OFFSET
0,3
COMMENTS
For n > 0, a(n) is the largest number k such that k^5 + n is divisible by k + n, or -1 if k is infinite. When k = n^5 - 2n, (k^5 + n)/(k + n) = n^20 - 9n^16 + 31n^12 - 49n^8 + 31n^4 - 1.
For n > 1 and j > 0, the largest number k such that k^(2j+1) + n is divisible by k + n is given by k = n^(2j+1) - 2n.
For n > 1, a(n) is also the largest number k such that k + n^5 is divisible by k + n. - Derek Orr, Oct 16 2014
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
FORMULA
From Wesley Ivan Hurt, Aug 17 2014: (Start)
G.f.: (-x+34*x^2+54*x^3+34*x^4-x^5) / (x-1)^6;
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6);
EXAMPLE
a(2) = 28. 28 is the largest number k such that k^5 + 2 is divisible by k + 2. The resulting integer is 573679 = 2^20 - 9*2^16 + 31*2^12 - 49*2^8 + 31*2^4 - 1.
MAPLE
MATHEMATICA
Table[n^5 - 2 n, {n, 0, 30}] (* Wesley Ivan Hurt, Aug 17 2014 *)
CoefficientList[Series[(-x + 34 x^2 + 54 x^3 + 34 x^4 - x^5)/(-1 + x)^6, {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 17 2014 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, -1, 28, 237, 1016, 3115}, 40] (* Harvey P. Dale, Apr 07 2017 *)
PROG
(PARI) vector(100, n, (n-1)^5 - 2*(n-1))
(Magma) [n^5-2*n: n in [0..35]]; // Vincenzo Librandi, Oct 17 2014
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Derek Orr, Aug 16 2014
STATUS
approved