1, 11, 21, 1211, 111221, 312211… The terms of this sequence get longer and longer without end, and yet no matter how far out you go, you will never find a single digit other than 1, 2, or 3. Not one 4. Not one 9. Why not?
Check out the channel! @polymathematic
First, the rule. This is the look-and-say sequence, and each term simply describes the one before it, read out loud. Start with 1. What do you see? One 1, so the next term is 11. Now you see two 1s, so you write 21. Then one 2 and one 1: 1211. Then one 1, one 2, two 1s: 111221. And so on, forever.
Here’s the key observation. Every term after the first is built out of pairs: a count, then the digit being counted. The counts always land in the odd positions, so call them O. The digits being counted always land in the even positions, so call them E. Every term is O E, O E, O E, for as many pairs as it takes to describe the term before.
So where could a 4 possibly come from? The E digits are just copied from the previous term, so if you don’t start with a 4, a 4 can never show up as an E. The very first 4 would have to be a count, which means the term before it would need four of the same digit in a row. So the whole question comes down to this: can you ever get four of the same digit in a row?
You can’t, and the reason is surprisingly tidy. Two neighboring E’s can never be the same digit. Suppose one pair said “two 1s” and the very next pair said “one 1.” Then the term before would have had three 1s in a row, and nobody would read that as “two 1s, one 1.” They’d just say “three 1s.” We always count the whole run, so neighboring E’s always have to be different.
Now try to build a run of four. Because the positions alternate O, E, O, E, any four digits in a row must contain two neighboring E’s, and those two are guaranteed to be different. Four of a kind is impossible. Three in a row, though, is fine, as long as only one E is involved: an O E O stretch where the count, the digit, and the next count all happen to match. That’s exactly what happens in 111221, which begins “one 1, one 2.” Both counts are 1, and the digit being counted is 1 too. Read it aloud and those three 1s collapse into the first 3 the sequence ever produces: 312211.
So you can never count four of anything, which means a 4 can never be written, which means there’s never a 4 to copy. Every term, forever, is made of 1s, 2s, and 3s.
#lookandsay #numbersequences #mathpuzzles
Watch more Math Videos:
Math Minis:
Math Minutes:
Number Sense:
MATHCOUNTS:
Follow Tim Ricchuiti:
TikTok:
Mathstodon:
Instagram:
Reddit:
Facebook:
source
We could start with another seed, like X and see how it propagates. X 1X 111X 311X 13211X 111312211X and observe that the seed symbol only appears at the very end.
some kid said i shit on this sequence
I'm gonna do a sequence called binary. 2 never appears
What if you have 111 ones
fun facte!
there are related sequences for different counting systems
binary:
1
11
101
111011
11110101
100110111011
…
trinary:
1
11
21
1211
111221
1012211
1110112221
101102110211
111021101221101221
…
roman numerals:
I
II
III
IIII
IVI
IIIVII
IIIIIVIII
VIIVIIII
IVIIIIVIVI
IIIVIVIIVIIIVII
…
you can even create your own by assigning different sets of symbols to different numbers:
1=A
2=B
3=AA
4=C
5=AB
…
A
AA
BA
ABAA
AAABBA
AAABBAA
AAABBBA
AAAAABAA
ABAABBA
…
Utterly Useless 😢
I’m guessing 6 will be 312211
i think if four of a number are consecutive in the sequence we will get a 4
“Bijective base 4 but there is no 4”ahh video
The most interesting is 4 and upward numbers won't come in the sequence which comprise only three numbers 1, 2, and 3❤❤👌✌👍
How about change the starting number
This reminds me of Kolakoski sequence, which I have still no idea how it works.
Is this sequence ever gonna loop, or is every element of it fully unique? I suspect the second option, but cannot formaly prove it right now.
It is also imposible to get 333 anywhere in the sequence. That's because we have to write those 3 in oeo, which means that we had to had 3 threes in previous row, wchich is impossible if we assume that this is the first 333 sequence in the whole sequence.
1 2 1 1. Why is it not 3 1 1 2 ? I thought we were saying what we saw?
Me who likes not following rules: 0 4-9’s and 0 0’s
I haven’t seen this sequence since I was young, I’ve been wondering what it meant.
its like were using quaternary
The opening statement and the title of this isn't very impressive. There are infinite sequences that match that criteria. I could just take a sequence starting with 1, then append a 0, then start over including the previous sequence but add one more 0 each time. It fits the criteria.
Trinary based
Hello
This doesn't seem like it serves any purpose, or has any function whatsoever.
I just recently discovered this sequence as well! Apparently this has a growth rate of an algebraic irrational from a 71-degree polynomial, that’s crazy
I didn't understand anything but I watched the entire video.
Just testing something and using the comments to type
4
14
1114
3114
132114
1113122114
311311222114
13211321322114
Ok if we start with a 4 the sequence is different but it always ends in 14. I assume this is true for any single positive digit so for the numbers from 4-9 its always that sequence with 1n at the end. Wait is that also true for any negative digit !?!?
-3
1-3
111-3
311-3
HOLY COW !!!!!!
I feel very uncomfortable how you use the strings of digits as if they stand for actual decimal numbers. Just look and say, damnit!
Term 4 is "1211", and Term 5 is "111221". Is there a variant of the look-and-say sequence in which Term 5 would be "3112", where all of the same digit (in this case, 1) are accounted for at once? Terms would be constructed by giving the count of 1s, then the count of 2s, of 3s, etc.
Going through a few terms like this:
1
11
21
1112
3112
211213
312213
212223
114213
We reach a 4 at Term 9.
Are we discovering trinary…
wait hear me out I know it's existed for so long but when I see it I think about it in terms of computer science
Not me arguing that it has to be stated that a count has to interrupted when another integer is encountered 😅
See, the way I do it I do get a 4.
1
1104
211014
So on and so forth
This was the first math proof I ever did, when I was about three years old. Obviously I didn't use formal mathematics, but… my dad would always give me this puzzle by giving me the sequence of numbers and asking me what comes next. He would always stop it before the first 3, so until that point, you only see 1s and 2s and you think that maybe the sequence only has 1s and 2s if you don't see the pattern, so it's hard to see that there will be a 3 next. That got me inspired to try and see how long it would take for a 4 to show up. I kept going for a long time, never seeing one, and then began to ask myself if it will ever happen. And then I worked through the proof – completely on my own – and deduced that it was impossible. Years later, I thought about trying the sequence in binary, so "2" would be written as "10", "3" would be written as "11", "4" would be written as "100", and so on. And under this version of the sequence, it is easy to get 4 to show up (though it is written as "100", of course).
if we start with a 9, will we see numbers 1 through 9 eventually?
What's the sequence for though
I feel like you teach me math better than my math teacher
From base 4 upwards it will be the same results/sequence?
cool
resist teh urge to say first
also shut up johnny im first