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%I #21 Dec 04 2016 10:27:07
%S 0,1,4,21,124,782,5144,34845,241196,1697498,12104872,87246770,
%T 634425752,4647805372,34267130928,254035385949,1892315106252,
%U 14155536314786,106288436980488,800753707211430,6050872882024520
%N G.f. satisfies A(x) = x*(1+A)^4/(1+A^2).
%H M. Bousquet-Mélou, <a href="http://arXiv.org/abs/math.CO/0501266">Limit laws for embedded trees</a>
%F G.f. (1-(1-8*x)^(1/4))/(1+(1-8*x)^(1/4))-1, a(n)=sum(m=1..n, m*sum(k=0..n-m(-1)^(n-m-k)*binomial(n+k-1,n-1)*sum(j=0..k, binomial(j,n-m-3*k+2*j)*binomial(k,j)*2^(2*n-2*m-5*k+3*j)*3^(-n+m+3*k-j))))/n, n>0, a(0)=0. - _Vladimir Kruchinin_, Dec 10 2011
%F a(n) ~ 2^(3*n-1)/(Gamma(3/4)*n^(5/4)) * (1 - 2*Gamma(3/4)/ (n^(1/4)*sqrt(Pi)) + 3*Gamma(3/4)^2/(sqrt(2*n)*Pi)). - _Vaclav Kotesovec_, Sep 16 2013
%F Conjecture: n*(n-1)*(n+1)*a(n) -12*n*(n-1)*(2*n-3)*a(n-1) +12*(n-1)*(16*n^2-64*n+65)*a(n-2) -16*(2*n-5)*(4*n-9)*(4*n-11)*a(n-3)=0. - _R. J. Mathar_, Nov 10 2013
%p A:= proc(n) option remember; if n=0 then 0 else convert(series(x* (1+A(n-1))^4/ (1+A(n-1)^2), x,n+1), polynom) fi end: a:= n-> coeff(A(n), x,n): seq(a(n), n=0..20); # _Alois P. Heinz_, Aug 23 2008
%t a[0]=0; a[n_] := Sum[m*Sum[(-1)^(n-m-k)*Binomial[n+k-1, n-1]*Sum[Binomial[j, n-m-3*k+2*j]*Binomial[k, j]*2^(2*n-2*m-5*k+3*j)*3^(-n+m+3*k-j), {j, 0, k}], {k, 0, n-m}], {m, 1, n}]/n; Table[a[n], {n, 0, 20}] (* _Jean-François Alcover_, Mar 30 2015, after _Vladimir Kruchinin_ *)
%o (Maxima)
%o a(n):=sum(m*sum((-1)^(n-m-k)*binomial(n+k-1,n-1)*sum(binomial(j,n-m-3*k+2*j)*binomial(k,j)*2^(2*n-2*m-5*k+3*j)*3^(-n+m+3*k-j),j,0,k),k,0,n-m),m,1,n)/n; (* _Vladimir Kruchinin_, Dec 10 2011 *)
%K nonn
%O 0,3
%A _Ralf Stephan_, Jan 21 2005
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