A330167 Length of the longest run of 1's in the ternary expression of n.

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

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

Table of n, a(n) for n=0..90.

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}]
Table[Max[Length/@Select[Split[IntegerDigits[n, 3]], MemberQ[#, 1]&]], {n, 0, 100}]/.(-\[Infinity]->0) (* Harvey P. Dale, Jan 07 2023 *)

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