%I #18 Dec 04 2019 18:19:41

%S 56,190,812,1672,4522,5278,16065,24244,25070,33915,39585,56252,80104,

%T 93496,102856,107156,140296,157586,220616,224536,316274,317205,365638,

%U 389732,423045,479655,546592,559845,596666,601312,696514,731962,1123605,1161508,1181895

%N Numbers n such that sigma(n) = 5*phi(n).

%C If p>2 and 2^p-1 is prime (a Mersenne prime) then 377*2^(p-2)*(2^p-1) is in the sequence (the proof is easy). So for n>1 377*2^(A000043(n)-2)*(2^A000043(n)-1) is in the sequence.

%H Amiram Eldar, <a href="/A136547/b136547.txt">Table of n, a(n) for n = 1..10000</a> (calculated using data from Jud McCranie, terms 1..1000 from Donovan Johnson)

%H Kevin A. Broughan and Daniel Delbourgo, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Broughan/broughan26.html">On the Ratio of the Sum of Divisors and Euler’s Totient Function I</a>, Journal of Integer Sequences, Vol. 16 (2013), Article 13.8.8.

%H Kevin A. Broughan and Qizhi Zhou, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Broughan/bro32.html">On the Ratio of the Sum of Divisors and Euler's Totient Function II</a>, Journal of Integer Sequences, Vol. 17 (2014), Article 14.9.2.

%e sigma(56) = 120 = 5*24 = 5*phi(56) so 56 is in the sequence.

%t Do[If[DivisorSigma[1,m]==5*EulerPhi[m],Print[m]],{m,1500000}]

%o (PARI) is(n)=sigma(n)==5*eulerphi(n) \\ _Charles R Greathouse IV_, May 09 2013

%Y Cf. A000043, A062699, A104900, A104901, A104902.

%K nonn

%O 1,1

%A _Farideh Firoozbakht_, Jan 29 2008, Jan 30 2008