A084781 G.f. A(x) satisfies A(x) = 1 + x*(1+x+x^2)*A(x)^2.

1, 1, 3, 10, 35, 132, 519, 2105, 8746, 37033, 159229, 693343, 3051290, 13550083, 60642857, 273248824, 1238567263, 5643738611, 25837579578, 118785766683, 548182891007, 2538522337214, 11792272546723, 54936210525388, 256603469498039, 1201486779137257

Offset: 0

Author

Paul D. Hanna, Jun 14 2003

Keywords

• nonn

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Programs

Magma

I:=[1,1,3,10,35]; [n le 5 select I[n] else (3*(n-2)*Self(n-1) + (7*n-18)*Self(n-2) + 12*(n-3)*Self(n-3) + 2*(4*n-15)*Self(n-4) + 2*(2*n-9)*Self(n-5))/n: n in [1..40]]; // G. C. Greubel, Jun 06 2023

Mathematica

a[n_]:= a[n]= Sum[Sum[a[i]a[j-i], {i,0,j}], {j, n-3, n-1}]; a[0]=1; Table[a[n], {n, 0, 30}]
Flatten[{1,Table[Sum[Sum[Sum[Binomial[j,n-3*k+2*j]*Binomial[k,j] *Binomial[-m+2*k-1,k-1]/k*m,{j,0,k}],{k,m,n}],{m,1,n}],{n,1,20}]}] (* Vaclav Kotesovec, Sep 17 2013 *)

Maxima

a(n):=sum((sum(((sum(binomial(j,n-3*k+2*j)*binomial(k,j),j,0,k))* binomial(-m+2*k-1,k-1))/k,k,m,n))*m,m,1,n); /* Vladimir Kruchinin, May 28 2011 */

SageMath

@CachedFunction
def a(n): # a = A084781
    if n==0: return 1
    else: return sum( sum( a(k)*a(j-k) for k in range(j+1) ) for j in range(n-3,n) )
[a(n) for n in range(41)] # G. C. Greubel, Jun 06 2023

Formula

a(0)=1; for n > 0, a(n) = Sum_{j=n-3..n-1} Sum_{i=0..j} a(i)*a(j-i). - Mario Catalani (mario.catalani(AT)unito.it), Jun 19 2003

G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x+x^2+x^3 (continued fraction); equivalently g.f. C(x+x^2+x^3) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011

a(n) = Sum_{m=1..n} Sum_{k=m..n} (Sum_{j=0..k} binomial(j,n-3*k+2*j) * binomial(k,j)) * (binomial(-m+2*k-1,k-1)/k) * m, for n > 0. - Vladimir Kruchinin, May 28 2011

Recurrence: (n+1)*a(n) = 3*(n-1)*a(n-1) + (7*n-11)*a(n-2) + 12*(n-2)*a(n-3) + 2*(4*n-11)*a(n-4) + 2*(2*n-7)*a(n-5). - Vaclav Kotesovec, Sep 17 2013

a(n) ~ 1/sqrt(3)*sqrt(-(1350 + 66*sqrt(131)*sqrt(3))^(2/3) - 48 + 21*(1350 + 66*sqrt(131)*sqrt(3))^(1/3))/((1350 + 66*sqrt(131)*sqrt(3))^(1/6)) * (((190 + 6*sqrt(393))^(2/3) + 28 + 4*(190 + 6*sqrt(393))^(1/3))/(190 + 6*sqrt(393))^(1/3)/3)^n / (n^(3/2)*sqrt(Pi)). - Vaclav Kotesovec, Sep 17 2013