%I
%S 1,2,4,6,10,16,18,22,28,30,42,46,52,58,66,70,78,82,
%T 100,102,106,126,130,136,138,148,150,162,166,172,178,
%U 190,196,198,210,222,226,228,238,250,256,262,268,270,282,292,306
%N Values of A023900(k) the only solutions of which have a single distinct prime factor.
%C Terms are equal to A023900(p) = A023900(p^2) = A023900(p^3) = ... with p prime, but is never equal to A023900(m*p) with m <> p.
%C abs(a(n)) + 1 is prime (A301590).
%C For n > 1, if and only if n can't be factored into 2*m factors, m > 0, distinct factors f > 1 where f + 1 is prime then n is a term.  _David A. Corneth_, Mar 25 2018
%e a(1) = 1 = A023900(2^m), m > 0.
%e a(2) = 2 = A023900(3^m), m > 0.
%e a(3) = 4 = A023900(5^m), m > 0.
%e a(4) = 6 = A023900(7^m), m > 0.
%e a(5) = 10 = A023900(11^m), m > 0.
%e a(6) = 16 = A023900(17^m), m > 0.
%e A023900(13) = 12 is not a term as A023900(42) = 12, and 42 is the product of three prime factors.
%e From _David A. Corneth_, Mar 25 2018: (Start)
%e 10 can't be factored in an even number of distinct factors f > 1 such that f + 1 is prime, so 10 is in the sequence.
%e 12 can be factored in an even number of distinct factors f > 1; 12 = 2 * 6 and both 2 + 1 and 6 + 1 are prime, hence 12 is not a term. (End)
%t Keys@ Select[Union /@ PrimeNu@ PositionIndex@ Array[DivisorSum[#, # MoebiusMu[#] &] &, 310], # == {1} &] (* _Michael De Vlieger_, Mar 26 2018 *)
%o (PARI) f(n) = sumdivmult(n, d, d*moebius(d));
%o isok(p, vp) = {for (k=p+1, p^21, if (f(k) == vp, return (0));); return (1);}
%o lista(nn) = {forprime(p=2, nn, vp = f(p); if (isok(p, vp), print1(vp, ", ")););} \\ _Michel Marcus_, Mar 23 2018
%Y Cf. A000040, A001055, A023900, A301590, A301591.
%K sign,easy
%O 1,2
%A _Torlach Rush_, Mar 19 2018
