Secret of row 10: a new visual key to ancient Pascalian puzzles
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- Опубліковано 14 чер 2024
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Today's video is about a recent chance discovery (2002) that provides a new beautiful visual key to some hidden self-similar patterns in Pascal's triangle and some naturally occurring patterns on snail shells. Featuring, Sierpinski's triangle, Pascal's triangle, some modular arithmetic and my giant pet snail shell.
Thank you very much to Marty for all his help with finetuning the script for this video and to Steve Humble and Erhard Behrends for making some photos available to me.
Enjoy :)
P.S.: The article I mentioned in this video is: Steve Humble, Erhard Behrends, ”Triangle Mysteries“, The Mathematical Intelligencer 35 (2), 2013, 10-15. There is also a followup article:
”Pyramid Mysteries“, The Mathematical Intelligencer 36 (3), 2014, 14 - 19.
And there is a book by Erhard Behrends that has a couple of chapters dedicated to this topic: The Math Behind the Magic: Fascinating Card and Number Tricks and How They Work: bookstore.ams.org/mbk-122/ :)
A Wolfram demonstration project that implements the 3-color game: demonstrations.wolfram.com/Tri...
Philip Smolen contributed this animation
www.trade-ideas.com/home/phil...
Someone pointed out these links to some code wars problems:
www.codewars.com/kata/coloure...
www.codewars.com/kata/5a331ea...
Juan Mir Pieras pointed out these earlier references:
mathcentral.uregina.ca/mp/arch... (Problem of the month June 2001)(mathcentral.uregina.ca/mp/arch...) The webpage says the problem is from "Crux Mathematicorum 27:3 (April 2001) pages 204-205 - it is problem 3 from the Ninth Annual Konhauser Problemfest (Carleton College, prepared by David Savitt and Russell Mann."
www.macalester.edu/mscs/stude...
cms.math.ca/crux/v27/n3/CRUXv...
Today's t-shirt: www.teepublic.com/en-au/poste...
I felt that I had to take a break from all the heavy algebra that dominated recent videos. Today’s video is about a fairly recent chance discovery of some very beautiful mathematics by the mathematician Steve Morton. After I read about this discovery in an article by Steve Humble and his colleague Erhard Behrens in the Mathematical Intelligencer it occurred to me that this beautiful mathematics can be used as a very nifty visual key to some hidden self-similar patterns in Pascal’s triangle. Today is the first time that I tell anybody about this insight. Enjoy and please let me know how this video worked for you :)
Thank you!
Wow, your content is amazing, and even more so, because it is actually original. How are we worthy?
Schöne Grüße aus Deutschland :)
@Mathologer: I tried with 9 colors. Each color has a number from 1 to 9. The result of the addition of 2 colors is determined by the addition of the digit until the result is a 1 digit number. For example color 3 + color 8 = color 2.
Then I tried the shortcut you proved by applying the rule 9^n+1 to find the row of the triangle but it did not work.
It works for length 2 for sure but not for length 10 and 82.
Do I make a mistake or the shortcut just work with 2 or 3 items ?
Mathologer - love your videos. Will this work for prime powers to get the special numbers?
@@alre9766 try -(3 +8) mod 9 = 7. Does it work then?
It seems algebra, geometry, modular arithmetic, and fractals are all the same thing wearing different hats.
How surprised would you be if I told you there is an area in math called Algebraic Geometry, where modular arithmetic is heavily used?
@@user-jc2lz6jb2e Not very, but it sounds like it has groups in it.
@@TomatoBreadOrgasm It does!
It's fractals all the way down.
@@user-jc2lz6jb2e Have you heard of Geometric Algebra? Different boat, but still profound! It uses Pascal's triangle
One of Mathologer best videos.
The great graphic design,
The clear explanation,
The general idea, all are simply perfect!
@@rosiefay7283 just watch on 2x speed if you feel it's too slow
@@valeriobertoncello1809 meanwhile I'm here pausing lol
"Did you get it? Without pausing the video?"
Yes!
"Or just guessing?"
Oh...
Mathologer: explains rules about two colours added -> the third one
me: ok, addition mod 3, easy
Mathologer: two same colours -> that colour
me: hmm
me: something something group theory???
Mathologer: -(a+b) mod 3
me: takes maths degree certificate off wall
I thought addition mod 3 at first too haha
I thought the same at first glance, but there is no identity element :s
@@PurpleCoin7777 yes, my thoughts exactly, immediately at 0:45, but a moment later also "what? that doesn't work! let's continue watching"
and finally exactly 19 minutes of being sidetracked(?) in interesting ways, i could continue my first thought at 19:45 "oh, that simple!" :-)
(but i have an excuse, i like mathematics, but have not studied it)
At first I was thinking of addition mod3 and then taking the result times 2. It works in mod3 but I guess it doesn't generalize to other modulos.
@@valeriobertoncello1809 Yeah, that's because in mod 3, 2 and -1 are the same.
The sliding magenta triangles at 26:04 made my jaw drop. The concept, while amazing as are most automata, wasn't hyper-noteworthy, but the idea of these little protein-like slider triangles infixing colors upon corner matching was just a revolutionary conceptualization for me. Now I want to reconsider all sorts of automata and the Wolfram/Cook Turing completeness in light of little slider machines.
8:40 Up to left-right symmetry and color permutations, there are only 10 top rows to check:
1 way to fill with 1 color (aaaa)
5 ways to fill with 2 colors (abbb, babb, aabb, abab, abba)
4 ways to fill with 3 colors (aabc, abac, abca, baac)
and the counts
(aaaa) -3
(abbb) - 12, (babb) - 12, (aabb) - 6, (abab) - 6, (abba) - 6
(aabc) - 12, (abac) - 12, (abca) - 6, (baac) - 6.
Giving the 81 triangles.
Wow that blew my mind that Pascal's triangle mod 2 gives the Sierpinski triangle, and mod 3 makes a similar pattern. I wrote a script to play with this more and it seems that mod any prime number it gives a nice Sierpinski-like pattern, and mod composite numbers give more messy overlapping patterns that relate to the factors. It's always amazing when seemingly disparate areas of math connect together in unexpected ways!
I noticed the same thing! This means that “special” numbers only exist when M = p is a prime number. Otherwise, zero rows (mod M) fail to exist, due to M being composite and generating extra patterns where the zeroes would be
Wow very cool
"Are there any other special numbers?" Well obviously 2 works :p
And 1 :D
And4
1 does not work
That’s exactly what I thought
@@dhu2056 Why not? If the same hexagon is all three corners, they're all the same color, which follows the rule
I'm here to brag about my bragging rights for choosing yellow arbitrarly
@oH well,lord! same
@oH well,lord! exactly
I guessed it by reading this comment
@@jogiff Big brain time
Me too! :-)
His shirt doesn't have a 27
How odd...
@@scudlee :)
@@Mathologer This was such an amazing video! This is really satisfying maths
But the number of numbers on his shirt is even!
It's a hint to the homework. 81/3 = 27
This video is a great example of why Mathologer is the BEST Mathematics channel all over UA-cam. Thank you Burkard, for making my day!!
really love how much production quality goes into this (all the animation programming, etc)!
Saturday morning fun with Mathologer! What a pleasant surprise! 😎
1 - trivial
1 1 - still trivial
*1 2 1 - the middle ones have two as their greatest common factor*
*1 3 3 1 - GCF = 3*
1 4 6 4 1 - GCF = 2, NOT 4
*1 5 10 10 5 1 - GCF = 5*
1 6 15 20 15 6 1 - GCF = 1
*1 7 21 35 35 21 7 1 - GCF = 7*
1 8 28 56 70 56 28 8 1 - GCF = 2
So, when P is prime, doing Pascal mod P makes the triangles of size P^k + 1 special because the GCF of the middle numbers in the P'th row is P itself
Yep only works for primes. For non primes you can't fine special numbers
Found the same!
Can you use this as a way to find prime numbers?
beautiful!
I did not know that
This is one of the best mathologer videos so far, incredible. I inspires to play with math.
this guy is the best. He is everything I ever wanted in a teacher.
11:00 - a proof by induction, my favorite. 🙂
Amazing video and I noticed a neat trick as an extension of this. If you have a number n, and want to know all mod(k), where k is all the numbers up to n, then take pascals triangle and extend out the top row by (n-1) zeros on the LHS and add an extra zero to the RHS after the 1 giving a row length of n+1 for the top row and that puts our value n, at the bottom position of a special triangle for mod(n) ... in fact you are at the bottom of many special triangles and can walk down each mod(k) by just cutting the top row off and our value in question will remain at the bottom of the new special triangle mod(k-1). Plus since the LHS is zero, the RHS is our value mod(k). Hope you find this trick neat and you really are inspiring and I hope to watch many more of your videos.
It's beautiful how complex behavior emerges from simple rules. Great video!
You miss so much to talk about this more!!
Do part 2 !!!!
I loved it as always. Didn't want it to end.
I love it when pure math connects to natural phenomena like these Pascalian triangles and Fibonacci numbers, both connected to the growth of shells.
Great video! From the beginning of the video I waited for cellular automaton to be mentioned. Still waiting :)
The snail shell thing is amazing.
Absolutely amazing!
blown away!
what a great presentation, Mathologer!!
You made my day!
That proof for width 10 triangles was absolutely beautiful!
Such puzzles and mathematical "mysteries" are like a concentrated food for always hungry young mathematical intellects - they're teaching them to mathematical beauty, elegance and aesthetics, and hone their logic, heuristic ability, pattern recognition, intuition (I saw the shortcut too, yesss!), making seemingly impossible connections between semantically distant objects or facts.
For not such young minds it is pure pleasure and recreational activity to tackle such problems.
So, thank you, sir, for making my day.
:)
@VibratorDefibulator I agree on the old side, speaking for myself. I also use these brain limbering exercises (as I absorb them) as posers to the young'uns i get the chance to interact with.
The colour rule also works for width 1. The cell at the foot of the side 1 triangle always is identical to both ends if the top row, and that satisfies the rule.
That presumably means that zero is a power of three ;)
I just won a lifetime subscription to Mathologer! So cool!
2 minutes into the video amd here's what's on my mind: the rule is the same as the average modulo 3, meaning (x+y)/2. If you assign a number to each color, and use this average, the answer will be the number for the corresponding color. For example: (1+0)/2 =1/2 = 2 (inverse of 2 modulo 3 is 2). Average of x and x will be x.
This also reminds me of tricolorability of knots. (Or whatever it's called. I forget)
that's clever. I think division by two is multiplication by -1. In the field of remainder classes mod 3, 1/2=-1. Ah, I see, you got that already (2=-1)
I think it was of maps (aka graphs) not knots but I'm not sure
@@sinom no, it's for knots. See here: en.m.wikipedia.org/wiki/Tricolorability
See the second rule in the rules section.
Fantastic video thank you for increasing my love of maths with every video you make
At last a Mathologer video that makes me feel special.
Are you a triangle with 3^n + 1 rows?
@@simo4875 Terve, kaima 🙋🏼♂️!
Pascal's triangle is so cool! There are all kinds of interesting patterns in it. This color thing is about the best I've seen.
Wow! I actually got it! In the time given, and for the right reason! I couldn't *justify* the reason, but, still kind of impressed with myself.
What better way to begin a lazy Sunday morning than with some simple and beautiful mathematics presented so eloquently?
You only need to check 27 since only the difference between the colors matter. So you can ignore the first one and use relative values to that one instead of absolute ones. This leads to 3^3=27 combinations.
Greetings from Germany :)
27 definitely suffice, but when you take symmetry into consideration you can get away with even less :)
Exercising my bragging right - yellow.
Ξενοφώντας Σούλης proti fora blepo ellina sta comments
I recently solved that problem on codewars and came up with the 10 rule myself (looking at the results of many different triangles). Helped me solve it fast enough
Have you got a reference ? :)
@@Mathologer www.codewars.com/kata/coloured-triangles
and this
www.codewars.com/kata/5a331ea7ee1aae8f24000175
@@nowifi4u thanks
@@nowifi4u That's great, thank you very much :)
I don't know if this video is beautifully awesome or awesomly beautiful...
Woah, I was able to guess the shortcut rule! Do I get bragging rights? 🙃
Sure :)
I guessed it too!
same, checked the previous examples afterwards to confirm this is always the case
I guessed the rule also! I don't care if I get the right to brag, I will anyway. I think I'll use it as a party trick.
Great job, you win a life subscription too. Just hit the button to claim your prize
I was *_INSTANTLY_* reminded of the Sierpiński triangle 😎.
🔺
🔺🔺
11:30 ach, das ist beinahe das Schönste, was ich an Erklärungen gesehen habe. Außerdem auch ist es so ein typischer "waaaas?" Moment
Great video. Want to see more videos on origami. Or about the "Golden ratio" for example with such a "Golden sword".
1 - 1(one way)
2=1+1 (2w)
3=1+1+1=1+2=2+1 (3w, prime numbers)
4=1+1+1+1=1+1+2=2+1+1=1+2+1=2+2 (5w, cquare number)
5=1+1+1+1+1=... (8w)
...
9=...=2+2+2+1+1+1 (..w, cquare number)
9-4=2+2+2+1+1+1-2-2=2+1+1+1=5 (prime number).
This one clicked so hard in my head that I ended up laughing out loud, and I was calling out answers before the video. My family must think I'm nuts. :)
The first thing I noticed is if you rotate the triangle so a different edge is on top, it still follows the rules.
That's because every group of three that are all touching are either all different colours, or all the same colour. Rotating doesn't change that.
For the width 4 triangles, there are 10 "essentially" different triangles, if your symmetries are permuting colors and reflections across the vertical axis. I used Burnside's lemma twice.
How many with rotations counting only once?
I just did it. With rotations, it's 6. Here they are:
1111
1112
1121
1122
1231
1213
Each digit represents a color, and each group of 4 digits represents a starting row. There are many other starting rows you can choose, but once you build the triangle, you will see that one of the other two sides matches one of these 6 patterns, once you adjust the colors or reflect.
@@user-jc2lz6jb2e do those 6 elements have a special Name in Group theory? How do calculate them?
@@cubicardi8011 no. These are just elements from a set that we're acting on with some bigger group. They don't form a groul themselves or anything like that.
The way I found them was use Burnside's lemma on the 81 starting rows for the triangles with permutations of colors. All 81 are fixed by the identity element, and then swapping two colors fixes 1 color (and we have 3 possible swaps), and swapping all 3 colors together fixes no triangle (we have 2 ways to do this). Burnside says we have (81+1+1+1+0+0)/6 = 14.
Then I just wrote them down and did the reflection part, and I got (14+6)/2 = 10. Then finally I just drew the possible triangles and saw which sides appear together, and got 6 different possible triangles.
@@user-jc2lz6jb2e Can you give the details for the Burnside calculations? I'm not familiar with that and I'm not sure how to count the permutations.
On a scale of 1-10 for how beautiful that recursive argument is, because 10 is just a special case of 4, the rating I give is 4.
??? what a strange comment :)
Army of Won -- Your comment deserves at least a '7,' too.
This is the best thing ever. On our last two COCI programming competitions we had triangulations of polygons (with 3 possible colors of edges) where each triangle had all sides differently colored.
Did they mention the art gallery theorem ? Its proof features a nice application of 3-colouring triangulations :)
We haven't heard of that, but I shall come up with a cool problem about it someday :)
For the trick to convert pascal's triangle to our game traingle, I originally thought of multiplication by 2.
Which is the same as multiplication by (-1) anyway since we are doing modular arithmetic here :)
That’s what I thought too!
But it turns out making them negative is more general, since it applies to any number of colors
In particular, notice that since we define color addition as c = -a - b (mod M), we can also rewrite this as a + b + c = 0 (mod M). This is exactly why the whole triangle is symmetric under rotations (in the sense that the addition still works from any direction), as mentioned towards the end of the video
Great! You go quite deep in this problem. I didn’t find the reduction argument as compelling as I would have liked and had to pause and think a fair bit.
Always love proofs that are related to mathematical induction. So simple, but also so beautiful...:)
I didn't realise at first why pascal triangles might be related (except the 2 objects add to one part), but when i saw the pattern animation of how the colours added; it struck me that we could interpret addition results through modular arithmetic. Simply brilliant.
Yes. Induction is a beautiful tool, indeed. 🙂
This video was really awesome and accessible. Really fun!
I Finally caught up with mathogoger!
8:40 This is probably the more straightforward Mathologer homework question ever. Let's start by defining an essentially different triangle. Let's say two triangles are essentially the same if we can get from one to the other by relabeling colors or by flipping across the vertical axis. Some have mentioned rotational symmetry but I'm not including that.
Now let's canonize the colorings: We'll relabel the colors so that the top left hex is always red and the next color as we go rightwards is always yellow. If there is no next color, we're OK because we know there's only one solid red triangle.
Let's also canonize the orientation: We'll also say that if we can flip the triangle before relabeling to get more red hexes on the left half of the top row after relabeling, we do. If we had larger triangles, we would need a more complex rule using yellow hexes as a tie-breaker, but triangles of 4 are too small to ever need that.
Now all we need to do is count the triangles that have red in their upper left, yellow next, and can't be flipped before relabeling to get more red hexes in the left half of the top row.
There are 10.
You can actually further reduce it to 6 by realizing that the way the 'addition' is defined still works whichever side of the triangle you choose as the top, so triangles related by rotational symmetry can also go in the same group.
EDIT: I see you actually mentioned rotational symmetry already but chose not to include it. I think 10 is really the correct answer anyway, since the rotational symmetry depends on the specific sum rule and wouldn't apply in general.
@@Reddles37 Thanks, but I said "some have mentioned rotational symmetry but I'm not including that".
A cool way to state the rule for that last 3D example is this: Assign the quaternion units to the four different colours, eg. red = 1, yellow = i, blue = j and green = k. Three colours now give the fourth by the rule that the product of the four associated quaternions must be real(1 or -1). Alternatively, C4 = |C1 C2 C3| .
Why is math so beautiful?
You deserve over a million subscribers!
I'm 34 and I just learned about Pascal's triangle yesterday.. Very interesting indeed.
Great video - I really learned a lot from this one!
Pefect, amazing, bravo!!! 👏👏👏
These triangleceptions made my day
Just from preliminary observation, it seems to me that there aren't any non-trivial solutions for composite numbers, but prime numbers do have them. Come to think of it, I seem to remember hearing that Pascal's triangle being used as a primality test for extremely large numbers. How interesting that these seemingly unrelated subjects might intersect.
It is weird watching a video of something I work with in my tool kit and having a book beside me with a very similar example. The book is, Artificial Life: An Overview, MIT Press, 1997.
My toolkit is my mind full of math and algorithms I use to make art. And that's why I absorb as much as I can from Mathologer. Thank you!
A variation I do is I would start as you did with N random integer numbers in a row of a given range.
The next row I would choose a selection of cells from the previous row relative to the cell in the current row that I am evaluating. (Best to think of a row as a ring, end connects to start) The values of the selected cells would go through a function that had modulus the output which would be mapped to a value to be given to the current cell. I would do this in Microsoft Excel for fun years ago.
Got the yellow right..! Love this episode, very interesting and something that i actually do understand..!:)
Last time I was this early Fermat's Last Theorem was still an unsolved problem
:)
Do you mean before Andrew Wiles solved it, or before Fermat did?
Fermat didn't. There is no other explanation than that he had an intuition of this, but his "truly marvelous" proof was faulty.
This one rocks! Very fun.
Reminds me of the matrix exponential. The power expansion of { e^(A) * e^(B) } for commutative matrices A & B can be rearranged to fit the binomial theorem. Fun!
3:20 ...wait so my gut instinct was RIGHT??? When I saw all the examples I thought "maybe the corners are related!" and it worked somehow???
Yeah me too
23:56 Challenge for M = 4+ colors
As far as I can tell, there are no special numbers for M = 4 colors. (In particular, the fractal for Pascal’s Triangle mod 4 is similar to Sierpinski’s Triangle, except that the huge “hole” triangle in the middle is filled in with a smaller copy of Sierpinski’s Triangle - Wolfram Alpha was a great help here. This means that we never get an “empty” row of zero coefficients, which would be required for our special trick)
However, when it comes to M = 5 colors, it looks like the first “empty” row of Pascal’s Triangle mod 5 occurs on the 6th row (which looks like 1, 5, 10, 10, 5, 1 before modding). That is, 6 is the smallest non-trivial special number for 5 colors
Notice that 6 = 5^1 + 1… Do we detect a pattern here? To find the next special number, we would have to jam together 5 rows of size-6 triangles, whose corners would form a super triangle of size-6 as well. The overall triangle would have a size of 5^2 + 1 = 26. Thus, the special numbers for 5 colors are of the form 5^n + 1
A similar non-existence issue occurs for M = 6 colors (as we saw for 4 colors). The rows of Pascal’s Triangle mod 6 where we might hope to find a gap, are instead filled in with smaller periodic patterns. I suspect this is caused by elements of Z/6Z that don’t have order 6…
In other words, I conjecture that special numbers for M colors exist if and only if M = p is prime. For instance, the 8th row of Pascal’s Triangle
1 7 21 35 35 21 7 1
consists entirely of multiples of 7 (aside from the 1’s at the end), which means that 8 is the first special number for M = 7 colors (and will ultimately lead to a special number formula of 7^n + 1)
Proof of special triangles iff M = p is prime:
First, we will prove the “reverse” direction. That is, let M = p be our given prime number. Then the (p+1)-th row of Pascal’s Triangle looks like:
1 p (p choose 2) … (p choose 2) p 1
In general, the entries of this row are of the form
(p choose k) = p! / [k! (p-k)!],
where k = 0 to p. For simplicity, let’s call the left side A, and the denominator on the right D:
A = p! / D, so that p! = A * D
Notice that p divides p!, so p must divide AD. However, since D = k! (p-k)! consists of factors strictly smaller than p (for k ≠ 0 or p), it is clear that p does not divide D. Thus, since p is prime and p divides AD, it must be that p divides A
In other words, A = (p choose k) = 0 (mod p), for all 0 < k < p (this excludes the ends of the row, where A = 1). This means that p+1 is a “special” number for M = p colors. In other words, if we define color addition as addition modulo p, then all the “middle” coefficients will cancel p rows down (since they will all be multiples of p), which means that triangles with p+1 rows will satisfy our trick of just adding the top 2 corners, as desired
(And ultimately, all triangles of size p^n + 1 will be special as well, which can easily be shown by following the same “super triangle” method we’ve used several times before)
Next, let’s prove the “forward” direction, by way of contraposition. Let our number of colors M be composite…
(Might come back to this later. The basic idea is that as you move along a row of Pascal’s Triangle, you multiply by (M-k+1) / k to find the next entry. At first, these entries will be multiples of M, and multiplying by (M-k+1) will never change that. However, if M is composite, then some k < M will divide it, and eventually break our multiple of M. That is, we will find some (M choose k) ≠ 0 mod M, which means that our coefficient cancellation trick does not work. We also need to think about how to extend this to other rows, and/or why we only consider the (M+1)-th row in the first place…)
For the 3D version, the fourth color d we put on top of three colors a, b and c is a⊕b⊕c where ⊕ is the nim-addition en.wikipedia.org/wiki/Nimber#Addition_and_multiplication_tables . We also have a⊕b⊕c⊕d = 0, which explains why rotating any such pyramid gives another pyramid of the same type.
Yes, basically, the rotation property works because it works at the level of the rules :)
Excellent as well as beautiful
how good is this argument? 10/10
On the pascels triangle thing, if you start from 1 past the bottom corner and go up, you'll notice that numbers inbetween two numbers are those numbers subtracted
17:11
mathologer-instead of 1+powers of 3, we take 1, can you guess it?
me-powers of 2(word to word exact)
mathologer-powers of 2.
me-OMG i am right.
btw I just watched a video of mathologer relating to reuleaux triangles and then I got hitched to this channel.I love mathematics and you are just increasing my affection for it.
Mistake @ 8:58: bottom row has 2 copies of:
Blue, Blue, Red, Yellow
and
Blue, Blue, Red, Blue
The ones missing are: Blue, Blue, Yellow, Yellow, and Blue, Blue, Yellow, Blue
The trick still works for these 2, of course.
Great video :)
"Time for a cat video?" lol
It's already a cat video. Obviously this represents the genetics of breeding cats of three different colours, and shows that after 10 generations the colour of a cat depends only on the colours of two of its ancestors.
I guessed it right! I am taking my free subscription! 😊
I was actually looking at the 2 corners but had no idea why. Just somehow felt right and consistent with the first example.
I actually almost used my initial example also as the challenge example (to see how many people would pick up on it :)
Great video! A masterpiece.
Yay I have lifetime subscription just by rewinding the video once and finding out the pattern and the answer in half a minute lol
Fractal patterns on Australian shell is fascinating, like the dots on the shell of a Chinese turtle that triggered the first 'magic square', 2000 bc. The turtle was found walking up the shore of river Lo, and the magic square gave rise to the mystery of Lo Shu.
By the way, I learned when I was a teen that if you take the rows of pascal's triangle, mod 2, not only do you have a fractal, but if you turn that into a digital sound stream, say go out 1024 elements then skip down and do the same, it plays a little tune all made up from the same note in different octaves and beats that are also at multiples of that frequency.
Yes, lots of fun games you can play with this idea :)
I played with this a while ago and found that if you colour Pascal's triangle with two colour according to whether the entries are zero mod N, if N is prime you get a regular pattern of triangles of size N. If N is not prime, you get a superposition of the patterns corresponding to the prime factors of N. So for smallish N you can often look at the pattern and get an idea of what its prime factors are.
@@Mathologer by the way, I think I was sort of generating Pascal's triangle at an angle or backwards. The actual rule was a 1024 bit shift register and a exor gate, I think the result is playing the Sierpinski triangle but starting with click then a pure tone the whole width of the shift register and up, working your way toward that ever smaller tip, then it starts over...
If you make the shift register something other than a power of two it sounds chaotic, things a bit like distorted guitar and down to noise.
I also tried mod3 when I did it in software, but I didn't find that as interesting.
I didn't make it up myself. I learned it from a binary clock that some college kid made and gave his dad. Every so often it would play this song and his dad was enough of a nerd to explain to 15 year old me how it worked. So I went home and programmed it into my TRS-80 model I in assembly language and it worked.
There’s 81 unique initial conditions. There’s 3 uniform arrangements that can be one category. Swapping colors changes does not how each one works. With both symmetries there are 6 families.
The arrangements can be color-swapped, rotated and reflected;
AAAA,
AAAB/BAAA/ABAB,
AABA/ABAA/ABBA,
AABB/AABC/ABCC,
ABAC/ABCB/ABBC,
& ABCA.
Very clean
The rules for 3D: if the base 3 are all different, pick the 4th. If one of the base’s colors is less than the others pick that one, if they’re all the same pick itself
I love this sort of thing. Excellent and even insightful video despite starting from such a simple rule, a deep dive into the "why" of things, as much of the best math starts.
Am I the only one who feels an odd moment of terror when I realize that sierpinski's triangle is going to show up again though? I've seen it in so many surprising places and yet can't understand why, so my guess is that my brain fears it as some intelligent unknown, like some lovecraftian entity lmao.
That was so cool!
I rate the argument 10/10 💖
9:17 Yeah; I spotted the pattern. The gap between the special numbers gets tripled, each time. Thus, the next special number, after 82, is: 82 + (3 * 54) = 82 + 162 = 244. The general formula, thus, is:
S(n) = S(1) + 2 * 3^(n-2) = 2 + 2 * 3^(n-2).
*EDIT:* Or, as Mathologer put it:
S(n) = 3^(n-1) + 1, which is, actually, a lot simpler form.
👍🏻😎
Yet another amazing video:)
Glad you like it :)
Knowing there is some trick to guessing quickly I looked back at the examples. I noticed that the top left and right corners added to the bottom one in each. Let's see if that hypothesis is right!
Loved it.
I actually tried to think of the most intelligent thing to do when he asked the question, so I took those 2 corners and added them together because it felt right. How the hell did I get that right off the bat?
I would say there are fourteen (14) essentially different coloring patterns. First we remove the 3 full colors, then the remaininng 78 can all be mirror reflected down to 39. add the 3 full colors back in for 42. But then I realized it doesn't matter which color is which, since the 'addition' of colors is circular so if you label the colors A B and C then a triangl starting w/ AABC will end in B but if you cycle the numbers round to CCAB then the bottom will also cycle to A proving another threefold symmetry and cutting 42 down to 14. ^-^ thanks for the challenge Mathologer, now to the rest of the video :P
-Tuonra
Getting close but you can still do a bit better :)
11:36 I’m gonna give the argument a solid 9. 😎👍🏻
Very nice vid!
One of the first things i noticed is that the entire triangle works with the same rules if rotated under either of its other 2 sides. That is, every colour of every row is determined by the colours in the row above, no matter the rotation.
That was fascinating, thank you. I wish I wasn't under the weather.
Reminds me a lot of sierpinski‘a triangle and its relation to pascal’s triangle
It's fascinating