can anyone explain this differently? "There exists a non-self-intersecting path starting from this node where N is the sum of the weights of the edges on that path. Multiple numbers indicate multiple paths that may overlap." Not quite catching how it relates to the numbers in the graph
@@ChrisTian-uw9tq You can follow any path and choose when to stop. The edge weights you pass (not the nodes) need to sum to the number (or one of the numbers) of the dark green node.
I've come to the comments section to write how happy I am to see Dr James Grime again on Numberphile and how much he's been missed, but I see everyone's done the same thing already!
I always remember triangular and tetrahedral numbers because of the song 12 Days of Christmas. If you interpret the lyrics as listing all gifts up to that point (including previous days), then the running total of gifts is the first twelve triangular numbers. If instead you interpret it as listing the gifts for only that day (i.e. the gifts from all previous days are given again, leading to, e.g., 12 partridges in 12 pear trees) the running total of gifts is the first 12 tetrahedral numbers.
@@shruggzdastr8-facedclown I think so. It took my a while to realize that since it isn’t used as profanity (or really at all) in my dialect of English.
@@JamesDavy2009 Honestly the two were so important that listing any one thing they did as an example feels like it can only ever understate their contribution. Even that formula is just one example of an expression that drops out of what is an entire mathematical framework that Euler pretty much constructed from scratch, and that entire framework is just scratching the surface of his contributions to mathematics.
Classic Numberphile with the OG presenter! ❤❤❤ I'm happy to see James again, being guest in other channels. Hopefully he'll upload new video in his own. 🤞🏼
James' closing comments are spot on. I was in high school (late 1980s for me; my brain is very middle-aged now) when I found the pattern of adding consecutive odd numbers to generate the square numbers, and then I figured out that the Nth level difference between consecutive N-dimensional numbers was N! (N factorial)... it's easiest to see this with the square/odd numbers, in which adding 2! starting at 1 generates the odd numbers. I found some hiccups in the first few iterations at each new power, but in general the pattern normalized at N^N.
My favorite tetrahedral number fact: The numbers along the finite diagonals of the multiplication table sum to the tetrahedral numbers. 1, 2+2, 3+4+3,4+6+6+4, ...
Fun fact: 2024's the only tetrahedral year all our lives~ And there's a book all about triangles coming out later this year! Seems like a triangle-y type of year~
True, but 2024 is also the only year we'll live through that is also a dodecahedral number and the first one since 1330. Every (3n + 1)th triangular number is the nth dodecahedral number.
It’s incredible that to this day, every episode gets its own special animation to make visualize the lesson in a delightful way. Stop motion!! Brady you animate so well!
I don't know exactly why but this is the most beautiful fundamental proof I've stumbled upon in Mathematics thus far. Thanks so much for making this video!
My favorite tetrahedral number is 4060. It is the 28th, and it is exactly 10 times bigger than the 28th triangular number which is 406. And 28 itself is a triangular number
Fun fact that i just found out: If you, let's say, color tetrahedral numbers in Pascal's triangle, you'll notice a straight line of colored numbers! This also works with triangular numbers and tentatopal numbers (numbers that can be arranged as a hupertetrahedron, though I'm not sure tentatopal is the correct name)
i've used triangular numbers to verify if a group of unique integers (in any order) was a gapless sequence or not. i was goofing around with some very basic arithmetic and i kept getting results that were oddly familiar. they turned out to be triangular numbers! around this time i had just been introduced to triangular numbers from numberphile! my specific use case was to determine if a set of years had gaps in it. turned out that there were much easier ways for me to do this programmatically with code, but i'm still proud of having such an epiphany as a non-mathematician. i have a working demo and explanation that i can link to, but i don't want this comment to go to spam jail! basically, the formula is this: `(max(set) * length(set)) - sum(set) = T(length(set) - 1)` where `T(n) = (n * (n + 1)) / 2`. `length` is the amount of entries in the `set` of unique integers.
It is a cool find and definitely works assuming the integers are unique, however, if you know the maximum you probably also know the minimum and thus max(set) - min(set) = length(set) - 1 is likely easier to check.
@@benjaminpedersen9548 lol of course i was overthinking it! it's funny because i did think of something like this but i must've forgotten to -1 from the length before i derailed and went on this magical journey. also, i almost immediately found another way to do this leveraging the native features of the programming language i was using. i ended up not using my original idea at all. but i won't let that take away the epiphany i got from this "discovery", however useless it may be. 🤣 thank you for the simplification!
11:33 this is exactly why I started working on the Collatz conjecture. I knew I'd learn a lot by thinking about the numbers and how they connect, and I was right.
The animations are great, but the synth effects i liked even more. Reminded me of those VHSes math teachers might put on in the 90s showing weird math ideas.
I think it makes sense to me that it doesn't require more than n n-gonal numbers. Here's my hand wavy intuition/psuedo-proof: Lemma: any sequence of n-gonal numbers starts as "1, n, ...". This is almost by definition: you start with 1, then add as many red checkers as it takes to make n sides. Of course, that's n checkers total. So the second number in the sequence is n. So now let's just keep adding checkers (start with 1, then 2, etc.) to see how to arrange them into at most n n-gonal numbers. If we add 1 checker, it might take 1 more n-gonal number. If we add 2, it might take 2 more n-gonal numbers (a 1 and another separate 1). Once we get to adding n more checkers, then it only needs 1 more n-gonal number, because those extra n checkers can be arranged into 1 "pile" (the lemma). So this shows that every n new checkers we add, it sort of collapses back down to one extra pile. Of course, that alone doesn't necessarily mean the "collapsing" keeps it under n piles *forever*, but it's some sort of intuition. I wonder how close this is to the real proof, if at all
Given that you seem to need at most 5 tetrahedral numbers to construct any number and at most nine cubes, it would seem that one would in general need at most n+1 3D-numbers to construct any number, where n is the number of vertexes the 3D-number has.
4:19 Funny: The first way that occurred to me was one you didn't mention. Seeing as 4|28, I divided it by 4, getting 7=4+1+1+1, then enlarged, getting 28=16+4+4+4.
I have encountered a lot of content on this channel where people have checked a conjecture up to a very large number but with no proof, i think it would be rather more useful to learn about all of the anomalies unproven conjectures which even after checking it up to very high numbers would eventually show something unexpected. Knowing about all of the anomalous unexpectancies would give one a good head start approaching any new theories.
The difference between the same (in order like the 5th pentagonal and the 5th hexagonal) pentagonal and hexagonal number is a triangular number and then the difference between the next pentagonal and hexagonal numbers is the next triangular number Same goes for square and pentagonal Triangular and square etc Very interesting
It might not be the main point of the video, but, I am really enjoying the sounds in the animations. Assuming these where created by the animator, this is really classy sound design, very buchla-esque / synthi etc. It really adds to a great video; always good to see James Grime. Like +1
What a coincidence, one of yesterday's videos on friends of the channel Cracking The Cryptic involved a puzzle where the solution path touched on tetrahedral numbers!
I imagine the way to prove the conjectures is through the jumps between singles. So for example with the triangle ones the jump from 1 to 3 is 2, 3 to 6 is 3, 6 to 10 is 4, 5 the next, 6 the next, you get the picture. Presumably the numbers between will only refer the the Ngonals that came before.
There are all these conjectures that hold true as far as we can calculate, but have not been definitively Proven. I wonder if there is a conjecture that was proven wrong by a hundred digit number or something like that. And yup, I see James, I click.
1. So if we name triangle-, square-, pentagonal- etc numbers as "plane" numbers; 2. And we have proof that we can write any whole number as sum of 1n of n-numbers for "plane" numbers; 3. Also we can name tetrahedral-, cube-, dodecahedral- etc numbers as "volume" numbers; Could there be relation between shape of plane and quantity of planes to describe how many "volume" numbers we need for different shape of volumes? Or something further beyond: relation between quantity of planes and volumes, and shape of these planes and volumes for description of "hyperspace" numbers?
I am new to this problem. What I see is that any way you attempt it, you will require 5 separate sequential logical axioms to describe the full body of any tetrahedron.
This all makes sense from very basic number theory, or more clearly, from the construction of the integers themselves. Normally the integers are constructed by incrementing them by some unit size, from some initial value. Using that operation, and the unit size of 1, and the initial value of 0, the positive integers can be constructed. Clearly if this process were altered, the resulting sequence would vary from that of the sequential positive whole numbers in some way. Like by changing the unit size, you can produce multiples of some integer, or you can produce a cyclical pattern of ratios, or you can produce a sequence of the harmonics of some irrational or transcendental number. Another way to change that process would be to adjust the algorithm, like by altering the unit size according to the step number. By incrementing the unit size by the step number, you get the triangular numbers. By incrementing the unit size by two times the step number, you get the square numbers. Likewise, by incrementing it by 3n, where n is the step number, you get the pentagonal numbers. In general, for polygonal number P with m sides and step count n, and initial value of 0, P(n) + 1 + n(m-2) = P(n+1) ... for triangular numbers, this produces the following: 0+1+0*1 = 1 1+1+1*1 = 3 3+1+2*1 = 6 6+1+3*1 = 10 ... and for squares it produces: 0+1+0*2 = 1 1+1+1*2 = 4 4+1+2*2 = 9 9+1+3*2 = 16 ... So, having said all that, since these are variations on the basic constructive sequence that builds all positive integers, it follows that they are going to have one of those altered sequences as a result, with characteristic gaps in the integers they produce. These gaps get larger, so they aren't some cyclical pattern produced by the beat ratio of some rational number. Also they still consist only of whole numbers, and so it doesn't produce the patterns made by transcendental or irrational numbers. So then, these are transformations applied to the basic integer construction operation that are both aperiodic, and that produce nothing but integers. Since they are aperiodic, and only produce integers, the outputs up to and including some step can all be combined in various ways to produce different sums. And since they themselves are comprised of elements that are aperiodic in origin (as opposed to a list of elements where each is the same distance apart), the number of possible sums from combining them grows nontrivially (if they were equidistant, there would be far fewer unique results, as more combinations would be redundant). I lack the rigor to connect the dots the rest of the way, it seems very intuitive that, because the gaps grow at the rate they grow, and because the list of unique elements from which to make sums also grows at a rate influenced by the nonperiodic growth of those gaps, those two would grow in balance, and that the gaps created by the sequence would never grow fast enough such that they could not be filled in using the same number of elements or less.
Another way to think of this, which makes it much more intuitive, is to think of this as a way to describe numerical systems where the base changes, such that each digit uses a different modular value.. The task then is to see how many bits are required to write each integer in this system. Like for triangular numbers, you get digits corresponding to 1s, 3s, 6s, 10s, 15s, etc... to match it to normal binary representation, with the largest values on the left, you'd get the mirror order of that though, so an array like such: [15s, 10s, 6s, 3s, 1s], and instead of each value in the array being a 1 or a 0, it'd be the count of how many of that number. I'm fairly sure some proof could be constructed to show that for any n-gon, you can create an n-ary numerical system using its n-gonal numbers, where all integers can be represented while keeping the total value of all elements in that array to n or less.
This is closely related to Waring's problem: What's the smallest number k such that every positive integer can be written as the sum of at most k n-th powers? For square numbers (n=2), the answer is 4. For cubes (n=3), it is 9, although 4 are enough for sufficiently large numbers. For n=4, you may need 19, but 16 are enough for large numbers. For n=5, it is 37, and it is conjectured that 6 may always be sufficient if the number is large enough. The general case (dimensions larger than 4) remains unsolved.
For regular convex polygons, the maximum combinations are the number of vertices. For regular convex polyhedrons, the maximum combinations are the number of vertices +1. For 4D convex regular polytopes, it's the number of vertices +2. For 5D convex regular polytopes, it's the number of vertices +3. And so on.
triangle numbers have 3 sides and you need 3 of them max. 2 sided numbers are a line of length 1 2 3 4 or 1 3 5 7 as you add one on each end? If it was 1 2 3 4 then it would never need 2 of them... ..love this.
Bug Byte puzzle from Jane Street at bit.ly/janestreet-bugbyte and programs at bit.ly/janestreet-programs (episode sponsor)
120 is also a triangular number that I am using in Pyramid Chess, a pyramid of 120 hexagons.
seems more like a bean dish puzzle!
can anyone explain this differently?
"There exists a non-self-intersecting path starting from this node where N is the sum of the weights of the edges on that path. Multiple numbers indicate multiple paths that may overlap." Not quite catching how it relates to the numbers in the graph
@@ChrisTian-uw9tq You can follow any path and choose when to stop. The edge weights you pass (not the nodes) need to sum to the number (or one of the numbers) of the dark green node.
@@Artaxo Then how is the pre-populated 31 meant to have its following edge filled to sum to 31 if max number allowed is 24?
A big applause for all the stop motion inserts and the clay balls and the discs! Wow ❤ I adore the clay Bollocks run Pollocks 😅
I wonder if it's actually stop motion or if it was just made to look like stop motion (like the Lego movie).
@@sergio_henriqueall real, moving little things around and taking photos
The stop motion is georgious. Appreciate the effort
By our man Pete 👍🏻
@@numberphile Thanks Pete! :)
@@numberphile Thanks Pete! :)
Pete ftw
Wait you didn’t just put an overhead camera on top of James’ paper and let him slowly move all the dots around then edited out the hands?
Very happy to see Dr Grime back on numberphile!
I've come to the comments section to write how happy I am to see Dr James Grime again on Numberphile and how much he's been missed, but I see everyone's done the same thing already!
Whoever animated this episode, you earned your paycheck.
Reminded me of an episode of Gumby.
For reference Lagrange actually proved any number is the sum of four squares. Which is why it is usually called Lagrange's four-square theorem.
I loved that game when I was in grade school
6:34 The Fermat-Haran Conjecture 😀
I always remember triangular and tetrahedral numbers because of the song 12 Days of Christmas. If you interpret the lyrics as listing all gifts up to that point (including previous days), then the running total of gifts is the first twelve triangular numbers. If instead you interpret it as listing the gifts for only that day (i.e. the gifts from all previous days are given again, leading to, e.g., 12 partridges in 12 pear trees) the running total of gifts is the first 12 tetrahedral numbers.
I think having that damn song stuck in my head in class was the reason I worked out the tetrahedral formula.
James is really Mr. Numberphile =D
I haven't watched the video yet, but I'm very excited about the combination of Numberphile, James Grime, and a specific large number.
A perfect storm
@@numberphile Superior highly perfect storm 😉
The animation / stop-motion is looking smooth as heck
EYPHKA! Delightful historical coincidence that you can still write this Greek word with Latin characters
ЕВРИКА
ΕΥΡΗΚΑ is not EYPHKA 🙂
@@jlljlj6991 I see what you did there
@@jlljlj6991 oh don't go splitting hairs
@@jlljlj6991I can’t tell the difference
I watch your videos, I don't understand anything about numbers, but I like your enthusiasm and your healthy joy, greetings from Chile
This guy makes maths ALOT more fun than when I was in school.
8:34 “I said «Pollock’s», you’ve heard me quite distinctly.”
😂
The only mathematician owned by a dog
Sounds like ☝️
Was he trying to make sure that people weren't mishearing him as saying "bollocks"?
@@shruggzdastr8-facedclown I think so. It took my a while to realize that since it isn’t used as profanity (or really at all) in my dialect of English.
Pollock's conjecture is bollocks.
Or, alternatively,, Pollock's conjecture is the dog's bollocks.
Gaus and Euler, the people who took a look at mathematics and went "that s***'s boring, but I can fix it."
The latter being the guy who gave us the base of the natural logarithm and the formula: e^πi + 1 = 0.
@@JamesDavy2009
Honestly the two were so important that listing any one thing they did as an example feels like it can only ever understate their contribution.
Even that formula is just one example of an expression that drops out of what is an entire mathematical framework that Euler pretty much constructed from scratch, and that entire framework is just scratching the surface of his contributions to mathematics.
@@JamesDavy2009 Many things in math are named after the second person who discovered them, because the first person was always Euler.
What wonderful video! As usual, perfect presentation by Mr. Grime and a generally very interesting topic 🤗
Thanks so much. 🙏
Glad you enjoyed it! Cheers.
Always waiting James' videos❤
Same 😅❤
Classic Numberphile with the OG presenter! ❤❤❤
I'm happy to see James again, being guest in other channels. Hopefully he'll upload new video in his own. 🤞🏼
James' closing comments are spot on. I was in high school (late 1980s for me; my brain is very middle-aged now) when I found the pattern of adding consecutive odd numbers to generate the square numbers, and then I figured out that the Nth level difference between consecutive N-dimensional numbers was N! (N factorial)... it's easiest to see this with the square/odd numbers, in which adding 2! starting at 1 generates the odd numbers. I found some hiccups in the first few iterations at each new power, but in general the pattern normalized at N^N.
I love maths! James adores it.
That joke about Fermat's margins is so granular, and I'm 100% here for it
Parker squares are the 21st-century version of Fermat margins.
My favorite tetrahedral number fact: The numbers along the finite diagonals of the multiplication table sum to the tetrahedral numbers. 1, 2+2, 3+4+3,4+6+6+4, ...
Fun fact: 2024's the only tetrahedral year all our lives~ And there's a book all about triangles coming out later this year! Seems like a triangle-y type of year~
True, but 2024 is also the only year we'll live through that is also a dodecahedral number and the first one since 1330. Every (3n + 1)th triangular number is the nth dodecahedral number.
That IS a fun fact!
For anyone curious 1771 was the last one and 2300 will be the next!
2024 is also the only dodecahedral number year we'll live through.
Loving the sound effects!
Has a very 70s animation vibe (or thereabouts) ✨
Great to have you back on Numberphile, James, and thanks for the video! And congrats on the ring 😉
I really appreciate the precision with saying (every time) that any POSITIVE WHOLE number :)
It’s incredible that to this day, every episode gets its own special animation to make visualize the lesson in a delightful way. Stop motion!! Brady you animate so well!
I don't know exactly why but this is the most beautiful fundamental proof I've stumbled upon in Mathematics thus far. Thanks so much for making this video!
This video, more than any other, reminded me of a Sesame Street episode brought to us by the number 343867.
My favorite tetrahedral number is 4060. It is the 28th, and it is exactly 10 times bigger than the 28th triangular number which is 406. And 28 itself is a triangular number
Also, the digits of all these numbers each add up to 10
I love James' little speech at the end of this
Fun fact that i just found out:
If you, let's say, color tetrahedral numbers in Pascal's triangle, you'll notice a straight line of colored numbers! This also works with triangular numbers and tentatopal numbers (numbers that can be arranged as a hupertetrahedron, though I'm not sure tentatopal is the correct name)
Numberphile's vid editor is probably my favorite person in the world that I don't know
I love that Gauss uses the same asterisk I his writings that I overuse today.
✺✺✺ I switched to the Sixteen pointed asterisk 😄 ✺✺✺
i've used triangular numbers to verify if a group of unique integers (in any order) was a gapless sequence or not. i was goofing around with some very basic arithmetic and i kept getting results that were oddly familiar. they turned out to be triangular numbers! around this time i had just been introduced to triangular numbers from numberphile!
my specific use case was to determine if a set of years had gaps in it. turned out that there were much easier ways for me to do this programmatically with code, but i'm still proud of having such an epiphany as a non-mathematician.
i have a working demo and explanation that i can link to, but i don't want this comment to go to spam jail!
basically, the formula is this: `(max(set) * length(set)) - sum(set) = T(length(set) - 1)` where `T(n) = (n * (n + 1)) / 2`. `length` is the amount of entries in the `set` of unique integers.
It is a cool find and definitely works assuming the integers are unique, however, if you know the maximum you probably also know the minimum and thus max(set) - min(set) = length(set) - 1 is likely easier to check.
@@benjaminpedersen9548 lol of course i was overthinking it! it's funny because i did think of something like this but i must've forgotten to -1 from the length before i derailed and went on this magical journey. also, i almost immediately found another way to do this leveraging the native features of the programming language i was using. i ended up not using my original idea at all. but i won't let that take away the epiphany i got from this "discovery", however useless it may be. 🤣
thank you for the simplification!
Numberphile is extra special with Dr. James.
Glad to see James Grime again!
11:33 this is exactly why I started working on the Collatz conjecture. I knew I'd learn a lot by thinking about the numbers and how they connect, and I was right.
Wow, I just noticed that for the square numbers you used square waves and so on. Pretty nice touch!!
I noticed that but i forgot what they were called :)
Nice!
I love (that) stop motion!
How did I miss a James Grime vid!
First things first: Click like!
Now let’s watch what this video is about…
Why why WHY is this so fascinating? It should be complicated, abstract and boring but it's interesting as heck and I don't know why
The animations are great, but the synth effects i liked even more. Reminded me of those VHSes math teachers might put on in the 90s showing weird math ideas.
I love how we can shine a light on an arbitrary number like 343,867 with this channel
Also always great seeing James in a video!
Good to see James Grimes again!🌞
My favourite banana! 🍌
I think it makes sense to me that it doesn't require more than n n-gonal numbers. Here's my hand wavy intuition/psuedo-proof:
Lemma: any sequence of n-gonal numbers starts as "1, n, ...". This is almost by definition: you start with 1, then add as many red checkers as it takes to make n sides. Of course, that's n checkers total. So the second number in the sequence is n.
So now let's just keep adding checkers (start with 1, then 2, etc.) to see how to arrange them into at most n n-gonal numbers. If we add 1 checker, it might take 1 more n-gonal number. If we add 2, it might take 2 more n-gonal numbers (a 1 and another separate 1). Once we get to adding n more checkers, then it only needs 1 more n-gonal number, because those extra n checkers can be arranged into 1 "pile" (the lemma). So this shows that every n new checkers we add, it sort of collapses back down to one extra pile.
Of course, that alone doesn't necessarily mean the "collapsing" keeps it under n piles *forever*, but it's some sort of intuition. I wonder how close this is to the real proof, if at all
today I was reminded about figurate numbers and went to read more about them. and now you release a video :D love this coincidence!
Amazing video as always, I’m glad with the stop-motion, can’t imagine how much work it took to make
GRIIIIIIME
I MISSED YOU, MAN
welcome back, singingbanana!
Fr
Tetrahedral numbers are the running partial sums of the sequence of triangular numbers. Somebody make something of that.
Given that you seem to need at most 5 tetrahedral numbers to construct any number and at most nine cubes, it would seem that one would in general need at most n+1 3D-numbers to construct any number, where n is the number of vertexes the 3D-number has.
4:19 Funny: The first way that occurred to me was one you didn't mention. Seeing as 4|28, I divided it by 4, getting 7=4+1+1+1, then enlarged, getting 28=16+4+4+4.
I have encountered a lot of content on this channel where people have checked a conjecture up to a very large number but with no proof,
i think it would be rather more useful to learn about all of the anomalies unproven conjectures which even after checking it up to very high numbers
would eventually show something unexpected.
Knowing about all of the anomalous unexpectancies would give one a good head start approaching any new theories.
Love that they still used the brown paper
Getting a new Singingbanana and a new Engineerguy video in one day, nay, within an hour of each other is *crazy*
Great animation. The kind of thing that hooks the kids.
That stop motion was pretty sweet!
James is the best
Fun Fact: 2024 is also a tetrahedron number. I think the side is 22
I like Brady's proof by pronouncement.🎉
The difference between the same (in order like the 5th pentagonal and the 5th hexagonal) pentagonal and hexagonal number is a triangular number and then the difference between the next pentagonal and hexagonal numbers is the next triangular number
Same goes for square and pentagonal
Triangular and square etc
Very interesting
thank you so much for your kindness and information
I’m a simple man, I see james, I click 🙌🏻😹
Cool. I solved the bug byte puzzle. Took me about 2 hours, but it was fun.
Did you use a computer?
The stacking sound effect is adorable
It might not be the main point of the video, but, I am really enjoying the sounds in the animations. Assuming these where created by the animator, this is really classy sound design, very buchla-esque / synthi etc. It really adds to a great video; always good to see James Grime. Like +1
5:20 also that was my conjecture :)
2:54 'try and go even further' sounds a lot like 'triangle even further' lol. was that intentional?
Just when we thought we had seen everything, Numberphile comes up with yet another 👍
Always with Fermat and those margins.
Great animation
As the number of people mentioning they are happy to see Dr Grime back approaches TREE(3), I'll just add my contribution!
Nth Tetrahedral number is the sum of Triangular numbers from 1to N !!!!
omg is the singingbanana
there's nothing like James Grime in a Numberphile video
What a coincidence, one of yesterday's videos on friends of the channel Cracking The Cryptic involved a puzzle where the solution path touched on tetrahedral numbers!
And the triangular number for nine appears in almost every video lol.
@@jimi02468 shh that's a secret
I imagine the way to prove the conjectures is through the jumps between singles. So for example with the triangle ones the jump from 1 to 3 is 2, 3 to 6 is 3, 6 to 10 is 4, 5 the next, 6 the next, you get the picture. Presumably the numbers between will only refer the the Ngonals that came before.
Your method and solution are so intresting!! Wish you the best 🙂
Now do it for all 4-dimensional pyramid-numbers 😄
hyper numbers
There are all these conjectures that hold true as far as we can calculate, but have not been definitively Proven. I wonder if there is a conjecture that was proven wrong by a hundred digit number or something like that.
And yup, I see James, I click.
I love the episodes with connection to Geometry.
my favorite presenter on Numberphile 👍👍 1:24
Cracking the cryptic mentioned a few days ago the concept of tetrahedron numbers. Nice coincidence 😃
1. So if we name triangle-, square-, pentagonal- etc numbers as "plane" numbers;
2. And we have proof that we can write any whole number as sum of 1n of n-numbers for "plane" numbers;
3. Also we can name tetrahedral-, cube-, dodecahedral- etc numbers as "volume" numbers;
Could there be relation between shape of plane and quantity of planes to describe how many "volume" numbers we need for different shape of volumes?
Or something further beyond: relation between quantity of planes and volumes, and shape of these planes and volumes for description of "hyperspace" numbers?
I am new to this problem. What I see is that any way you attempt it, you will require 5 separate sequential logical axioms to describe the full body of any tetrahedron.
For the longest time I thought Grimey was the host of Numberphile as he was in so many videos, until I saw Brady!
That stop go animation must have taken ages to do. Good job Brady and his elves
James is back!!!
Give that man a wider margin!
Oh Cauchy, always ruining my life by being a better mathematician than I could ever dream of aspiring to be
What a beautiful video. Thank you.
This all makes sense from very basic number theory, or more clearly, from the construction of the integers themselves. Normally the integers are constructed by incrementing them by some unit size, from some initial value. Using that operation, and the unit size of 1, and the initial value of 0, the positive integers can be constructed.
Clearly if this process were altered, the resulting sequence would vary from that of the sequential positive whole numbers in some way. Like by changing the unit size, you can produce multiples of some integer, or you can produce a cyclical pattern of ratios, or you can produce a sequence of the harmonics of some irrational or transcendental number. Another way to change that process would be to adjust the algorithm, like by altering the unit size according to the step number. By incrementing the unit size by the step number, you get the triangular numbers. By incrementing the unit size by two times the step number, you get the square numbers. Likewise, by incrementing it by 3n, where n is the step number, you get the pentagonal numbers.
In general, for polygonal number P with m sides and step count n, and initial value of 0, P(n) + 1 + n(m-2) = P(n+1) ... for triangular numbers, this produces the following:
0+1+0*1 = 1
1+1+1*1 = 3
3+1+2*1 = 6
6+1+3*1 = 10
...
and for squares it produces:
0+1+0*2 = 1
1+1+1*2 = 4
4+1+2*2 = 9
9+1+3*2 = 16
...
So, having said all that, since these are variations on the basic constructive sequence that builds all positive integers, it follows that they are going to have one of those altered sequences as a result, with characteristic gaps in the integers they produce. These gaps get larger, so they aren't some cyclical pattern produced by the beat ratio of some rational number. Also they still consist only of whole numbers, and so it doesn't produce the patterns made by transcendental or irrational numbers. So then, these are transformations applied to the basic integer construction operation that are both aperiodic, and that produce nothing but integers. Since they are aperiodic, and only produce integers, the outputs up to and including some step can all be combined in various ways to produce different sums. And since they themselves are comprised of elements that are aperiodic in origin (as opposed to a list of elements where each is the same distance apart), the number of possible sums from combining them grows nontrivially (if they were equidistant, there would be far fewer unique results, as more combinations would be redundant).
I lack the rigor to connect the dots the rest of the way, it seems very intuitive that, because the gaps grow at the rate they grow, and because the list of unique elements from which to make sums also grows at a rate influenced by the nonperiodic growth of those gaps, those two would grow in balance, and that the gaps created by the sequence would never grow fast enough such that they could not be filled in using the same number of elements or less.
Another way to think of this, which makes it much more intuitive, is to think of this as a way to describe numerical systems where the base changes, such that each digit uses a different modular value.. The task then is to see how many bits are required to write each integer in this system. Like for triangular numbers, you get digits corresponding to 1s, 3s, 6s, 10s, 15s, etc... to match it to normal binary representation, with the largest values on the left, you'd get the mirror order of that though, so an array like such: [15s, 10s, 6s, 3s, 1s], and instead of each value in the array being a 1 or a 0, it'd be the count of how many of that number. I'm fairly sure some proof could be constructed to show that for any n-gon, you can create an n-ary numerical system using its n-gonal numbers, where all integers can be represented while keeping the total value of all elements in that array to n or less.
This is closely related to Waring's problem: What's the smallest number k such that every positive integer can be written as the sum of at most k n-th powers?
For square numbers (n=2), the answer is 4. For cubes (n=3), it is 9, although 4 are enough for sufficiently large numbers. For n=4, you may need 19, but 16 are enough for large numbers. For n=5, it is 37, and it is conjectured that 6 may always be sufficient if the number is large enough. The general case (dimensions larger than 4) remains unsolved.
I wonder what the sequence looks like for tetra, cube, etc. 5, 9, ... and what the formula looks like, that produces infinite terms of that sequence.
There are only 5 regular polyhedrons so it's not clear to me what the "infinite sequence" would even be?
For regular convex polygons, the maximum combinations are the number of vertices.
For regular convex polyhedrons, the maximum combinations are the number of vertices +1.
For 4D convex regular polytopes, it's the number of vertices +2.
For 5D convex regular polytopes, it's the number of vertices +3. And so on.
@KrelianLoke which of these are proven? I thought he said in the video that it's only a conjecture for tetrahedral numbers
Don't worry about the inappropriate bots in the comments section, I will report them and hopefully UA-cam will see the reports.
Ive never been this early to a numberphile
me neither
Welcome to the party
same
triangle numbers have 3 sides and you need 3 of them max. 2 sided numbers are a line of length 1 2 3 4 or 1 3 5 7 as you add one on each end? If it was 1 2 3 4 then it would never need 2 of them...
..love this.
The Katamari speaking sound effects are perfect