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Primes p such that there is at least one integer x satisfying p=x*tau(x)-1 where tau(x)=A000005(x), the number of divisors of x.
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%I #9 Dec 01 2013 08:27:09

%S 3,5,11,13,23,31,37,59,61,71,73,79,83,103,107,131,139,151,157,167,191,

%T 193,227,239,263,269,277,307,311,313,347,379,397,421,431,443,449,457,

%U 479,487,491,499,541,563,571,587,593,613,619,631,643,661,673,701,719

%N Primes p such that there is at least one integer x satisfying p=x*tau(x)-1 where tau(x)=A000005(x), the number of divisors of x.

%e 6*tau(6)=24 hence 23 is in the sequence.

%o (PARI) lista(nn) = {forprime(p=2, nn, for (n=1, p, if (n*numdiv(n)-1 == p, print1(p, ", "); break;);););} \\ _Michel Marcus_, Dec 01 2013

%Y Cf. A079449.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Jan 13 2003