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Numbers k such that phi(k)/4 is a prime, where phi is the Euler totient function.
1

%I #20 Mar 08 2020 01:39:49

%S 13,15,16,20,21,24,25,26,28,29,30,33,36,42,44,50,53,58,66,69,92,106,

%T 138,141,149,173,177,188,236,249,269,282,293,298,317,321,332,346,354,

%U 389,428,498,501,509,537,538,557,586,634,642,653,668,681,716,773,778,789

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

%C 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)

%C 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

%H Amiram Eldar, <a href="/A180642/b180642.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) >> n log^2 n. - _Charles R Greathouse IV_, Feb 04 2013

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

%t Select[Range[800],PrimeQ[EulerPhi[#]/4]&] (* _Harvey P. Dale_, Feb 11 2015 *)

%o (PARI) is(n)=n=eulerphi(n);n%4==0 && isprime(n/4) \\ _Charles R Greathouse IV_, Feb 04 2013

%o (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

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

%K nonn

%O 1,1

%A _Carmine Suriano_, Sep 14 2010