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.

A134163 1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2.

Original entry on oeis.org

1, 187, 1531, 5977, 16441, 36811, 71947, 127681, 210817, 329131, 491371, 707257, 987481, 1343707, 1788571, 2335681, 2999617, 3795931, 4741147, 5852761, 7149241, 8650027, 10375531, 12347137, 14587201, 17119051, 19966987, 23156281, 26713177
Offset: 0

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Author

Artur Jasinski, Oct 10 2007

Keywords

Comments

A000541(n) is divisible by A000537(n) if and only n is congruent to 1 mod 3 (see A016777).

Crossrefs

Programs

  • Magma
    [1 + 12*n + 81*n^3 + n*(105*n+ 81*n^3)/2: n in [0..30]]; // Vincenzo Librandi, May 09 2011
  • Maple
    A134163:=n->1 + 12*n + 81*n^3 + n*(105*n + 81*n^3)/2: seq(A134163(n), n=0..30); # Wesley Ivan Hurt, Oct 23 2014
  • Mathematica
    Table[(3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6, {n, 0, 100}] (* or *) Table[Sum[k^7, {k, 1, 3n + 1}]/Sum[k^3, {k, 1, 3n + 1}], {n, 0, 100}]

Formula

a(n) = (3(3n + 1)^4 + 6(3n + 1)^3 - (3n + 1)^2 - 4 (3n + 1) + 2)/6.
a(n) = ( sum_{k=1..3n+1} k^7 ) / ( sum_{k=1..3n+1} k^3 ).
G.f.: (1+182*x+606*x^2+182*x^3+x^4)/(1-x)^5. - R. J. Mathar, Nov 14 2007
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, Oct 23 2014