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A284259
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a(n) = number of distinct prime factors of n that are < the square of smallest prime factor of n, a(1) = 0.
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10
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0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2
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OFFSET
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1,6
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LINKS
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FORMULA
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a(n) = Sum_{p|n, p prime and < spf(n)^2} sign(p), where spf(n) (A020639) gives the smallest prime factor of n, and sign(p) = A057427(p) = 1 for all p.
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EXAMPLE
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For n = 4 = 2*2, the prime factor 2 is less than 2^2, and it is counted only once, thus a(4) = 1.
For n = 45 = 3*3*5, both prime factors 3 and 5 are less than 3^2, thus a(45) = 2.
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MATHEMATICA
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Table[If[n == 1, 0, Count[#, d_ /; d < First[#]^2] &@ FactorInteger[n][[All, 1]]], {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 len(primefactors(n))
def A(n):
for i in primefactors(n):
if i>min(primefactors(n))**2: 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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Cf. A284262 (where obtains first time value n, also positions of records).
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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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