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A180642
Numbers k such that phi(k)/4 is a prime, where phi is the Euler totient function.
1
13, 15, 16, 20, 21, 24, 25, 26, 28, 29, 30, 33, 36, 42, 44, 50, 53, 58, 66, 69, 92, 106, 138, 141, 149, 173, 177, 188, 236, 249, 269, 282, 293, 298, 317, 321, 332, 346, 354, 389, 428, 498, 501, 509, 537, 538, 557, 586, 634, 642, 653, 668, 681, 716, 773, 778, 789
OFFSET
1,1
COMMENTS
Apparently the sequence is infinite, but I have no proof. There are many n-ples of consecutives: (15,16)-(20,21)-(24,25,26)-(537,538)-(1436,1437)-...-(30236-30237)
This sequence is infinite if and only if there are infinitely many primes of the form 2p+1 or 4p+1 with prime p. - Charles R Greathouse IV, Feb 04 2013
LINKS
FORMULA
a(n) >> n log^2 n. - Charles R Greathouse IV, Feb 04 2013
EXAMPLE
a(5) = 21 since pi(21)/4 = 12/4 = 3 is prime.
MATHEMATICA
Select[Range[800], PrimeQ[EulerPhi[#]/4]&] (* Harvey P. Dale, Feb 11 2015 *)
PROG
(PARI) is(n)=n=eulerphi(n); n%4==0 && isprime(n/4) \\ Charles R Greathouse IV, Feb 04 2013
(PARI) is(n)=if(n<51, n=eulerphi(n); n%4==0 && isprime(n/4), my(v=[3, 4, 6]); for(i=1, #v, if(n%(2*v[i])==v[i]&&gcd(n/v[i], v[i])==1&&isprime(n/v[i])&&isprime(eulerphi(n)/4), return(1))); if(n%4==2, n/=2); n%4==1&&isprime(n)&&isprime(n\4)) \\ Charles R Greathouse IV, Feb 04 2013
CROSSREFS
Cf. A000010, A065966 (phi(k)/2 is prime), A090866 (subsequence of primes).
Sequence in context: A043702 A278978 A212908 * A176717 A070603 A329183
KEYWORD
nonn
AUTHOR
Carmine Suriano, Sep 14 2010
STATUS
approved