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

Author

Joshua Oliver, Dec 04 2019

Keywords

• nonn
• base

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.

Examples

			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
		

Links

• Wikipedia, Ternary numeral system.

Crossrefs

Cf. A003462, A007089, A062756, A330036, A330166, A330168.

Equals zero iff n is in A005823.

Programs

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 *)

Formula

a(A003462(n)) = a((3^n-1)/2) = n.

a(n) = 0 iff n is in A005823.