A294647 Differential variant of the Kolakoski sequence, with initial terms 1, 1.

1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1

Offset

1,3

Comments

Conjecture: the density of 1's is equal to 1/2.

The differential Kolakoski sequence a(n) starting with 1, 1, 2, 2, 2, 1, 2, ... is an infinite sequence of numbers {1, 2} such that the sequence D(n) = |a(n+1) - a(n)| = 0, 1, 0, 0, 1, 1, ... of numbers {0, 1} generates, by taking run lengths of its terms, the initial sequence a(n).

To construct the sequence, start with the pair of digits (a(1), a(2)) = (1, 1) that generates D(1) = a(2) - a(1) = 0. The first run in D of length a(1) = 1 consists of just that single 0, so the next term in D should be 1. Therefore, require that the next run in D to consist only of a(3) - a(2) = 1, giving 01 as the initial part of D. The following run is a(4) - a(3), a(5) - a(4) = 00 with two identical digits, having length 2. So, the first three terms of the sequence are 1, 1, 2.

+------+-------------------------------------------------------------------

| a(n) | 1 1 2 2 2 1 2 2 2 1 1 1 2 1 1 1 2

+------+-------------------------------------------------------------------

| D(n) | 0 1 0 0 1 1 0 0 1 0 0 1 1 0 0 1

+------+-------------------------------------------------------------------

\-----/ \-----/ \-----/ \-----/ \-----/ \-----/

| a(n) | 1 1 2 2 2 1 2 2 2 1

-------+------------------------------------------------------------------

Links

Andrey Zabolotskiy, Table of n, a(n) for n = 1..20000

Wikipedia, Kolakoski sequence

Maple

nn:=100:a:=array(1..151):d:=array(1..151):
a[1]:=1:a[2]:=1:a[3]:=2:d[1]:=0:d[2]:=1:d[3]:=0:j:=2:
for n from 3 to nn-1 do:
  if a[n]= 2 and d[j]=1
   then
    j:=j+1:d[j]:=0:j:=j+1:d[j]:=0:
   else
  if a[n]= 2 and d[j]=0
   then
    j:=j+1:d[j]:=1:j:=j+1:d[j]:=1:
   else
  if a[n]=1 and d[j]=1
   then
    j:=j+1:d[j]:=0:
     else
  if a[n]=1 and d[j]=0
   then
    j:=j+1:d[j]:=1:
   else
   fi:fi:fi:fi:
   for k from 1 to j do:
   s:=a[k]+d[k]:
   if s=3 then
   a[k+1]:=1:
   else
   a[k+1]:=s:fi:
   od:od:
print(a):print(d):

Prog

(Python)
a = [1]
for n in range(100):
    a += [3-a[-1] if n%2 else a[-1], a[-1]][:a[n]]
print(a)
# Andrey Zabolotskiy, Nov 20 2017

Crossrefs

Cf. A000002.

Sequence in context: A139394 A240128 A125829 * A077099 A276337 A160385

Adjacent sequences: A294644 A294645 A294646 * A294648 A294649 A294650

Keyword

nonn

Author

Michel Lagneau, Nov 06 2017

Status

approved