|
%I #11 Sep 08 2022 08:45:49
%S 5,13,35,37,61,73,157,193,277,313,397,421,455,457,541,613,661,665,673,
%T 733,757,877,997,1085,1093,1153,1201,1213,1237,1295,1321,1381,1453,
%U 1621,1657,1753,1873,1933,1993,2017,2137,2169,2341,2473,2557,2593,2797,2857
%N Numbers n such that phi(n) = 2*phi(n+1).
%C Theorem: A prime p is in the sequence iff 1/2*(p+1) is prime.
%C Proof: If both numbers p & 1/2*(p+1) are prime then phi(p)=p-1=2*(p-1)/2
%C 2*(1/2*(p+1)-1)=2*phi(1/2*(p+1)), 1/2*(p+1) is odd so phi(1/2*(p+1))=
%C phi(p+1) hence phi(p)=2*phi(p+1), namely p is in the sequence.
%C Also if p is a prime term of the sequence
%C then phi(p)=2*phi(p+1) so
%C p-1=2*phi(p+1) or phi(p+1)=1/2*(p+1)-1 hence 1/2*(p+1)is prime.
%H Ray Chandler, <a href="/A171262/b171262.txt">Table of n, a(n) for n = 1..10000</a>
%F phi(35)=2*12=2*phi(35+1), so 35 is in the sequence.
%t Select[Range[2900],EulerPhi[ # ]==2EulerPhi[ #+1]&]
%o (Magma) [n: n in [1..3*10^3] | EulerPhi(n) eq 2*EulerPhi(n+1)]; // _Vincenzo Librandi_, Apr 14 2015
%Y Cf. A005383, A171271.
%K nonn,easy
%O 1,1
%A _Farideh Firoozbakht_, Feb 23 2010
|
