The hardest "What comes next?" (Euler's pentagonal formula) 

Mathologer: What does a partition have to do with a pentagon (aside from beginning with "p"). Me: *blinding flash of insight* They both end in "n"!!!

"We do real math, which means we prove things" *squashes pentagon into a house shape"

I like how this guy laughs at his own presentation. It tells me he is really enjoying himself and I like to see people who are.

Best part of this was realizing how I can use the logic to solve 3 of my yet unsolved ProjectEuler problems! Awesome video!

3 b 1b and mathologer are the gifts of gods to all the math lovers around the globe

mathologer, coloring numbers green and orange: "and now the pattern should be really obvious to you!" me, extraordinarily colorblind: oh my god am I bad at math what's going on

"What comes next?" Me: Almighty Lagrange's interpolation

Pure magic Burkhard. I went though this video in detail with my gr. 9 students this week. Curriculum be damned..! It’s so fun to see them light up when understanding. I hope they appreciate that very intense math concepts are made accessible to math neophytes thanks to your phenomenal animations and eloquence. Very Much appreciated by me at the very least.

As a 50 year old man, I may have done better in maths if I had teachers like you! Thank you for your simplification of complex maths. :-)

This is literally magic, the video kept getting more and more interesting (and complicated) and I more and more amazed

Ramanujan and Euler is everywhere ... And I love it ... ❤️

YES! The guy with the towel hat!! I've always always wondered about this image of Euler, and what he was wearing on top of his head! Lol!!

I am so blown away.. I was never a big math guy though I did use a lot of geometry and right angle trig in my construction life.. But now here in my old age (68) I see the amazement of math laid out before me. The wonder that a few of my fiends had talked about but I could not see.. Oh to take this knowledge back 50 years and do it all over again... What fun it would have been.. Thank you my friend for giving me a taste of the fun and joy my old friends had in their day.. They are gone now but I remember.. thank you!

"I made it to the very end."

I like when this guy laughs, he sounds like he really loves what he does and gives good vibes

If there is ever a mathematical hall of fame; I sure hope you and your entire shirt collection is inducted. Thank you for your contribution to math, and sharing the knowledge!

The hardest "What comes next?" is the year 2020

Dear Mathologer, Seeing your video this morning has brightened my day so incredibly much. Your videos allow me to transcend my body(have pain) and live in a world of pure mathematics. Please never stop. - Your fan and student.

Love the video, Partition numbers are what got me so interested in OEIS. I was hoping you were going to go into A008284 which is kind of a transformation of Pascal's triangle but spits out the partition numbers.

I made it to the very end. Can't say I fully understand Euler's Pentagonal Formula, but I'm happy to know it exists and that you have visually given me enough to feel I've discovered a new facet of the universe today. Thank you!

Ramanujan may have been The Man Who Knew Infinity, but Mathologer is the Man Who Made Infinity Long Videos About Them :)

What a mathematician! Whatever problem you approach on math, Euler has done something there.

I love you Mathologer. Really. Few other channels dare to dive into such a level of details. And even when it gets too complicated for a video, we at least get the main intuition. Love it!

i always wanted to dig into partitions but never got around to it. Thank you for outlying it and making it so easy to follow! Euler used to be my favourite as well, that dude was amazing. Good job Mathologer, keep it up

16:08 - Challenge Accepted: Firstly, by 666th partition number, do you count the first 1 (from 0) as the first? If so: 11393868451739000294452939 If 666th is the one associated with 666 then: 11956824258286445517629485

I just watched the first 5 minutes and have to really compliment the way you present you material. It's inspiring how you structure it in a way that makes it engaging. The "tricking" shows how important it is to really check what's going and that's what math is all about :)

Interesting story. I have severe anxiety and ones I get an anxiety attack there is no way for me to take my mind off it. Until I discovered math. When I feel my anxiety sneaking up on me I watch math problems. Hours later I realize not only have I forgotten about my anxiety but I am also getting better in math and even enjoying it. Weird how my brain works.

Your channel is life changing, no other way to put it, how come no one in the education system explained these formula like you do

Those "complete the sequence" questions are my pet peave. The thing is, *any* number can continue *any* sequence, and there will be a formula (a polynomial; actually, infinitely many polynomials) to produce the resulting new sequence. That type of question is routinely used in school tests and intelligence tests, but what it really tests for is a kind of learned bias toward small integers.

I made it to the very end! And I actually followed everything you presented, cuz by the time you got to the p(n)(O-E) setup, I bursted aloud: "Some are zero, and the others will be pentagonal exceptions alternating between 1 and -1!!!" I felt sheepishly proud, but really, it was only obvious because the previous 47 minutes were presented so masterfully by you!

Excellent exploration of an extremely fascinating number sequence (or rather sequence of sequences). A pleasure to watch as always!

probably the best and most mindblowing maths video ive ever watched Really excited to try to read about it on my own thx for the links in the description :D

I said the first pattern should continue with 31. I didn't expect you to add the "evenly spaced" criterion.

I made it to the very end... ...and I liked it. I know a decent bit of recreational math and most Mathologer videos contain "something old, something new, something borrowed, something blue". But this one - apart from the concept of the partition numbers - open a new part of the math world. Thanks Burkard for coming up with these amazing and very followable adventures! 👍

I made it to the very end... ;) and thanks for always making me smile whenever I see one of your videos pop up in my subscriptions!

❤️ I admire your "ease" of presentation on all videos and topics (and the great visualizations)...

«Whoever has trouble with this pattern should change channels now»...hahaha. Mathologer, you always manage to make Maths fun and funny at the same time

Wow, I’m not even a number theory fan in general, but this was incredible! Thank you so much for this video, really appreciate it!

I am enjoying this very much - since you ask. I have needed to rewind very often and sometimes play at half speed and am bewildered for most of the time but eventually it comes across. I could never get on with Real Math, still don't in very many ways but your methods are enjoyable and interesting. Thank you.

You’re so much fun and it’s so fun to see you have fun with your presentations!

11:54 “It’s always 2 pluses, followed by 2 minuses, followed by 2 pluses and so on” How do you just expect me to know where this sequence is going? Hahaha

Haha, was just wondering why you didn't cover partitions and then seen this. Very interesting and an intriguing topic with the contributions of several important people like Euler and of course Ramanujan .

To the very end. Thank you for the many gifts you have given me and many others in your videos. You see the intuition and are able to help others like me see. Thank You

25:08 "Where does he enter the picture?" Right there, on the left!

The pentagonal numbers for negative n are also the numbers of cards you need to build a n-story house of cards.

I just have watched 50 minutes straight of man taking about various partitions in math. That has to be magic of some sort.

Absolutely marvellest mathologer video... I am returning to watch this from time to time and always find myself lerning something more...

I made it to the very end ! I'm so hooked to your channel and 3Blue1Brown, then there is that Matefacil one in Spanish with tons of exercises detailed as never seen before, I think that the three channels complement each other very well. Who would say 20 years ago that math was going to be my Hobby. Thanks a lot.

This video is a prime example of how maths is like a never ending rabbit hole that you can keep going down, never running out of new things to discover. Marvelous. Also I made it to the very end.

13:46 I was so annoyed that the pattern was that simple. I had worked out a completely different, more complicated pattern. The sums of the differences were always factors of the double position number they surrounded. (e.g. the position numbers surrounding 2 were 1 and 3, which sum to 4, which is double the original position number). Furthermore, the number needed to multiply the sum to get to double the position number went 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, etc. It was a pattern, just a way more complicated one. Now I'm going to have to try to prove that they are equivalent patterns, which may be quite difficult.

THIS is what I love about mathematics. The puzzles may seem impossible, but a shift in point of view brings everything into focus... or perhaps bring much (but mot everything) into focus. There is beauty in it. It is beguiling and can lead a person on as far as they are willing to follow.

I am so pleased I could follow the maths all the way through. Usually I am very lost and give up about halfway into increasingly complex Margologer videos.

just wanna compliment the pacing of this video. first time i didnt have to pause/rewind to absorb, except when you prompted me to for the last chapter, which was when i was planning to take a break anyway lol

Everyone: maths is boring :( Mathsloger : let me take care of it. ;) Btw your videos are very interesting and full of knowledge...... Love from india 🇮🇳❤❤

As I watch your videos I feel more and more amazed! I already thought that maths is beautiful..., but now I am sure of it. Thank you!

I finished the video! The reasoning is so beautiful! Thank you for that :) Can't wait your video on Galois Theory :D

I couldn't concentrate after chapter 3 because all I could think about was that amazing modified machine!

I enjoy partitions as one of the many studies in mathematics that can get mind-numbingly complicated, but starts from a place an elementary school student can understand. Amazing.

Awesome content as always. He is as good communicator to Mathematics as Richard Feynman was to Physics. I've just applied for admission to a master's in Mathematics, in good part inspired by this channel.

This is the first video I am seeing on this channel. I am really passionate about coding(c++) and maths and I like to combine the two to get some not so useful results, but its fun. This is like a dream come true channel for me. Thank you mr. Mathologer.

I had somehow never heard of partitions before the other day when I watched “The Man Who Knew Infinity.” Now it’s seems like I’m seeing them everywhere! This video’s release had good timing!

I made it to the very end. Partitions are amazing. My first introduction to them was through Ferrer diagrams and then later again with generating functions. It was nice to see yet another connection with pentagonal numbers. That one was new to me.

Dear Mathologer all cuadratic numbers are the sum of Two triangular numbers, for example: 4=3+1, 9=3+6, 16=6+10, in general C(n)=T(n-1) + T(n) AND we may to Write the penthagonal numbers how the sum: P(n) = 2 T(n-1) + T(n). So 35 = 2(10)+15...etc.

This gave me goosebumps and a dizzy head - but I made it to the end - thank you for opening this up for us.

27:05 When Mathologer became a physicist.

Teacher math test will be easy. Math test: 2:14

I made it to the very end! I'm glad I did and learned lots and lots and lots of interesting number theory. Thanks Mathologer team!

I made it to the very end. I had my last formal maths lesson around 50 years ago - all I can say is if maths teachers had been as engaging as you in the 60s and 70s, lessons would have been more understandable and much more enjoyable. With your help, I'm beginning to plug some of the gaping chasms in my mathematical knowledge. Thank you.

"Where does Ramanujan fit into the picture?" Everywhere...

"I made it to the very end and this is really it for today until next time"

Lovely mathematics being expounded here! This is perhaps my favorite Mathologer video yet.

I made it to the very end. Oh it would have been so much fun to be Euler working out these patterns. The computer loves the patterns even without putting it to formula or proving it, so much of the extra heavy lifting is necessarily satisfying only to mathematicians and hardcore mental gymnasts such as Ramanujan. It is such hard work they went through to prove their observations. Thank you for retelling and showing it.

This has blown my mind. This is now my favorite Mathologer video, as I can actually follow along with it to the end.

This is an awesome video! I didn't know this version of the pentagonal number theorem, and it's a lot more intuitive than multiplying out lots of generating functions. Really enjoyed every minute of it.

I made it to the very end! After woke up from a deep slumber for the 7th time. The formula for the partition number serie at the halfway mark is the killer. I lasted no more than a minute after seeing that. Had to rewind and start again, and then the next time I lasted 2 minutes past it. Truely magical.

Brilliant video - best @mathologer one I've seen to date!

I made it to the very end! Love the visual proof as always. My first guess at the second "what comes next" is 1, 3, 8, 21, 55, 144, just the Fib numbers.

Excelent video! For the last problem (here I call it PPS partition product sum) you can show that the x number in the sequence is: PPS(x)=1*PPS(x-1)+2*PPS(x-2)+...+(x-1)*PPS(1)+x*PPS(0) We take that PPS(0)=1 to make the formula simetric instead of adding a fixed x So the sequence is 1, 3, 8, 21, 55, 144, 377, 987

Thanks a lot for this video. I don't recall another Mathologer video that so baffled me in its underlying relations.

Thank you so much for sharing the knowledge & explaining it with everyone.

that machine in chapter 3 can also find perfects (if black=red, then red=perfect). this proves that there are no perfect primes. thank you for coming to my ted talk

Taking a shot at the "Multiplication Partition" problem at the end... Empirically, the numbers seem to follow the pattern F(2n), where F(n) is the Fibonacci function. So the pattern is every-other Fibonacci number, henceforth called the "Skiponacci sequence". I will prove the hypothesis that the sum of products generated from partitions of the number n follows the Skiponacci sequence. Here, S(n) is the Skiponacci function, and P(n) is the partition-product-sum function. Firstly, analizing the stacks of equations, we can utilize the "recursion" mentioned early in the video. Looking at the example given for n=4, we see that the products are: 4 3 * 1 1 * 3 2 * 2 2 * 1 * 1 1 * 2 * 1 1 * 1 * 2 1 * 1 * 1 * 1 We focus on the products ending with "1", and removing the "* 1" we see: 3 2 * 1 1 * 2 1 * 1 * 1 Oh look, its the products for n=3 ! Looking at the products ending with "2" and removing it: 2 1 * 1 Its the products for n=2. The pattern is becoming clearer. Looking at the remaining products: 4 1 * 3 The "1 * 3" clearly follows the pattern, being 3 times the n=1 product. The 4 sticks out, but for now its easy to write it off as just "n". The final formula for this pattern is: P(n) = P(n-1) + 2P(n-2) + ... + (n-1)P(1) + n The reason for this formula makes sense. The "recursion" is because the partition products that are multiplied by 2 are made from partitions that are 2 less than n. Hence the "+2" in the partition list becoming a "*2", giving us the 2*P(n-2) part of the equation. Now how does this fit into the Skiponacci sequence? It becomes clearer if we write the terms out into a pyramid. For instance, for n=5, the answer is the sum of these numbers. Each row has n copies of P(n-1), except the last row, which is written as n "1"s, for the "+n" term. 21 8 8 3 3 3 1 1 1 1 1 1 1 1 1 To aid in making sense of this, here is the pyramid for n=4: 8 3 3 1 1 1 1 1 1 1 Notice the recursion? The n=5 pyramid contains the n=4, just with the extra diagonal. This makes sense, since every time n increases by 1, each P(n-k) factor's coefficient increases by 1 (and the "+n" term increases by 1, naturally). All this means that this equation holds: P(n) - P(n-1) = P(n-1) + P(n-2) + ... + P(1) + 1. This gives us a neater equation for P(n) if you add P(n-1) to both sides, but for now lets test our hypothesis and replace P(n) with S(n). S(n) - S(n-1) = S(n-1) + S(n-2) + ... + S(1) + 1 The left side is easy to simplify, because S(n) = F(2n) S(n) - S(n-1) F(2n) - F(2n-2) F(2n-1) For the right side, we can recursively replace the two right-most elements with another fibonacci number, until we are left with F(2n-1) S(n-1) + S(n-2) + ... + S(1) + 1 F(2n-2) + F(2n-4) + ... + F(4) + F(2) + F(1) F(2n-2) + F(2n-4) + ... + F(6) + F(4) + F(3) F(2n-2) + F(2n-4) + ... + F(8) + F(6) + F(5) ... F(2n-2) + F(2n-4) + F(2n-5) F(2n-2) + F(2n-3) F(2n-1) This leaves us with this equation, which is obviously true: F(2n-1) = F(2n-1) Therefore, because we were able to replace P(n) with S(n) in our equation, we showed that P(n) = S(n). QED. Also I did the math and found that the general equation for S(n) and P(n): S(n) = 2/sqrt(5) * sinh(2 * ln((1+sqrt(5))/2) * n) This is a long way of saying I think the next number is 55 :)

Amazing Video, as usual. I love how you delve into the details of the proof. Some further motivation on "why partitions are interesting" would be welcome at the beginning of the video, especially for viewers (like myself) whose training is more in applied mathematic, physics, engineering, and computer science. In spite of not understanding the significance of partitions, I followed your reasoning with delight. "If the journey is enjoyable, the destination may be less important"

Love your work!! Beautiful & Inspiring stuff :)

Every video is so damn interesting and explained incredibly well. Words can't explain how thankful I am to have found a channel like yours.

The problem at the end is extremely interesting. Changing the sum to product is called "norm of a partition" (Sills-Schneider 2019). There are very few papers on this very subject. Thus, the sum of norms is quite intriguing to ponder upon.

When I was in High School, I was very much into math; including being rated #1 in the state Math Team League. Unfortunately, the guidance counselors claimed that the only job path in math was being an accountant which was not the math that interested me, so I was side tracked. They did direct me to MIT, but because of their guidance (or mis-guidance) I didn't really get into math at MIT and ended up in my second choice CS (at the time I was there, the 1970s, the saying was "no matter what degree you get from MIT, you'll probably work in computers"). I would like to thank you for rekindling my earlier obsession. I'm now too old to really concentrate enough to follow everything, but watching your videos and seeing the results and your incredible enthusiasm for it reminds me of what it was like. I did actually get the next number from the thumbnail right, but my pattern didn't continue beyond that. But then I saw your final comment and realized I was actually almost on the right path, just needed one more tweak.

This has to be one of the most beautiful videos I've watched on the internet.

Every Mathologer video is a path like I know this... I can understand this... This is interesting... What do you mean by that... Wtf

14:23 i always fail these "pause and figure out" so it was amazingly satisfying to finally get one right! Great video as always.

Best channel for discovering math beauty. J'adore.

Unbelievably good video ! My mind is genuinely blown 😊👍

I made it to the very end As for what comes next in the 1,3,8,21,... series. My immediate guess is that it looks like the Fibonacci series, except you omit every other term. By that logic the next one would be 55. *(edit, I think I've solved it, this comment and its reply shows my solution, don't look at it if you don't want to spoil it for yourself)* Now as for actually giving it some thought... We could go by the opening and closing gaps idea at 4:00 in the video. Appending a new block, we can get the partitions plus a disconnected block at the far right, or with a connected block at the far right. In other words, adding an additional x1 or alternatively increasing the last factor by 1. How can we account for how this? We could track the sums of the products whose last factor are specific values. i.e., 1 -> [1: 1], 2 -> [1: 1, 2: 2], 3 -> [1: 3, 2: 2, 3: 3] 4 -> [1: 8, 2: 6, 3: 3, 4: 4] Which is kind of easier to reason about. - The 1: value of the next iteration is always going to be the total sum of the previous iteration. - The 2: value of the next iteration is always going to be the 1: value of the previous iteration, times 2. - The k: value of the next iteration is always going to be the (k-1): value of the previous iteration, times k/(k-1) Hm. Well this is what I thought of so far anyways. On that front. Supposing the pattern is every other fibonnaci number, given the last two values in the sequence p,q the next one should be 3q - p. Which you can derive without too much difficulty. p = a, b, q = a+b, a+2b, 2a+3b ==> 2a+3b = 3q - p. So somehow it has to be tied in with that recurrence to get an inductive proof. (assuming that it does fit the pattern).

Amazing as always. Love it when you can start with a concept which is easy to formulate like ways of summing to an integer and end up needing e, pi, infinity and derivatives to express a general solution. Makes you appreciate how interconnected maths really are.

Truly excellent video as always. Very enjoyable.

The chapters on this video are incredibly helpful. Sometimes you need to rewatch just a small segment before moving on.

I made it up to the end due to your amazing sense of humor!!!

I found interesting about the first seaquence that if we delete the condition that the n points has to be equally distancied on the circle then we have the Moser's circle problem then with n=6 we found the sequence 1,2,4,8,16,31 and not 30. Moreover the formula is much more simply with an apparently more diffcult problem since it is given by binom(n,4) + binom(n,2) + 1 = max #{ number of region created by the edges}.

"I made it to the very end". Thank you for the presentation. I enjoyed it thoroughly.

For the question at 24:10; I started with function f(n) = (n(3n-1))/2. From there, I found the values of f(-k) and f(k), where k is some real number (obviously we're only interested in integers, but it doesn't hurt to be more general). When you subtract the first from the second, you get a difference of k, which is what we wanted! Then, after -k, you jump up to k+1, so I found the value for f(k+1). I then subtracted f(-k) from this new value, and when you simplify and cancel, you end up with 2k+1, which will compute all the odd numbers. Again, that's just what we wanted!