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%I #14 Jun 17 2018 06:20:05
%S 5,7,7,5,13,7,5,17,5,19,5,13,5,5,7,11,7,5,5,5,13,5,7,31,5,5,5,5,5,5,
%T 13,5,5,5,5,5,7,5,5,5,5,5,7,7,5,5,5,5,5,11,5,5,5,5,5,5,5,5,5,7,5,5,5,
%U 7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,7,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5
%N Number of divisors of square of true prime powers arising in A065405.
%F a(n) = A000005(A065405(n)^2). If A065405(n) = q^c, a prime-power, then sigma(q^(2c)) = A000203(q^(2c)) = (-1 + q^(2c+1))/(q-1) = (-1 + q^A000005(A065405(n)^2))/(q-1) also a prime, from A065403.
%e n = 3125, tau(nn) = 11, sigma(nn) = 12207031 = (5^(tau(nn)) - 1)/4 = A065403(16) is also a prime.
%o (PARI) { n=0; for (m=1, 10^9, if (isprime(m), next); x=sigma(m^2); if (isprime(x), a=numdiv(m^2); write("b065773.txt", n++, " ", a); if (n==100, return)) ) } \\ _Harry J. Smith_, Oct 30 2009
%Y Cf. A000005, A065405, A000203, A065771, A065772, A025475.
%K nonn
%O 1,1
%A _Labos Elemer_ and _Robert G. Wilson v_, Nov 19 2001
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