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A058340
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Primes p such that phi(x) = p-1 has only 2 solutions, namely x = p and x = 2p.
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17
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11, 23, 29, 31, 47, 53, 59, 67, 71, 79, 83, 103, 107, 127, 131, 137, 139, 149, 151, 167, 173, 179, 191, 197, 199, 211, 223, 227, 229, 239, 251, 263, 269, 271, 283, 293, 307, 311, 317, 331, 347, 359, 367, 373, 379, 383, 389, 419, 431, 439, 443, 463, 467, 479
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OFFSET
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1,1
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COMMENTS
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Two solutions, p and 2p, exist for all odd primes p; primes in sequence have no other solutions.
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
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LINKS
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FORMULA
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EXAMPLE
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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.
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MAPLE
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filter:= n -> isprime(n) and nops(numtheory:-invphi(n-1))=2:
select(filter, [seq(i, i=3..10000, 2)]); # Robert Israel, Aug 12 2016
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MATHEMATICA
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Take[Rest@ Keys@ Select[KeySelect[KeyMap[# + 1 &, PositionIndex@ Array[EulerPhi, 10^4]], PrimeQ], Length@ # == 2 &], 54] (* Michael De Vlieger, Dec 29 2017 *)
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CROSSREFS
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Cf. A000010, A000040, A005385, A006093, A058339, A066071, A066072, A066073, A066074, A066075, A066076, A066077, A066080, A138537.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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