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A276310 G.f. A(x) satisfies: x = A(x)-2*A(x)^2-2*A(x)^3.

%I A276310 #20 Oct 15 2019 12:21:59
%S A276310 1,2,10,60,404,2912,21984,171600,1373680,11215776,93039648,781936896,
%T A276310 6643741440,56973685760,492482782208,4286561051904,37536888622848,
%U A276310 330471001126400,2923338431270400,25970490200202240,231607762146309120,2072719382680535040
%N A276310 G.f. A(x) satisfies: x = A(x)-2*A(x)^2-2*A(x)^3.
%H A276310 Michael De Vlieger, <a href="/A276310/b276310.txt">Table of n, a(n) for n = 1..1023</a>
%H A276310 Elżbieta Liszewska, Wojciech Młotkowski, <a href="https://arxiv.org/abs/1907.10725">Some relatives of the Catalan sequence</a>, arXiv:1907.10725 [math.CO], 2019.
%H A276310 Thomas M. Richardson, <a href="http://arxiv.org/abs/1609.01193">The three 'R's and the Riordan dual</a>, arXiv:1609.01193 [math.CO], 2016.
%F A276310 G.f.: Series_Reversion(x - 2*x^2 - 2*x^3).
%F A276310 Conjecture: 3*n*(n-1)*a(n) -13*(n-1)*(2*n-3)*a(n-1) -3*(3*n-5)*(3*n-7)*a(n-2)=0. - _R. J. Mathar_, Sep 17 2016
%F A276310 a(n) ~ (13 + 5*sqrt(10))^(n - 1/2) / (2^(5/4) * 5^(1/4) * sqrt(Pi) * n^(3/2) * 3^(n - 1/2)). - _Vaclav Kotesovec_, Aug 22 2017
%e A276310 G.f.: A(x) = x + 2*x^2 + 10*x^3 + 60*x^4 + 404*x^5 + 2912*x^6 + 21984*x^7 +...
%e A276310 Related expansions.
%e A276310 A(x)^2 = x^2 + 4*x^3 + 24*x^4 + 160*x^5 + 1148*x^6 + 8640*x^7 + 67296*x^8 +...
%e A276310 A(x)^3 = x^3 + 6*x^4 + 42*x^5 + 308*x^6 + 2352*x^7 + 18504*x^8 +...
%e A276310 where x = A(x) - 2*A(x)^2 - 2*A(x)^3.
%t A276310 Rest[CoefficientList[InverseSeries[Series[x - 2*x^2 - 2*x^3, {x, 0, 20}], x],x]] (* _Vaclav Kotesovec_, Aug 22 2017 *)
%o A276310 (PARI) {a(n)=polcoeff(serreverse(x - 2*x^2 - 2*x^3 + x^2*O(x^n)), n)}
%o A276310 for(n=1, 30, print1(a(n), ", "))
%Y A276310 Cf. A250886, A276314, A276315, A276316.
%K A276310 nonn,easy
%O A276310 1,2
%A A276310 _Tom Richardson_, Aug 29 2016