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

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, 13, 15, 5, 7, 29, 15, 23, 11, 11, 11, 11, 3, 15, 5, 35, 19, 23, 21, 22, 15, 13, 7, 15, 11, 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

Offset: 1

Author

Altug Alkan, Aug 07 2016

Keywords

• nonn

Comments

From Robert Israel, Aug 09 2016: (Start)

a(n) <= A038700(n) if n >= 4, since sigma(k^k) == 0 (mod n) if k is an odd prime == -1 (mod n).

If n is prime and n-2 is squarefree, then a(n) <= n-2 since sigma((n-2)^(n-2)) == 0 (mod n).

Conjecture: a(n) <= n-2 for all n > 15, but a(n) = n-2 for infinitely many n. (End)

Examples

			a(11) = 19 because sigma(19^19) is divisible by 11.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000203, A038700, A062727, A275800.

Programs

Maple

N:= 200: # to get a(1)..a(N)
S:= {$1..N}:
for k from 1 while S <> {} do
  v:= numtheory:-sigma(k^k);
  F:= select(t -> v mod t = 0, S);
  for n in F do A[n]:= k od:
  S:= S minus F;
od:
seq(A[n],n=1..N); # Robert Israel, Aug 09 2016

PARI

a(n) = {my(k=1); while(sigma(k^k) % n != 0, k++); k; }