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%I #17 Jan 24 2020 14:11:21
%S 0,0,1,0,0,1,1,1,2,0,0,1,0,0,1,1,1,2,1,1,1,1,1,1,2,2,3,0,0,1,0,0,1,1,
%T 1,2,0,0,1,0,0,1,1,1,2,1,1,1,1,1,1,2,2,3,1,1,1,1,1,1,1,1,2,1,1,1,1,1,
%U 1,1,1,2,2,2,2,2,2,2,3,3,4,0,0,1,0,0,1,1,1,2,0
%N Length of the longest run of 2's in the ternary expression of n.
%C All numbers appear in this sequence. Numbers of the form 3^n-1 (A024023(n)) have n 2's in their ternary expression.
%C The longest run of zeros possible in this sequence is 2, as the last digit of the ternary expression of the integers cycles between 0, 1, and 2, meaning that at least one of three consecutive numbers has a 2 in its ternary expression.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ternary_numeral_system">Ternary numeral system</a>.
%F a(A024023(n)) = a(3^n-1) = n.
%F a(n) = 0 iff n is in A005836.
%e For n = 74, the ternary expression of 74 is 2202. The length of the runs of 2's in the ternary expression of 74 are 2 and 1, respectively. The larger of these two values is 2, so a(74) = 2.
%e n [ternary n] a(n)
%e 0 [ 0] 0
%e 1 [ 1] 0
%e 2 [ 2] 1
%e 3 [ 1 0] 0
%e 4 [ 1 1] 0
%e 5 [ 1 2] 1
%e 6 [ 2 0] 1
%e 7 [ 2 1] 1
%e 8 [ 2 2] 2
%e 9 [ 1 0 0] 0
%e 10 [ 1 0 1] 0
%e 11 [ 1 0 2] 1
%e 12 [ 1 1 0] 0
%e 13 [ 1 1 1] 0
%e 14 [ 1 1 2] 1
%e 15 [ 1 2 0] 1
%e 16 [ 1 2 1] 1
%e 17 [ 1 2 2] 2
%e 18 [ 2 0 0] 1
%e 19 [ 2 0 1] 1
%e 20 [ 2 0 2] 1
%t Table[Max@FoldList[If[#2==2,#1+1,0]&,0,IntegerDigits[n,3]],{n,0,90}]
%Y Cf. A007089, A024023, A081603, A330036, A330166, A330167.
%Y Equals zero iff n is in A005836.
%K nonn,base
%O 0,9
%A _Joshua Oliver_, Dec 04 2019