A216490 G.f. satisfies A(x) = 1 + x*A(x)^2 + x^3*A(x)^5.

1, 1, 2, 6, 21, 78, 302, 1210, 4979, 20913, 89284, 386308, 1690221, 7465594, 33243970, 149080710, 672682035, 3051859515, 13913105076, 63704186436, 292825392338, 1350782681404, 6251139672412, 29014088205700, 135029529235367, 629978080093921, 2945889534054758

Offset: 0

Author

Paul D. Hanna, Sep 07 2012

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 21*x^4 + 78*x^5 + 302*x^6 + 1210*x^7 +...
Related expansions:
A(x)^2 = 1 + 2*x + 5*x^2 + 16*x^3 + 58*x^4 + 222*x^5 + 880*x^6 + 3588*x^7 +...
A(x)^5 = 1 + 5*x + 20*x^2 + 80*x^3 + 330*x^4 + 1391*x^5 + 5950*x^6 +...
Given (1) A(x) = 1 + x*A(x)^2 + x^3*A(x)^5,
suppose (2) A(x) = 1/A(-x*A(x)^3),
then substituting x in (1) with -x*A(x)^3 yields:
1/A(x) = 1 - x*A(x)^3/A(x)^2 - x^3*A(x)^9/A(x)^5,
which illustrates that (2) is consistent with (1).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A112806.

Programs

Mathematica

nmax=20; aa=ConstantArray[0,nmax];aa[[1]]=1; Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[1+x*AGF^2+x^3*AGF^5-AGF,x,j]==0,koef][[1]];aa[[j]]=koef/.sol[[1]],{j,2,nmax}]; Flatten[{1,aa}] (* Vaclav Kotesovec, Sep 10 2013 *)

PARI

{a(n)=local(A=1+x);for(i=1,n,A=1+x*A^2+x^3*A^5 +x*O(x^n));polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f. satisfies: A(x) = 1/A(-x*A(x)^3); note that the Catalan function C(x) = 1 + x*C(x)^2 also satisfies this condition: C(x) = 1/C(-x*C(x)^3).

Recurrence: 8*n*(2*n+1)*(4*n+1)*(4*n+3)*(18857*n^5 - 123695*n^4 + 307075*n^3 - 357745*n^2 + 193668*n - 38880)*a(n) = (29662061*n^9 - 194572235*n^8 + 475490846*n^7 - 510273230*n^6 + 169643069*n^5 + 97131085*n^4 - 80249496*n^3 + 9135180*n^2 + 4205520*n - 777600)*a(n-1) - 3*(13463898*n^9 - 108514077*n^8 + 347609938*n^7 - 547442512*n^6 + 395255792*n^5 - 25515983*n^4 - 129824028*n^3 + 55362252*n^2 + 3442320*n - 3888000)*a(n-2) + 5*(5*n - 12)*(5*n - 11)*(5*n - 9)*(5*n - 8)*(18857*n^5 - 29410*n^4 + 865*n^3 + 9880*n^2 - 1092*n - 720)*a(n-3). - Vaclav Kotesovec, Sep 10 2013

a(n) ~ c*d^n/n^(3/2), where d = 1/768*((2775409885 + 28964352 * sqrt(8502))^(2/3) + 1573*(2775409885 + 28964352 * sqrt(8502))^(1/3) + 829273)/(2775409885 + 28964352 * sqrt(8502))^(1/3) = 4.952774592083496... is the root of the equation -3125 + 2142*d - 1573*d^2 + 256*d^3 = 0 and c = 0.3472109760934755295223550512446412412267... - Vaclav Kotesovec, Sep 10 2013

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * binomial(2*n-k+1,n-2*k)/(2*n-k+1). - Seiichi Manyama, Aug 28 2023