The Kolakoski Sequence - Numberphile
video description
since the quantity of /those/ quantities (which we are adding and subtracting) is the same sequence, we can drop all 2's since it will be either 2-2 or 1-1 or -2+2 or -1+1, leaving us with a sequence of infinitely many 1's in a row, describing either a single 2 or a single 1.
since this is a sequence of infinitely many 1's in a row, considering how many 2's were skipped is important.
we get the sequence (describing the quantity of 2's or 1's in chain before a different number, starting with 1's):
1 1 1 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 which translates to the signs, 1 means you flip the sign, 0 means you don't for the alternating 1 2 1 2 1 2. sequence. this can evolve further into a proof, I just don't yet know how
Date: 2022-04-08
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Comments and reviews: 9
DarkStorn
It seem weird to me that -no-one knows if there is more 1s than 2s-
When you try to generate this sequence, you quickly realize there is only one (that uses 1, 2 and starts with 1)
Start with 1, which means there is one 1, so the next is 2, which means there are two of them, then you put 1, and since 3rd number is 2 there is two 1s, then it is one 2, one 1, two 2, .
So at any point, you put (more or less) randomly 1 or 2 and (with the same randomness) you put one or two of them. So yes, it can be hard to predict how many of each number has been used at any point, but the infinite series must go to 50/50 by rules of probability.
(Maybe that's what he meant? That there is no way to tell which number is more common at given point, other than generating the sequence to that point)
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It seem weird to me that -no-one knows if there is more 1s than 2s-
When you try to generate this sequence, you quickly realize there is only one (that uses 1, 2 and starts with 1)
Start with 1, which means there is one 1, so the next is 2, which means there are two of them, then you put 1, and since 3rd number is 2 there is two 1s, then it is one 2, one 1, two 2, .
So at any point, you put (more or less) randomly 1 or 2 and (with the same randomness) you put one or two of them. So yes, it can be hard to predict how many of each number has been used at any point, but the infinite series must go to 50/50 by rules of probability.
(Maybe that's what he meant? That there is no way to tell which number is more common at given point, other than generating the sequence to that point)
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Farhan
Some people are wondering how it's generated, and since I just realised, I'll explain. You start with 1. What does this tell you? It means the first chain of numbers must be 1 long, in other words, it's already over, and the next number needs to be different. Let's go with 2.
Now we have 1, 2.
What does that tell us? The second chain must have length 2. Since it currently has length 1, we need another 2.
1, 2, 2.
The second chain must now be over, and since we added a 2, the third chain must also be 2 long, and must consist of 1s.
1, 2, 2, 1, 1.
Things start to spiral out of control, as we end up with far more information than necessary at any given step.
This line of reasoning could probably be used to show that the sequence needn't ever terminate.
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Some people are wondering how it's generated, and since I just realised, I'll explain. You start with 1. What does this tell you? It means the first chain of numbers must be 1 long, in other words, it's already over, and the next number needs to be different. Let's go with 2.
Now we have 1, 2.
What does that tell us? The second chain must have length 2. Since it currently has length 1, we need another 2.
1, 2, 2.
The second chain must now be over, and since we added a 2, the third chain must also be 2 long, and must consist of 1s.
1, 2, 2, 1, 1.
Things start to spiral out of control, as we end up with far more information than necessary at any given step.
This line of reasoning could probably be used to show that the sequence needn't ever terminate.
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Joey
I don't remember exactly how I had done it but I coincidentally discovered this sequence myself about a year ago while playing around with the Fibonaccis and primes. I was exploring number theory as a way to find rhythmically pleasing sequences and found this to be one that worked well for drum patterns if I remember correctly. I wish I could recall exactly what steps I took but yea- it's amazing what you can find from just fiddling about!
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I don't remember exactly how I had done it but I coincidentally discovered this sequence myself about a year ago while playing around with the Fibonaccis and primes. I was exploring number theory as a way to find rhythmically pleasing sequences and found this to be one that worked well for drum patterns if I remember correctly. I wish I could recall exactly what steps I took but yea- it's amazing what you can find from just fiddling about!
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joseph
When something is this obvious I submit that there is a difference between _knowing_ something and bring able to -prove- something.
I'm really not impressed with the idea that dropping the -one- from the beginning defines a separate sequence. This video didn't have enough meat to stand alone. I'm sure there are a dozen equally trivial sequences or patterns that could have stirred up some more interesr
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When something is this obvious I submit that there is a difference between _knowing_ something and bring able to -prove- something.
I'm really not impressed with the idea that dropping the -one- from the beginning defines a separate sequence. This video didn't have enough meat to stand alone. I'm sure there are a dozen equally trivial sequences or patterns that could have stirred up some more interesr
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Nathan
I wonder if there are names for the processes of intending to generate a sequence from the first sequence, to list our the second sequence side by side with the first sequence, to find out the similarities of the first and second sequences, and the discovery that the second sequence repeats the first sequence. I wonder if there are any proper terminology to label these processes.
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I wonder if there are names for the processes of intending to generate a sequence from the first sequence, to list our the second sequence side by side with the first sequence, to find out the similarities of the first and second sequences, and the discovery that the second sequence repeats the first sequence. I wonder if there are any proper terminology to label these processes.
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Isaac
Kolakoski sequence with 3s:
1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 2, 3, 1, 1, 2, 2, 2.
3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 2, 3, 1, 1, 2, 2, 2,
3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1 3, 3, 1, 2, 3,
3, 3, 3, 2, 2, 2, 1, 1, 1,
3, 3, 3,
3,
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Kolakoski sequence with 3s:
1, 3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 2, 3, 1, 1, 2, 2, 2.
3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1, 3, 3, 1, 2, 3, 1, 1, 2, 2, 1, 2, 3, 3, 3, 2, 2, 2, 3, 1, 1, 2, 2, 2,
3, 3, 3, 2, 2, 2, 1, 1, 1, 2, 2, 1, 1 3, 3, 1, 2, 3,
3, 3, 3, 2, 2, 2, 1, 1, 1,
3, 3, 3,
3,
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Stephen
The Wikipedia article states that there are an infinite number of these sequences and gives the -1, 3-, -2, 3-, -1, 2, 3-, -3, 1, 2-, and -2, 1, 3, 1- sequences as examples. Fascinating article. there are sets that generate their brothers, that generate them. It is possible to generate them from infinite integer sets, probably trivial if you think about it.
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The Wikipedia article states that there are an infinite number of these sequences and gives the -1, 3-, -2, 3-, -1, 2, 3-, -3, 1, 2-, and -2, 1, 3, 1- sequences as examples. Fascinating article. there are sets that generate their brothers, that generate them. It is possible to generate them from infinite integer sets, probably trivial if you think about it.
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Uny
The answer to the question of are there more 1s or 2s in the sequence is that they are both the same amount, which is infinite. My proof being that if one ends the sequence finitely and does the whole n=1 2n=2 thing for a while eventually it will just become a series of 1s, which will just repeat forever, therefore the sequence must be infinite.
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The answer to the question of are there more 1s or 2s in the sequence is that they are both the same amount, which is infinite. My proof being that if one ends the sequence finitely and does the whole n=1 2n=2 thing for a while eventually it will just become a series of 1s, which will just repeat forever, therefore the sequence must be infinite.
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KedarOthort
-Are there more 1's or 2's? - Neither. The cycle has 3 of each then starts over, so even if you repeat the cycle ad nauseum, you'd always have the same multiple of 3 for both numbers, so there's not more of one than the other. Only way to change that would be to stop mid-sequence.
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-Are there more 1's or 2's? - Neither. The cycle has 3 of each then starts over, so even if you repeat the cycle ad nauseum, you'd always have the same multiple of 3 for both numbers, so there's not more of one than the other. Only way to change that would be to stop mid-sequence.
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