OFFSET
1,1
COMMENTS
Two solutions, p and 2p, exist for all odd primes p; primes in sequence have no other solutions.
Conjecture: if q > 7 is in A005385, then q is in the sequence. - Thomas Ordowski, Jan 04 2017
Proof of conjecture: q'=(q-1)/2 is an odd prime > 3. If phi(x)=2q', which has 2-adic order 1 but is not a power of 2, there must be exactly one odd prime r dividing x. We could also have a factor of 2 (but no higher power, which would contribute more 2's to phi(x)). If x = r^e or 2r^e, then phi(x) = (r-1) r^(e-1). For this to be 2q' one possibility is r-1 = 2 and r^(e-1)=q', but then q'=r=3, ruled out by q > 7. The only other possibility is r-1=2q' and e=1, which makes r=q and x=q or 2q. - Robert Israel, Jan 04 2017
Information from Carl Pomerance: It is known that almost all primes (in the sense of relative asymptotic density) are in the sequence. - Thomas Ordowski, Jan 08 2017
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
FORMULA
a(n) ~ n log . - Charles R Greathouse IV, Nov 18 2022
EXAMPLE
For p=2, phi(x)=1 has only two solutions, but they are 1 and 2, not 2 and 4, so 2 is not in the sequence.
MAPLE
filter:= n -> isprime(n) and nops(numtheory:-invphi(n-1))=2:
select(filter, [seq(i, i=3..10000, 2)]); # Robert Israel, Aug 12 2016
MATHEMATICA
Take[Rest@ Keys@ Select[KeySelect[KeyMap[# + 1 &, PositionIndex@ Array[EulerPhi, 10^4]], PrimeQ], Length@ # == 2 &], 54] (* Michael De Vlieger, Dec 29 2017 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Dec 14 2000
EXTENSIONS
Edited by Ray Chandler, Jun 06 2008
STATUS
approved