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A124830
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Number of distinct prime factors of A055932(n).
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3
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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
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OFFSET
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1,4
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LINKS
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FORMULA
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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