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A275391 Least k such that n divides sigma(k^k) (k > 0).

%I A275391 #38 Aug 10 2016 00:54:06
%S A275391 1,3,5,3,3,5,2,3,5,3,19,11,11,5,15,7,15,5,11,3,5,19,10,11,7,11,17,11,
%T A275391 13,15,5,7,29,15,23,11,11,11,11,3,15,5,35,19,23,21,22,15,13,7,15,11,
%U A275391 23,17,19,11,11,13,28,15,11,5,5,15,15,29,21,15,65,23,34,11,4,11,29,11,39,11,23,7,17
%N A275391 Least k such that n divides sigma(k^k) (k > 0).
%C A275391 From _Robert Israel_, Aug 09 2016: (Start)
%C A275391 a(n) <= A038700(n) if n >= 4, since sigma(k^k) == 0 (mod n) if k is an odd prime == -1 (mod n).
%C A275391 If n is prime and n-2 is squarefree, then a(n) <= n-2 since sigma((n-2)^(n-2)) == 0 (mod n).
%C A275391 Conjecture: a(n) <= n-2 for all n > 15, but a(n) = n-2 for infinitely many n. (End)
%H A275391 Robert Israel, <a href="/A275391/b275391.txt">Table of n, a(n) for n = 1..10000</a>
%e A275391 a(11) = 19 because sigma(19^19) is divisible by 11.
%p A275391 N:= 200: # to get a(1)..a(N)
%p A275391 S:= {$1..N}:
%p A275391 for k from 1 while S <> {} do
%p A275391   v:= numtheory:-sigma(k^k);
%p A275391   F:= select(t -> v mod t = 0, S);
%p A275391   for n in F do A[n]:= k od:
%p A275391   S:= S minus F;
%p A275391 od:
%p A275391 seq(A[n],n=1..N); # _Robert Israel_, Aug 09 2016
%o A275391 (PARI) a(n) = {my(k=1); while(sigma(k^k) % n != 0, k++); k; }
%Y A275391 Cf. A000203, A038700, A062727, A275800.
%K A275391 nonn
%O A275391 1,2
%A A275391 _Altug Alkan_, Aug 07 2016