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Reversion of x - x^2 - 2*x^5.
0

%I #18 Apr 07 2019 19:05:09

%S 1,1,2,5,16,56,204,759,2880,11132,43732,174122,700952,2847840,

%T 11661592,48080811,199433880,831649380,3484523460,14662036550,

%U 61931353880,262503848400,1116179957160,4759795460550,20351410848288,87229181620152,374722175164232,1613115479264852,6957700944802160,30064406772108544

%N Reversion of x - x^2 - 2*x^5.

%C For the reversion of x - a*x^2 - b*x^5 (a!=0, b!=0) we have a(n) = Sum_{j=0..floor((n-1)/3)} a^(n-4*j-1)*b^j*binomial(n-3*j-1,j)*binomial(2*n-3*j-2,n-1)/n, n > 0.

%H Vladimir Kruchinin, <a href="http://arxiv.org/abs/1211.3244">The method for obtaining expressions for coefficients of reverse generating functions</a>, arXiv:1211.3244 [math.CO], 2012.

%F a(n) = Sum_{j=0..floor((n-1)/3)} 2^j*binomial(n-3*j-1,j)*binomial(2*n-3*j-2,n-1)/n, n > 0.

%o (Maxima)

%o a(n):=sum(2^j*binomial(n-3*j-1,j)*binomial(2*n-3*j-2,n-1),j,0,(n-1)/3)/n;

%K nonn

%O 1,3

%A _Vladimir Kruchinin_, May 28 2011