Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #16 Sep 08 2024 10:02:22
%S 1,3,17,121,965,8247,73841,683713,6493145,62898859,619079889,
%T 6173490857,62239144525,633323532783,6496052173665,67093423506049,
%U 697181754821297,7283521984427283,76455801614169809,806004056649062937,8529783421905380629,90584730265930813607
%N G.f. satisfies A(x) = (1 + x*A(x)) * (1 + 2*x*A(x)^2).
%C The radius of convergence of g.f. A(x) is r = 0.08774268876242660659654020... with A(r) = 2.04748732367111203761312028274219344812311691... where y=A(r) satisfies 6*y^3 - 14*y^2 + 4*y - 1 = 0.
%C r = 1/(((40465 + 387*sqrt(129))^(2/3) + 1174 + 34*(40465 + 387*sqrt(129))^(1/3)) / (40465+387*sqrt(129))^(1/3)/9). - _Vaclav Kotesovec_, Sep 17 2013
%H Vincenzo Librandi, <a href="/A216314/b216314.txt">Table of n, a(n) for n = 0..200</a>
%H R. Bacher, <a href="http://arxiv.org/abs/math/0409050">On generating series of complementary plane trees</a> arXiv:math/0409050 [math.CO]
%F G.f. A(x) satisfies:
%F (1) A(x) = exp( Sum_{n>=1} x^n*A(x)^n/n * Sum_{k=0..n} C(n,k)^2 * 2^(n-k)/A(x)^k ).
%F (2) A(x) = (1/x) * Series_Reversion( x*(1 - 2*x - 2*x^2)/(1+x) ).
%F (3) A(x) = Sum_{n>=0} A028859(n) * x^n * A(x)^n, where g.f. of A028859 = (1+x)/(1-2*x-2*x^2).
%F The formal inverse of the g.f. A(x) is (sqrt(1-4*x+12*x^2) - (1+2*x))/(4*x^2).
%F a(n) = [x^n] ( (1+x)/(1-2*x-2*x^2) )^(n+1) / (n+1).
%F Recurrence: 3*n*(n+1)*(43*n-76)*a(n) = n*(1462*n^2 - 3315*n + 1274)*a(n-1) + (86*n^3 - 324*n^2 + 523*n - 330)*a(n-2) + (n-2)*(2*n-5)*(43*n-33)*a(n-3)
%F a(n) ~ 1/516*sqrt(86)*sqrt((1448486261 + 1803807*sqrt(129))^(1/3)*((1448486261 + 1803807*sqrt(129))^(2/3) + 1280110 + 1118*(1448486261 + 1803807*sqrt(129))^(1/3)))/(1448486261 + 1803807*sqrt(129))^(1/3) * (((40465 + 387*sqrt(129))^(2/3) + 1174 + 34*(40465 + 387*sqrt(129) )^(1/3)) / (40465+387*sqrt(129))^(1/3)/9)^n / (n^(3/2)*sqrt(Pi)). - _Vaclav Kotesovec_, Sep 17 2013
%F a(n) = Sum_{k=0..n} 2^k * binomial(n+k+1,k) * binomial(n+k+1,n-k) / (n+k+1). - _Seiichi Manyama_, Sep 08 2024
%e G.f.: A(x) = 1 + 3*x + 17*x^2 + 121*x^3 + 965*x^4 + 8247*x^5 + 73841*x^6 +...
%e Related expansions.
%e A(x)^2 = 1 + 6*x + 43*x^2 + 344*x^3 + 2945*x^4 + 26398*x^5 + 244615*x^6 +...
%e A(x)^3 = 1 + 9*x + 78*x^2 + 696*x^3 + 6399*x^4 + 60321*x^5 + 580316*x^6 +...
%e where A(x) = 1 + A(x)*(1+2*A(x))*x + 2*A(x)^3*x^2.
%e The g.f. also satisfies the series:
%e A(x) = 1 + 3*x*A(x) + 8*x^2*A(x)^2 + 22*x^3*A(x)^3 + 60*x^4*A(x)^4 + 164*x^5*A(x)^5 + 448*x^6*A(x)^6 +...+ A028859(n)*x^n*A(x)^n +...
%e The logarithm of the g.f. equals the series:
%e log(A(x)) = (1*2 + 1/A(x))*x*A(x) + (1*2^2 + 2^2*2/A(x) + 1/A(x)^2)*x^2*A(x)^2/2 +
%e (1*2^3 + 3^2*2^2/A(x) + 3^2*2/A(x)^2 + 1/A(x)^3)*x^3*A(x)^3/3 +
%e (1*2^4 + 4^2*2^3/A(x) + 6^2*2^2/A(x)^2 + 4^2*2/A(x)^3 + 1/A(x)^4)*x^4*A(x)^4/4 +
%e (1*2^5 + 5^2*2^4/A(x) + 10^2*2^3/A(x)^2 + 10^2*2^2/A(x)^3 + 5^2*2/A(x)^4 + 1/A(x)^5)*x^5*A(x)^5/5 +...
%e Explicitly,
%e log(A(x)) = 3*x + 25*x^2/2 + 237*x^3/3 + 2361*x^4/4 + 24203*x^5/5 + 252757*x^6/6 + 2674185*x^7/7 + 28567105*x^8/8 +...+ L(n)*x^n/n +...
%e where L(n) = [x^n] (1+x)^n/(1-2*x-2*x^2)^n.
%t CoefficientList[1/x * InverseSeries[Series[x*(1 - 2*x - 2*x^2)/(1+x),{x,0,20}],x],x] (* _Vaclav Kotesovec_, Sep 17 2013 *)
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=(1 + x*A)*(1 + 2*x*(A+x*O(x^n))^2)); polcoeff(A, n)}
%o (PARI) {a(n)=polcoeff( (1/x)*serreverse( x*(1-2*x-2*x^2)/(1+x +x*O(x^n))), n)}
%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1, n, A=exp(sum(m=1, n, sum(j=0, m, binomial(m, j)^2*2^(m-j)/A^j)*x^m*A^m/m))); polcoeff(A, n)}
%o for(n=0, 31, print1(a(n), ", "))
%Y Cf. A007863, A215661, A215654, A028859, A364374.
%K nonn
%O 0,2
%A _Paul D. Hanna_, Sep 03 2012