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Multiplicative with a(3^k) = 0, a(p^(2k)) = 0 and a(p^(2k+1)) = 1 if p == 1 (mod 3), and a(p^(2k)) = 1 and a(p^(2k+1)) = 0, if p == -1 (mod 3).
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%I #17 Jan 13 2025 01:34:29

%S 1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,

%T 0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,0,0,1,0,

%U 0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,1

%N Multiplicative with a(3^k) = 0, a(p^(2k)) = 0 and a(p^(2k+1)) = 1 if p == 1 (mod 3), and a(p^(2k)) = 1 and a(p^(2k+1)) = 0, if p == -1 (mod 3).

%C This differs from A070563 for the first time at n=2401.

%H Antti Karttunen, <a href="/A374053/b374053.txt">Table of n, a(n) for n = 1..100000</a>

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

%t f[p_, e_] := If[Mod[p, 3] == 1, Mod[e, 2], 1 - Mod[e, 2]]; f[3, e_] := 0; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Jan 13 2025 *)

%o (PARI) A374053(n) = { my(f=factor(n)); prod(i=1, #f~, if(0==(f[i,1]%3), 0, if(1==(f[i, 1]%3), f[i, 2]%2, (1+f[i, 2])%2))); };

%Y Cf. A070563, A353815.

%K nonn,mult

%O 1

%A _Antti Karttunen_ after _Christian G. Bower_'s conjectured formula for A070563, Jul 03 2024