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G.f. satisfies: A( A(x) - 4*A(x)^2 ) = x.
3

%I #52 Mar 09 2023 09:00:39

%S 0,1,2,12,96,880,8720,90752,975936,10737152,120093056,1360051456,

%T 15556087296,179424700416,2084953411584,24393551634432,

%U 287204585508864,3400978267127808,40480500900446208,484006813958356992,5810240353159839744,70001749695581061120

%N G.f. satisfies: A( A(x) - 4*A(x)^2 ) = x.

%C First negative term is a(45).

%H Alois P. Heinz, <a href="/A213422/b213422.txt">Table of n, a(n) for n = 0..200</a>

%H Dmitry Kruchinin and 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 G.f. satisfies: A(A(x)) = (1 - sqrt(1-16*x))/8.

%F G.f. satisfies: A(A(x)) = x + 4*A(A(x))^2.

%F a(n) = T(n,1), T(n,m) = 1/2*(4^(n-m)*m/n*C(2*n-m-1,n-1)-sum(i=m+1..n-1, T(n,i)*T(i,m))), n>m, T(n,n)=1. [_Dmitry Kruchinin_, Dec 02 2012]

%e G.f.: A(x) = x + 2*x^2 + 12*x^3 + 96*x^4 + 880*x^5 + 8720*x^6 + 90752*x^7 +...

%e where

%e A(A(x)) = x + 4*x^2 + 32*x^3 + 320*x^4 + 3584*x^5 + 43008*x^6 + 540672*x^7 +...+ A000108(n-1)*4^(n-1)*x^n +...

%e The series reversion of the g.f. A(x) begins:

%e A(x) - 4*A(x)^2 = x - 2*x^2 - 4*x^3 - 16*x^4 - 80*x^5 - 432*x^6 - 2304*x^7 -...

%t max = 21; a[0] = 0; a[1] = 1; f[x_] := Sum[a[n]*x^n, {n, 0, max}]; se = Series[f[f[x]] - (1 - Sqrt[1 - 16*x])/8 , {x, 0, max}]; coes = CoefficientList[se, x]; sol = Solve[Thread[coes == 0]]; Table[a[n], {n, 1, max}] /. sol // First (* _Jean-François Alcover_, Feb 19 2013, from 1st formula *)

%t T[0, 1]=0; T[n_, n_]=1; T[n_, m_]:= T[n, m]= 1/2*(4^(n-m)* m/n * Binomial[2*n-m-1, n-1] - Sum[T[n, i]*T[i, m], {i, m+1, n-1}]);

%t a[n_] := T[n, 1];

%t Table[a[n], {n, 0, 21}] (* _Jean-François Alcover_, Jan 11 2018, after _Dmitry Kruchinin_ *)

%o (PARI) {a(n)=local(A, B, F); A=x+2*x^2; if(n<1, 0, for(i=0, n, B=subst(A,x,A+x*O(x^n)); B=x+4*B^2; F=serreverse(A+x*O(x^n)); A=(A+subst(B, x, F))/2); polcoeff(A, n, x))}

%o for(n=1,31,print1(a(n),", "))

%o (Maxima)

%o T(n,m):=if n=m then 1 else 1/2*(4^(n-m)*m/n*binomial(2*n-m-1,n-1) -sum(T(n,i) *T(i,m), i,m+1,n-1));

%o makelist(T(n,1),n,1,10); [_Dmitry Kruchinin_, Dec 02 2012]

%o (SageMath)

%o @CachedFunction

%o def T(n,k):

%o if (k<0 or k>n): return 0

%o elif (n==0): return 0

%o elif (k==n): return 1

%o else: return 2^(2*n-2*k-1)*(k/(2*n-k))*binomial(2*n-k, n) - (1/2)*sum( T(n, n-j-1)*T(n-j-1, k) for j in range(n-k-1) )

%o def A213422(n): return T(n,1)

%o [A213422(n) for n in range(31)] # _G. C. Greubel_, Mar 08 2023

%Y Cf. A000108, A107700, A307103.

%K sign

%O 0,3

%A _Paul D. Hanna_, Jun 29 2012