%I #22 Feb 21 2020 15:02:03
%S 1,1,1,2,3,6,12,25,52,113,247,548,1226,2770,6298,14419,33183,76760,
%T 178327,415960,973693,2286781,5386573,12723097,30127465,71506140,
%U 170081575,405359177,967899981,2315131955,5546597838,13308818691,31979667219,76947325788
%N Number of quaternary rooted identity (distinct subtrees) trees with n nodes.
%C It is not known if these trees have the asymptotic form C rho^{-n} n^{-3/2}, whereas the identity binary trees, A063895, do, see the Jason P. Bell et al. reference.
%H Alois P. Heinz, <a href="/A116380/b116380.txt">Table of n, a(n) for n = 1..1000</a>
%H Jason P. Bell, Stanley N. Burris and Karen A. Yeats, <a href="https://arxiv.org/abs/math/0512432">Counting Rooted Trees: The Universal Law t(n) ~ C rho^{-n} n^{-3/2}</a>, arXiv:math/0512432 [math.CO], 2005-2006.
%F G.f. satisfies: A(x) = x(1 + A(x) + A(x)^2/2-A(x^2)/2 + A(x)^3/6-A(x)A(x^2)/2+A(x^3)/3 + A(x)^4/24-A(x)^2A(x^2)/4+A(x)A(x^3)/3+A(x^2)^2/8-A(x^4)/4), that is A(x) = x(1+Set_{<=4}(A)(x)).
%p A:= proc(n) option remember; local T; if n<=1 then x else T:= unapply(A(n-1), x); convert(series(x* (1+T(x)+ T(x)^2/2- T(x^2)/2+ T(x)^3/6- T(x)*T(x^2)/2+ T(x^3)/3+ T(x)^4/24- T(x)^2* T(x^2)/4+ T(x)* T(x^3)/3+ T(x^2)^2/8- T(x^4)/4), x,n+1), polynom) fi end: a:= n-> coeff(A(n),x,n): seq(a(n), n=1..40); # _Alois P. Heinz_, Aug 22 2008
%t A[n_] := A[n] = If[n <= 1, x, T[y_] = A[n-1] /. x -> y; Normal[Series[y*(1+T[y]+T[y]^2/2-T[y^2]/2+T[y]^3/6-T[y]*T[y^2]/2+T[y^3]/3+T[y]^4/24-T[y]^2*T[y^2]/4+T[y]*T[y^3]/3+T[y^2]^2/8-T[y^4]/4), {y, 0, n+1}]] /. y -> x]; a[n_] := Coefficient[A[n], x, n]; Table[a[n], {n, 1, 40}] (* _Jean-François Alcover_, Feb 13 2014, after Maple *)
%o (C) #include <ginac/ginac.h> using namespace GiNaC; int main(){ int i, order=40; symbol x("x"); ex T; for (i=0; i<order; i++) T = (x+x*(T + pow(T,2)/2 - T.subs(x==pow(x,2))/2 + pow(T,3)/6 - T*T.subs(x==pow(x,2))/2 + T.subs(x==pow(x,3))/3 + pow(T,4)/24 - pow(T,2)*T.subs(x==pow(x,2))/4 + T*T.subs(x==pow(x,3))/3 + pow(T.subs(x==pow(x,2)),2)/8 - T.subs(x==pow(x,4))/4)).series(x, i+3); for (i=1; i<=order; i++) std::cout << T.coeff(x, i) << ", "; }
%Y Cf. A004111, A063895, A116379.
%K nonn
%O 1,4
%A _Karen A. Yeats_, Feb 06 2006