A153396 - OEIS

A153396

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.

1, 1, 5, 32, 228, 1726, 13587, 109923, 907499, 7609898, 64609346, 554108863, 4792190298, 41739160686, 365746143064, 3221723465187, 28509044813580, 253295607463902, 2258539046009268, 20203103111671575, 181242298665210280

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

LINKS

Table of n, a(n) for n=0..20.

Alexander Burstein and Louis W. Shapiro, Pseudo-involutions in the Riordan group, arXiv:2112.11595 [math.CO], 2021.

FORMULA

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.

G.f. satisfies: A(x) = 1 + x*G(x)^3*A(x)^2 where G(x) is the g.f. of A002293.

G.f. satisfies: A(x/F(x)^2) = F(x*F(x)) where F(x) is the g.f. of A000108.

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.

G.f. satisfies: A(-x*A(x)^9) = 1/A(x). - Alexander Burstein, Apr 14 2020

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

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

EXAMPLE

G.f.: A(x) = F(x*G(x)^3) = 1 + x + 5*x^2 + 32*x^3 + 228*x^4 +... where

F(x) = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 +...

F(x)^2 = 1 + 2*x + 5*x^2 + 14*x^3 + 42*x^4 + 132*x^5 + 429*x^6 +...

G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...

G(x)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 +...

G(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 612*x^4 + 4389*x^5 +...

G(x)^4 = 1 + 4*x + 22*x^2 + 140*x^3 + 969*x^4 + 7084*x^5 +...

A(x)^2 = 1 + 2*x + 11*x^2 + 74*x^3 + 545*x^4 + 4228*x^5 +...

G(x)^3*A(x)^2 = 1 + 5*x + 32*x^2 + 228*x^3 + 1726*x^4 + 13587*x^5 +...

MATHEMATICA

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 *)