Ok, so after digging a little bit in the sequence, I wanted to share a bit of what I’ve found. I started having in mind to stop when the numbers from 1 to 10 would have appeared but it took me a bit longer than I thought. I finally got a bit further and got the first 252 numbers of the sequence. (I’ve done this on paper, no programming, so it’s possible I failed it at some point) Here are the 56 numbers that appeared in order : 0, 1, 2, 6, 5, 4, 3, 9, 14, 15, 17, 11, 8, 42, 20, 32, 18, 7, 31, 33, 56, 19, 37, 46, 23, 21, 25, 52, 13, 62, 40, 36, 16, 27, 10, 92, 51, 131, 39, 12, 44, 34, 97, 72, 41, 78, 24, 105, 107, 167, 61, 26, 22, 127, 28 and 29. One thing that I found funny with this sequence is that is has the tendency to quickly come back to a number that newly appeared. For exemple when the 9 shows up for the first time, it takes only 3 steps to appear again. Same for 7 and 31. 5, 6, 18 are taking 5 steps to appear a 2nd time, 107 takes 28 steps, etc. But it doesn’t happen for every number, like for 14 that takes 131 steps to appear a 2nd time, but takes 4 steps to appear a 3rd time. ^^ 17 didn’t appear a second time for me even though it comes pretty early in the sequence. It’s hard to find coherence in there but it’s strange to see more often that not new numbers reappearing pretty quickly even though there are still lot of numbers that haven’t appeared yet. The second thing that surprise me a bit is the frequency of new numbers appearing, only takes about 4,5 steps (the longest chain of numbers between two 0s I’ve found is 8 numbers long (found it 2 times)) Thought it would take a bit longer but it’s pretty rare that a new number takes more than 6 steps to appear. But like I said, I only checked the 250 first numbers so I don’t know if it grows up, shrinks or stay pretty much the same if you go further and further. I usually don’t really dig into that kind of stuff, mostly I listen to the video and continue my way elsewhere, but this time my curiosity hasn’t been fulfilled enough, so here I am writing this :p It was worth the try. Thanks Numberphile o/
I just realized that adding 0 as the next term when there's a number you haven't seen before, isn't as arbitrary as I first thought: It's really just in agreement with the rule of writing down "how far back it occurred last time". When it's never occurred before, the last time it occurred was _right now,_ zero steps ago, so we add a zero. Awesome :D
Nice logic isn't it? Does it mean that the sequence can only start with 0 and no other number? Also the rule seems to mean that 0 can only occur twice in succession at the beginning of the sequence or immediately after the first number n if n is allowed to be non-0.
@@chrisg3030 I don't see any reason you couldn't start with something else than 0. Then the sequence will be different depending on which number we start with. A funny thought: Since we don't know whether every number will eventually appear, let's say that m is a number that never appears in the sequence. Then if you start with m, only then will you get the same sequence as the one where you just start with 0 :D
Similarly there's no reason why you couldn't add something else, say 2, when there's a number you haven't seen before. So 0 2 2 1 2 2 1 3 2 3 2 2 1 6 ... The original rule says "Add n when the number last occurred n places back", so when it's new - that is last occurred 0 places back - you add 0. With my variant it's the same but with the exception of 0 places back, in which case you add 2 but still use 2 for 2 places back as well. We still seem to get the same kind of sequence though (except in the case of 1, if we add 1 for new as well as 1 place back we just get endless 1's). Please check.
One thing that can be proven about the sequence is that VE(n) < n for n > 0 (since the entire sequence has length n+1, the most number of moves back it could take is n, but VE(0)=0 and VE(1)=0, so you'll never go all the way back to VE(0) and thus VE(n) < n). So yeah, f(n) = n seems like a fairly good approximation of the growth of the sequence, but it is also an absolute upper bound on the sequence.
But isn't there a sense in which any sequence obeying a rule is self-referencing? Let's express the rule for the van Eck as "Add n when the current term last appears n places back". So if the current term is 1 and it last appeared 6 places back then we add 6. If the current term is 6 and it last appeared 0 places back (in other words it's never appeared before) we add 0. Now let's change that rule a tad: "Add n when the current term FIRST appears n places back". If we start with 0 we go on 0 1 0 3 0 5 0 7 0 9 0 11 ..., a both boringly regular and not apparently self referencing sequence, even though our defining rule makes sound like it should be. But in my example the first place of appearance of a term is never going to stop being just that, whereas the latest place of appearance of a van Eck term can change quite frequently. So perhaps we should talk instead of term-index variant and invariant sequences.
@@chrisg3030 i think the important distinction for a self-referential sequence" is when a series checks something other than the ordinality of a previous term. If you do something with the number other than use how big it is, it feels like using a meta-property of the sequence itself
I would change the definition of Van Eck's sequence. The sequence doesn't begin from 0 necessarily. Then it is only 0-sequence but it can be N-sequence as well. Then the Van Eck's sequence family was created.
That's interesting, actually. They're related-if you start the sequence at n, it will look identical to the 0 sequence up to the first instance of n in the sequence, at which point it change completely. And the first different number will be much higher than anything around it, which could affect the shape of the large-scale triangle-my wild guess says its slope wouldn't change but its height would jump up at that point. Now I want to find out.
'Oh come on! How can you not know how fast it grows? Surely that's easy to prove! We just... okay maybe we.... what if....' *Three hours later* 'Alright, you win this round...'
It does feel like there is a provable lower bound using the repeating argument described in the video. But it is probably super low, logarithmic in n or something.
@7:06 4 ways. specifically (+,+,+,+), (+,-,+,+), (-,+,-,-) (-,-,-,-) I found this by the following logic chain: 1. 81 is already divisible by 3, therefore we only need to manipulate the pluses and minuses to preserve this property. 2. 9 is also divisible by 3, therefore it doesn't matter if it is added or subtracted, it will not change the remainder after division by 3. 3. 31, 13 and 4 are each numbers of the form 3x+1, therefore for the purposes of determining whether their sum will be divisible by 3, we need only concern ourselves with the '1' part. 4. the only way to add or subtract 3 1s to each other in any combination and end up with a number that is divisible by 3 is if either all of them are subtracted (-1-1-1=-3) or all of them are added (1+1+1=3), therefore, the first, third and fourth sign must match each other. 5. (4) combined with (2) implies that the second sign can be either plus or minus and the remaining ones must match each other but be either plus or minus and any such combination will work, this means we have 2*2=4 combinations
Or how about the roman version, it starts like this I II III IIII IVI IIIVII IIIIIVIII VIIVIIII IVIIIIVIVI IIIVIVIIVIIIVII IIIIIVIIIVIIIIVIIIIIVIII VIVIIIIIVIVIIVVIIVIIII IVIIIVVIIVIIIVIIIIIVIIIIVIVI
Answer to the daily challenge problem: 4. It is a modular arithmetic question. 81 is divisible by 3 and so is 9. The other numbers each are divisible by 3 with a remainder of 1. All three of those must have either a plus or minus sign. But it must be the same sign for all three. Then thr nine can take a plus or minus and it is independent of the other one. So you have 2 independent choices with 2 options each. 2x2=4.
Brilliant question: Mod 3, the question is: 0 ( ) 1 ( ) 0 ( ) 1 ( ) 1 Where ( ) should be + or -. The maximum value of the expression is 3 and the minimum is -3, occurring when all the signs are + and - respectively (except for the sign before the 0, which can be either). This yields 2×2=4 possibilities. 0 cannot be achieved since the parity of the expression must be odd.
So I saw the title and clicked on the video, and I just glanced at the description for maybe a few hundred milliseconds, and I saw OEIS mentioned, and I thought "oh, nice, they've got the Sloane's entry for it." Then I watched the video and realized that they've also got *Sloane.* :-D
Is it starting to rain? Afraid so. Is this going to hurt? Afraid so. Are we out of coffee? Afraid so. Is the car totaled? Afraid so. Will this leave a scar? Afraid so. Hotel? Trivago.
I'm going to answer on a new comment, cause I find the answer interesting by itself, to someone who remarked that if the sequence started with 1,1,... then the sequence would be periodic. The statement is true, but with this set of rules, the first number determines the sequence, and 1,1 is not a valid start for a sequence. In other words, all sequences generated with this rule start by x,0,... . However, we can actually verify that there are at least 2 such sequences that are "profoundly" different (i.e. one is not a subsequence of the other): 0,0,1,0,2,0,2,2,1,... and 1,0,0,1,3,0,3,2,0,3,3,1,8,0,... ("0,0" is a subsequence that appears exactly once on each sequence). A "not profoundly different" sequence would be: -1,0,0,1,0,2,... , if we allow for x to be a negative integer. With this I just realized that if 0,0,... does take all the positive integer values, then it might be "easy" to prove that x,0,... is a "profoundly different" sequence from y,0,... iff x!=y and both are natural numbers. Looking at it in the other way, if there's a value z that's not part of the sequence 0,0,... , then z,0,... is not "profoundly different" from 0,0,... .
Agreed. There are some more interesting sequences with modified rules: Add 1 to any new number. Subtract 1 from the number following a zero. That sequence looks just .. loopy. Very interesting.
@@Ashebrethafe, haha... "factorial". Funny how programmers have managed to decide on ways to type 'not equal' and understand eachother eg '!=', 'neq', '>
7:02 Interesting question, I'm thinking 4? 31 mod 3 = 13 mod 3 = 4 mod 3 = 1 and 81 mod 3 = 9 mod 3 = 0 so if the result is divisible by 3 (ie. result mod 3 = 0) the signs in front of 31, 13, & 4 can be + or - but they must match. Then the sign in front of 9 can then be + or - so that makes 2*2=4 combinations.
This series is amazing. Not intuitive, sort of alternating and unsolved. Reminds me of the 3n+1 problem, but in a more interesting and (probably also easier to solve) way
Really shouldn't be that surprising. At any nth term x, x cannot be larger than n, because that would mean you would have to look back an amount of steps larger than the total amount of steps you have taken. Therefore, since the maximum value of x is equal to the value of n, drawing a line through all the peaks should give a line that approximately maps to y=x, or a slope of 1.
@@firstlast8858 That's only half an argument. You've only explained why the slope cannot be above 1, not why it should be near 1. Indeed, since the sequence starts with 0, the maximum value of x is less than n. Further, it is easy to assume that x grows slower than n, so it isn't immediately evident that the slope would remain near 1.
@@BainesMkII Hmm, as soon as a number is "used" to look back to, it will never be used again. So eventually all the starting numbers must get "used" up. I wonder how fast the consecutive used-up numbers progresses right, because that could limit HOW MUCH less than 1 the slope is.
5:44 "there might be other copies of z in the period" Okay, but what if there aren't? I guess the same reasoning still holds, but it confused me when he didn't complete cover both possible cases, felt like a loose end.
When he draws the first period that is the earliest point at which the period can start. So if there was no other z in the period then x would have to be the length of the period (it can't be any more than that length). But that means that we could draw the period one step earlier, because another z would have to occur just before the first period, and that's a contradiction. So z must occur somewhere else within the period. It's essentially the same argument but with x=a. Sloane generalises by looking at the first z in the period, wherever that may be.
Well if there wasn't, you could just redefine the period to be two iterations of the pattern instead which would mean you'd have two copies of z and the proof would still follow.
I don't understand the final step of the proof. Specifically, the statement at 6:23 confused me. He didn't push back the beginning of the period; he demonstrated that there is a z in the previous period.
which means there always has to be a period before the one you're looking at right now, so the periods could never start due to a bootstrapping paradox since the sequence has a beginning.
I don't know why he made it so complex.. So lets say a period looks like C D E ... L M N. Since the second period starts with C this means the length of the period is C and before the first period there has to be a N. But this means the period now looks like N C D E ... L M and so on, so there can only be a repeating pattern if it goes all the way to the beginning. The problem here is there can't be any 0's in a period because zeroes are only used when it's a new number and a repetitive sequence can't have anything new. The sequence has by design 0s so periodicity is not possible.
2:17 there is a poem out there that does follow this pattern of "don't know" at the end kinda, "Vietnam" by Wisława Szymborska. it's a sad poem, but it's really nice :)
If you guys are interested in playing with this sequence, I wrote some javascript code that you can use to generate terms quite easily: function van_eck(terms){ function find_index_in_array_from_back(arr, i){ for(var c = arr.length-1; c >= 0; c--){ if(arr[c] == i){ return c; } } return -1; } var s = [0]; var s_1 = 0; for(var c = 0; c < terms; c++){ var index = find_index_in_array_from_back(s, s_1); var distance_back = s.length - index; s.push(s_1); if(index >= 0){ s_1 = distance_back; }else{ s_1 = 0; } } return [s, s_1]; } In terms of playing with it, you can, for example: console.log(Array.from(new Set(van_eck(100000)[0])).sort((a,b)=> a - b)) You can see all the unique numbers within the fist 100000 terms of the sequence. By matching up the numbers with the indexes in the output, we can see that all the numbers up to somewhere in the 1500s are included in this number of terms (as well as several numbers beyond, but EVERY whole number up to there is included). If we do: console.log(Array.from(new Set(van_eck(1000000)[0])).sort((a,b)=> a - b)) Every number up to somewhere in the 8000s is included, and many more beyond. Anyway, that's just one idea, you can of course do whatever you want. I had some fun playing around with the sequence, so if you want to play with it, the code is there for you, just do CTRL+i in chrome (or bring up developer tools in any browser) go over to the console, paste it in, and away you go!
We also know that the nth number cant be larger than n, because there arent more than n steps before n. Therefore the fastest way for the sequence to grow is linearly. it could still be root of n or log n, but n^2 or 2^n are ruled out.
About the demonstration "there might be some z's in the middle" and after thag absumption proving a contradiction seems weak, why add a z inside which is the same the last number of the period, and instead not take x directly (or z and then the a is x)
I have a hunch (meaning I have no idea how to demonstrate anything) that this is related to the primes. Why? Because each time a new number enters, the sequence produces a zero. Just like, when you do an Eratosthenes sieve, you write off any ‘’new’’ factor you encounter.
I love numbers and how they relate with each other. I never heard of this. Has anyone ever programmed a computer to see how far you can go? What I am fond of saying is, "The more I learn, the less I don't know!" (Or realize I don't know.)
5:44 A claim is made that “there might be a copy of z” within the range of one of the periods, after which it is proven that this sequence grows endlessly. However, I do not see why there is a guarantee that any such copy of z ever has to appear between period markers x and z. Can anyone clarify?
Ohh gosh, that's an amazing sequence!And there are lots of questions rising: Does the sequence has infinite non zero terms? how often does each term appear? Does each positive integer appear in there? Can we find an algebraic expression for it? In order to find the n-th term, do we really need to know all the previous terms? So many questions, i love it!
6:30 i think the animation is a little bit misleading. all that the proof says is that x is not the first number of the period but the z before the first x is. Which leads to a contradiction to the assumption that x is the first number of the period. Of course you can repeat it with every number before that, but that is not what Neil Sloane says and is not neccessary. (He says: The period began one step earlier)
In case anyone else was confused by the last step of his proof, it is stated a little more clearly in the OEIS: The periodic part does not contain any zeros. Suppose the period has length p, and starts at term r, with a(r)=x, ..., a(r+p-1)=z, a(r+p)=x, ... There is another z after q
What I don't follow is why for it to be bounded it has to be periodic. I get the M^M possible permutations of M numbers 1-M. The sequence will have to repeat blocks at some point, but pi has that same restriction (albeit with single numbers and not blocks). There's only 10 numbers available there and it manages to be non-periodic and carry on infinitely. Why could this sequence not 'settle' to repeating 10 or more blocks of M numbers in a similar non-periodic way? EDIT: And of course I realise immediately after you could never repeat 2 blocks or this leads to the same logic as the original proof, that this will lead to a periodic part one step back etc. And I assume you can only rearrange the blocks a finite number of ways before being forced to either repeat 2 or repeat the order.
@@lyndenw2240 Assuming the sequence is bounded, the next number in the sequence is always determined by at most the past M numbers. So if a chunk of M numbers occurs a second time, the number immediately following that chunk will be the same the first and second times. Successive application of the same argument shows that the whole sequence would be periodic. Because there are only finitely many (M*M) possible chunks of length M, some chunk is guaranteed to repeat.
Well, I don't know if the slope is actually 1, because I'm getting no numbers in the sequence at all that are more than the position in the sequence, that is to say, X>Y as a general rule. If we graph the known points of increase, to get the maximum values, you have (1,0)(3,1)(5,2)(10,6)(24,9)(30,14).and (56,20) That starts at a slope of 0.5, and slows to a rate as low as 0.23, leaving an expected value of Y (the highest number that you should get if you randomly choose an X value) of less than X/2 Edit: I might have messed up. I picked the points with the highest X values in a group, when I should have picked the lowest x values Fixing issues above.
That would be an illegal starting pair by the definition of the sequence. If you start with a 1, the next number has to be 0. Note that the sequence as-shown doesn't start with 0,0 but rather just 0 and proceeds from there.
@@MattStum Definition of sequence is the mechanism by which new numbers are added The starting sequence is free parameters that allow to generate different strings of same ruleset
I can't prove it grows linearly, but it is quite simple to prove limsup a(n)/√n ≥ 1 Proof: Whenever a(n)=0, either there have been √n zeros in the sequence, thus √n new distinct numbers (and at least one bigger than √n), or there have been less than √n zeros in the sequence, and thus there is at least one gap between two zeros which is at least √n wide. Even though this is very far from linear, I haven't seen any lower bound yet, so let's start here. And go back to find a better one!
I would consider it linear growth. The value at a(n) is always less than n. In my reasoning this rules out exponential growth. I would certainly like to prove that the growth approaches { y = 0.809 x }.
I don’t really understand the part from 4:11 until 5:15. If i’m not mistaken, he explains that if a sequence could have a highest number, that would mean the sequence has to be periodic. But I don’t get why that is. Would somebody be kind enough to help me understand? :) thanks!
If you begin with 1, 1, and then employ this rule thereafter, you generate a periodic sequence. Are there any other prefixes for which the Van Eck rule generates a periodic sequence?
The answer is 4 You have 2 numbers divisible by 3, and three numbers divisible by 3 ONLY if added together or subtracted from one another; so the 2 that are divisible by 3 can be added together or subtracted from each other and the other 3 can be added to or subtracted from each result... 4 possible answers to the +/- question
To print the first 1000 numbers with python 🐍 nMAX=1000 L=[0,0] n=2 while n0 and b==0): if L[a]==x: b=1 a=a-1 if a==0: L.append(0) print(0) else: L.append(n-2-a) print(n-2-a) n=n+1
It appears that in the first million terms, the smallest number not to appear in the sequence is 8,756. The largest number to appear more than once in the first million terms is 815,746.... which is the 929,837th term (following 33,801, which is the 114,120th and 929,836th term) and the 943,716th term (following 23,927, which is the 128,006th and 943,722nd term).
This is what I think he means: Suppose M is the maximum number of that particular sequence and it occurs as the pth term within the sequence. Then the number of possible sub sequences between the (p-M)th term to the (p-1) term is M to the M as there are M terms in that subsequence and for each of those terms, there are M possible values (0 to M-1) that a term could take
What if the periodic part is a repetition of 2 (or more) M sized blocks? Then there is no longer guaranteed to be a z at the end of the block preceding the ...,z,a,... (which is what lead to finding a periodic part one element prior and the contradiction). I'm also quite interested by what you can say doesn't happen. As other comments have pointed out (albeit as initial conditions), you couldn't find 1,1 anywhere in the sequence, this would imply there's a 1 immediately before, and 1 before that eventually leading to contradiction. Leading to the idea there's a set of sub-sequences that would never be seen.
Brady: what do we know about this sequence?
Neil Sloane: nothing.
Brady: great! Let's make a video about it!
We know how to make it.
Truttle1: what do we know about this programming language?
ais523: nothing.
Truttle1: great! Let's make a video about it!
@@anawesomepet But why outside of realizing it, (the sequence) do we need or use it?
??
@@official-obamaQuite rare finding a esolang enjoyer on random place
This man is a legend. I could listen to him talk about numbers forever
Definitely an enjoyable video!
You mean professor Farnsworth???
You mean guy who spouts the same boring sequence stuff all the time, all in a comedy accent?
By “forever”, do you mean ℵ₀ seconds or something greater, like, say, ℵ₁ seconds?
false.
"boy, that's a really great sequence" my favourite kind of person
??
"X to Z" mathematicians favourite drug
Not in the UK. There's not much of a market here for ecsta-zed.
@@rogerkearns8094 is this the reasoning behind zedd's name?
You are a better drug.
cursive Z, nonetheless. That's the strong stuff.
I once did x to zee and almost ended up zed
"Boy, that's a really great sequence!"
_better do math to it before anyone else_
that's a really great sequence you got there!
be a shame if someone...
*did* *math* *to* *it*
I know it is
666 likes.
I saw this right when he said it
- Did you do anything fun this weekend?
- Yeah
- Yeah? What?
- 5:42
Man that's some clever stuff
Amazing
Don't get it :/
@@kim15742 same
Kim, "X to Z" sounds like "ecstasy".
Ok, so after digging a little bit in the sequence, I wanted to share a bit of what I’ve found.
I started having in mind to stop when the numbers from 1 to 10 would have appeared but it took me a bit longer than I thought. I finally got a bit further and got the first 252 numbers of the sequence.
(I’ve done this on paper, no programming, so it’s possible I failed it at some point)
Here are the 56 numbers that appeared in order : 0, 1, 2, 6, 5, 4, 3, 9, 14, 15, 17, 11, 8, 42, 20, 32, 18, 7, 31, 33, 56, 19, 37, 46, 23, 21, 25, 52, 13, 62, 40, 36, 16, 27, 10, 92, 51, 131, 39, 12, 44, 34, 97, 72, 41, 78, 24, 105, 107, 167, 61, 26, 22, 127, 28 and 29.
One thing that I found funny with this sequence is that is has the tendency to quickly come back to a number that newly appeared. For exemple when the 9 shows up for the first time, it takes only 3 steps to appear again. Same for 7 and 31.
5, 6, 18 are taking 5 steps to appear a 2nd time, 107 takes 28 steps, etc.
But it doesn’t happen for every number, like for 14 that takes 131 steps to appear a 2nd time, but takes 4 steps to appear a 3rd time. ^^ 17 didn’t appear a second time for me even though it comes pretty early in the sequence.
It’s hard to find coherence in there but it’s strange to see more often that not new numbers reappearing pretty quickly even though there are still lot of numbers that haven’t appeared yet.
The second thing that surprise me a bit is the frequency of new numbers appearing, only takes about 4,5 steps (the longest chain of numbers between two 0s I’ve found is 8 numbers long (found it 2 times)) Thought it would take a bit longer but it’s pretty rare that a new number takes more than 6 steps to appear. But like I said, I only checked the 250 first numbers so I don’t know if it grows up, shrinks or stay pretty much the same if you go further and further.
I usually don’t really dig into that kind of stuff, mostly I listen to the video and continue my way elsewhere, but this time my curiosity hasn’t been fulfilled enough, so here I am writing this :p
It was worth the try.
Thanks Numberphile o/
Numberphile: Don't know
Me: * Gets spooked *
vsauce sound
funny
@@alveolate Moon Men
Has the spook
I just realized that adding 0 as the next term when there's a number you haven't seen before, isn't as arbitrary as I first thought: It's really just in agreement with the rule of writing down "how far back it occurred last time". When it's never occurred before, the last time it occurred was _right now,_ zero steps ago, so we add a zero. Awesome :D
Nice logic isn't it? Does it mean that the sequence can only start with 0 and no other number? Also the rule seems to mean that 0 can only occur twice in succession at the beginning of the sequence or immediately after the first number n if n is allowed to be non-0.
@@chrisg3030 I don't see any reason you couldn't start with something else than 0. Then the sequence will be different depending on which number we start with. A funny thought: Since we don't know whether every number will eventually appear, let's say that m is a number that never appears in the sequence. Then if you start with m, only then will you get the same sequence as the one where you just start with 0 :D
Similarly there's no reason why you couldn't add something else, say 2, when there's a number you haven't seen before. So 0 2 2 1 2 2 1 3 2 3 2 2 1 6 ... The original rule says "Add n when the number last occurred n places back", so when it's new - that is last occurred 0 places back - you add 0. With my variant it's the same but with the exception of 0 places back, in which case you add 2 but still use 2 for 2 places back as well. We still seem to get the same kind of sequence though (except in the case of 1, if we add 1 for new as well as 1 place back we just get endless 1's). Please check.
I immediately want to extend this to the negative and imaginary numbers
This needs more up votes.
One thing that can be proven about the sequence is that VE(n) < n for n > 0 (since the entire sequence has length n+1, the most number of moves back it could take is n, but VE(0)=0 and VE(1)=0, so you'll never go all the way back to VE(0) and thus VE(n) < n). So yeah, f(n) = n seems like a fairly good approximation of the growth of the sequence, but it is also an absolute upper bound on the sequence.
2:58 now that's some genuine enthusiasm, love it.
i was just thinking the same thing and looking for a comment about that. Warms my heart that people noticed
@@Lyle-xc9pg I felt tickled when he said it that way! Neil is the best
??
@@Triantalex the “yeahhhhh… I think it’s lovely”. Really has some genuine expression to it.
"oooh that's a really great sequince, let me analyze it before anyone else does" I'm gonna go with things only a mathematician would say for 500
Suddenly, Jeopardy.
Very farnsworth
Sequence*
This guy is so obsessed with weird series
I can relate. I wanted to analyze it myself before watching the rest of the video :)
Dr Sloane has such a relaxing voice and his love for sequences just radiates from him.
This is my new favorite sequence. I love self-descriptive sequences.
Nice , same (they're kinda like storing information about themselves)
This is my new favorite sequence because it's interesting, and also because my last name is part of the name!
Reminds me of the Recaman sequence (Numberphile vid), also dependent on whether a number is new or not.
But isn't there a sense in which any sequence obeying a rule is self-referencing?
Let's express the rule for the van Eck as "Add n when the current term last appears n places back". So if the current term is 1 and it last appeared 6 places back then we add 6. If the current term is 6 and it last appeared 0 places back (in other words it's never appeared before) we add 0.
Now let's change that rule a tad: "Add n when the current term FIRST appears n places back". If we start with 0 we go on 0 1 0 3 0 5 0 7 0 9 0 11 ..., a both boringly regular and not apparently self referencing sequence, even though our defining rule makes sound like it should be.
But in my example the first place of appearance of a term is never going to stop being just that, whereas the latest place of appearance of a van Eck term can change quite frequently. So perhaps we should talk instead of term-index variant and invariant sequences.
@@chrisg3030 i think the important distinction for a self-referential sequence" is when a series checks something other than the ordinality of a previous term. If you do something with the number other than use how big it is, it feels like using a meta-property of the sequence itself
Love Neil Sloane videos on Numberphile. Non convential maths at its very best.
I love these self-referencing number sequences. Reminds me of the Kolakoski sequence.
I could listen to him listing the sequence like he did in the first minute for hours
There's extra footage, right? _Please_ tell me there's extra footage.
I know right. I was immediately checking the description for the bonus video.
dont know
Don't know ;)
don't know 🤔
dont know
This is brilliant, it's so simple to think up, yet it's not been submitted before and so unpredictable. I really enjoyed this sequence.
I love it when Sloane is on the channel. His database inspired me to choose a maths major. I'm so excited for it!!
I would change the definition of Van Eck's sequence. The sequence doesn't begin from 0 necessarily. Then it is only 0-sequence but it can be N-sequence as well. Then the Van Eck's sequence family was created.
That's interesting, actually. They're related-if you start the sequence at n, it will look identical to the 0 sequence up to the first instance of n in the sequence, at which point it change completely. And the first different number will be much higher than anything around it, which could affect the shape of the large-scale triangle-my wild guess says its slope wouldn't change but its height would jump up at that point. Now I want to find out.
I did the graphing and I can't seem to find any patterns other than that initial outlier.
1-sequence: 1 0 0 1 3 0 3 2 0 3 3 1 8 0 5 0 2 9 0 3 9 3 2 6 0 6 2 4 0 4 2 4 2 2 1 23 0 8 25 0 3 19 0 3 3 1 11 0...
'Oh come on! How can you not know how fast it grows? Surely that's easy to prove! We just... okay maybe we.... what if....'
*Three hours later*
'Alright, you win this round...'
It does feel like there is a provable lower bound using the repeating argument described in the video. But it is probably super low, logarithmic in n or something.
@7:06
4 ways. specifically (+,+,+,+), (+,-,+,+), (-,+,-,-) (-,-,-,-)
I found this by the following logic chain:
1. 81 is already divisible by 3, therefore we only need to manipulate the pluses and minuses to preserve this property.
2. 9 is also divisible by 3, therefore it doesn't matter if it is added or subtracted, it will not change the remainder after division by 3.
3. 31, 13 and 4 are each numbers of the form 3x+1, therefore for the purposes of determining whether their sum will be divisible by 3, we need only concern ourselves with the '1' part.
4. the only way to add or subtract 3 1s to each other in any combination and end up with a number that is divisible by 3 is if either all of them are subtracted (-1-1-1=-3) or all of them are added (1+1+1=3), therefore, the first, third and fourth sign must match each other.
5. (4) combined with (2) implies that the second sign can be either plus or minus and the remaining ones must match each other but be either plus or minus and any such combination will work, this means we have 2*2=4 combinations
I love this guys enthusiasm.
Explaining a sequence with a totally unrelated poem. Love it!!
This is fascinating- it reminds me of John Conway's Look-&-Say Sequence.
The self describing sequence?
Like 0 , 10 , 1110, 3110, ... ??
Reminds me of Recaman, self-descriptive and also depends on whether a number is new or not, except you can't use it if not.
However this sequence gets boring if you have the 2 starting numbers be 1,1.
@@livedandletdie The second term is a lie, and we all know that you can derive anything from a false premise. :p
Or how about the roman version, it starts like this
I
II
III
IIII
IVI
IIIVII
IIIIIVIII
VIIVIIII
IVIIIIVIVI
IIIVIVIIVIIIVII
IIIIIVIIIVIIIIVIIIIIVIII
VIVIIIIIVIVIIVVIIVIIII
IVIIIVVIIVIIIVIIIIIVIIIIVIVI
Please keep us updated on this sequence, this is fascinating.
His smile at 3:50 says it all :)
Sloane is so relaxing to listen to.
Answer to the daily challenge problem:
4. It is a modular arithmetic question. 81 is divisible by 3 and so is 9. The other numbers each are divisible by 3 with a remainder of 1. All three of those must have either a plus or minus sign. But it must be the same sign for all three. Then thr nine can take a plus or minus and it is independent of the other one. So you have 2 independent choices with 2 options each. 2x2=4.
This channel is the channel which aided me to do very well in Mathematics, and is the channel responsible for my uprising interest in this subject!
I love this sequence, everytime I think it's going to repeat itself it doesn't.
Seriously, I keep seeing repeated patterns in it, but they're always in different sections and separated.
This guy is great. Love his enthusiasm.
There is something so calming about the way he basks in these sequences.
2:40 when your crush sends you their bionicle collection
I miss bionicle
Lunar arithmetic*
damn we are evrywhere. all hail bonkles
Brilliant question:
Mod 3, the question is:
0 ( ) 1 ( ) 0 ( ) 1 ( ) 1
Where ( ) should be + or -.
The maximum value of the expression is 3 and the minimum is -3, occurring when all the signs are + and - respectively (except for the sign before the 0, which can be either). This yields 2×2=4 possibilities. 0 cannot be achieved since the parity of the expression must be odd.
Would definitely like to see if there's any progress on this sequence
Don't know.
So I saw the title and clicked on the video, and I just glanced at the description for maybe a few hundred milliseconds, and I saw OEIS mentioned, and I thought "oh, nice, they've got the Sloane's entry for it." Then I watched the video and realized that they've also got *Sloane.* :-D
he always reminds me of Professor Farnsworth. I love it!
I see it. Now I can't unsee it.
Listen to this sequence in the library, it is amazing.
Neil: The obvious questions are…
Me: What set of circumstance led to someone creating such an arbitrary set of rules.
Boredom, probably
Well, pretty much all of math arose from bored people creating arbitrary sets of rules, and then figuring out what they did.
Creativity, folks.
Someone looking for an interesting sequence to submit to the number sequence encyclopedia.
@@letao12 the rules might be arbitrary but the relationships enable spaceflight
Is it starting to rain? Afraid so. Is this going to hurt? Afraid so. Are we out of coffee? Afraid so. Is the car totaled? Afraid so. Will this leave a scar? Afraid so. Hotel? Trivago.
GOSH DARNIT
Is this comment bland? Afraid so. Should this meme be left to die? Afraid so.
Is the comment above true? Afraid so.
i hate this
hotel? afraid so
r? afraid so
afraid so? afraid so
so afraid? afraid so
are you afraid so? afraid so
deez nuts? gottem
I'm going to answer on a new comment, cause I find the answer interesting by itself, to someone who remarked that if the sequence started with 1,1,... then the sequence would be periodic. The statement is true, but with this set of rules, the first number determines the sequence, and 1,1 is not a valid start for a sequence. In other words, all sequences generated with this rule start by x,0,... . However, we can actually verify that there are at least 2 such sequences that are "profoundly" different (i.e. one is not a subsequence of the other): 0,0,1,0,2,0,2,2,1,... and 1,0,0,1,3,0,3,2,0,3,3,1,8,0,... ("0,0" is a subsequence that appears exactly once on each sequence).
A "not profoundly different" sequence would be: -1,0,0,1,0,2,... , if we allow for x to be a negative integer.
With this I just realized that if 0,0,... does take all the positive integer values, then it might be "easy" to prove that x,0,... is a "profoundly different" sequence from y,0,... iff x!=y and both are natural numbers. Looking at it in the other way, if there's a value z that's not part of the sequence 0,0,... , then z,0,... is not "profoundly different" from 0,0,... .
Agreed.
There are some more interesting sequences with modified rules:
Add 1 to any new number. Subtract 1 from the number following a zero. That sequence looks just .. loopy. Very interesting.
This looked wrong at first -- then I realized that x!=y was supposed to be "x is not equal to y", not "x factorial is equal to y".
@@Ashebrethafe, haha... "factorial".
Funny how programmers have managed to decide on ways to type 'not equal' and understand eachother eg '!=', 'neq', '>
eq is the true way to write not equal for mathematicians
@@JNCressey I just use a custom keyboard layout that allows me to type symbols like ≠ ;)
7:02 Interesting question, I'm thinking 4?
31 mod 3 = 13 mod 3 = 4 mod 3 = 1
and
81 mod 3 = 9 mod 3 = 0
so if the result is divisible by 3 (ie. result mod 3 = 0) the signs in front of 31, 13, & 4 can be + or - but they must match. Then the sign in front of 9 can then be + or - so that makes 2*2=4 combinations.
Numberphile is my favourite channel
This series is amazing. Not intuitive, sort of alternating and unsolved. Reminds me of the 3n+1 problem, but in a more interesting and (probably also easier to solve) way
I love you neil sloane for oeis, it is very handy for an amateurish mathematician like me
I could sit and watch an animation showing each number getting added and counting the spaces back for ages, it's hypnotic and pleasing
the slope roughly equalling 1 is kinda blowing my mind.
Really shouldn't be that surprising. At any nth term x, x cannot be larger than n, because that would mean you would have to look back an amount of steps larger than the total amount of steps you have taken. Therefore, since the maximum value of x is equal to the value of n, drawing a line through all the peaks should give a line that approximately maps to y=x, or a slope of 1.
@@firstlast8858 Doesn't your argument show that the slope should be "less than or equal to 1", rather than "equal to 1" ?
@@firstlast8858 That's only half an argument. You've only explained why the slope cannot be above 1, not why it should be near 1. Indeed, since the sequence starts with 0, the maximum value of x is less than n. Further, it is easy to assume that x grows slower than n, so it isn't immediately evident that the slope would remain near 1.
@@BainesMkII Hmm, as soon as a number is "used" to look back to, it will never be used again. So eventually all the starting numbers must get "used" up. I wonder how fast the consecutive used-up numbers progresses right, because that could limit HOW MUCH less than 1 the slope is.
@@BobStein my guess, based on the first 173 numbers of the sequence is about 1/10
I guess there will never be an end to learning about these number sequences that make me think "well I could have thought of that"
I have watched this video a few times now and absolutely enjoy this video! This is now one of my favorite sequences, it's so delightful! 😀
The animation at 2:47 is pure magic. Also, YES, love this guy.
Anyone here from advent of code?
Yes! I was hoping there's be clever ways to speed up generation of the sequence haha. Seems I'll be running it for a few more hours yet!
@@petermarsh4578 There is a way to speed it up. Think about how you store your generated numbers and how you look them up.
This one took me particularly long to work out. I can't actually remember how I managed it.
5:44 "there might be other copies of z in the period"
Okay, but what if there aren't? I guess the same reasoning still holds, but it confused me when he didn't complete cover both possible cases, felt like a loose end.
Well there has to be at least one, because z is defined to be the last number in the period.
When he draws the first period that is the earliest point at which the period can start. So if there was no other z in the period then x would have to be the length of the period (it can't be any more than that length). But that means that we could draw the period one step earlier, because another z would have to occur just before the first period, and that's a contradiction. So z must occur somewhere else within the period. It's essentially the same argument but with x=a. Sloane generalises by looking at the first z in the period, wherever that may be.
Well if there wasn't, you could just redefine the period to be two iterations of the pattern instead which would mean you'd have two copies of z and the proof would still follow.
I was wondering the same, but:
there's at least one z (in the end of the period) and the argument also works for it.
I don't understand the final step of the proof. Specifically, the statement at 6:23 confused me. He didn't push back the beginning of the period; he demonstrated that there is a z in the previous period.
which means there always has to be a period before the one you're looking at right now, so the periods could never start due to a bootstrapping paradox since the sequence has a beginning.
I don't know why he made it so complex..
So lets say a period looks like C D E ... L M N. Since the second period starts with C this means the length of the period is C and before the first period there has to be a N. But this means the period now looks like N C D E ... L M and so on, so there can only be a repeating pattern if it goes all the way to the beginning. The problem here is there can't be any 0's in a period because zeroes are only used when it's a new number and a repetitive sequence can't have anything new. The sequence has by design 0s so periodicity is not possible.
2:17 there is a poem out there that does follow this pattern of "don't know" at the end kinda, "Vietnam" by Wisława Szymborska.
it's a sad poem, but it's really nice :)
MOAR OF THIS GUY PLS
this is a super cool sequence, I hope one day someone else wants to talk to this channel about discoveries made about it!
If you guys are interested in playing with this sequence, I wrote some javascript code that you can use to generate terms quite easily:
function van_eck(terms){
function find_index_in_array_from_back(arr, i){
for(var c = arr.length-1; c >= 0; c--){
if(arr[c] == i){
return c;
}
}
return -1;
}
var s = [0];
var s_1 = 0;
for(var c = 0; c < terms; c++){
var index = find_index_in_array_from_back(s, s_1);
var distance_back = s.length - index;
s.push(s_1);
if(index >= 0){
s_1 = distance_back;
}else{
s_1 = 0;
}
}
return [s, s_1];
}
In terms of playing with it, you can, for example:
console.log(Array.from(new Set(van_eck(100000)[0])).sort((a,b)=> a - b))
You can see all the unique numbers within the fist 100000 terms of the sequence. By matching up the numbers with the indexes in the output, we can see that all the numbers up to somewhere in the 1500s are included in this number of terms (as well as several numbers beyond, but EVERY whole number up to there is included).
If we do:
console.log(Array.from(new Set(van_eck(1000000)[0])).sort((a,b)=> a - b))
Every number up to somewhere in the 8000s is included, and many more beyond.
Anyway, that's just one idea, you can of course do whatever you want.
I had some fun playing around with the sequence, so if you want to play with it, the code is there for you, just do CTRL+i in chrome (or bring up developer tools in any browser) go over to the console, paste it in, and away you go!
We also know that the nth number cant be larger than n, because there arent more than n steps before n. Therefore the fastest way for the sequence to grow is linearly. it could still be root of n or log n, but n^2 or 2^n are ruled out.
Bless this man
2:15 accidental poetry by Neil Sloane
About the demonstration "there might be some z's in the middle" and after thag absumption proving a contradiction seems weak, why add a z inside which is the same the last number of the period, and instead not take x directly (or z and then the a is x)
Love this guy’s explanations
Another great video. Thanks for producing this extremely engaging material.
He's wearing a Pink Floyd shirt! One more reason he's a badass.
hahah he wore a jimi hendrix shirt in another episode! a true beast
You worship the establishment too much
@@StefanReich You worship my root chakra too much
3:51 for a DSOTM T-shirt. What a legend Neil Sloane is
@@InzaneFlippers my favorite
Neil Sloane is the piper at the gates of dawn.
I have a hunch (meaning I have no idea how to demonstrate anything) that this is related to the primes.
Why?
Because each time a new number enters, the sequence produces a zero. Just like, when you do an Eratosthenes sieve, you write off any ‘’new’’ factor you encounter.
I love numbers and how they relate with each other. I never heard of this. Has anyone ever programmed a computer to see how far you can go?
What I am fond of saying is, "The more I learn, the less I don't know!" (Or realize I don't know.)
Sequence:
Boring, boring, boring, ohmygoodnesswhathappenedthere
seems to have a slope 1 because it is bounded by a line with slope 1 (at position n, the furthest back you could have seen a number was n steps ago)
5:44 A claim is made that “there might be a copy of z” within the range of one of the periods, after which it is proven that this sequence grows endlessly. However, I do not see why there is a guarantee that any such copy of z ever has to appear between period markers x and z. Can anyone clarify?
I wish I had as much enthusiasm as this guy explaining math
Fascinating! I have never seen anything quite like this before!
This was a fun programming challenge. Created an algorithm to compute n values in linear time!
Van Eck: You know nothing, Neil Sloane XD
Adrian Pietkiewicz Neil: afraid so :(
It's 1:17 AM me right now. Some might say that this video was my night's watch.
@@galgrunfeld9954
I couldn't Clegane on it fast enough.
Ohh gosh, that's an amazing sequence!And there are lots of questions rising:
Does the sequence has infinite non zero terms? how often does each term appear? Does each positive integer appear in there? Can we find an algebraic expression for it? In order to find the n-th term, do we really need to know all the previous terms?
So many questions, i love it!
Definitely want to see more on this sequence
2:51 Analyzing A Sequence Before Somebody Else Does PRANK *GONE WRONG* *COPS CALLED* *OMG*
Love the sequence,
Love the proof,
Love the Pink Floyd shirt!!
This man is 80 years old. Incredible.
6:30 i think the animation is a little bit misleading. all that the proof says is that x is not the first number of the period but the z before the first x is. Which leads to a contradiction to the assumption that x is the first number of the period. Of course you can repeat it with every number before that, but that is not what Neil Sloane says and is not neccessary. (He says: The period began one step earlier)
I’m a simple man. I see Neil Sloan, I like.
I saw “sequence” and knew it would be Neil!
In case anyone else was confused by the last step of his proof, it is stated a little more clearly in the OEIS:
The periodic part does not contain any zeros.
Suppose the period has length p, and starts at term r, with a(r)=x, ..., a(r+p-1)=z, a(r+p)=x, ... There is another z after q
What I don't follow is why for it to be bounded it has to be periodic. I get the M^M possible permutations of M numbers 1-M. The sequence will have to repeat blocks at some point, but pi has that same restriction (albeit with single numbers and not blocks). There's only 10 numbers available there and it manages to be non-periodic and carry on infinitely. Why could this sequence not 'settle' to repeating 10 or more blocks of M numbers in a similar non-periodic way?
EDIT: And of course I realise immediately after you could never repeat 2 blocks or this leads to the same logic as the original proof, that this will lead to a periodic part one step back etc. And I assume you can only rearrange the blocks a finite number of ways before being forced to either repeat 2 or repeat the order.
@@lyndenw2240 Assuming the sequence is bounded, the next number in the sequence is always determined by at most the past M numbers. So if a chunk of M numbers occurs a second time, the number immediately following that chunk will be the same the first and second times. Successive application of the same argument shows that the whole sequence would be periodic. Because there are only finitely many (M*M) possible chunks of length M, some chunk is guaranteed to repeat.
"Boy that's a great sequence, I'm gonna analyze that before anyone does"
Always gets me
This man is an excellent teller.
5:42
Extasy?
That’s not how you spell ecstasy
"X to Z."
Well, I don't know if the slope is actually 1, because I'm getting no numbers in the sequence at all that are more than the position in the sequence, that is to say, X>Y as a general rule.
If we graph the known points of increase, to get the maximum values, you have (1,0)(3,1)(5,2)(10,6)(24,9)(30,14).and (56,20)
That starts at a slope of 0.5, and slows to a rate as low as 0.23, leaving an expected value of Y (the highest number that you should get if you randomly choose an X value) of less than X/2
Edit: I might have messed up. I picked the points with the highest X values in a group, when I should have picked the lowest x values Fixing issues above.
2:32 - Sloane claims to be an editor at OEIS. What is he hiding? The fact he is actually the founder
As a Brazilian, I'm really curious to know what are those "Brazil" books in the background.
Sequence that starts from 2 numbers - "1,1" - can be periodic
@@mxmdabeast6047 "sequence that starts"
That would be an illegal starting pair by the definition of the sequence. If you start with a 1, the next number has to be 0. Note that the sequence as-shown doesn't start with 0,0 but rather just 0 and proceeds from there.
@@MattStum
Definition of sequence is the mechanism by which new numbers are added
The starting sequence is free parameters that allow to generate different strings of same ruleset
NoName the rule is if you haven’t seen the number before then you write a 0. You haven’t seen 1 before so the sequence starts 1,0,0,1,3,...
Can "1,1" ever appear anywhere in the sequence?
I can't prove it grows linearly, but it is quite simple to prove limsup a(n)/√n ≥ 1
Proof: Whenever a(n)=0, either there have been √n zeros in the sequence, thus √n new distinct numbers (and at least one bigger than √n), or there have been less than √n zeros in the sequence, and thus there is at least one gap between two zeros which is at least √n wide.
Even though this is very far from linear, I haven't seen any lower bound yet, so let's start here. And go back to find a better one!
I would consider it linear growth. The value at a(n) is always less than n. In my reasoning this rules out exponential growth. I would certainly like to prove that the growth approaches { y = 0.809 x }.
I don’t really understand the part from 4:11 until 5:15. If i’m not mistaken, he explains that if a sequence could have a highest number, that would mean the sequence has to be periodic. But I don’t get why that is. Would somebody be kind enough to help me understand? :) thanks!
If you begin with 1, 1, and then employ this rule thereafter, you generate a periodic sequence. Are there any other prefixes for which the Van Eck rule generates a periodic sequence?
The answer is 4
You have 2 numbers divisible by 3, and three numbers divisible by 3 ONLY if added together
or subtracted from one another; so the 2 that are divisible by 3 can be added together or subtracted from each other and the other 3 can be added to or subtracted from each result... 4 possible answers to the +/- question
me clicking on a video about sequences: :)
Me seeing its neil sloane: :D
I just admire him so much!
To print the first 1000 numbers with python 🐍
nMAX=1000
L=[0,0]
n=2
while n0 and b==0):
if L[a]==x:
b=1
a=a-1
if a==0:
L.append(0)
print(0)
else:
L.append(n-2-a)
print(n-2-a)
n=n+1
thanks, stranger whom i never met before. you truly are a genius. and also sexy.
It appears that in the first million terms, the smallest number not to appear in the sequence is 8,756. The largest number to appear more than once in the first million terms is 815,746.... which is the 929,837th term (following 33,801, which is the 114,120th and 929,836th term) and the 943,716th term (following 23,927, which is the 128,006th and 943,722nd term).
4:50 Why would there be only m to the M possible terms?
This is what I think he means:
Suppose M is the maximum number of that particular sequence and it occurs as the pth term within the sequence. Then the number of possible sub sequences between the (p-M)th term to the (p-1) term is M to the M as there are M terms in that subsequence and for each of those terms, there are M possible values (0 to M-1) that a term could take
What if the periodic part is a repetition of 2 (or more) M sized blocks? Then there is no longer guaranteed to be a z at the end of the block preceding the ...,z,a,... (which is what lead to finding a periodic part one element prior and the contradiction).
I'm also quite interested by what you can say doesn't happen. As other comments have pointed out (albeit as initial conditions), you couldn't find 1,1 anywhere in the sequence, this would imply there's a 1 immediately before, and 1 before that eventually leading to contradiction. Leading to the idea there's a set of sub-sequences that would never be seen.