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
'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.
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.
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.
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 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
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.
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', '>
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.
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
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
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!
But how do we know that 1 1 doesn't appear at some as yet unexplored reaches of the sequence, with the result that indeed all subsequent terms are 1? I guess because 1 only ever appears after two consecutive identical terms, so the next term must always be bigger than 1 since that 1 must have last appeared at least 3 places back, or 0 places back (if new) in which case the next term will be 0. So you can never get two consecutive 1s in the first place. Another interesting question is can more than two consecutive identical terms ever appear?
I have a solution for you. Perform the sequence on base 2. Look at the number of steps it takes to complete each number of digits. Notice as you near completion a subset of digits, you begin a new subset of larger digits. Let A be the set of integers in base 2 that have 1 digit. Let B be the set of integers in base 2 that have 2 digits.... Let N be the set of integers in base 2 that have N+1 digits. I conjecture that as van eck sequence completes the N set after m steps, the m+1 step is a number in the N+1 set and the last number in N is within m+1 to m+k steps. Where m+k steps nears completion of the N+1 subset. That is to say, as the sequence completes a subset of integers, it begins the next larges subset and within the completion of that subset is the last integer that completes the prior subset. I suppose you could order the numbers in N subsets by difficulty or some other ordering other than size. I also surmise that if its true in base 2, then each subsequent base it will also be true as we don't change anything about the numbers only the number of symbols we have in each subset of digits.
I have a suggestion: This is called The five sticks problem Each stick is valued 10, 20, 30, 40 and 50 You can not repeat sticks or made an earlier group of sticks that has existed, for instance: 10, 30 and 40 You can not make 30 40 and 10 cause its the same But you could do 10, 30 Or any subgroup (is not like Tree(3)) The question is: In how many ways you can get each result of the adittion of all points worthed each stick? Each result is done like this: Case X: 20, 30 and 50 20 + 30 + 50 = 100 So 100 could be done like taht, but also 50, 40 and 10 and others... I'd love to see a video of that problem, thanks! ☺️
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.)
Since the Nth term can never be larger than N, we at least know it cant grow faster than linearly over the long term. I've established an upper bound on it's growth! :D *pats self on back ironically*
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).
The proof that the sequence isn't cyclic only works if there is a first copy of the last number. Although Mr Sloane may know, for me, the question "What is the guarantee for there to be another copy of the last number?" The result is "Don't know"
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.
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!
You will never have two ones in a row or else the sequence will repeat ones forever. I don’t think that’s possible though because in order to get a one you need to numbers in a row, and then the one after that can’t be a one. It would have to be at least a three.
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
VanSickle's Conjecture is that every discovery in mathematics will eventually have an application outside of mathematics. I'm wondering what Van Eck's sequence will be used for. (Proof of VanSickle's Conjecture is left as an exercise for the student.)
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 }.
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?
Okay so I was wondering if I could code this, and here it is (in JavaScript): /** * Get a van Eck sequence * * @param {Array} s - The seqeunce. * @param {Number} n - The number that was calculated by the previous itteration * and is to be added to the sequence. * @return {Array} */ const vanEck = (s = [], n = 0) => { // Get the previous index of n in the sequence s. let previousIndex = s.lastIndexOf(n); // Push n into the sequence. s.push(n); // Get the length of the sequence. let l = s.length; // We want to do a maximum of 5000 itterations. if (l > 100) return s; // If n did not occur in s, we want to pass 0 as an argument, else we want to // pass the steps to the previous occurence to the current occurence. return (previousIndex === -1) ? vanEck(s, 0) : vanEck(s, l - (previousIndex + 1)); }
Really nice! Here's a quick and dirty Python 3 implementation: def vaneck(start=0): vdict = {} last = start while True: # lazily generate the entire, infinitely long list yield last vdict = {key: val+1 for key, val in vdict.items()} try: vdict[last], last = 0, vdict[last] # order matters! except KeyError: vdict[last] = last = 0 van = vaneck() # you can specify a starting value for the generalized sequence. For instance: van = vaneck(1) numofterms = 1000 # increase to generate more of the list for n, v in zip(range(numofterms), van): print(n, v)
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.
I knew at a certain point that I really like this man. Not only is he a great maths enthusiast and genius, he also seems to like the dark side of the moon album by Pink Floyd by looking at his shirt. That sealed the deal for me.
Excel formula to generate the sequence if you want play. put 0 in cells A1 and A2, copy formula in A3 and copy down as far as you like. =IF(COUNTIF($A$1:A1,A2)>0,ROW(A2)-AGGREGATE(14,6,ROW($A$1:A1)/($A$1:A1=A2),1),0)
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.
"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
Numberphile: Don't know
Me: * Gets spooked *
vsauce sound
funny
@@alveolate Moon Men
Has the spook
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.
"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 :)
"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
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
Dr Sloane has such a relaxing voice and his love for sequences just radiates from him.
'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.
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
Love Neil Sloane videos on Numberphile. Non convential maths at its very best.
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...
I love it when Sloane is on the channel. His database inspired me to choose a maths major. I'm so excited for it!!
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.
Please keep us updated on this sequence, this is fascinating.
I could listen to him listing the sequence like he did in the first minute for hours
he always reminds me of Professor Farnsworth. I love it!
I see it. Now I can't unsee it.
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
Sloane is so relaxing to listen to.
I love this guys enthusiasm.
Explaining a sequence with a totally unrelated poem. Love it!!
Would definitely like to see if there's any progress on this sequence
Don't know.
This guy is great. Love his enthusiasm.
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.
MOAR OF THIS GUY PLS
I could sit and watch an animation showing each number getting added and counting the spaces back for ages, it's hypnotic and pleasing
2:40 when your crush sends you their bionicle collection
I miss bionicle
Lunar arithmetic*
damn we are evrywhere. all hail bonkles
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"
Listen to this sequence in the library, it is amazing.
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
2:15 accidental poetry by Neil Sloane
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.
Numberphile is my favourite channel
Hello advent of code folks :)
There is something so calming about the way he basks in these sequences.
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 ≠ ;)
Neil Sloane is the piper at the gates of dawn.
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.
Love this guy’s explanations
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.
Don't know ♥
All the teen titans except Robin: ugh ugh UGH!!!
Robin: PURE UNDERSTANDING (or something like that)
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
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 feel that this sequence and its name could be made into a book in the writing style of Dr.Seuss.
We have seen a zero before.
We saw it one minute ago!
And when you've seen the number before,
the next term is how far back you saw it, y'know?
this is a super cool sequence, I hope one day someone else wants to talk to this channel about discoveries made about it!
I feel like this is another video which is going to inspire a person to “solve” this sequence.
I saw “sequence” and knew it would be Neil!
Definitely want to see more on this sequence
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.
That is such a poetic sequence.
Love the sequence,
Love the proof,
Love the Pink Floyd shirt!!
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!
I guess we can also say that two consecutive 1s never appear because once '11' appears all subsequent terms are 1.
But how do we know that 1 1 doesn't appear at some as yet unexplored reaches of the sequence, with the result that indeed all subsequent terms are 1? I guess because 1 only ever appears after two consecutive identical terms, so the next term must always be bigger than 1 since that 1 must have last appeared at least 3 places back, or 0 places back (if new) in which case the next term will be 0. So you can never get two consecutive 1s in the first place. Another interesting question is can more than two consecutive identical terms ever appear?
Neil Sloan playlist!
I have a solution for you.
Perform the sequence on base 2. Look at the number of steps it takes to complete each number of digits. Notice as you near completion a subset of digits, you begin a new subset of larger digits.
Let A be the set of integers in base 2 that have 1 digit. Let B be the set of integers in base 2 that have 2 digits.... Let N be the set of integers in base 2 that have N+1 digits.
I conjecture that as van eck sequence completes the N set after m steps, the m+1 step is a number in the N+1 set and the last number in N is within m+1 to m+k steps. Where m+k steps nears completion of the N+1 subset.
That is to say, as the sequence completes a subset of integers, it begins the next larges subset and within the completion of that subset is the last integer that completes the prior subset.
I suppose you could order the numbers in N subsets by difficulty or some other ordering other than size. I also surmise that if its true in base 2, then each subsequent base it will also be true as we don't change anything about the numbers only the number of symbols we have in each subset of digits.
I have a suggestion:
This is called
The five sticks problem
Each stick is valued 10, 20, 30, 40 and 50
You can not repeat sticks or made an earlier group of sticks that has existed, for instance: 10, 30 and 40
You can not make 30 40 and 10 cause its the same
But you could do 10, 30
Or any subgroup (is not like Tree(3))
The question is: In how many ways you can get each result of the adittion of all points worthed each stick?
Each result is done like this:
Case X: 20, 30 and 50
20 + 30 + 50 = 100
So 100 could be done like taht, but also 50, 40 and 10 and others...
I'd love to see a video of that problem, thanks! ☺️
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.)
The less that is known about a sequence the happier Neil is. 🙂
There’s a less interesting sequence known as the Xan Acks sequence.
J J it’s just a series of Z’s
I love you both, guys.
The video is about Van Eck's sequence. What are you talking about?
@PUN not genius - The video is about Van Eck’s sequence. What are YOU talking about
Since the Nth term can never be larger than N, we at least know it cant grow faster than linearly over the long term. I've established an upper bound on it's growth! :D *pats self on back ironically*
Fascinating! I have never seen anything quite like this before!
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).
The proof that the sequence isn't cyclic only works if there is a first copy of the last number. Although Mr Sloane may know, for me, the question "What is the guarantee for there to be another copy of the last number?" The result is "Don't know"
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 wish I had as much enthusiasm as this guy explaining math
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!
The animation at 2:47 is pure magic. Also, YES, love this guy.
I love simplicity of this sequence.
You will never have two ones in a row or else the sequence will repeat ones forever. I don’t think that’s possible though because in order to get a one you need to numbers in a row, and then the one after that can’t be a one. It would have to be at least a three.
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
This man is 80 years old. Incredible.
That is a great sequence.
VanSickle's Conjecture is that every discovery in mathematics will eventually have an application outside of mathematics. I'm wondering what Van Eck's sequence will be used for. (Proof of VanSickle's Conjecture is left as an exercise for the student.)
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 }.
we want more of neil!
This guy really likes his integer sequences.
Another great video. Thanks for producing this extremely engaging material.
As a Brazilian, I'm really curious to know what are those "Brazil" books in the background.
I’m a simple man. I see Neil Sloan, I like.
2:43 when you see a sequence
2:51 Analyzing A Sequence Before Somebody Else Does PRANK *GONE WRONG* *COPS CALLED* *OMG*
Brilliant sequence
This man is an excellent teller.
MOAR NEIL
Surprisingly interesting.
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?
Has my brain just exploded? Don't know.
why is nobody talking about the fact that this man is writing down book names ON THE PAGES
Okay so I was wondering if I could code this, and here it is (in JavaScript):
/**
* Get a van Eck sequence
*
* @param {Array} s - The seqeunce.
* @param {Number} n - The number that was calculated by the previous itteration
* and is to be added to the sequence.
* @return {Array}
*/
const vanEck = (s = [], n = 0) => {
// Get the previous index of n in the sequence s.
let previousIndex = s.lastIndexOf(n);
// Push n into the sequence.
s.push(n);
// Get the length of the sequence.
let l = s.length;
// We want to do a maximum of 5000 itterations.
if (l > 100) return s;
// If n did not occur in s, we want to pass 0 as an argument, else we want to
// pass the steps to the previous occurence to the current occurence.
return (previousIndex === -1) ? vanEck(s, 0) : vanEck(s, l - (previousIndex + 1));
}
Question is... What's its application?
_I can already hear the surge of "Don't know"s flooding in._
Really nice! Here's a quick and dirty Python 3 implementation:
def vaneck(start=0):
vdict = {}
last = start
while True:
# lazily generate the entire, infinitely long list
yield last
vdict = {key: val+1 for key, val in vdict.items()}
try:
vdict[last], last = 0, vdict[last] # order matters!
except KeyError:
vdict[last] = last = 0
van = vaneck()
# you can specify a starting value for the generalized sequence. For instance: van = vaneck(1)
numofterms = 1000 # increase to generate more of the list
for n, v in zip(range(numofterms), van):
print(n, v)
:o using errors to control logic flow. You weren't lying when you said dirty.
@@JNCressey As far as I know this is actually common practice in Python? But sure, yeah
I don't understand this video at all but keep up the great work!
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.
Lovely!
I knew at a certain point that I really like this man. Not only is he a great maths enthusiast and genius, he also seems to like the dark side of the moon album by Pink Floyd by looking at his shirt. That sealed the deal for me.
Excel formula to generate the sequence if you want play. put 0 in cells A1 and A2, copy formula in A3 and copy down as far as you like.
=IF(COUNTIF($A$1:A1,A2)>0,ROW(A2)-AGGREGATE(14,6,ROW($A$1:A1)/($A$1:A1=A2),1),0)
Neil's videos are amongst the most interesting on Numberphile but his sheets of brown paper have got to be the messiest.