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

1,4

Links

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

Formula

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A106698 A097848 A308964 * A291598 A287820 A191373

Adjacent sequences: A124827 A124828 A124829 * A124831 A124832 A124833

Keyword

nonn

Author

Franklin T. Adams-Watters, Nov 09 2006

Status

approved