OFFSET
1,2
COMMENTS
Given three consecutive numbers x=n-2, y=n-1 and z=n, the sum over all products is x*y*z + x*y + x*z + y*z + x + y + z = n^3 - n - 1 = a(n). - J. M. Bergot, Aug 25 2011
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
For n > 1, a(n) = floor(n^6/(n^3+n+1)). - Gary Detlefs, Feb 10 2010
G.f.: x*(-1 + 9*x - 3*x^2 + x^3) / (x-1)^4. - R. J. Mathar, Aug 28 2011
a(-n) = -A061600(n). - Bruno Berselli, Aug 29 2011
E.g.f.: (-1 + 6*x + 6*x^2 + x^3)*exp(x) = -E(0) where E(k) = 1 - 6*x/(1 - x/(1 + x - x/(6 + x - 6/(1 - x^2*(k+1)/E(k+1) )))); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 09 2013
MATHEMATICA
a = {}; Do[AppendTo[a, x^3 - x - 1], {x, 1, 100}]; a
PROG
(Magma) [n^3-n-1: n in [1..50]]; // Vincenzo Librandi, Aug 29 2011
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Artur Jasinski, Dec 26 2006
STATUS
approved