login
Numbers n such that (phi(n) + 1) | sigma(n + 1), where phi is Euler's totient function A000010.
5

%I #34 Jan 01 2019 06:59:42

%S 2,4,16,25,170,256,264,1920,9384,26664,65536,263040,437760,1057800,

%T 2038648320

%N Numbers n such that (phi(n) + 1) | sigma(n + 1), where phi is Euler's totient function A000010.

%C 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

%p with(numtheory): select(n->modp(sigma(n+1),phi(n)+1)=0,[$1..10000]); # _Muniru A Asiru_, Dec 10 2018

%t Select[Range[10000], Mod[DivisorSigma[1, #+1], EulerPhi[#] +1] == 0 &] (* _G. C. Greubel_, Dec 10 2018 *)

%o (PARI) is(n)=!(sigma(n+1)%(eulerphi(n)+1)) \\ _Charles R Greathouse IV_, Sep 25 2012

%Y Cf. A000010 (phi), A000203 (sigma).

%K nonn,more

%O 1,1

%A _Robert G. Wilson v_

%E Corrected by _Charles R Greathouse IV_, Sep 25 2012

%E Incorrect a(1) = 1 removed and a(12)-a(14) added by _Sean A. Irvine_, Dec 10 2018

%E a(15) from _Amiram Eldar_, Dec 10 2018