login
A015775
Numbers n such that (phi(n) + 1) | sigma(n + 1), where phi is Euler's totient function A000010.
5
2, 4, 16, 25, 170, 256, 264, 1920, 9384, 26664, 65536, 263040, 437760, 1057800, 2038648320
OFFSET
1,1
COMMENTS
For 0 <= k <= 4, p = 2^(2^k) + 1 is a (Fermat) prime, so sigma(p) = p + 1 = 2*(2^(2^k-1) + 1) and phi(2^(2^k)) = 2^(2^k-1), so we have sigma(p) = 2*(phi(p-1) + 1) and n = p-1 = 2^(2^k) is in the sequence. For k = 5 this is no more the case. - M. F. Hasler, Dec 10 2018
MAPLE
with(numtheory): select(n->modp(sigma(n+1), phi(n)+1)=0, [$1..10000]); # Muniru A Asiru, Dec 10 2018
MATHEMATICA
Select[Range[10000], Mod[DivisorSigma[1, #+1], EulerPhi[#] +1] == 0 &] (* G. C. Greubel, Dec 10 2018 *)
PROG
(PARI) is(n)=!(sigma(n+1)%(eulerphi(n)+1)) \\ Charles R Greathouse IV, Sep 25 2012
CROSSREFS
Cf. A000010 (phi), A000203 (sigma).
Sequence in context: A292369 A357917 A153665 * A330582 A104258 A143904
KEYWORD
nonn,more
EXTENSIONS
Corrected by Charles R Greathouse IV, Sep 25 2012
Incorrect a(1) = 1 removed and a(12)-a(14) added by Sean A. Irvine, Dec 10 2018
a(15) from Amiram Eldar, Dec 10 2018
STATUS
approved