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 A330167 Length of the longest run of 1's in the ternary expression of n. 3
 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 3, 2, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 3, 4, 3, 2, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 2, 3, 2, 1, 1, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS All numbers appear in this sequence. Numbers of the form (3^n-1)/2 (A003462(n)) have n 1's in their ternary expression. 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 1 in its ternary expression. LINKS Wikipedia, Ternary numeral system. FORMULA a(A003462(n)) = a((3^n-1)/2) = n. a(n) = 0 iff n is in A005823. EXAMPLE For n = 43, the ternary expression of 43 is 1121. The length of the runs of 1's in the ternary expression of 43 are 2 and 1, respectively. The larger of these two values is 2, so a(43) = 2.    n [ternary n] a(n)    0 [        0] 0    1 [        1] 1    2 [        2] 0    3 [      1 0] 1    4 [      1 1] 2    5 [      1 2] 1    6 [      2 0] 0    7 [      2 1] 1    8 [      2 2] 0    9 [    1 0 0] 1   10 [    1 0 1] 1   11 [    1 0 2] 1   12 [    1 1 0] 2   13 [    1 1 1] 3   14 [    1 1 2] 2   15 [    1 2 0] 1   16 [    1 2 1] 1   17 [    1 2 2] 1   18 [    2 0 0] 0   19 [    2 0 1] 1   20 [    2 0 2] 0 MATHEMATICA Table[Max@FoldList[If[#2==1, #1+1, 0]&, 0, IntegerDigits[n, 3]], {n, 0, 90}] CROSSREFS Cf. A003462, A007089, A062756, A330036, A330166, A330168. Equals zero iff n is in A005823. Sequence in context: A098055 A344739 A092111 * A307776 A341027 A050317 Adjacent sequences:  A330164 A330165 A330166 * A330168 A330169 A330170 KEYWORD nonn,base AUTHOR Joshua Oliver, Dec 04 2019 STATUS approved

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Last modified June 19 00:04 EDT 2021. Contains 345125 sequences. (Running on oeis4.)