A124830 Number of distinct prime factors of A055932(n).

0, 1, 1, 2, 1, 2, 1, 2, 2, 3, 1, 2, 2, 2, 3, 1, 2, 3, 2, 2, 3, 1, 2, 3, 2, 3, 2, 4, 2, 3, 1, 3, 2, 3, 2, 3, 2, 4, 2, 3, 3, 2, 1, 3, 2, 3, 4, 2, 3, 3, 2, 3, 4, 2, 3, 3, 2, 1, 4, 3, 2, 3, 4, 2, 3, 3, 2, 4, 3, 2, 3, 4, 2, 3, 4, 3, 2, 1, 4, 3, 3, 2, 5, 3, 3, 4, 2, 3, 3, 2, 4, 3, 2, 4, 3, 4, 2, 3, 3, 4, 3, 2, 3, 1, 4

Offset: 1

Author

Franklin T. Adams-Watters, Nov 09 2006

Keywords

• nonn

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000 (First 1000 terms from G. C. Greubel.)

Crossrefs

Cf. A055932, A001221, A124829, A124831, A061394.

Programs

Mathematica

PrimeNu /@ Select[Range[4000], ! MemberQ[Function[f, ReplacePart[Table[0, {PrimePi[f[[-1, 1]]]}], #] &@ Map[PrimePi@ First@ # -> Last@ # &, f]]@ FactorInteger@ #, 0] &] (* Michael De Vlieger, Feb 02 2017 *)
A055932[n_] := Module[{f = Transpose[FactorInteger[n]][[1]]}, f == {1} || f == Prime[Range[Length[f]]]]; PrimeNu[Select[Range[2000], A055932]] (* G. C. Greubel, May 11 2017 *)

Python

from sympy import nextprime, primefactors
def a053669(n):
    p = 2
    while True:
        if n%p!=0: return p
        else: p=nextprime(p)
def ok(n): return True if n==1 else a053669(n)>max(primefactors(n))
print([len(primefactors(n)) for n in range(1, 10001) if ok(n)]) # Indranil Ghosh, May 11 2017

Formula

a(n) = A001221(A055932(n)).