cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A200753 G.f. satisfies: A(x) = 1 + (x-x^2)*A(x)^3.

Original entry on oeis.org

1, 1, 2, 6, 22, 89, 381, 1694, 7744, 36168, 171831, 827814, 4034589, 19857262, 98555324, 492710856, 2478897620, 12541604830, 63768192378, 325674039636, 1669922290311, 8593644472017, 44369362778645, 229767801472366, 1193126351099007, 6211253430642091
Offset: 0

Views

Author

Paul D. Hanna, Nov 21 2011

Keywords

Comments

S. Corteel et al. ask whether this sequence also gives the number of inversion sequences avoiding the pattern 102. - Michel Marcus, Oct 26 2015
Concerning the previous comment: This was proved by Mansour and Shattuck. - Eric M. Schmidt, Jul 18 2017

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 22*x^4 + 89*x^5 + 381*x^6 + ...
Related expansion:
A(x)^3 = 1 + 3*x + 9*x^2 + 31*x^3 + 120*x^4 + 501*x^5 + 2195*x^6 + ...
where a(2) = 3 - 1; a(3) = 9 - 3; a(4) = 31 - 9; a(5) = 120 - 31; ...
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^k*Binomial[n-k,k]*Binomial[3*(n-k),n-k]/(2*(n-k)+1),{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Aug 19 2013 *)
  • PARI
    {a(n)=local(A=1+x);for(i=1,n,A=1+(x-x^2)*A^3+x*O(x^n));polcoeff(A,n)}
    
  • PARI
    {a(n)=polcoeff((1/x)*serreverse( x*(1+x^2 + sqrt((1+x^2)^2 - 4*x +x^2*O(x^n)))/2 ),n)}
    
  • PARI
    {a(n)=sum(k=0,n\2,(-1)^k*binomial(n-k, k)*binomial(3*(n-k),n-k)/(2*(n-k)+1))}

Formula

a(n) = Sum_{k=0..[n/2]} (-1)^k * C(n-k, k) * C(3*(n-k), n-k) / (2*(n-k)+1).
G.f.: A(x) = G(x-x^2) where G(x) = 1 + x*G(x)^3 is the g.f. for A001764.
G.f.: A(x) = (1/x)*Series_Reversion( x*(1+x^2 + sqrt((1+x^2)^2 - 4*x))/2 ).
G.f.: A(x) = (1 - x^2*A(x)^3) / (1 - x*A(x)^2).
Conjecture: 2n*(2n+1)*a(n) -(13n-7)(3n-2)*a(n-1) +4(29n^2-87n+67)*a(n-2) +9(-15n^2+69n-80)*a(n-3) +6(3n-8)(3n-10)*a(n-4)=0. - R. J. Mathar, Nov 22 2011
Recurrence: 2*(n-1)*n*(2*n+1)*a(n) = (n-1)*(31*n^2 - 27*n + 6)*a(n-1) - 6*(9*n^3 - 27*n^2 + 22*n - 2)*a(n-2) + 3*n*(3*n-7)*(3*n-5)*a(n-3). - Vaclav Kotesovec, Aug 19 2013
a(n) ~ sqrt(18*sqrt(33)-66) * ((27+3*sqrt(33))/8)^n/(16*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Aug 19 2013