OFFSET
1,1
COMMENTS
Given S = (n^n+k^k)/(n+k) (here k = 2), when k = 2^m for some m > 0, there are significantly less values of n that make S an integer. For k=3, see A242883.
a(15) > 210000.
Equivalently, (-2)^n + 4 == 0 (mod n + 2). - Robert Israel, Jun 10 2014
Odd terms are A033984(2..infinity) - 2. - Robert Israel, Jun 10 2014
EXAMPLE
(5^5+2^2)/(5+2) = 3129/7 = 447 is an integer. Thus 5 is a member of this sequence.
MAPLE
filter:= proc(n) (-2)&^n + 4 mod (n+2) = 0 end proc;
select(filter, [$1..10^6]); # Robert Israel, Jun 10 2014
PROG
(PARI) for(n=1, 10^5, s=(n^n+2^2)/(n+2); if(floor(s)==s, print(n)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Derek Orr, May 25 2014
EXTENSIONS
a(16)-a(34) from Robert Israel, Jun 10 2014
STATUS
approved