A180642 Numbers k such that phi(k)/4 is a prime, where phi is the Euler totient function.

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

Amiram Eldar, Table of n, a(n) for n = 1..10000

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

Adjacent sequences: A180639 A180640 A180641 * A180643 A180644 A180645

Keyword

nonn

Author

Carmine Suriano, Sep 14 2010

Status

approved