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.

A178384 A (-1,1) Somos-4 sequence associated with the elliptic curve y^2 + y = x^3 + x.

Original entry on oeis.org

1, 1, 1, 3, -2, 11, -29, 21, -305, -764, -3761, -26829, 20827, -1044667, 7336774, -34981779, 829881529, 4656917815, 116074261249, 2133710863224, 4261714316929, 871401830149817, -15861891538169783, 387559539627947379, -20207945101587735626, -195471056819748264101
Offset: 0

Views

Author

Paul Barry, May 26 2010

Keywords

Comments

Hankel transform of the sequence with g.f. 1/(1-x^2/(1-x^2/(1-3x^2/(1+(2/9)x^2/(1-(33/4)x^2/(1-...)))))) where 1,3,-2/9,33/4,... are the x-coordinates of the multiples of (0,0).

Crossrefs

Programs

  • Haskell
    a178384 n = a178384_list !! n
    a178384_list = [1, 1, 1, 3] ++
       zipWith div (foldr1 (zipWith subtract) (map b [1..2])) a178384_list
       where b i = zipWith (*) (drop i a178384_list) (drop (4-i) a178384_list)
    -- Reinhard Zumkeller, Sep 15 2014
    
  • Magma
    I:=[1,1,1,3]; [n le 4 select I[n] else (-Self(n-1)*Self(n-3)+Self(n-2)^2)/Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 24 2015
    
  • Mathematica
    RecurrenceTable[{a[n] == (-a[n-1]*a[n-3] +a[n-2]^2)/a[n-4], a[0] == 1, a[1] == 1, a[2] == 1, a[3] == 3}, a, {n, 0, 30}] (* G. C. Greubel, Sep 18 2018 *)
  • PARI
    m=30; v=concat([1,1,1,3], vector(m-4)); for(n=5, m, v[n] = (-v[n-1]*v[n-3] + v[n-2]^2)/v[n-4]); v \\ G. C. Greubel, Sep 18 2018
    
  • PARI
    {a(n) = n++; sign(n) * (-1)^(n\2+n+1) * subst( elldivpol( ellinit([0, 0, 1, 1, 0]), abs(n)), x, 0)}; /* Michael Somos, Sep 14 2024 */

Formula

a(n) = (-a(n-1)*a(n-3)+a(n-2)^2)/a(n-4), n>3.
From Michael Somos, Jan 11 2015: (Start)
a(n) = (-1)^n * a(-2-n) for all n in Z.
0 = a(n)*a(n+4) + a(n+1)*a(n+3) - a(n+2)*a(n+2) for all n in Z.
0 = a(n)*a(n+5) + a(n+1)*a(n+4) - 3*a(n+2)*a(n+3) for all n in Z. (End)