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.

Showing 1-2 of 2 results.

A083586 Binomial transform of A083580.

Original entry on oeis.org

0, 1, 5, 22, 96, 419, 1819, 7824, 33302, 140317, 585873, 2427146, 9989068, 40886295, 166600967, 676368388, 2737758594, 11055063953, 44553867901, 179280877950, 720513441080, 2892804682891, 11605214879475, 46528078138232
Offset: 0

Views

Author

Paul Barry, May 01 2003

Keywords

Crossrefs

Cf. A083580.

Programs

  • Mathematica
    LinearRecurrence[{11,-43,69,-36},{0,1,5,22},30] (* Harvey P. Dale, Jul 19 2015 *)

Formula

a(n) = 2*4^n/3 -5/12 -n*3^(n-1)/2 -3^n/4.
O.g.f.: x*(1-6*x+10*x^2)/((-1+x)(-1+4*x)(-1+3*x)^2). - R. J. Mathar, Apr 02 2008
E.g.f.: exp(x)*(8*exp(3*x) - 3*exp(2*x)*(1 + 2*x) - 5)/12. - Stefano Spezia, Apr 12 2025

A083579 Generalized Jacobsthal numbers.

Original entry on oeis.org

0, 1, 1, 4, 8, 19, 39, 82, 166, 337, 677, 1360, 2724, 5455, 10915, 21838, 43682, 87373, 174753, 349516, 699040, 1398091, 2796191, 5592394, 11184798, 22369609, 44739229, 89478472, 178956956, 357913927, 715827867, 1431655750, 2863311514
Offset: 0

Views

Author

Paul Barry, May 01 2003

Keywords

Crossrefs

Cf. A083580.

Programs

  • GAP
    a:=[0,1,1,4];; for n in [5..40] do a[n]:=3*a[n-1]-a[n-2]-3*a[n-3] +2*a[n-4]; od; a; # G. C. Greubel, May 24 2019
  • Magma
    I:=[0,1,1,4]; [n le 4 select I[n] else 3*Self(n-1)-Self(n-2) - 3*Self(n-3)+2*Self(n-4): n in [1..40]]; // G. C. Greubel, May 25 2019
    
  • Mathematica
    LinearRecurrence[{3,-1,-3,2}, {0,1,1,4}, 40] (* G. C. Greubel, May 25 2019 *)
  • PARI
    concat(0, Vec(x*(1-2*x+2*x^2)/(1-3*x+x^2+3*x^3-2*x^4) + O(x^40))) \\ G. C. Greubel, May 25 2019
    
  • Sage
    (x*(1-2*x+2*x^2)/(1-3*x+x^2+3*x^3-2*x^4)).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, May 24 2019
    

Formula

a(n) = (2^(n+3) - 5*(-1)^n - 3*(2*n+1))/12.
a(n+2) = a(n+1) + 2*a(n) + n, a(0)=0, a(1)=1.
G.f.: x*(1 - 2*x + 2*x^2)/(1 - 3*x + x^2 + 3*x^3 - 2*x^4). - Colin Barker, Jan 16 2012
Showing 1-2 of 2 results.