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Parity of number of distinct primes dividing n (function omega(n)) parity of A001221.
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%I #31 Jun 02 2017 00:39:41

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

%C a(p^r) = 1 for all primes p and all exponents r>0. - _Tom Edgar_, Mar 22 2015

%H G. C. Greubel, <a href="/A092248/b092248.txt">Table of n, a(n) for n = 1..5000</a>

%H <a href="/index/Ch#char_fns">Index entries for characteristic functions</a>

%F 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 a(n) = A000035(A001221(n)). - _Michel Marcus_, Mar 22 2015

%F a(n) = A268411(A156552(n)). - _Antti Karttunen_, May 30 2017

%e For n = 1, 0 primes divide 1 so a(1)=0.

%e For n = 2, there is 1 distinct prime dividing 2 (itself) so a(2)=1.

%e For n = 4 = 2^2, there is 1 distinct prime dividing 4 so a(4)=1.

%e For n = 5, there is 1 distinct prime dividing 5 (itself) so a(5)=1.

%e For n = 6 = 2*3, there are 2 distinct primes dividing 6 so a(6)=0.

%t Table[Boole[OddQ[PrimeNu[n]]], {n, 1, 100}] (* _Geoffrey Critzer_, Feb 16 2015 *)

%o (PARI) for (i=1,200,if(Mod(omega(i),2)==0,print1(0,","),print1(1,",")))

%o (Python)

%o from sympy import primefactors

%o def a(n): return 0 if n==1 else 1*(len(primefactors(n))%2==1) # _Indranil Ghosh_, Jun 01 2017

%Y Cf. A001221, A268411.

%K easy,nonn

%O 1,1

%A Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Feb 19 2004

%E Offset corrected by _Reinhard Zumkeller_, Oct 03 2008