n-m)*Sum[Binomial[j, n - 3*m + 2*j]*Binomial[m, j], {j, 0, m}] - Sum[t[n, k]*t[k, m], {k, m+1, n-1}]); t[n_, n_] = 1; Table[t[n, 1], {n, 1, 27}] (* Jean-François Alcover, Feb 22 2013 *)

Maxima

T(n,m):=if n=m then 1 else 1/2*(2^(n-m)*sum(binomial(j,n-3*m+2*j)*binomial(m,j),j,0,m)-sum(T(n,k)*T(k,m),k,m+1,n-1));
makelist((T(n,1)),n,1,10);

Formula

a(n)=T(n,1), 2*T(n,m)= 2^(n-m) *sum_{j=0..m} binomial(j,n-3*m+2*j) *binomial(m,j) -sum_{k=m+1..n-1} T(n,k)*T(k,m), n>m, T(n,n)=1.