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 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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) LINKS Robert Israel, Table of n, a(n) for n = 1..10000 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: seq(A[n], n=1..N); # Robert Israel, Aug 09 2016 PROG (PARI) a(n) = {my(k=1); while(sigma(k^k) % n != 0, k++); k; } CROSSREFS Cf. A000203, A038700, A062727, A275800. Sequence in context: A135514 A251754 A225581 * A092553 A112755 A239278 Adjacent sequences: A275388 A275389 A275390 * A275392 A275393 A275394 KEYWORD nonn AUTHOR Altug Alkan, Aug 07 2016 STATUS approved

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Last modified June 18 21:09 EDT 2024. Contains 373487 sequences. (Running on oeis4.)