Integers k such that A071089(k) = k.
From Robert Israel, Mar 27 2018: (Start)
No more terms below 10^7.
Heuristically, the probability that k is a term is 1/prime(k) ~ 1/(k log k).
Since Sum_{k>=2} 1/(k log(k)) diverges, there should be infinitely many terms. However, the sum diverges very slowly, so terms may be very sparse: approximately log(log(k)) terms <= k. (End)
No more terms below 10^9.  Michel Marcus, Mar 28 2018
No more terms below 1.44*10^12.  Giovanni Resta, Apr 06 2018
