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%I #42 Mar 11 2018 17:41:11
%S 2,3,2,5,1,7,2,3,7,11,1,13,5,6,2,17,1,19,7,4,9,23,1,5,15,3,5,29,4,31,
%T 2,8,19,2,1,37,17,14,7,41,6,43,9,6,21,47,1,7,7,18,15,53,1,14,5,16,31,
%U 59,4,61,29,4,2,8,10,67,19,20,0,71,1,73,39,6,17
%N Binary XOR of prime divisors of n.
%C The sequence of indices of zeros begins: 70, 140, 280, 350, 490, 560, 646, 700, 980, 1120, 1292, 1400, 1750, 1798, 1960, 2145.
%H Alois P. Heinz, <a href="/A293212/b293212.txt">Table of n, a(n) for n = 2..20000</a>
%F a(n)=n iff n is a prime.
%e a(6) = a(24) = 2 XOR 3 = 1.
%e a(2145) = 3 XOR 5 XOR 11 XOR 13 = 0.
%p a:= proc(n) local d, r; r:=0; for d in numtheory
%p [factorset](n) do r:= Bits[Xor](r, d) od; r
%p end:
%p seq(a(n), n=2..100); # _Alois P. Heinz_, Mar 09 2018
%o (PARI) a(n) = my(vp = factor(n)[,1]~, k=0); for (i=1, #vp, k = bitxor(k, vp[i])); k; \\ _Michel Marcus_, Feb 05 2018
%Y Cf. A000040, A072594, A178910.
%K nonn,base
%O 2,1
%A _Alex Ratushnyak_, Feb 04 2018