login
Numbers n such that phi(n) = 2*phi(n+1).
7

%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