A284258 a(n) = number of distinct prime factors of n that are > the square of smallest prime factor of n, a(1) = 0.
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 2, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 2, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1
Offset: 1
Keywords
Examples
For n = 50, 2*5*5, the prime factor > 2^2 is 5, which is counted only once, thus a(50) = 1. For n = 70, 2*5*7, the prime factors > 2^2 are 5 and 7, thus a(70) = 2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10001
Crossrefs
Programs
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Mathematica
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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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))); for(n=1, 150, print1(omega(a(n)),", ")) \\ Indranil Ghosh, after David A. Corneth, Mar 24 2017
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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(a(n)) for n in range(1, 151)]) # Indranil Ghosh, Mar 24 2017
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Scheme
(define (A284258 n) (A001221 (A284254 n)))