OFFSET
1,1
COMMENTS
Theorem: A prime p is in the sequence iff 1/2*(p+1) is prime.
Proof: If both numbers p & 1/2*(p+1) are prime then phi(p)=p-1=2*(p-1)/2
2*(1/2*(p+1)-1)=2*phi(1/2*(p+1)), 1/2*(p+1) is odd so phi(1/2*(p+1))=
phi(p+1) hence phi(p)=2*phi(p+1), namely p is in the sequence.
Also if p is a prime term of the sequence
then phi(p)=2*phi(p+1) so
p-1=2*phi(p+1) or phi(p+1)=1/2*(p+1)-1 hence 1/2*(p+1)is prime.
LINKS
Ray Chandler, Table of n, a(n) for n = 1..10000
FORMULA
phi(35)=2*12=2*phi(35+1), so 35 is in the sequence.
MATHEMATICA
Select[Range[2900], EulerPhi[ # ]==2EulerPhi[ #+1]&]
PROG
(Magma) [n: n in [1..3*10^3] | EulerPhi(n) eq 2*EulerPhi(n+1)]; // Vincenzo Librandi, Apr 14 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Farideh Firoozbakht, Feb 23 2010
STATUS
approved