OFFSET
0,3
COMMENTS
For n > 1, these are also the largest positive integers k such that k + n divides k^3 + n^2. For all n > 1 and p > 1, the largest positive integer k such that k + n divides k^p + n^(p-1) is given by k = n^p - (-n)^(p-1) - n. Here, p = 3. - Derek Orr, Aug 13 2014
LINKS
Derek Orr, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: -x*(1-6*x-x^2)/(1-x)^4. - Bruno Berselli, Jul 27 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Wesley Ivan Hurt, Aug 13 2014
E.g.f.: exp(x)*x*(x^2 + 2*x - 1). - Stefano Spezia, Apr 15 2022
MAPLE
a:=n->sum(-1+sum(1+sum(1, i=2..n), j=2..n), k=1..n): seq(a(n), n=0..44); # Zerinvary Lajos, Dec 22 2008
MATHEMATICA
lst={}; Do[AppendTo[lst, n^3-n^2-n], {n, 0, 5!}]; lst
Table[n^3-n^2-n, {n, 0, 100}] (* Wesley Ivan Hurt, Oct 06 2013 *)
LinearRecurrence[{4, -6, 4, -1}, {0, -1, 2, 15}, 50] (* Harvey P. Dale, Sep 08 2024 *)
PROG
(PARI) vector(100, n, (n-1)^3-(n-1)^2-(n-1)) \\ Derek Orr, Aug 13 2014
(Magma) [n^3-n^2-n : n in [0..50]]; // Wesley Ivan Hurt, Aug 13 2014
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Vladimir Joseph Stephan Orlovsky, Nov 20 2008
EXTENSIONS
Offset changed by Bruno Berselli, Jul 27 2012
STATUS
approved