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

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, 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, 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

Offset: 1

Author

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

Keywords

• easy
• nonn

Comments

a(p^r) = 1 for all primes p and all exponents r>0. - Tom Edgar, Mar 22 2015

Examples

			For n = 1, 0 primes divide 1 so a(1)=0.
For n = 2, there is 1 distinct prime dividing 2 (itself) so a(2)=1.
For n = 4 = 2^2, there is 1 distinct prime dividing 4 so a(4)=1.
For n = 5, there is 1 distinct prime dividing 5 (itself) so a(5)=1.
For n = 6 = 2*3, there are 2 distinct primes dividing 6 so a(6)=0.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..5000
• Index entries for characteristic functions

Crossrefs

Cf. A001221, A268411.

Programs

Mathematica

Table[Boole[OddQ[PrimeNu[n]]], {n, 1, 100}] (* Geoffrey Critzer, Feb 16 2015 *)

PARI

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

Python

from sympy import primefactors
def a(n): return 0 if n==1 else 1*(len(primefactors(n))%2==1) # Indranil Ghosh, Jun 01 2017

Formula

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).

a(n) = A000035(A001221(n)). - Michel Marcus, Mar 22 2015

a(n) = A268411(A156552(n)). - Antti Karttunen, May 30 2017

Extensions

Offset corrected by Reinhard Zumkeller, Oct 03 2008