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

Author

Carmine Suriano, Sep 14 2010

Keywords

• nonn

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

Examples

			a(5) = 21 since pi(21)/4 = 12/4 = 3 is prime.
		

Links

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

Crossrefs

Cf. A000010, A065966 (phi(k)/2 is prime), A090866 (subsequence of primes).

Programs

Mathematica

Select[Range[800],PrimeQ[EulerPhi[#]/4]&] (* Harvey P. Dale, Feb 11 2015 *)

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

Formula

a(n) >> n log^2 n. - Charles R Greathouse IV, Feb 04 2013