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.

A084782 G.f.: A(x) = 1 + x*A(x)^2/(1-x-x^2).

Original entry on oeis.org

1, 1, 3, 11, 42, 168, 696, 2965, 12915, 57276, 257787, 1174597, 5407854, 25119663, 117579351, 554053049, 2626184688, 12513029640, 59898952650, 287931365692, 1389297316104, 6726449251539, 32668497856323, 159114598216251
Offset: 0

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Author

Paul D. Hanna, Jun 14 2003

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( 2/(Sqrt((x^2+5*x-1)/(x^2+x-1)) + 1) )); // G. C. Greubel, Jun 07 2023
    
  • Mathematica
    CoefficientList[Series[2/(Sqrt[(x^2+5*x-1)/(x^2+x-1)]+1), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 24 2012 *)
  • Maxima
    a(n):=sum(sum(binomial(i,n-k-i)*binomial(k+i-1,k-1),i, ceiling((n-k)/2),n-k)*1/(k+1)*binomial(2*k,k),k,1,n) /* Vladimir Kruchinin, Sep 15 2010 */
    
  • SageMath
    @CachedFunction
    def a(n): # a = A084782
        if n<2: return 1
        else: return sum( sum( a(k)*a(j-k) for k in range(j+1) )*fibonacci(n-j) for j in range(n) )
    [a(n) for n in range(41)] # G. C. Greubel, Jun 07 2023

Formula

a(0) = a(1) = 1; for n>1, a(n) = Sum_{j=0..n-1} Fibonacci(n-j)*( Sum_{i=0..j} a(i)*a(j-i) ). - Mario Catalani (mario.catalani(AT)unito.it), Jun 18 2003
a(n) = Sum_{k=1..n} (Sum_{i=ceiling((n-k)/2)..n-k} binomial(i,n-k-i) *binomial(k+i-1,k-1) * C(k) ), C(k) - Catalan numbers A000108. - Vladimir Kruchinin, Sep 15 2010
G.f.: 1/(1-z/(1-z/(1-z/(...)))) where z=x/(1-x-x^2) (continued fraction); more generally g.f. C(x/(1-x-x^2)) where C(x) is the g.f. for the Catalan numbers (A000108). - Joerg Arndt, Mar 18 2011
G.f.: 2/(sqrt((x^2+5*x-1)/(x^2+x-1)) + 1). - Vladimir Kruchinin, Oct 11 2011
Recurrence: (n+1)*a(n) = 3*(2*n-1)*a(n-1) - 3*(n-2)*a(n-2) - 3*(2*n-7) * a(n-3) - (n-5)*a(n-4). - Vaclav Kotesovec, Oct 24 2012
a(n) ~ 29^(1/4)*((5+sqrt(29))/2)^n/(2*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 24 2012