OFFSET
1,1
COMMENTS
f(x) = sum of prime factors with multiplicity, so that f(1500) = 2+2+3+5+5+5 = 22.
Numbers k such that A002217(k) = 3. - Andrew Howroyd, Sep 15 2019
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..10000
MATHEMATICA
okQ[n_]:=Module[{lst=NestList[Total[Times@@@FactorInteger[#]]&, n, 2]}, !PrimeQ[First[lst]] &&!PrimeQ[lst[[2]]]&&First[lst]!=4&&lst[[2]]!=4&&(PrimeQ[Last[lst]]||Last[lst]==4)]; Select[Range[400], okQ] (* Harvey P. Dale, Mar 23 2011 *)
PROG
(PARI) sopfr(n)={my(f=factor(n)); sum(i=1, #f~, f[i, 1]*f[i, 2])}
ok(n)={forstep(k=2, 1, -1, n=sopfr(n); if(n==4||isprime(n), return(k==1))); 0}
select(ok, [1..500]) \\ Andrew Howroyd, Sep 14 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved