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.

A115391 a(0)=0; then a(4*k+1)=a(4*k)+(4*k+1)^2, a(4*k+2)=a(4*k+1)+(4*k+3)^2, a(4*k+3)=a(4*k+2)+(4*k+2)^2, a(4*k+4)=a(4*k+3)+(4*k+4)^2.

Original entry on oeis.org

0, 1, 10, 14, 30, 55, 104, 140, 204, 285, 406, 506, 650, 819, 1044, 1240, 1496, 1785, 2146, 2470, 2870, 3311, 3840, 4324, 4900, 5525, 6254, 6930, 7714, 8555, 9516, 10416, 11440, 12529, 13754, 14910, 16206, 17575, 19096, 20540, 22140, 23821, 25670, 27434, 29370
Offset: 0

Views

Author

Pierre CAMI, Mar 15 2006

Keywords

Comments

Probable answer to the riddle in A115603.
Partial sums of the squares of the terms of A116966.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2,-1,0,2,-4,2,0,-1,2,-1},{0,1,10,14,30,55,104,140,204,285,406},50] (* Harvey P. Dale, Jul 01 2020 *)

Formula

G.f.: x*(4*x^7-3*x^6+8*x^5+7*x^4+12*x^3-5*x^2+8*x+1) / ((x-1)^4*(x+1)^2*(x^2+1)^2). - Colin Barker, Jul 18 2013
a(n) = (2*n+1)*(2*n*(n+1)+3*(1+cos(n*Pi)-2*cos(n*Pi/2)))/12. - Luce ETIENNE, Feb 01 2017

Extensions

More terms from Stefan Steinerberger, Mar 31 2006
Entry revised by Don Reble, Apr 06 2006
More terms from Colin Barker, Jul 18 2013
Offset adapted to definition by Georg Fischer, Jun 18 2021