%I #22 Sep 08 2022 08:45:57
%S 1,1,3,12,50,224,1054,5121,25509,129591,668811,3496740,18481512,
%T 98585788,530068840,2869725800,15630429306,85589391884,470905310206,
%U 2601941245750,14432082902820,80328808797750,448527122885700,2511672193514250
%N Reversion of x-x^2-x^3-2*x^4
%C For the reversion of x - a*x^2 - b*x^3 - c*x^4 (a!=0, b!=0, c!=0) we have
%C a(n) = sum(k=1,n-1, (sum(j=0..k, a^(-n+3*k-j+1)*b^(n-3*k+2*j-1)*c^(k-j)*binomial(j,n-3*k+2*j-1)*binomial(k,j)))*binomial(n+k-1,n-1))/n, n>1, a(1)=1.
%H Vincenzo Librandi, <a href="/A191242/b191242.txt">Table of n, a(n) for n = 1..200</a>
%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(k=1..n-1, (sum(j=0..k, binomial(j,n-3*k+2*j-1)*2^(k-j)*binomial(k,j)))*binomial(n+k-1,n-1))/n, n>1, a(1)=1.
%t a[1] = 1; a[n_] := Sum[Sum[Binomial[j, n - 3k + 2j - 1]*2^(k - j)* Binomial[k, j], {j, 0, k}]*Binomial[n + k - 1, n - 1], {k, 1, n - 1}]/n;
%t Array[a, 24] (* _Jean-François Alcover_, Jul 23 2018 *)
%o (Maxima)
%o a(n):=sum((sum(binomial(j,n-3*k+2*j-1)*2^(k-j)*binomial(k,j),j,0,k))*binomial(n+k-1,n-1),k,1,n-1)/n;
%o (PARI) x='x+O('x^66); /* that many terms */
%o Vec(serreverse(x-x^2-x^3-2*x^4)) /* show terms */ /* _Joerg Arndt_, May 28 2011 */
%o (Magma) [&+[Binomial(i,n-3*k+2*i-1)*2^(k-i)*Binomial(k,i)*Binomial(n+k-1,n-1)/n: k in [0..25], i in [0..n]]: n in [1..25]]; // _Vincenzo Librandi_, Jul 23 2018
%K nonn
%O 1,3
%A _Vladimir Kruchinin_, May 28 2011