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A242870 Numbers n such that (n^n-2^2)/(n-2) is an integer.

%I A242870 #19 Apr 24 2016 08:49:45
%S A242870 1,3,4,6,8,14,20,22,38,44,56,62,86,102,110,128,158,164,182,222,254,
%T A242870 296,302,326,344,380,422,470,488,502,542,590,622,662,686,758,782,822,
%U A242870 884,902,974,1028,1094,1102,1136,1262,1316,1334,1406,1460,1502,1622,1766,1808
%N A242870 Numbers n such that (n^n-2^2)/(n-2) is an integer.
%C A242870 a(n) is even for all n > 2. 1 and 3 are members of this sequence because (n^n-2^2)/(n-2) becomes (2^2-n^n) and (n^n-2^2), respectively, which are both integers.
%C A242870 Given the term (n^n-k^k)/(n-k) (here, k=2), whenever k = 2^m for some m, there are significantly fewer data values within a given range of numbers. See A242871 for k=3.
%C A242870 These are also numbers n such that (2^n-n^2)/(n-2) is an integer.
%H A242870 Robert Israel, <a href="/A242870/b242870.txt">Table of n, a(n) for n = 1..10000</a>
%e A242870 (4^4-2^2)/(4-2) = 252/2 = 126 is an integer. Thus, 4 is a member of this sequence.
%p A242870 filter:= proc(n) (n&^n - 4) mod (n-2) = 0 end proc;
%p A242870 select(filter, [1,$3..1000]); # _Robert Israel_, May 25 2014
%t A242870 Join[{1},Select[Range[3,2000],IntegerQ[(#^#-4)/(#-2)]&]] (* _Harvey P. Dale_, Apr 24 2016 *)
%o A242870 (PARI) for(n=1,2500,if(n!=2,s=(n^n-2^2)/(n-2);if(floor(s)==s,print(n))))
%Y A242870 Cf. A242871.
%K A242870 nonn
%O A242870 1,2
%A A242870 _Derek Orr_, May 24 2014