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A213422
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G.f. satisfies: A( A(x) - 4*A(x)^2 ) = x.
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3
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0, 1, 2, 12, 96, 880, 8720, 90752, 975936, 10737152, 120093056, 1360051456, 15556087296, 179424700416, 2084953411584, 24393551634432, 287204585508864, 3400978267127808, 40480500900446208, 484006813958356992, 5810240353159839744, 70001749695581061120
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OFFSET
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0,3
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COMMENTS
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First negative term is a(45).
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LINKS
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FORMULA
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G.f. satisfies: A(A(x)) = (1 - sqrt(1-16*x))/8.
G.f. satisfies: A(A(x)) = x + 4*A(A(x))^2.
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]
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EXAMPLE
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G.f.: A(x) = x + 2*x^2 + 12*x^3 + 96*x^4 + 880*x^5 + 8720*x^6 + 90752*x^7 +...
where
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 +...
The series reversion of the g.f. A(x) begins:
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 -...
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MATHEMATICA
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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[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}]);
a[n_] := T[n, 1];
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PROG
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(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))}
for(n=1, 31, print1(a(n), ", "))
(Maxima)
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));
(SageMath)
@CachedFunction
def T(n, k):
if (k<0 or k>n): return 0
elif (n==0): return 0
elif (k==n): return 1
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) )
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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