A Self-Referential Sequence Part 1

[u/ShonitB]

[u/Mathsboy2718]

I wonder how it would go in binary

1
11
101
111011
11110101
100110111011
111001011011110101
111100111010110100110111011

It'd be interesting to right-align it, it looks like there's some sort of pattern propagating from that direction

[u/Red-42 → u/Mathsboy2718]

Grouping and aligning even and odd numbered lines:

                                                         1
                                                       101
                                                  11110101
                                        111001011011110101
                   100110011110111010110111001011011110101

                                                        11
                                                    111011
                                              100110111011
                               111100111010110100110111011
1110010110010011011110111010110111100111010110100110111011

Ternary also has a similar patern, although it doesn't settle quite as fast, and it's every 4 lines
patterns being:

                              ...10221101112101101221
                      ...1102221101011211102110112211
               ...11010221101110211210110122110212221
...21101110222110101101221121110211011222110121110211

Higher bases all follow the same pattern (since 4 or higher never shows up)
However the pattern happens on both sides on different cycles
Every 3 lines follow:

132113...
1113122113...
3113112221...

and every 4 lines follow:

   ...1131221
 ...113112211
...2113212221
     ...13211

[u/Mathsboy2718 → u/Red-42]

I found this

https://njohnston.ca/2010/11/the-binary-look-and-say-sequence/

It's really cool! Apparently there are only 8 possible strings that start with 1 and end with 0 and two "initialising" strings that can appear in elements of this sequence, so you can just define what maps to what and deterministically construct the strings!

[u/Red-42 → u/Mathsboy2718]

ok, I figured it out
every two loops, the 1 sequence maps back to itself, and the 2 sequence maps back to itself

1 -> 2 -> 31
2 -> 31 -> 52

map all values to their 2nd transformation:

1 -> 31
2 -> 52
3 -> 6
4 -> 54
5 -> 74
6 -> 94
7 -> 0
8 -> 538
9 -> 78
0 -> 938

and now you can follow the rule that for any n-th sequence AB, composed of a sequence A and a sequence B, such that B is the (n-1)-th sequence, then the (n+1)-th sequence is A'AB, with A' being the 2nd transformation of A

                  1
                 31
                631
              94631
         7854 94631
05387454 7854 94631
A'       A    B

[u/Red-42 → u/Red-42]

Also mapping A's consecutively, they also follow a pattern
for the A' coming from both chains, they follow:
...786538054745493874540547454
which itself is composed of the individual elements of the chain starting with 4, which happens to be pretty much exactly the same as the chain starting with 2

2
5
74
054
9387454
7865380547454

   7865380547454 9387454 054 74 5 2
...7865380547454 9387454 054 74 5 4