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%I #13 Sep 28 2018 15:31:01
%S 2,3,4,5,7,8,9,11,13,16,17,19,20,21,23,25,27,29,31,32,33,37,41,43,44,
%T 47,49,53,57,59,60,61,64,67,68,69,71,73,79,81,83,85,89,93,97,101,103,
%U 105,107,109,112,113,116,121,125,127,128,129,131,133,137,139
%N Numbers whose average of prime factors is prime.
%C Prime factors counted with multiplicity. - _Harvey P. Dale_, Sep 28 2018
%H Harvey P. Dale, <a href="/A316525/b316525.txt">Table of n, a(n) for n = 1..1000</a>
%e 60 = 2*2*3*5 has average of prime factors (2+2+3+5)/4 = 3, which is prime, so 60 belongs to the sequence.
%t Select[Range[100],PrimeQ[Mean[If[#==1,{},Flatten[Cases[FactorInteger[#],{p_,k_}:>Table[p,{k}]]]]]]&]
%t Select[Range[200],PrimeQ[Mean[Flatten[Table[#[[1]],#[[2]]]&/@ FactorInteger[ #]]]]&] (* _Harvey P. Dale_, Sep 28 2018 *)
%o (PARI) isok(n) = {my(f=factor(n)); iferr(isprime(sum(k=1, #f~, f[k,1]*f[k,2])/sum(k=1, #f~, f[k,2])), E, 0);} \\ _Michel Marcus_, Jul 06 2018
%Y Subsequence of A078175.
%Y Cf. A000040, A000607, A056768, A071321, A100118, A316219, A316313.
%K nonn
%O 1,1
%A _Gus Wiseman_, Jul 05 2018