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A124830 Number of distinct prime factors of A055932(n). 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,4

LINKS

G. C. Greubel, Table of n, a(n) for n = 1..1000

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 1 if n==1 else a053669(n)>max(primefactors(n))

print [len(primefactors(n)) for n in xrange(1, 10001) if ok(n)] # Indranil Ghosh, May 11 2017

CROSSREFS

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

Sequence in context: A324117 A106698 A097848 * A291598 A287820 A191373

Adjacent sequences:  A124827 A124828 A124829 * A124831 A124832 A124833

KEYWORD

nonn

AUTHOR

Franklin T. Adams-Watters, Nov 09 2006

STATUS

approved

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Last modified October 17 06:08 EDT 2019. Contains 328106 sequences. (Running on oeis4.)