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A132900 Colored Motzkin paths where each of the steps has three possible colors.

%I A132900 #35 Feb 05 2024 10:52:03
%S A132900 1,3,18,108,729,5103,37179,277749,2119203,16435305,129199212,
%T A132900 1027098306,8243181351,66698502705,543507899346,4456368744804,
%U A132900 36738955831707,304354824214977,2532328310730798,21152326520189628,177310026608555619,1491097815365481477
%N A132900 Colored Motzkin paths where each of the steps has three possible colors.
%H A132900 Vincenzo Librandi, <a href="/A132900/b132900.txt">Table of n, a(n) for n = 0..200</a>
%F A132900 G.f.: (1-3*x-sqrt(1-6*x-27*x^2))/(18*x^2).
%F A132900 G.f. is the reversion of x/(1+3*x+9*x^2).
%F A132900 a(n) = 3^n * A001006(n).
%F A132900 a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*C(k)*3^(n-2k)*3^k*3^k, where C(n) = A000108(n).
%F A132900 a(n) = (1/(2*Pi))*Integral_{x=-3..9} x^n*sqrt(27 + 6x - x^2)/9.
%F A132900 Conjecture: (n+2)*a(n) - 3*(2*n+1)*a(n-1) + 27*(1-n)*a(n-2) = 0. - _R. J. Mathar_, Nov 14 2011
%F A132900 a(n) ~ 3^(2*n+3/2)/(2*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Oct 20 2012
%F A132900 G.f.: 1/G(x), with G(x) = 1-3*x-9*x^2/G(x) (Jacobi continued fraction). - _Nikolaos Pantelidis_, Feb 01 2023
%F A132900 From _Peter Bala_, Feb 02 2024: (Start)
%F A132900 G.f.: 1/(1 + 3*x)*c(3*x/(1 + 3*x))^2 = 1/(1 - 9*x)*c(-3*x/(1 - 9*x))^2, where c(x) = (1 - sqrt(1 - 4*x))/(2*x) is the g.f. of the Catalan numbers.
%F A132900 a(n) = 3^n *Sum_{k = 0..n} (-1)^(n+k)*binomial(n,k)*Catalan(k+1).
%F A132900 a(n) = 9^n * Sum_{k = 0..n} (-3)^(-k)*binomial(n,k)*Catalan(k+1). (End)
%p A132900 seq(9^n * simplify(hypergeom([3/2, -n], [3], 4/3)), n = 0..20); # _Peter Bala_, Feb 04 2024
%t A132900 CoefficientList[Series[(1-3*x-Sqrt[1-6*x-27*x^2])/(18*x^2), {x, 0, 20}], x] (* _Vaclav Kotesovec_, Oct 20 2012 *)
%o A132900 (PARI) my(x='x+O('x^50)); Vec((1-3*x-sqrt(1-6*x-27*x^2))/(18*x^2)) \\ _G. C. Greubel_, Mar 21 2017
%Y A132900 Cf. A001006, A129400.
%K A132900 easy,nonn
%O A132900 0,2
%A A132900 _Paul Barry_, Sep 04 2007