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A092248 Parity of number of distinct primes dividing n (function omega(n)) parity of A001221.

%I A092248 #31 Jun 02 2017 00:39:41
%S A092248 0,1,1,1,1,0,1,1,1,0,1,0,1,0,0,1,1,0,1,0,0,0,1,0,1,0,1,0,1,1,1,1,0,0,
%T A092248 0,0,1,0,0,0,1,1,1,0,0,0,1,0,1,0,0,0,1,0,0,0,0,0,1,1,1,0,0,1,0,1,1,0,
%U A092248 0,1,1,0,1,0,0,0,0,1,1,0,1,0,1,1,0,0,0,0,1,1,0,0,0,0,0,0,1,0,0,0,1,1,1,0,1
%N A092248 Parity of number of distinct primes dividing n (function omega(n)) parity of A001221.
%C A092248 a(p^r) = 1 for all primes p and all exponents r>0. - _Tom Edgar_, Mar 22 2015
%H A092248 G. C. Greubel, <a href="/A092248/b092248.txt">Table of n, a(n) for n = 1..5000</a>
%H A092248 <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>
%F A092248 If omega(n) is even then a(n) = 0 else a(n) = 1. By convention, a(1) = 0. (Also because A001221(1) = 0 is an even number too).
%F A092248 a(n) = A000035(A001221(n)). - _Michel Marcus_, Mar 22 2015
%F A092248 a(n) = A268411(A156552(n)). - _Antti Karttunen_, May 30 2017
%e A092248 For n = 1, 0 primes divide 1 so a(1)=0.
%e A092248 For n = 2, there is 1 distinct prime dividing 2 (itself) so a(2)=1.
%e A092248 For n = 4 = 2^2, there is 1 distinct prime dividing 4 so a(4)=1.
%e A092248 For n = 5, there is 1 distinct prime dividing 5 (itself) so a(5)=1.
%e A092248 For n = 6 = 2*3, there are 2 distinct primes dividing 6 so a(6)=0.
%t A092248 Table[Boole[OddQ[PrimeNu[n]]], {n, 1, 100}] (* _Geoffrey Critzer_, Feb 16 2015 *)
%o A092248 (PARI) for (i=1,200,if(Mod(omega(i),2)==0,print1(0,","),print1(1,",")))
%o A092248 (Python)
%o A092248 from sympy import primefactors
%o A092248 def a(n): return 0 if n==1 else 1*(len(primefactors(n))%2==1) # _Indranil Ghosh_, Jun 01 2017
%Y A092248 Cf. A001221, A268411.
%K A092248 easy,nonn
%O A092248 1,1
%A A092248 Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Feb 19 2004
%E A092248 Offset corrected by _Reinhard Zumkeller_, Oct 03 2008