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%I #8 Dec 01 2013 08:27:18
%S 2,5,7,11,13,23,41,47,59,61,73,83,89,107,109,137,157,167,179,193,227,
%T 229,233,241,263,271,277,313,337,347,349,359,373,379,383,409,433,449,
%U 457,461,467,479,503,563,569,587,709,719,733,809,821,839,853,857,863
%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 12*tau(12)=72 hence 73 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. A079448.
%K nonn
%O 1,1
%A _Benoit Cloitre_, Jan 13 2003
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