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%I #21 Nov 21 2024 22:25:40
%S 2,5,8,128,2144,4808,12872,14168,33672,40367,45992,116192,185768,
%T 186824,271208,426008,524288,601352,612768,673661,755792,990407,
%U 996032,1697607,1878368,2073125,2262752,4325960,4810808,6331808,6462647,6707352,7527197,7559477
%N Numbers n such that (n^n+2^2)/(n+2) is an integer.
%C 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.
%C a(15) > 210000.
%C Equivalently, (-2)^n + 4 == 0 (mod n + 2). - _Robert Israel_, Jun 10 2014
%C Odd terms are A033984(2..infinity) - 2. - _Robert Israel_, Jun 10 2014
%H Robert Israel, <a href="/A242875/b242875.txt">Table of n, a(n) for n = 1..100</a>
%e (5^5+2^2)/(5+2) = 3129/7 = 447 is an integer. Thus 5 is a member of this sequence.
%p filter:= proc(n) (-2)&^n + 4 mod (n+2) = 0 end proc;
%p select(filter, [$1..10^6]); # _Robert Israel_, Jun 10 2014
%o (PARI) for(n=1,10^5,s=(n^n+2^2)/(n+2);if(floor(s)==s,print(n)))
%Y Cf. A242883.
%K nonn
%O 1,1
%A _Derek Orr_, May 25 2014
%E a(16)-a(34) from _Robert Israel_, Jun 10 2014