This sequence is not the only self-describing sequence. It's the blueprint for a whole family of self-describing sequences. {1,3} 1,3,3,3,1,1,1,3,3,3,1,3,1,3,3,3,1,1,1,3,3,3,1,3,1... {1,2,3} 1,2,2,3,3,1,1,1,2,2,2,3,1,2,3,3,1,1,2,2,3,3,3,1,2... {1,2,1,3,1,4...} 1,2,2,1,1,3,1,4,4,4,1,5,5,5,5,1,1,1,1,6,6,6,6,1,7,7,7,7,7...
The sequence is special because the nth term describes the length of the nth run (it looks like both start counting from one). In your first example of 1s and 3s the 12th term is a 3, but the 12th run is is a single 3 (so of length 1), and so it doesn't have the special property that the one in this video has. I didn't check the others, but my guess is you'd be able to find a term that doesn't work.
@@wantsomechocolate OP merely made an error in transcribing the {1,3} sequence, at the point you identified, where indeed the 12th term is 3 and the 12th run (beginning at the 24th term) *_should be_* three 3s. So obviously it should be 1,3,3,3,1,1,1,3,3,3,1,3,1,3,3,3,1,1,1,3,3,3,1,3,3,3,1... OP did not make any transcription errors in the other examples they gave. You should have thought harder before making your claims. No, the property that "the nth term describes the nth run" is *_not unique_* to the sequence based on {1,2}. The exact same process can be generalized, as OP alludes, to construct a self describing sequence with that property on the basis of _any_ "seed" sequence of positive integers where adjacent seed terms are always different. {3,1,4} 3,3,3,1,1,1,4,4,4,3,1,4,3,3,3,3,1,1,1,1,4,4,4,4,3,3,3,1,4,4,4,4,3,3,3,1,1,1,4,4,4,3,3,3,1,4,3,1,... Now, *_among sequences with this property,_* the sequence based on {1,2} is arguably "simplest" by some complexity metric or "earliest" by a lexicographic metric, but it's not totally unique.
The Vurp Sequence: 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11... The number that each number repeats in the sequence is that number in the sequence.
The Kolakoski sequence was named after William Kolakoski (1944-97), who discussed it in 1965, but subsequent research has revealed that it first appeared in a paper by Rufus Oldenburger in 1939.
Yeah, and Rufus stole it from my grandpa, who created the sequence in 1938. It’s no more Oldenburger’s sequence than it is Kolakoski’s. It’s Peepaw’s sequence, and that’s all there is to it. You need to put some respeck on his name.
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!
There is another self-describing sequence called Golomb's sequence. It contains all positive integers instead of just alternating between runs of 1 and runs of 2. The sequence is 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, ...
Your question "are there more 1's or 2's?" reminds me of another question. Are there more primes that are sums of squares or more primes that are not sums of squares. Primes 2, 5, 13, 17, ... are sums of squares. Primes 3, 7, 11, 19, ... are not sums of squares. Which sequence of primes dominates?
1:20 This sequence isn't the only sequence with the property of self-describing consecutive similar terms, though it (with and without the first 1) is the only one consisting of only 1s and 2s. Take any positive integer and repeat it itself number of times, then take another number and repeat it the second-term number of times, and so on.
I got curious about the name Kolakoski now. It sounds very much like a Finnish name. "Kola" means a snow plough in modern Finnish (or cola as in Coke or Pepsi) but it has meant a tool for lifting or shifting things in general, and "koski" means river rapids. However, the Population Register Centre name statistics page doesn't find any occurrences of it. That means there has been no Finnish citizen by that name alive since the mid-1960s, so presumably there has been no one born with that name since the early 1900s. On the other hand, there seem to be a number of people with the name in the USA, mostly in the Pennsylvania-Minnesota area, which is where most of the Finnish emigrants in the late 19th and early 20th century settled. This is also where William Kolakoski came from. I can find one single place in Finland where the name Kolakoski occurs. In Eura, there is a bay in the lake Pyhäjärvi called Kolakoskenlahti ("the Kolakoski bay"). There is a small stream that runs into the bay, which I suspect might be locally called Kolakoski, but it's too small to have a name on the map. A nearby road is also called Kolakoskentie (the Kolakoski road). This area seems to be called Kolakoski in some official papers of the city of Eura. But somehow Kolakoski doesn't strike me as a particularly western name either. I also found a reference to some rapids named Kolakoski in the river Luiro in Lapland. But this information was in a badly scanned and OCR'd copy of the book Lappmarken by the Jacob Fellmann (published 1906). This turned out to be an incorrectly scanned Kotakoski, which is a relatively common name, it seems. There are plenty of those in Finland, and I found another OCR mistake like that as well. Maybe the whole name Kolakoski in the USA is derived from a misspelling of Kotakoski at some point in the past? :)
This is cool because it is self-producing. If you need to know the next number in the set, you can figure it out according to how much you know so far. You can sit and produce this yourself for as long as you like and make it more and more fractally.
if you take the sequence and do all odd terms minus all even terms, this tells you how many ones there are more than twos, (for the one that starts with 1) since this describes (for odd terms) the quantity of ones, and (for even) the quantity of twos. 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
1:25 To clarify, those are the only two sequences IF you only use 1s and 2s. You can make this sort of sequence with any numbers you please. For example, 2 and 3 give you 223322233322332233322233322332223332233222333222332233222333222 et cetera. 1/2/3 gives you 12233111222312331122333122333111231122 and so on. Or using ascending integers gets you the Golomb sequence, 1223344455566667777888899999 et cetera.
He says this and the sequence without the leading 1 are the only such sequences, but I think this only applies if you're talking about sequences of only 1 and 2. I messed around a bit with a sequence that also allows 3's in the mix and it works in the same way. Pretty cool, it took me a couple minutes longer than I'd like to admit to understand how to keep generating the sequence further hahaha
Freshman year my calc professor asked us to identify how this sequence was being generated and I figured it out. I never knew what the name of it was though!
This is the song that doesn't end. Yes, it goes on and on my friend! Some people started singing it not knowing what it was, And they'll continue singing it forever just because This is the song that doesn't end...
For anyone curious, this will generate a "kolakoski" sequence of length n with x integers (1 through x). def kolakoski(x, n): seq = [1] next_number = 2 for i in range(n): temp = seq[i]*[next_number] seq += temp if next_number == x: next_number = 1 else: next_number += 1 seq = [seq[0]] + [2] + seq[1:] return seq Edit: Python language, fixed to splice in the [2] that needed manual fixing (yes... it's lazy coding... I'm an analyst not a programmer ;)
I'm guessing this video only addresses sequences containing {1,2}. The Wikipedia page on Kolakoski sequences show several other examples with other digits, including yours.
The Kolakoski sequence's definition say that it can only consist 1 and 2, but there are a lots of self-generating sequences that have "Kolakoski sequence's property", your sequence is one of them.
It actually does have the same properties. I think the speaker just made a mistake when he said that it is the ONLY sequence that displays these properties (and the sequence without the first 1). There are actually an INFINITE number of sequences like this, like the one you commented. It's just that the Kolakoski sequence is the ONLY sequence that only uses the numbers 1 and 2.
I think if you look, all of his are sequential (i.e. a 2 always follows a 1 and a 3 always follows a 2). So the encoding still works assuming you apply that extra rule (which is also applicable, but trivial, for Kolakoski)
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.
What do you mean it's the only one? The Wikipedia page on this sequence has loads, and states that there are infinitely many, even chains of sequences which describe each other.
And even if the script said that it's only of sequences composed of 1's and 2's, the fact that there are only two such sequences isn't interesting at all. You pick a starting symbol, that determines run length, and there's no choice of transition symbols. Very poor script compared to the usual accurate and accessible content.
For the letter density question (help welcome): There seems to be only 4 possibilities to generate the next sequence. If you're 'at 1' and 'see 1' write down '2'. @1 see2 write22. @2 see1 write1. @2 see2 write 11. This generating algorithm promises to make the ratio of 1s to 2s asymptotic to 0.5. Seems you can start with anything, it'll move to 0.5 eventually.
A very fine continuation of the visualisation of math aspect last shown on the collatz conjecture. This video should have been sponsored by circlespace, though.
in a classic case of our brains seeing patterns they are trained on, if you move through the plot interpreting it as binary values, substituting 0 for 2 and rotating it 90 degrees, you get the (integer) values 0,0,1,2,3,4,4,7,8,8,9,14,33,32,32,35,34,57,66,67,64,64....
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.
+Numberphile, You assert (1:18) that there are only 2 sequences with the property of listing its own runs: Kolakoski and Kolakoski-sans-1. But here's another that does: 2 2 3 3 2 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 2 3 3 3... And another that does: 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 2 3 3 3 2 2... You can just start with any pair of positive numbers and form a sequence with this property, so there are infinitely many of them.
Here's a procedure to generate a sequence with the property from any pair of positive integers: A = your input # first positive integer. B = your input # second positive integer, A != B AB = A define a function 'switch' such that{ switch(A) = B switch(B) = A } term count = 0 run count = 0 sequence(0) = A loop forever{ L = sequence(run count) # get the run length from the sequence # write into sequence a run length L of whichever A or B is currently stored in AB repeat L times{ sequence(term count) = AB term count = term count + 1 } run count = run count + 1 AB = switch(AB) # end of run, so switch whether we're writing in A or B }
If you take this sequence and apply it to the number of penguins born every year and divide it by the speed of sound in a standard Olympic swimming pool it makes absolutely no sense whatsoever.
Huh? Assuming you have a definition of "apply a sequence to a concrete number", yes it does. It's just units of penguin/distance. I mean, it's utterly useless, but it still makes _sense_.
According to random websites and graphs I found using google, there are 42 million penguins on earth. Penguins live on average 15-20 years. Therefore, the minimum amount of penguins born each year to sustain such population is 42000000/17,5 = 2,4 MPenguins/year or 0,0761 Penguins/s. The temperature of an olympic swimming pool is required to be 25-28 degrees Celsius. The speed of sound in water at 25 degrees C is about 1485 m/s. If we divide number of penguins per year by speed of sound in olympic pool water we get the Penguin Coefficient = 5,125 * 10^-5 Penguins/m. It will vary each year depending on the amount of penguins in the world. Now, if we apply the Kolakoski sequence to this coefficient we get: 5.125, 10.25, 10.25, 5.125, 5.125, 10.25... [10*-5 Penguins/m] Please correct me if my calculations were wrong.
This can go with 1,2,3 as well. Something like: 1,2,2,3,3,2,2,2,1,1,1,...etc. First round will produce: 1,2,2,3,3,etc. Second round will produce: 1,2,2,..etc. third round will produce 1,2,..etc.
Those technically don't work, as they consist of 3s as well as 2s and 1s. One of the rules is that they should contain 1s and 2s. Gosh I hate wikipedia.
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.
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?)
The first number describes itself, and the second number describes itself and the third number. After that, the numbers don't describe themselves, they describe what's going on way out to the right, so you're reading somewhere near the start of the list (say the nth number) and writing somewhere far to the right (around the 3n/2 th number if there are as many ones as twos). You'll never get a contradiction because as you read down the list, you never get any new information about what you've already written, only about what to add to the end.
11 and 22 are both balanced in the next layer. So if one layer had a larger proportion of 2s, that influence would be corrected in the sequel, and same for pairs of 1s occurring too frequently. Alternately: imagine a sizable region with a significant imbalance. What predecessor layer produced it? You quickly run out of options.
Reminds me of the "Figure Figure" sequence from Godel, Escher, Bach. 1, 3, 7, 12, 18, 26, 35, 45, 56, 69,.... Numberphile should track down Hofstadter to do that video! But then again, there's so many videos you could do with Hofstadter...
You can go in the other direction: "second" iteration: 1,2,2,1,1 => "previous" iteration: 1,2,2,1,1,2,1... => "preprevious": 1,2,2,1,1,2,1,2,2,1 and so on I guess.
@Daneca B: I think you know if you take into account two levels of the sequence: the initial one, and the first referential one. both of the levels have to start with the same numbers (if they don't, they're not the same sequence); so, if you start with 1, the referential level has to start with 1. 1... 1... Now you have to change to another number (if you place another 1, the referential level would start with 2 because there are two 1's: 11... 2...) Remember, both sequences have to be the same; therefore, because the next number has to be 2 (this sequence only has 1's and 2's), that means that there will have to be two 2's because the referential level says so: 122... 12... [the 1 shows that there's one 1, the 2 signifies that there are two 2's] And you continue like this for infinity
you can think of it like a self-grown to-do list, since you know every new number describes how many of a number appear, start with 1, this tells you it must break at 1 copy of the first number (1), then you get 1 2, knowing the sequence must be 1 number (1) and 2 more (2 2) you get 1 2 [2], repeat. 1 2 2 [1 1], 1 2 2 1 1 [2 1], 12 2 1 1 2 1 [2 2 1] each [] tells you what to apphend
I would love to see a video on the binary carry sequence/ruler sequence since I have used it in a few problems I worked on recently. I'm not sure what interesting things there are to say about it but I find it fascinating just because I discovered it by myself and used it for calculations!
there's definitely the same number of one's and twos as you approach infinite integers. the base is 122112. if you look, there is an even number of 1s and 2s. this pattern repeats regardless
It was so calming and satisfying to see someone colour it in. And also, shouldn't you be able to kind of work out a new pattern using the colouribg method. You start from the center and repeat that pattern each row again. I haven't tried it yet, but I am not sure, what could go wrong
"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.
Would be more interesting as a binary number sequence of {0,1}'s being how many to add at each step-thus recognizably limited to 1|2's, whereas trinary would be 1|2|3's, etc. And then you'd have 0↔0|1, 1↔00|01|10|11, etc.
There's something very dragon curvish about that sequence. If you make the 1s and 2s into 90° turns left and right (or vice versa), it seems to start giving some kind of pattern. There obviously can't be any closed loops, as you'd need sequences of 3 for that to happen, but the shape doesn't seem to disappear off into the distance or try to cut across itself. Are there any drawings on line showing the shape an extended sequence might give?
they will make a video about how many mathematical secrets and facts are there, then a video about how many videos can you make about how many mathematical secrets and facts are there, then a video about how many videos can you make about how many videos can you make about how many mathematical secrets and facts are there...
Well, when Numberphile catches up to where mathematical progress is today, at that point maths will have progressed a bit further. When Numberphile catches up to where maths is at that point, maths will then have gotten a tiny bit further again. And so on and so on. Therefore endless content!
+Gyrocop But wait! For Numberphile to get to the mathematical progress of today, it must first get halfway here; to get halfway to the mathematical progress of today, it must first get a quarter of the way here. In fact, to get anywhere, they must do infinitely many things first! Since they can't move along any progress, there can _never be another numberphile video_! Oh noes!
Actually the full sequence that repeats itself isn't clearly mentioned in this video, it's longer than that: 1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,2,1,2,2,1,1
I feel like there's a lot of info missing from this video. What do we know about the sequence? How do you generate it? Is there a formula? Does it repeat?
The Look and Say Sequence is similar, but rather than being self describing its members are sequences themselves, generated by counting the consecutive occurences of a number in the previous one. Interestingly the numbers never get bigger than 3. It's listed here: oeis.org/A005150
So it doesn't repeat? That would have been cool to note near the beginning of the video. I only caught that indirectly since you do not know if there are more ones than twos.
Why is the question of whether there are more 1s or 2s unknown? I mean isn't it fairly obvious that both of them are countably infinite and therefore the same number of each?
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.
Does the Kolakoski sequence encompass all finite sequences of 1's and 2's? Or less specifically, is 0,12211... (continuing the sequence) transcendental?
I wonder... to solve whether there are more ones than twos, couldn't the pattern be reversed engineered from the single length description to the infinite description?
Welcome vsauce viewer! That's not what it means this case. Here, he is asking for the limit of the ratio of 1s to 2s in the finite subsequences as their length tends to infinity.
Isn't it possible to make this kind of sequence for any pair of numbers? For example, 133311133313133311133313331333111333131333111333... Has this same property of being descriptive of its own string, it just ends up more complicated
That uncolored blank circular pattern looks like something I've seen in Halo or another sci-fi game. Is it? Like on an interactive terminal of some kind? No? Idk but it looks familiar...
Hi numberphile I really enjoyed the video and so did my family who aren´t into maths at all! I have a question, doesn´t this work too? 2,2,3,3,2,2,2,3,3,3,2,2,3,3,2,2,3,3,3,2,2,2,3,3,3,2,2,3,3,2,2,2,3,3,3,... Thanks for posting!
can a computer generate this sequence? It would have to somehow think ahead? "if I put a 2 here, I would need to put a second 2 so that to have two 2's, then I would need to put two 1's to justify the second 2, that means I would have to put one 2 next and one 1 after that, which leaves me with how many 2's next?..
TFW you have an advertisement for a web site designer in front of a YT video supported by a DIFFERENT web site designer.... (I got an ad for GoDaddy's Go Central service in the ad before the video played...)
Why does this sequence not work? Aren't there supposed to be only 2 of them? 122331112223123311223331223331112311223331112223112233311122231233111231122333111222312... I'm sure, I'm missing something, but as I see it right now, you could construct any number of such sequences with any set of base numbers. You just have to use some arbitrary rule to pick the next number in your sequence, when you run out.
This sequence is not the only self-describing sequence. It's the blueprint for a whole family of self-describing sequences.
{1,3}
1,3,3,3,1,1,1,3,3,3,1,3,1,3,3,3,1,1,1,3,3,3,1,3,1...
{1,2,3}
1,2,2,3,3,1,1,1,2,2,2,3,1,2,3,3,1,1,2,2,3,3,3,1,2...
{1,2,1,3,1,4...}
1,2,2,1,1,3,1,4,4,4,1,5,5,5,5,1,1,1,1,6,6,6,6,1,7,7,7,7,7...
And 1
The sequence is special because the nth term describes the length of the nth run (it looks like both start counting from one). In your first example of 1s and 3s the 12th term is a 3, but the 12th run is is a single 3 (so of length 1), and so it doesn't have the special property that the one in this video has. I didn't check the others, but my guess is you'd be able to find a term that doesn't work.
@@wantsomechocolate
OP merely made an error in transcribing the {1,3} sequence, at the point you identified, where indeed the 12th term is 3 and the 12th run (beginning at the 24th term) *_should be_* three 3s.
So obviously it should be
1,3,3,3,1,1,1,3,3,3,1,3,1,3,3,3,1,1,1,3,3,3,1,3,3,3,1...
OP did not make any transcription errors in the other examples they gave. You should have thought harder before making your claims.
No, the property that "the nth term describes the nth run" is *_not unique_* to the sequence based on {1,2}.
The exact same process can be generalized, as OP alludes, to construct a self describing sequence with that property on the basis of _any_ "seed" sequence of positive integers where adjacent seed terms are always different.
{3,1,4}
3,3,3,1,1,1,4,4,4,3,1,4,3,3,3,3,1,1,1,1,4,4,4,4,3,3,3,1,4,4,4,4,3,3,3,1,1,1,4,4,4,3,3,3,1,4,3,1,...
Now, *_among sequences with this property,_* the sequence based on {1,2} is arguably "simplest" by some complexity metric or "earliest" by a lexicographic metric, but it's not totally unique.
The Vurp Sequence:
1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11...
The number that each number repeats in the sequence is that number in the sequence.
That's already called the golomb sequence
The Kolakoski sequence was named after William Kolakoski (1944-97), who discussed it in 1965, but subsequent research has revealed that it first appeared in a paper by Rufus Oldenburger in 1939.
Yeah, and Rufus stole it from my grandpa, who created the sequence in 1938. It’s no more Oldenburger’s sequence than it is Kolakoski’s. It’s Peepaw’s sequence, and that’s all there is to it. You need to put some respeck on his name.
Colorkoski sequence
Futfan genius
Oldenburger sequence.
??
My computer froze on 0:15 and I was just sitting there for a solid 20 seconds waiting for him to stop teasing me
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!
simply beautiful
*non parker square video
Splendid haha simply delicious
1 1 was a racehorse. 2 2 was 1 2. When 1 1 1 1 race, 2 2 1 1 2.
and then 1 1 8 2 2 4 dinner
That's numberwang!
Let's rotate the board!
Yes I know that
that took me awhile
1:18 "there are only two sequences that have this property"
What about
2 2 3 3 2 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 2 3 3 3 2 2 3 3
There is another self-describing sequence called Golomb's sequence. It contains all positive integers instead of just alternating between runs of 1 and runs of 2. The sequence is 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 9, ...
Your question "are there more 1's or 2's?" reminds me of another question. Are there more primes that are sums of squares or more primes that are not sums of squares. Primes 2, 5, 13, 17, ... are sums of squares. Primes 3, 7, 11, 19, ... are not sums of squares. Which sequence of primes dominates?
There's an answer it's relatively 50-50 but the sum of squares is smaller in the long run somewhere on the order of log x
1:20 This sequence isn't the only sequence with the property of self-describing consecutive similar terms, though it (with and without the first 1) is the only one consisting of only 1s and 2s. Take any positive integer and repeat it itself number of times, then take another number and repeat it the second-term number of times, and so on.
I got curious about the name Kolakoski now. It sounds very much like a Finnish name. "Kola" means a snow plough in modern Finnish (or cola as in Coke or Pepsi) but it has meant a tool for lifting or shifting things in general, and "koski" means river rapids. However, the Population Register Centre name statistics page doesn't find any occurrences of it. That means there has been no Finnish citizen by that name alive since the mid-1960s, so presumably there has been no one born with that name since the early 1900s.
On the other hand, there seem to be a number of people with the name in the USA, mostly in the Pennsylvania-Minnesota area, which is where most of the Finnish emigrants in the late 19th and early 20th century settled. This is also where William Kolakoski came from.
I can find one single place in Finland where the name Kolakoski occurs. In Eura, there is a bay in the lake Pyhäjärvi called Kolakoskenlahti ("the Kolakoski bay"). There is a small stream that runs into the bay, which I suspect might be locally called Kolakoski, but it's too small to have a name on the map. A nearby road is also called Kolakoskentie (the Kolakoski road). This area seems to be called Kolakoski in some official papers of the city of Eura. But somehow Kolakoski doesn't strike me as a particularly western name either.
I also found a reference to some rapids named Kolakoski in the river Luiro in Lapland. But this information was in a badly scanned and OCR'd copy of the book Lappmarken by the Jacob Fellmann (published 1906). This turned out to be an incorrectly scanned Kotakoski, which is a relatively common name, it seems. There are plenty of those in Finland, and I found another OCR mistake like that as well.
Maybe the whole name Kolakoski in the USA is derived from a misspelling of Kotakoski at some point in the past? :)
His name sounds very Polish to me.
Albert Sosnowski I'd agree
But the Kola peninsula is in Russia
But yea he is probably Polish
He was American born in Pennsylvania
Do a video on 2 + 2. You guys will probably figure out how to make it interesting
that's 5. it's a great "meme".
And I bet there would still be heated arguments in the comments 🤣
2+2 = 2*2 = 2^2 etc.
This is cool because it is self-producing. If you need to know the next number in the set, you can figure it out according to how much you know so far. You can sit and produce this yourself for as long as you like and make it more and more fractally.
Kowalski is part of Madagascar's penguins
Kolakoski, analysis.
"There is beauty in simplicity" - Hanzo Shimada
Endless forms, most beautiful !! Thanks again to Numberphile channel !!
if you take the sequence and do all odd terms minus all even terms, this tells you how many ones there are more than twos, (for the one that starts with 1) since this describes (for odd terms) the quantity of ones, and (for even) the quantity of twos.
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
I watch this channel because i hope somehow their smartness will soak through the screen into me.
1:25 To clarify, those are the only two sequences IF you only use 1s and 2s. You can make this sort of sequence with any numbers you please. For example, 2 and 3 give you 223322233322332233322233322332223332233222333222332233222333222 et cetera. 1/2/3 gives you 12233111222312331122333122333111231122 and so on. Or using ascending integers gets you the Golomb sequence, 1223344455566667777888899999 et cetera.
He says this and the sequence without the leading 1 are the only such sequences, but I think this only applies if you're talking about sequences of only 1 and 2. I messed around a bit with a sequence that also allows 3's in the mix and it works in the same way. Pretty cool, it took me a couple minutes longer than I'd like to admit to understand how to keep generating the sequence further hahaha
Freshman year my calc professor asked us to identify how this sequence was being generated and I figured it out. I never knew what the name of it was though!
Wow, that's pretty cool! My favorite types of math are calculus, number theory, and topology, this fits right in!
This is the song that doesn't end.
Yes, it goes on and on my friend!
Some people started singing it not knowing what it was,
And they'll continue singing it forever just because
This is the song that doesn't end...
3:38 almost looks like Enterprise hull plating pattern.
2:35 almost looks like some strangely shaped piano keys.
For anyone curious, this will generate a "kolakoski" sequence of length n with x integers (1 through x). def kolakoski(x, n):
seq = [1]
next_number = 2
for i in range(n):
temp = seq[i]*[next_number]
seq += temp
if next_number == x:
next_number = 1
else:
next_number += 1
seq = [seq[0]] + [2] + seq[1:]
return seq
Edit: Python language, fixed to splice in the [2] that needed manual fixing (yes... it's lazy coding... I'm an analyst not a programmer ;)
1, 2, 2, 3, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 3, 3, 1, 1, 2, 2, 3, 3, 3, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 3, 1, 1, 2, 2, 3, 3, 3, 1, 1, 1, 2, 2, 2...
So you're saying this sequence doesn't have the properties of the Kolakoski sequence? In which way?
I'm guessing this video only addresses sequences containing {1,2}. The Wikipedia page on Kolakoski sequences show several other examples with other digits, including yours.
The Kolakoski sequence's definition say that it can only consist 1 and 2, but there are a lots of self-generating sequences that have "Kolakoski sequence's property", your sequence is one of them.
It actually does have the same properties. I think the speaker just made a mistake when he said that it is the ONLY sequence that displays these properties (and the sequence without the first 1).
There are actually an INFINITE number of sequences like this, like the one you commented. It's just that the Kolakoski sequence is the ONLY sequence that only uses the numbers 1 and 2.
Maybe because what you wrote doesn't repeat the same sequence of numbers over and over again?
I think if you look, all of his are sequential (i.e. a 2 always follows a 1 and a 3 always follows a 2). So the encoding still works assuming you apply that extra rule (which is also applicable, but trivial, for Kolakoski)
I saw a lecture about this from Douglas Hofstadter. It's a very Hoftsadteresque sequence.
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.
One of the rules is that it must consist of 1s and 2s. In this way of thinking about it, those are technically not correct.
Watched a 5 minute video on mathematics, tried it out on paper, learned something new and amazing 😎
What do you mean it's the only one? The Wikipedia page on this sequence has loads, and states that there are infinitely many, even chains of sequences which describe each other.
See the full-length colouring video: th-cam.com/video/a-6oOVdU7Oo/w-d-xo.html
Numberphile make sure you pin your own comment so it stays at the top so people see it btw ^-^
That's wrong. There are infinite sequences with this property, but these two are the only ones with only 1s and 2a.
And even if the script said that it's only of sequences composed of 1's and 2's, the fact that there are only two such sequences isn't interesting at all. You pick a starting symbol, that determines run length, and there's no choice of transition symbols. Very poor script compared to the usual accurate and accessible content.
For the letter density question (help welcome): There seems to be only 4 possibilities to generate the next sequence. If you're 'at 1' and 'see 1' write down '2'. @1 see2 write22. @2 see1 write1. @2 see2 write 11. This generating algorithm promises to make the ratio of 1s to 2s asymptotic to 0.5. Seems you can start with anything, it'll move to 0.5 eventually.
It looks like a 'Binary / Grey Code encoder disc'
A very fine continuation of the visualisation of math aspect last shown on the collatz conjecture. This video should have been sponsored by circlespace, though.
I instantly figured it out.
It's about how many matching numbers there are in a sequance
"Like Kolakoski, in Vanishing Point." - Bobby Gillespie
I'M SO PROUD OF MYSELF FOR HAVING GUESSED WHEN HE WROTE IT DOWN THE 1ST TIME
in a classic case of our brains seeing patterns they are trained on, if you move through the plot interpreting it as binary values, substituting 0 for 2 and rotating it 90 degrees, you get the (integer) values 0,0,1,2,3,4,4,7,8,8,9,14,33,32,32,35,34,57,66,67,64,64....
Interesting and worthwhile video.
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.
Thanks for this blissfully short video
one of my favourite sequences is self describing
1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, ...
Student of Conway?
+Numberphile,
You assert (1:18) that there are only 2 sequences with the property of listing its own runs: Kolakoski and Kolakoski-sans-1.
But here's another that does:
2 2 3 3 2 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 2 3 3 3...
And another that does:
3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 3 3 3 2 2 2 3 3 3 2 2 3 3 2 2 2 3 3 3 2 2...
You can just start with any pair of positive numbers and form a sequence with this property, so there are infinitely many of them.
Here's a procedure to generate a sequence with the property from any pair of positive integers:
A = your input # first positive integer.
B = your input # second positive integer, A != B
AB = A
define a function 'switch' such that{
switch(A) = B
switch(B) = A
}
term count = 0
run count = 0
sequence(0) = A
loop forever{
L = sequence(run count) # get the run length from the sequence
# write into sequence a run length L of whichever A or B is currently stored in AB
repeat L times{
sequence(term count) = AB
term count = term count + 1
}
run count = run count + 1
AB = switch(AB) # end of run, so switch whether we're writing in A or B
}
Super interesting topic, I've never heard of this before!
If you take this sequence and apply it to the number of penguins born every year and divide it by the speed of sound in a standard Olympic swimming pool it makes absolutely no sense whatsoever.
JB So you think
According to Benford's law, the number you get by doing this would be likely (probability>50%) to start with the digit 1, 2 or 3.
+JB Love it! Maybe you'll end up with the digital code for a flat earth video.
Huh? Assuming you have a definition of "apply a sequence to a concrete number", yes it does. It's just units of penguin/distance. I mean, it's utterly useless, but it still makes _sense_.
According to random websites and graphs I found using google, there are 42 million penguins on earth. Penguins live on average 15-20 years.
Therefore, the minimum amount of penguins born each year to sustain such population is 42000000/17,5 = 2,4 MPenguins/year or 0,0761 Penguins/s.
The temperature of an olympic swimming pool is required to be 25-28 degrees Celsius. The speed of sound in water at 25 degrees C is about 1485 m/s.
If we divide number of penguins per year by speed of sound in olympic pool water we get the Penguin Coefficient = 5,125 * 10^-5 Penguins/m. It will vary each year depending on the amount of penguins in the world.
Now, if we apply the Kolakoski sequence to this coefficient we get:
5.125, 10.25, 10.25, 5.125, 5.125, 10.25... [10*-5 Penguins/m]
Please correct me if my calculations were wrong.
Another thing which is quite interesting is that the negative space shapes the golden ratio.
This can go with 1,2,3 as well. Something like: 1,2,2,3,3,2,2,2,1,1,1,...etc. First round will produce: 1,2,2,3,3,etc. Second round will produce: 1,2,2,..etc. third round will produce 1,2,..etc.
Those technically don't work, as they consist of 3s as well as 2s and 1s. One of the rules is that they should contain 1s and 2s.
Gosh I hate wikipedia.
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.
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?)
Has Numberphile done a video on the number thing what goes
1
11
21
1211
111221
312211
etc?
TaliesinMyrddin i think it's called the look and say sequence, there might be a video with John Conway on the topic
Yes, featuring John Conway himself. watch?v=ea7lJkEhytA
Cheers!
Aaand the first comment I see is one from myself 2 years ago.
Whoops...
TaliesinMyrddin Grade A idiot!
Has it been proven that one can actually build this sequence up to infinity without getting a "contradiction"?
I can prove it but can't explain it to you
The first number describes itself, and the second number describes itself and the third number. After that, the numbers don't describe themselves, they describe what's going on way out to the right, so you're reading somewhere near the start of the list (say the nth number) and writing somewhere far to the right (around the 3n/2 th number if there are as many ones as twos).
You'll never get a contradiction because as you read down the list, you never get any new information about what you've already written, only about what to add to the end.
Proportions and ratios go a long way for solving cosmic distance and distribution. Mostly prime reductions.
11 and 22 are both balanced in the next layer. So if one layer had a larger proportion of 2s, that influence would be corrected in the sequel, and same for pairs of 1s occurring too frequently. Alternately: imagine a sizable region with a significant imbalance. What predecessor layer produced it? You quickly run out of options.
Reminds me of the "Figure Figure" sequence from Godel, Escher, Bach. 1, 3, 7, 12, 18, 26, 35, 45, 56, 69,.... Numberphile should track down Hofstadter to do that video! But then again, there's so many videos you could do with Hofstadter...
how do you know what number comes next in the initial series?
You can go in the other direction:
"second" iteration: 1,2,2,1,1
=>
"previous" iteration: 1,2,2,1,1,2,1...
=>
"preprevious": 1,2,2,1,1,2,1,2,2,1 and so on I guess.
@Daneca B: I think you know if you take into account two levels of the sequence: the initial one, and the first referential one.
both of the levels have to start with the same numbers (if they don't, they're not the same sequence); so, if you start with 1, the referential level has to start with 1.
1...
1...
Now you have to change to another number (if you place another 1, the referential level would start with 2 because there are two 1's:
11...
2...)
Remember, both sequences have to be the same; therefore, because the next number has to be 2 (this sequence only has 1's and 2's), that means that there will have to be two 2's because the referential level says so:
122...
12... [the 1 shows that there's one 1, the 2 signifies that there are two 2's]
And you continue like this for infinity
you can think of it like a self-grown to-do list, since you know every new number describes how many of a number appear, start with 1, this tells you it must break at 1 copy of the first number (1), then you get 1 2, knowing the sequence must be 1 number (1) and 2 more (2 2) you get 1 2 [2], repeat. 1 2 2 [1 1], 1 2 2 1 1 [2 1], 12 2 1 1 2 1 [2 2 1] each [] tells you what to apphend
I would love to see a video on the binary carry sequence/ruler sequence since I have used it in a few problems I worked on recently. I'm not sure what interesting things there are to say about it but I find it fascinating just because I discovered it by myself and used it for calculations!
I tried writing the sequence myself, was kinda easy but also very satisfying 😀
there's definitely the same number of one's and twos as you approach infinite integers. the base is 122112. if you look, there is an even number of 1s and 2s. this pattern repeats regardless
That's what I first thought but you can see at 3:50 the sequence does change
It was so calming and satisfying to see someone colour it in. And also, shouldn't you be able to kind of work out a new pattern using the colouribg method. You start from the center and repeat that pattern each row again. I haven't tried it yet, but I am not sure, what could go wrong
It seems pretty obvious that there are the same number of 1s and 2s in the sequence. Each grouping of 1,2,2,1,1,2 is three and three.
The graph at 2:35 reminds me of some of the Kadish Tolesa puzzles from Ages Beyond Myst.
"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.
There is no cycle. At least, not one of length six. If you've found a longer one, apply to get your paper published :)
Would be more interesting as a binary number sequence of {0,1}'s being how many to add at each step-thus recognizably limited to 1|2's, whereas trinary would be 1|2|3's, etc. And then you'd have 0↔0|1, 1↔00|01|10|11, etc.
There exists an O(n^0.63) algorithm to calculate the prefix sum for the general Kolakoski sequence
Sombra would be proud of this skycode.
??
Forseti scrolled to comments to see if anyone else thought this too :). Loved that ARG
¿Quién es Sombra?
its a Overwatch character
¿Quién es Overwatch?
I just suddenly tuned in when he said "This is a coloring book."
There's something very dragon curvish about that sequence. If you make the 1s and 2s into 90° turns left and right (or vice versa), it seems to start giving some kind of pattern. There obviously can't be any closed loops, as you'd need sequences of 3 for that to happen, but the shape doesn't seem to disappear off into the distance or try to cut across itself. Are there any drawings on line showing the shape an extended sequence might give?
I wonder when will numberphile tell us every mathematical secrets and facts etc. Until there is nothing left.........
they will make a video about how many mathematical secrets and facts are there, then a video about how many videos can you make about how many mathematical secrets and facts are there, then a video about how many videos can you make about how many videos can you make about how many mathematical secrets and facts are there...
Well, when Numberphile catches up to where mathematical progress is today, at that point maths will have progressed a bit further. When Numberphile catches up to where maths is at that point, maths will then have gotten a tiny bit further again. And so on and so on. Therefore endless content!
Information theory tells us that for everything we learn, we discover roughly 10 more things we don't know.
+Haiku Metzger No it most certainly does not.
+Gyrocop But wait! For Numberphile to get to the mathematical progress of today, it must first get halfway here; to get halfway to the mathematical progress of today, it must first get a quarter of the way here. In fact, to get anywhere, they must do infinitely many things first! Since they can't move along any progress, there can _never be another numberphile video_! Oh noes!
Actually the full sequence that repeats itself isn't clearly mentioned in this video, it's longer than that:
1,2,2,1,1,2,1,2,2,1,2,2,1,1,2,1,1,2,2,1,2,1,1,2,1,2,2,1,1
I feel like there's a lot of info missing from this video. What do we know about the sequence? How do you generate it? Is there a formula? Does it repeat?
The Look and Say Sequence is similar, but rather than being self describing its members are sequences themselves, generated by counting the consecutive occurences of a number in the previous one. Interestingly the numbers never get bigger than 3. It's listed here: oeis.org/A005150
Here it is on Numberphile with John Conway: th-cam.com/video/ea7lJkEhytA/w-d-xo.html
Shayka ! I think there is already a video on this?
Awesome, should've known Numberphile couldn't have missed that
Shayka ! Sorry, did not realise numberphile already replied...
U are first
So it doesn't repeat? That would have been cool to note near the beginning of the video. I only caught that indirectly since you do not know if there are more ones than twos.
For some reason, I started transcribing the Kolakoski Sequence as a drum rudiment. I need to use it on my drum pad and see how it works.
Why is the question of whether there are more 1s or 2s unknown? I mean isn't it fairly obvious that both of them are countably infinite and therefore the same number of each?
Heterilogical nouns are in math too
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.
Y'all should do a video over the golomb sequence
Does the Kolakoski sequence encompass all finite sequences of 1's and 2's?
Or less specifically, is 0,12211... (continuing the sequence) transcendental?
3,3,3,2,2,2,1,1,1,3,3,2,2,1,1,3,2,1,3,3,3,2,2,2,1,1
Same thing, but with 3->2->1
thanks.I like sequences.And oeis.
If the sequence is infinite, won't the ones and twos both just be countably infinite?
thanks again for the great video!
It's the simplest form for the building blocks of reality.
I wonder... to solve whether there are more ones than twos, couldn't the pattern be reversed engineered from the single length description to the infinite description?
Yes, there are the same number of 1s and 2s: aleph nought of each of them
Welcome vsauce viewer!
That's not what it means this case. Here, he is asking for the limit of the ratio of 1s to 2s in the finite subsequences as their length tends to infinity.
You could do infinity, infinity, infinity etc
Isn't it possible to make this kind of sequence for any pair of numbers?
For example, 133311133313133311133313331333111333131333111333...
Has this same property of being descriptive of its own string, it just ends up more complicated
If you mapped these to 1s and 0s, I'd like to see the binary numbers that would result, expressed in base ten.
David Learmonth pick 0.7945071927794792762403624156360456462... then multiply it by powers of two and round it down
That's the kolakoski constant ;)
That uncolored blank circular pattern looks like something I've seen in Halo or another sci-fi game. Is it? Like on an interactive terminal of some kind? No? Idk but it looks familiar...
Hi numberphile I really enjoyed the video and so did my family who aren´t into maths at all!
I have a question, doesn´t this work too?
2,2,3,3,2,2,2,3,3,3,2,2,3,3,2,2,3,3,3,2,2,2,3,3,3,2,2,3,3,2,2,2,3,3,3,...
Thanks for posting!
can a computer generate this sequence? It would have to somehow think ahead? "if I put a 2 here, I would need to put a second 2 so that to have two 2's, then I would need to put two 1's to justify the second 2, that means I would have to put one 2 next and one 1 after that, which leaves me with how many 2's next?..
a significant difference from Fibonacci.. sequence.. but what will the pattern look like of fibonacci no... i wnna see the colouring...
Ther is something we need to talk about... Square Space.
I love this. Numberline described by its own numberline :D. So what is the numberline tyen exactly?
So I assume then that the pattern doesn't repeat? Feels like there are limited permutations.
It doesn't repeat, which is why they don't know the proportion of 1s and 2s.
Talk about question 3 from IMO. That hard one
TFW you have an advertisement for a web site designer in front of a YT video supported by a DIFFERENT web site designer.... (I got an ad for GoDaddy's Go Central service in the ad before the video played...)
Why does this sequence not work? Aren't there supposed to be only 2 of them?
122331112223123311223331223331112311223331112223112233311122231233111231122333111222312...
I'm sure, I'm missing something, but as I see it right now, you could construct any number of such sequences with any set of base numbers. You just have to use some arbitrary rule to pick the next number in your sequence, when you run out.
2:56
I bet Vihart would love to paint that!
Is this sequence related to the golumn sequence?