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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
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
Sequence in context: A106698 A097848 A308964 * A291598 A287820 A191373
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 19 09:40 EDT 2024. Contains 370981 sequences. (Running on oeis4.)