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A284257
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a(n) = number of prime factors of n that are < the square of smallest prime factor of n (counted with multiplicity), a(1) = 0.
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9
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0, 1, 1, 2, 1, 2, 1, 3, 2, 1, 1, 3, 1, 1, 2, 4, 1, 3, 1, 2, 2, 1, 1, 4, 2, 1, 3, 2, 1, 2, 1, 5, 1, 1, 2, 4, 1, 1, 1, 3, 1, 2, 1, 2, 3, 1, 1, 5, 2, 1, 1, 2, 1, 4, 2, 3, 1, 1, 1, 3, 1, 1, 3, 6, 2, 2, 1, 2, 1, 1, 1, 5, 1, 1, 3, 2, 2, 2, 1, 4, 4, 1, 1, 3, 2, 1, 1, 3, 1, 3, 2, 2, 1, 1, 2, 6, 1, 1, 2, 2, 1, 2, 1, 3, 3, 1, 1, 5, 1, 1, 1, 4, 1, 2, 2, 2, 2, 1, 2, 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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EXAMPLE
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For n = 45 = 3*3*5, all prime factors 3, 3 and 5 are less than 3^2, thus a(45) = 3.
For n = 120 = 2*2*2*3*5, the prime factors less than 2^2 are 2*2*2*3, thus a(120) = 4.
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MATHEMATICA
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Table[If[n == 1, 0, Count[#, d_ /; d < First[#]^2] &@ Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]], {n, 120}] (* Michael De Vlieger, Mar 24 2017 *)
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PROG
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(PARI) A(n) = if(n<2, return(1), my(f=factor(n)[, 1]); for(i=2, #f, if(f[i]>f[1]^2, return(f[i]))); return(1));
a(n) = if(A(n)==1, 1, A(n)*a(n/A(n)));
(Python)
from sympy import primefactors
def Omega(n): return 0 if n==1 else Omega(n//min(primefactors(n))) + 1
def A(n):
pf = primefactors(n)
if pf: min_pf2 = min(pf)**2
for i in pf:
if i > min_pf2: return i
return 1
def a(n): return 1 if A(n)==1 else A(n)*a(n//A(n))
print([Omega(n//a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 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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