A060582 If the final digit of n in base 3 is the same as a([n/3]) then this digit, otherwise a(n)= mod 3-sum of these two digits, with a(0)=0.

0, 2, 1, 1, 0, 2, 2, 1, 0, 2, 1, 0, 0, 2, 1, 1, 0, 2, 1, 0, 2, 2, 1, 0, 0, 2, 1, 1, 0, 2, 2, 1, 0, 0, 2, 1, 0, 2, 1, 1, 0, 2, 2, 1, 0, 2, 1, 0, 0, 2, 1, 1, 0, 2, 2, 1, 0, 0, 2, 1, 1, 0, 2, 1, 0, 2, 2, 1, 0, 0, 2, 1, 0, 2, 1, 1, 0, 2, 2, 1, 0, 2, 1, 0, 0, 2, 1, 1, 0, 2, 1, 0, 2, 2, 1, 0, 0, 2, 1, 0, 2, 1, 1, 0, 2

Offset: 0

Author

Henry Bottomley, Apr 04 2001

Keywords

• base
• nonn

Examples

			a(4)=0 since a([4/3])=a(1)=2, (4 mod 3)=1 and 3-2-1=0. a(5)=2 since a([5/3])=a(1)=2 and (5 mod 3)=2.
		

Links

• Index entries for sequences related to final digits of numbers

Crossrefs

Cf. A060588.

Formula

a(n) =(-a([n/3])-n) mod 3 =A060583(n) mod 3.