A218251 G.f. satisfies A(x) = (1 + x*A(x))^2 * (1 + x^3*A(x)).

1, 2, 5, 15, 48, 160, 550, 1937, 6954, 25355, 93633, 349490, 1316397, 4997306, 19100278, 73440718, 283876092, 1102466529, 4299673200, 16832894330, 66127276201, 260595497227, 1029913570587, 4081124171097, 16211144100379, 64539011439944, 257474646313530

Offset: 0

Author

Paul D. Hanna, Oct 24 2012

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 15*x^3 + 48*x^4 + 160*x^5 + 550*x^6 +...
where
A(x) = 1 + (2+x^2)*x*A(x) + (1+2*x^2)*x^2*A(x)^2 + x^5*A(x)^3.
		

Links

• Charles H. Conley and Valentin Ovsienko, Quiddities of polygon dissections and the Conway-Coxeter frieze equation, arXiv:2107.01234 [math.CO], 2021.

Crossrefs

Cf. A218250, A182053.

Programs

Mathematica

nmax=20; aa=ConstantArray[0,nmax]; aa[[1]]=2; Do[AGF=1+Sum[aa[[n]]*x^n,{n,1,j-1}]+koef*x^j; sol=Solve[Coefficient[(1 + x*AGF)^2 * (1 + x^3*AGF) - 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); for(i=1, n, A=(1+x*A)^2*(1+x^3*A)+x*O(x^n)); polcoeff(A, n)}
for(n=0,30,print1(a(n),", "))

Formula

Recurrence: (n+2)*(n+3)*(1241*n^5 - 8896*n^4 + 14395*n^3 + 17632*n^2 - 50640*n + 20520)*a(n) = - 6*(n+2)*(1201*n^4 - 9868*n^3 + 26581*n^2 - 24270*n + 2340)*a(n-1) + 2*(12410*n^7 - 64140*n^6 - 41011*n^5 + 524724*n^4 - 340939*n^3 - 550044*n^2 + 232560*n + 81000)*a(n-2) - 6*(2482*n^7 - 16551*n^6 + 12327*n^5 + 105521*n^4 - 209527*n^3 + 39268*n^2 + 134496*n - 70920)*a(n-3) + 2*(4964*n^7 - 40548*n^6 + 79541*n^5 + 175950*n^4 - 881383*n^3 + 1128540*n^2 - 373392*n - 118152)*a(n-4) + 6*(2482*n^7 - 23997*n^6 + 57469*n^5 + 92361*n^4 - 533975*n^3 + 581508*n^2 - 19896*n - 133272)*a(n-5) + 60*(n-5)*(2*n - 7)*(n^3 - 34*n^2 + 132*n - 144)*a(n-6) - 2*(n-6)*(2*n - 9)*(1241*n^5 - 2691*n^4 - 8779*n^3 + 19851*n^2 - 1570*n - 5748)*a(n-7). - Vaclav Kotesovec, Sep 10 2013

a(n) ~ c*d^n/n^(3/2), where d = 4.2142983943967634... is the root of the equation 4 - 12*d^2 - 8*d^3 + 12*d^4 - 20*d^5 + d^7 = 0 and c = 2.164253883870... - Vaclav Kotesovec, Sep 10 2013

a(n) = Sum_{k=0..floor(n/3)} binomial(n-2*k,k) * binomial(2*n-4*k+3,n-3*k+1)/(2*n-4*k+3). - Seiichi Manyama, Aug 28 2023