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%I #61 Apr 12 2023 11:00:41
%S 1,1,4,25,190,1606,14506,137089,1338790,13403950,136846144,1419257434,
%T 14911016596,158363649640,1697452010230,18338919413425,
%U 199496184219910,2183299541440150,24021874198331080,265559590979820910,2948253066186839140,32857382497018933060
%N Generalized Catalan numbers C(3; n).
%C a(n+1) = Y_{n}(n+1) = Z_{n}, n >= 0, in the Derrida et al. 1992 reference (see A064094) for alpha=3, beta =1 (or alpha=1, beta=3).
%C Hankel transform is A060722. - _Paul Barry_, Jan 30 2009
%D S. Ramanujan, Modular Equations and Approximations to pi, pp. 23-39 of Collected Papers of Srinivasa Ramanujan, Ed. G. H. Hardy et al., AMS Chelsea 2000. See page 39, equation (50).
%H Fung Lam, <a href="/A064063/b064063.txt">Table of n, a(n) for n = 0..925</a>
%H J. Abate, W. Whitt, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL14/Whitt/whitt6.html">Brownian Motion and the Generalized Catalan Numbers</a>, J. Int. Seq. 14 (2011) # 11.2.6, example section 3.
%H Paul Barry, <a href="https://www.emis.de/journals/JIS/VOL22/Barry3/barry422.html">Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles</a>, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
%H S. B. Ekhad, M. Yang, <a href="http://sites.math.rutgers.edu/~zeilberg/tokhniot/oMathar1maple12.txt">Proofs of Linear Recurrences of Coefficients of Certain Algebraic Formal Power Series Conjectured in the On-Line Encyclopedia Of Integer Sequences</a>, (2017)
%F G.f.: (1+3*x*c(3*x)/2)/(1+x/2) = 1/(1-x*c(3*x)) with c(x) g.f. of Catalan numbers A000108.
%F a(n) = Sum_{m=0..n-1} (n-m)*binomial(n-1+m, m)*(3^m)/n.
%F a(n) = (-1/2)^n * (1 - 3*Sum_{k=0..n-1} C(k)*(-6)^k, n >= 1, a(0) = 1, with C(n) = A000108(n) (Catalan).
%F a(n) = Sum_{ k=0..n} A059365(n, k)*3^(n-k). - _Philippe Deléham_, Jan 19 2004
%F Given the semi-axes a,b of an ellipse, then Ramanujan gave the highly accurate formula for the perimeter p = Pi((a+b) + (3(a-b)^2)/(10(a+b) + sqrt(a^2 + 14ab + b^2))). If we let h = ((a-b)/(a+b))^2, then (p/(Pi(a+b))-1)/4 = (3/4)* h/(10 + sqrt(4 - 3*h)) = 1*(h/16) + 1*(h/16)^2 + 4*(h/16)^3 + 25*(h/16)^4 + ... . - _Michael Somos_, Apr 11 2007
%F G.f.: 1/(1-x/(1-3x/(1-3x/(1-3x/(1-.... = 1/(1-x-3*x^2/(1-6*x-9*x^2/(1-6*x-9*x^2/(1-.... (continued fractions). - _Paul Barry_, Jan 30 2009
%F G.f.: 6/(5+sqrt(1-12*x)). - _Harvey P. Dale_, Mar 11 2011
%F From _Gary W. Adamson_, Jul 12 2011: (Start)
%F a(n) = upper left term in M^n, M = the infinite square production matrix:
%F 1, 1, 0, 0, 0, 0, ...
%F 3, 3, 3, 0, 0, 0, ...
%F 3, 3, 3, 3, 0, 0, ...
%F 3, 3, 3, 3, 3, 0, ...
%F 3, 3, 3, 3, 3, 3, ...
%F ... (End)
%F D-finite with recurrence: 2*n*a(n) + (-23*n+36)*a(n-1) + 6*(-2*n+3)*a(n-2) = 0. - _R. J. Mathar_, Dec 03 2012 (Formula verified and used for computations. - _Fung Lam_, Mar 05 2014)
%F a(n) ~ 3^(n+1) * 4^n / (25*n^(3/2)*sqrt(Pi)). - _Vaclav Kotesovec_, Mar 05 2014
%F a(n) = hypergeometric([1-n, n], [-n], 3) for n>0. - _Peter Luschny_, Nov 30 2014
%t CoefficientList[Series[6/(5+Sqrt[1-12 x]),{x,0,50}],x] (* _Harvey P. Dale_, Mar 11 2011 *)
%o (PARI) a(n)=if(n<0,0,polcoeff(serreverse((x-2*x^2)/(1+x)^2+O(x^(n+1))),n)) \\ _Ralf Stephan_, Jun 12 2004
%o (PARI) {a(n)= if(n<1, n==0, polcoeff( serreverse( (x-2*x^2)/ (1+x)^2 +x*O(x^n)), n))} /* _Michael Somos_, Apr 11 2007 */
%o (Sage)
%o def a(n):
%o if n==0: return 1
%o return hypergeometric([1-n, n], [-n], 3).simplify()
%o [a(n) for n in range(24)] # _Peter Luschny_, Nov 30 2014
%o (Magma) R<x>:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 6/(5+Sqrt(1-12*x)) )); // _G. C. Greubel_, May 02 2019
%Y Cf. A064062 (C(2; n)).
%K nonn,easy
%O 0,3
%A _Wolfdieter Lang_, Sep 13 2001
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