OFFSET
1,1
COMMENTS
Other solutions of this equation are in A001838.
Also, numbers k such that 2^(k-1)-1 is prime. Proof: If x = 2^k-2, phi(x)+2 = phi(x+2) <==> phi(2^k-2)+2 = phi(2^k) <==> phi(2(2^(k-1)-1)) + 2 = 2^k(1-1/2) <==> phi(2)*phi(2^(k-1)-1)+2 = 2^(k-1) <==> phi(2^(k-1)-1) = 2^(k-1)-2 if y = 2^(k-1)-1. We have phi(y) = y-1 <==> y=2^(k-1)-1 is prime. Therefore a(n) = A000043(n)+1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 19 2004
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..48 (terms 1..47 from Ivan Panchenko)
FORMULA
a(n) = A000043(n) + 1. - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 19 2004
EXAMPLE
phi(2^18-2)+2 = 131072 = phi(2^18), so 18 is in the sequence.
MATHEMATICA
Flatten[Position[EulerPhi[2^# - 2] + 2 == EulerPhi[2^# ] & /@ Range[1, 250], True]] (* Vit Planocka *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jud McCranie, Dec 24 1999
EXTENSIONS
a(39)-a(44) from Ivan Panchenko, Apr 11 2018
STATUS
approved