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%I #29 Mar 19 2024 17:22:58
%S 0,1,2,0,1,2,0,2,1,0,0,0,0,1,2,0,2,1,0,0,0,0,2,1,0,1,2,0,0,0,0,0,0,0,
%T 0,0,0,0,0,0,1,2,0,2,1,0,0,0,0,2,1,0,1,2,0,0,0,0,0,0,0,0,0,0,0,0,0,2,
%U 1,0,1,2,0,0,0,0,1,2,0,2,1,0,0,0,0,0,0,0,0,0,0,0
%N Product of digits of (n written in base 3) mod 3.
%C a(A032924(n)) = 1 or 2. For n >= 1, a(A032924(n)) - 1 = A309953(A032924(n)) mod 3 - 1 = A010059(n+1).
%F a(n) = A309953(n) mod 3.
%F a(A081605(n)) = 0.
%e n = 5: 5_10 = 12_3 thus a(5) = 1*2 mod 3 = 2.
%e n = 8: 8_10 = 22_3 thus a(8) = 2*2 mod 3 = 1.
%t a[n_] := Mod[Times @@ IntegerDigits[n, 3], 3]; Array[a, 100, 0] (* _Amiram Eldar_, Mar 18 2024 *)
%o (Python)
%o from functools import reduce
%o from sympy.ntheory import digits
%o def A371222(n): return reduce(lambda a,b: a*b%3,digits(n,3)[1:],1) # _Chai Wah Wu_, Mar 19 2024
%Y Cf. A007089, A010059, A032924, A081605, A309953, A371281 (base 10).
%K nonn,base
%O 0,3
%A _Ctibor O. Zizka_, Mar 18 2024