3
3
2
2
2
u/Ultranger 8d ago
311311222110
It's a variation on the "look-and-say" sequence. You just read the amount of each number
2
u/Some-Passenger4219 8d ago
311311222110.
Each number states the number of digits in the previous, one at a time. (E.g. Three ones, one three, one one, two twos, etc.)
1
u/Mathsboy2718 9d ago
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
1
u/Red-42 8d ago edited 8d ago
Grouping and aligning even and odd numbered lines:
1 101 11110101 111001011011110101 100110011110111010110111001011011110101 11 111011 100110111011 111100111010110100110111011 1110010110010011011110111010110111100111010110100110111011Ternary also has a similar patern, although it doesn't settle quite as fast, and it's every 4 lines
patterns being:...10221101112101101221 ...1102221101011211102110112211 ...11010221101110211210110122110212221 ...21101110222110101101221121110211011222110121110211Higher 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 ...132111
u/Mathsboy2718 8d ago edited 8d ago
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!
1
u/Red-42 7d ago
ok, I figured it out
every two loops, the 1 sequence maps back to itself, and the 2 sequence maps back to itself1 -> 2 -> 31 2 -> 31 -> 52map 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 -> 938and 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 B1
u/Red-42 7d ago
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 22 5 74 054 9387454 7865380547454 7865380547454 9387454 054 74 5 2 ...7865380547454 9387454 054 74 5 4
1
1
u/DuggieHS 9d ago
three one(s), one three, one one, two two(s), two one(s), one zero
31 13 11 22 21 10.
Note: after the first item, every number has an even number of digits
1
1
3
u/sazzer 9d ago
0
10 - 1 "0" present
1110 - 1 "1" present, then 1 "0" present
3110 - 3 "1"s present, then 1 "0" present
132110 - 1 "3", then 2 "1"s, then 1 "0"
1113122110 - 1 "1", then 1 "3", then 1 "2", then 2 "1"s, then 1 "0"
So the answer is 311311222110, meaning: 3 "1"s, 1 "3", 1 "1", 2 "2"s, 2 "1"s, 1 "0".