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%I #9 Nov 23 2022 15:41:03
%S 8,9,14,20,24,26,27,38,44,62,68,74,105,112,116,125,126,134,150,160,
%T 180,188,192,195,208,212,216,218,231,234,243,254,275,278,314,330,332,
%U 343,352,356,362,396,398,422,428,465,483,490,496,548,558,575,588,609,614
%N Numbers n such that the sum of prime factors of n (counted with multiplicity) is 3 times a prime.
%C The numbers A100118(n)^3 are in the sequence.
%H Harvey P. Dale, <a href="/A213016/b213016.txt">Table of n, a(n) for n = 1..1000</a>
%F sopfr(n) = 3*p, p prime.
%e 44 is in the sequence because 44 = 2^2 * 11 => sum of prime factors = 2*2+11 = 15 = 3*5 where 5 is prime.
%p with(numtheory):A:= proc(n) local e, j; e := ifactors(n)[2]: add (e[j][1]*e[j][2], j=1..nops(e)) end: for m from 1 to 3000 do: if type(A(m)/3,prime)= true then printf(`%d, `,m):else fi:od:
%t L = {}; Do[ww = Transpose[FactorInteger[k]]; w = ww[[1]].ww[[2]]; If[PrimeQ[w/3], AppendTo[L, k]], {k, 2, 1000}]; L
%t Select[Range[700],PrimeQ[Total[Times@@@FactorInteger[#]]/3]&] (* _Harvey P. Dale_, Nov 23 2022 *)
%Y Cf. A001414, A100118, A213015.
%K nonn
%O 1,1
%A _Michel Lagneau_, Jun 02 2012
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