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%I #19 Dec 23 2021 03:05:47
%S 1,1,5,32,228,1726,13587,109923,907499,7609898,64609346,554108863,
%T 4792190298,41739160686,365746143064,3221723465187,28509044813580,
%U 253295607463902,2258539046009268,20203103111671575,181242298665210280
%N G.f.: A(x) = F(x*G(x)^3) where F(x) = G(x/F(x)^2) = 1 + x*F(x)^2 is the g.f. of A000108 (Catalan) and G(x) = F(x*G(x)^2) = 1 + x*G(x)^4 is the g.f. of A002293.
%H Alexander Burstein and Louis W. Shapiro, <a href="https://arxiv.org/abs/2112.11595">Pseudo-involutions in the Riordan group</a>, arXiv:2112.11595 [math.CO], 2021.
%F a(n) = Sum_{k=0..n} C(2k+1,k)/(2k+1) * C(4n-k,n-k)*3k/(4n-k) for n>0 with a(0)=1.
%F G.f. satisfies: A(x) = 1 + x*G(x)^3*A(x)^2 where G(x) is the g.f. of A002293.
%F G.f. satisfies: A(x/F(x)^2) = F(x*F(x)) where F(x) is the g.f. of A000108.
%F G.f. satisfies: A(x/H(x)) = F(x*H(x)^2) where H(x) = 1 + x*H(x)^3 is the g.f. of A001764 and F(x) is the g.f. of A000108.
%F G.f. satisfies: A(-x*A(x)^9) = 1/A(x). - _Alexander Burstein_, Apr 14 2020
%F Recurrence: 243*(n-1)*n*(n+1)*(3*n - 5)*(3*n - 4)*(3*n - 2)*(3*n - 1)*(147456*n^6 - 1998336*n^5 + 11209920*n^4 - 33294250*n^3 + 55173779*n^2 - 48321229*n + 17452260)*a(n) = 72*(n-1)*n*(3*n - 5)*(3*n - 4)*(127401984*n^9 - 2045067264*n^8 + 14240360448*n^7 - 56278911936*n^6 + 138595592064*n^5 - 219567715966*n^4 + 222542820712*n^3 - 138190518059*n^2 + 47259501167*n - 6683489400)*a(n-1) - 48*(n-1)*(16307453952*n^12 - 351459606528*n^11 + 3428587929600*n^10 - 20001961205760*n^9 + 77643945578496*n^8 - 211031837008384*n^7 + 411217026027200*n^6 - 577827896836090*n^5 + 579810023200127*n^4 - 403994885007838*n^3 + 184802213339825*n^2 - 49548085570200*n + 5838168798000)*a(n-2) + 128*(2*n - 5)*(4*n - 11)*(4*n - 9)*(8*n - 23)*(8*n - 21)*(8*n - 19)*(8*n - 17)*(147456*n^6 - 1113600*n^5 + 3430080*n^4 - 5488810*n^3 + 4779029*n^2 - 2123685*n + 369600)*a(n-3). - _Vaclav Kotesovec_, Feb 22 2015
%F a(n) ~ (256/27)^n / n^(5/4) * (3^(1/4)*sqrt(EllipticK(1/sqrt(2)))/(2*Pi)^(3/4) - sqrt(3/(2*Pi))/n^(1/4) + (2/(3*Pi))^(1/4) / sqrt(EllipticK(1/sqrt(2)))/n^(1/2)), where EllipticK(1/sqrt(2)) = A093341 = GAMMA(1/4)^2/(4*(Pi)^(1/2)) = 1.85407467730137191843385... (= EllipticK[1/2] in Mathematica). - _Vaclav Kotesovec_, Feb 22 2015
%e G.f.: A(x) = F(x*G(x)^3) = 1 + x + 5*x^2 + 32*x^3 + 228*x^4 +... where
%e F(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 +...
%e F(x)^2 = 1 + 2*x + 5*x^2 + 14*x^3 + 42*x^4 + 132*x^5 + 429*x^6 +...
%e G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...
%e G(x)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 +...
%e G(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 612*x^4 + 4389*x^5 +...
%e G(x)^4 = 1 + 4*x + 22*x^2 + 140*x^3 + 969*x^4 + 7084*x^5 +...
%e A(x)^2 = 1 + 2*x + 11*x^2 + 74*x^3 + 545*x^4 + 4228*x^5 +...
%e G(x)^3*A(x)^2 = 1 + 5*x + 32*x^2 + 228*x^3 + 1726*x^4 + 13587*x^5 +...
%t Join[{1},Table[Sum[Binomial[2k+1,k]/(2k+1) Binomial[4n-k,n-k]3 k/(4n-k), {k,0,n}],{n,20}]] (* _Harvey P. Dale_, Feb 09 2015 *)
%o (PARI) {a(n)=if(n==0,1,sum(k=0,n,binomial(2*k+1,k)/(2*k+1)*binomial(4*(n-k)+3*k,n-k)*3*k/(4*(n-k)+3*k)))}
%Y Cf. A000108, A001764, A002293; A153395, A153397, A093341.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 15 2009
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