A306556 Integers that appear as (unreduced) numerators of segment endpoints when a ternary Cantor set is created.

0, 1, 2, 3, 6, 7, 8, 9, 18, 19, 20, 21, 24, 25, 26, 27, 54, 55, 56, 57, 60, 61, 62, 63, 72, 73, 74, 75, 78, 79, 80, 81, 162, 163, 164, 165, 168, 169, 170, 171, 180, 181, 182, 183, 186, 187, 188, 189, 216, 217, 218, 219, 222, 223, 224, 225, 234, 235, 236, 237, 240, 241, 242, 243

Offset: 1

Author

Dan Dima, Feb 23 2019

Keywords

• nonn
• frac
• easy

Comments

Nonnegative integers whose ternary representation contains only digits 0 and 2 except for at most a single digit 1 that is followed only by 0's.

Nonnegative integers that can be written in base 3 using only 0's and 2's, allowing the use of the "decimal" point (.) and replacing ....10..0(.) by ....02..2(.)2222...

Note that fractions are not reduced.

List of integers in the closure of the ternary Cantor set under multiplication by 3. The closure is the union of the translated ternary Cantor sets spanning [a(1), a(2)], [a(3), a(4)], [a(5), a(6)], ... . - Peter Munn, Jul 09 2019

Examples

			On 1st step we have [0,1/3] U [2/3,3/3] so we get a(1)=0, a(2)=1, a(3)=2, a(4)=3.
On 2nd step we have [0,1/9] U [2/9,3/9] U [6/9,7/9] U [8/9,9/9] so we get in addition a(5)=6, a(6)=7, a(7)=8, a(8)=9.
		

Links

• Georg Cantor, Über unendliche, lineare Punktmannigfaltigkeiten V" [On infinite, linear point-manifolds (sets), Part 5]. Mathematische Annalen (in German). (1883) 21: 545-591.
• Paul du Bois-Reymond, Der Beweis des Fundamentalsatzes der Integralrechnung, Mathematische Annalen (in German), (1880), 16, footnote on p. 128.
• Eric Weisstein's World of Mathematics, Cantor Set
• Wikipedia, Cantor set
• Index entries for 3-automatic sequences.

Crossrefs

Cf. A005823, A054591, A055247, A191106, A191108.

Programs

PARI

A306556(n) = {sm=0;while(n>1,ex=floor(log(n)/log(2));if(n-2^ex==0,sm=sm+3^(ex-1),sm=sm+2*3^(ex-1));n=n-2^ex);return(sm)}

PARI

a(n) = n--; fromdigits(binary(n>>1),3)*2 + (n%2); \\ Kevin Ryde, Apr 23 2021

Formula

a(1)=0, a(2)=1;

a(2^n) = 3^(n-1) for n >= 1;

a(2^n+k) = 2*3^(n-1) + a(k) for 1 <= k <= 2^n.

From Peter Munn, Jul 09 2019: (Start)

a(2n-1) = A005823(n) = A191106(n)-1.

a(2n) = A191106(n) = A005823(n)+1.

a(2n-1) = (A055247(2n-1)-1)/3.

a(2n) = (A055247(2n) +1)/3.

a(2n-1) = (A191108(n)-1)/2.

a(2n) = (A191108(n)+1)/2.

(End)