A126420 a(n) = n^3 - n - 1.

-1, 5, 23, 59, 119, 209, 335, 503, 719, 989, 1319, 1715, 2183, 2729, 3359, 4079, 4895, 5813, 6839, 7979, 9239, 10625, 12143, 13799, 15599, 17549, 19655, 21923, 24359, 26969, 29759, 32735, 35903, 39269, 42839, 46619, 50615, 54833, 59279, 63959, 68879, 74045, 79463

Offset: 1

Author

Artur Jasinski, Dec 26 2006

Keywords

• sign
• easy

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).

Crossrefs

Cf. A002327, A116581.

Programs

Magma

[n^3-n-1: n in [1..50]]; // Vincenzo Librandi, Aug 29 2011

Mathematica

a = {}; Do[AppendTo[a, x^3 - x - 1], {x, 1, 100}]; a

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