OFFSET
1,1
COMMENTS
f(x) = sum of prime factors with multiplicity, so that f(1500) = 2+2+3+5+5+5 = 22.
Getting 4 is rare (only 1 such instance in the first 5,000 terms). - Harvey P. Dale, Nov 19 2023
LINKS
Harvey P. Dale, Table of n, a(n) for n = 1..5000
MATHEMATICA
fi[n_]:=Total[Flatten[Table[#[[1]], #[[2]]]&/@FactorInteger[n]]]; bpQ[n_]:=Boole[PrimeQ[NestList[ fi[#]&, n, 8]]] =={0, 0, 0, 0, 0, 0, 0, 0, 1}||Nest[fi[#]&, n, 7]==4; Select[Range[11000], bpQ] (* Harvey P. Dale, Nov 19 2023 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved