A171262 Numbers n such that phi(n) = 2*phi(n+1).

5, 13, 35, 37, 61, 73, 157, 193, 277, 313, 397, 421, 455, 457, 541, 613, 661, 665, 673, 733, 757, 877, 997, 1085, 1093, 1153, 1201, 1213, 1237, 1295, 1321, 1381, 1453, 1621, 1657, 1753, 1873, 1933, 1993, 2017, 2137, 2169, 2341, 2473, 2557, 2593, 2797, 2857

Offset: 1

Author

Farideh Firoozbakht, Feb 23 2010

Keywords

• nonn
• easy

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

Crossrefs

Cf. A005383, A171271.

Programs

Magma

[n: n in [1..3*10^3] | EulerPhi(n) eq 2*EulerPhi(n+1)]; // Vincenzo Librandi, Apr 14 2015

Mathematica

Select[Range[2900],EulerPhi[ # ]==2EulerPhi[ #+1]&]

Formula

phi(35)=2*12=2*phi(35+1), so 35 is in the sequence.