

A275391


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


2



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
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OFFSET

1,2


COMMENTS

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 n2 is squarefree, then a(n) <= n2 since sigma((n2)^(n2)) == 0 (mod n).
Conjecture: a(n) <= n2 for all n > 15, but a(n) = n2 for infinitely many n. (End)


LINKS



EXAMPLE

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


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:


PROG

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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



