Why do prime numbers make these spirals? (2019)

https://www.3blue1brown.com/lessons/prime-spirals

by nivethan • 3 years ago

372 100 3 years ago

zmmmmm• 3 years ago

Bit of a let down when they reveal that plotting integers in general results in the same spiral pattern as primes. So naturally primes as a subset of integers also produce spirals.

Seems the title is a bit misleading!

desertraven2• 3 years ago

A bit misleading I agree. I suppose since prime numbers DO make these spirals, it's not inaccurate.

smitty1e2• 3 years ago

We expect patterns from quotidian integers. The seeming chaos of primes, in contrast, brings the intrigue.

Bytewave81• 3 years ago

It's referencing the Math Stack Exchange question which asked the same question: https://math.stackexchange.com/questions/885879/meaning-of-r...

crdrost• 3 years ago

I don't know, for me the fact that this gives you a visualization of a totient is really interesting.

Another really interesting idea is to deliberately change the thing which you are rationally approximating; you don't have to rationally approximate π if you don't want to, that's just if you make steps of 1 radian. Make steps of q radians and you get the denominators for rational approximations of q/π.

This is used in the golden spiral algorithm[1] to evenly-ish distribute points on a sphere, we choose the most irrational number q/π = φ, the golden ratio. Since all of its rational approximations suck, the spirals are as inoffensive as they can be.

1. https://stackoverflow.com/questions/9600801/evenly-distribut...

subroutine• 3 years ago

Something less interesting is testing what the polar plot looks like if you plot the angle in degrees instead of radians. Or, like in this plot, where I defined 10 degrees as exactly one complete turn around the circle:

https://i.ibb.co/59k5dRT/polarplot.png

Slightly more interesting is what the primes look like when one complete turn is defined as 1, 2, 3... 30 units:

https://i.ibb.co/F0GvrSY/polarplot2.png

onethought2• 3 years ago

The whole point of the video is that asking these kinds of questions even if at first they seem kind of silly or arbitrary often lead to really interesting observations. In this case: Dirichlet's theorem on arithmetic progressions [0]

[0] https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arith...

laserbeam2• 3 years ago

I found it insightful, not a let down. It says a lot about human nature and our biases towards spotting patterns that may be incomplete or inaccurate.

euroderf2• 3 years ago

In the absence of reliable telepathy, there is indeed survival value in spotting patterns in incomplete information.

desertraven2• 3 years ago

What's the commentary on prime numbers currently? Is it expected that there is some unfound pattern to them? Or will they always remain elusive?

timetraveller26• 3 years ago

Yes

frogpelt• 3 years ago

No.

Taywee• 3 years ago

Maybe.

C-x_C-f• 3 years ago

Elusiveness is subjective, but there's certainly plenty of patterns to be found (though, to be fair, what constitutes a pattern might also be subjective). You should check out Terence Tao's blog [0], he's not only one of the most prominent mathematicians in the field but he's also excellent at explaining things. These slides [1] for example don't require any background in math.

[0] https://terrytao.wordpress.com

[1] https://www.math.ucla.edu/~tao/preprints/Slides/primes.pdf

smackeyacky• 3 years ago

Prime numbers are important for cryptography - if there is a pattern to them it might have an effect on what we think is secure or not.

https://www.abc.net.au/news/science/2018-01-20/how-prime-num...

mathgenius• 3 years ago

The answer to these questions is the Riemann hypothesis. John Baez just wrote a related paper about "motives" for beginners (undergrads?) [1]. It's worth a read even if you squint at the equations like i do, there's plenty of interesting commentary as well.

[1] https://math.ucr.edu/home/baez/motives.pdf

harles• 3 years ago

It could use a (2019) in the title. Interesting, but not recent.

irrational2• 3 years ago

Does it matter for math articles? It’s not like the math has changed in 4 years.

dylan604• 3 years ago

Something has to define when new math came to be /s

mypornaccount• 3 years ago

I double checked, 1+1 is still 2 in ios 16.3.1

edgineer• 3 years ago

There is an aspect of novelty when it comes to YouTube videos

thrdbndndn• 3 years ago

Does it matter for math articles

It matters for all the articles, so yes.

laserbeam2• 3 years ago

It has less to do with math, and more to do with the conventions of posting to HN where people expect dates for non-recent links. There's nothing wrong with them otherwise.

m3kw9• 3 years ago

I can also make all integers look like sine waves sin(x)

mlyle• 3 years ago

Better divide x by something or you get a pretty crummy sine.

Or multiply by 19.

quickthrower2• 3 years ago

Or use the high-school sin. (Take's degrees as input)

b33j0r• 3 years ago

y = (-1)**x

-1 is a circle, brah!

MayeulC2• 3 years ago

Tangentially, this is directly related to the article, as x*sin(x) is the x-component of plotting (x,x) in polar coordinates: with more precision (with x a real), you'd obtain a single spiral, but with integers where 2pi≈6, they appear as six.

Plotting primes p like this is plotting p*e^(ip), but the spirals are an interesting observation. You can straighten them out by adding a pi factor to the polar coordinate (so iside the sin...) but that's less interesting.

gnarbarian• 3 years ago

I suspect most functions would make a spiral on a polar graph if you zoomed out far enough.

onethought2• 3 years ago

Well, except as the video points out - primes don't make spiral when you zoom out far enough, they make a "rays" pattern... which then if you zoom out even further becomes a much gentler spiral.

jameshart2• 3 years ago

And the fact that pi is irrational means you will continue zooming out getting ever closer to rays, but never actually find rays - you’ll always have a slight spiral.

smitty1e2• 3 years ago

These patterns are certainly beautiful, but they don't have a hidden, divine message about primes.

As far as we know, for now.

jacknews• 3 years ago

Yeah, it's clearly someone's fingerprint:

https://i.imgur.com/yFMPmtW.png

rarx• 3 years ago

a factor, plus one. maybe? so it’s the pair of primes in that short sequence

Scubabear68• 3 years ago

I’m hardly a math genius, but isn’t this basically because he’s using polar coordinates, where nearly any trend is going to look like curves and spirals?

hn_throwaway_993• 3 years ago

Hence, the video...

CyberDildonics• 3 years ago

Hence the video what? It's polar coordinates making the spiral, not prime numbers.

hn_throwaway_993• 3 years ago

I thought this video was fantastic. Just a great intro and example of how "playing" with numbers and visualizations can help you understand deeper, more profound concepts.

For commenters saying "well, doesn't anything in polar coordinates end up like spirals?", do yourself a favor and watch the video. He says as much pretty early on that yes, any plotting of the integers like that will look like spirals, but goes into excellent detail in my opinion explaining why the specific patterns you see with prime numbers are as they are.

westcort2• 3 years ago

One of the thrills of studying the primes is the discovery that all primes greater than 3 are of the form 6k+1 or 6k-1. And for primes greater than 2, all primes are of the form 4k+1 or 4k-1. It is something that is commonly rediscovered by students, and that new independent finding was quite exciting for me.

The reasoning, which is in the article here, is that you can make any whole number you wish if the number is of the form 6k+1, 6k-1, 6k+2, 6k-2, 6k+3, or 6k-3. But you cannot make primes with the numbers of the form 6k+2, 6k-2 (they would always have to be divisible by 2), and you cannot makes primes with numbers of the form 6k+3, 6k-3 because they are always divisible by 3. So what are you left with? All primes >3 must of the form 6k+1 or 6k-1. And that factor 6 is just a bit less than 2 pi (a full turn in radians) so you get spirals from the offset. They are also a pixel or two off, but that is imperceptible.

The same logic is a nice exercise to apply to the problem of why primes >2 can only be of the form 4k+1 or 4k-1. Apply the same logic as above.

pcmaffey• 3 years ago

Yes, it was such a thrill to discover this in college (not a math major). I formulated it as all primes >2 are “factors” of 1 and 5 in a base 6 notation.

MayeulC2• 3 years ago

Well, factor is clearly the wrong technical term here, you probably meant the last digit? Like the last digit (remainder of modulo 10 division: p%10 in C-notation) of a prime number (p>5) in base 10 can be {1,3,7,9}.

And yes, in base 6 that becomes 1 and 5, as p%6 = 1 or 5, equivalent to p=6k±1. Interesting observation, thanks :)

CGamesPlay• 3 years ago

This is neat and I never noticed this. For 5, I guess the linear component would be (2*3*5)k, but it isn't as interesting or useful because the constant component would be +/- 1, 7, 11, or 13. This method feels like it's basically a "higher order prime sieve".

[append] Oh, and because this pattern (of 1 always being one of the constants) carries out for arbitrarily large linear coefficients, that also explains the "twin prime" phenomenon: https://www.youtube.com/watch?v=QKHKD8bRAro

oluckyman• 3 years ago

A “higher order prime sieve” indeed: https://en.wikipedia.org/wiki/Sieve_of_Pritchard

tie_• 3 years ago

Isn't 4k+1/4k-1 rule trivial though?

If we take any number K=N*4 divisible by 4 and >2, that'd be an even number by definition. The two closest odd numbers on either side would be (K-3), (K-1), (K+1), (K+3). As it happens (K+3) is the same as K(-1) for the next N, and (K-3) is the same as (K+1) for the previous N. So _all_ odd numbers follow this rule.

What "4k+1 or 4k-1" says in a roundabout way is that all prime numbers (>2) are odd, which isn't much of a surprise.

naniwaduni3• 3 years ago

So is the 6k±1 rule: 6k and 6k±2 are all even, 6k±3 is divisible by 3. You can extend this further: all primes greater than 5 must take one of the forms 30k±1, 30k±7, 30k±11, 30k±13. This is much less exciting, but ... suggestive. (No, not that suggestion, that one isn't actually true.)

For a certain point of view, most of math is trivial corollaries.

(Proof: check.)

waitforit• 3 years ago

Somehow I find 30k +/- 1, 7, 11, 13 more pleasing. It's 8 numbers and the primality of each "group" can be encoded in a byte.

bheadmaster• 3 years ago

I have discovered the same thing while practicing prime sieve algorithms back in the day. Such properties of primes are quite useful for optimizing both speed and memory of sieve algorithms.

More generally, if you take the first n primes p_1, ..., p_n and define P=p_1*...*p_n, then all primes bigger than P be in the form of P*k +- a, such that a < P and GCD(P, a) = 1. In case of n=2 (P=2*3=6), there is this nice property that the only such a are 1 and 5 (which are equivalent), but in principle, the same can be done for any n. It's just that the set of all a has the size equal to Euler's totient function of P, which grows pretty fast as n increases.

For example, if n = 3, then P = 2*3*5 = 30, so all prime numbers bigger than 30 have to be in the form of 30k +- 1, 30k +- 7, 30k +- 11, 30k +- 13, 30k +- 17, 30k +- 19, 30k +- 23 or 30k +- 29 (notice that half of these are equivalent to the other half and can be ommitted). It is interesting that in this particular case, all a are either 1 or a prime less than P: I don't think that property holds for all n, though.

naniwaduni3• 3 years ago

Indeed it's not, it's just anything coprime to P. This explains why there are always ±1s, and in particular why e.g. 210±1/210±209 are viable, even though 209=11*19.

nh23423fefe• 3 years ago

https://en.wikipedia.org/wiki/Wheel_factorization

hinkley• 3 years ago

You can also approximate pi as 22/7, which explains some of the other patterns that appear farther out.

ihatepython• 3 years ago

You can also approximate pi as 22/7

Except in Indiana, where the legal value of pi is 3.2 by mandate.

qikInNdOutReply• 3 years ago

Its the dimensionality? 2nd dimension, 3rd dimension and there permutation shadows? So by the mathematical principal, there is only one interesting permutation, and thats the "natural" one within the first dimension, aka 1 and 2s.

kfrzcode3• 3 years ago

imperceptible

isn't that sort of cheating, when it comes to math?

It's "almost" the thing just not the thing

WastingMyTime892• 3 years ago

Nah, it opens a whole new area of research. Everything is a great opportunity for new fun in maths.

You just have to properly define what you mean by "imperceptible" and you are good to go. Now you can look for a bounded approximation. See how it changes as things get bigger. See if it can be improved.

Bounds and approximation are generally a very useful tool. Properties simplifications and working on sub parts can also yield interesting results.

Generally I would say that if you can’t prove something, trying something close but simpler is nearly always a good idea.

Traubenfuchs• 3 years ago

What's k?

Hendrikto• 3 years ago

An integer.

WastingMyTime892• 3 years ago

A weird way of writing n?

More seriously, it’s an integer variable. By convention, letters from the middle of the alphabet are used for them (generally n then k).

Here, the commenter uses k because that’s what’s used in the article and that’s what’s used because n is already used to designate the class in the definition of a residue class.

quietbritishjim• 3 years ago

Trivial correction:

6k+3, or 6k-3.

These are the same collection of numbers, you meant to put 6k instead of one of them :-)

kazinator3• 3 years ago

the discovery that ...

Wouldn't the vast majority of those studying primes learn this from their textbook?

Note that 6k-1 is the same as 6k + 5. By writing that way, we can focus in positive representations of the modulo 6 congruence.

6k + 0 can't be prime, it's divisible by 6, yielding k

6k + 1 might be prime: we cannot rule it out by division.

6k + 2 cannot be prime, it's divisible by 2, yielding 3k + 1.

6k + 3 cannot be prime, it's divisible by 3, yielding 2k + 1

6k + 4 cannot be prime, it's divisible by 2, yielding 3k + 2

6k + 5 might be prime again.

That covers all cases of the modulo 6 congruence.

Thus only 6k + 1 and 6k + 5 can possibly be prime.

This is trivial fluff, only a smidgeon more clever than "all primes greater than 2 are of the form 2k + 1".

Another line of reasoning:

If a number N is divisible by 6, then it is even. This means that N + 2 and N + 4 are also even. Thus none of those numbers are prime.

If a number N is divisible by 6, it is also divisible by 3. This means that N + 3 is also divisible by 3. Thus, it cannot be prime.

That leaves N + 1 and N + 5, whose divisibility doesn't relate to 6.

Arnt2• 3 years ago

My textbook, at least, spent its space on the important axioms, theorems and corollaries. There were some easily-rediscovered results there, but mostly its pages described stuff that wasn't trivial to me.

And that's why it's one of the five books I have kept in the decades since.

leoff• 3 years ago

all primes greater than 3 are of the form 6k+1 or 6k-1

what's the value in using this "formula"? We could also keep extending this rule, and say that all primes greater than 5 are of the form 30k±1, 30k±7, 30k±11, or 30k±13. Or go further by multiplying coefficient of X with the next primes

Vvector• 3 years ago

30k±1 = 6x±1 where x = 5k and so on. Why use four formulas when one will suffice?

jameshart2• 3 years ago

This simple fact blows up into a bunch of even more surprising things that are true of all primes when you follow the algebra to its logical conclusions. Like:

Since this means every pair of twin primes > 3 must be separated by a multiple of 6, so they can be written as 6k+1, 6k-1. That means the product of any pair of twin primes will be of the form 36k^2-1.

In other words take any pair of twin primes, multiply them together, add 1, divide by 36, you are guaranteed to get a perfect square. E.g. 11*13=143, +1=144, /36=4 =2^2.

Or (and this one’s actually a little more complicated because it gets kind of casewise) you can show that the square of any prime (>3) is either one more or one less than a multiple of 24 (which is the product of the 6 and the 4 from the 6k and 4k rules)

eternalban• 3 years ago

That is because 6 is a priomorial. For any primorial p', k*p' +/- 1 will result in a number relatively prime to p', some of which are absolute primes.

The key to understanding primes is in relative primes and reduced residue sets. All patterns in (higher) primes (absolute) are generated by the members of RRS of smaller primes. This includes the clusters, such as twins, triple, quadruple, ..., primes. RRSs also hint [imo] at intimate connection between complex numbers and primes.

https://en.wikipedia.org/wiki/Primorial

https://en.wikipedia.org/wiki/Coprime_integers

https://en.wikipedia.org/wiki/Reduced_residue_system

https://en.wikipedia.org/wiki/Root_of_unity

csense• 3 years ago

Let's think like a programmer implementing the sieve of Eratosthenes.

You can think of the sieve of Eratosthenes as starting with an infinite string of 1 bits, then for each prime p, you AND it with a periodic infinite string that's all 1's except for 0's at multiples of p. Then repeat until you get to sqrt(sieve_size), with the next p being the next 1 bit in the string.

AND is associative so you could alternatively AND together several of the periodic strings first, then you get a string whose period is the product of the periods.

Could you use that to optimize? Yes. For example if your big sieve is 1GB, naively you'd need to do four 1GB passes to knock out four consecutive primes, say 11, 13, 17, 19. But you can instead calculate the combined action of those primes by doing four passes over a (much smaller!) bit vector of size 11x13x17x19, then apply that in one pass over the main 1GB sieve. (I guess you'd want to tune the max size of the small bit vector based on your cache size.)

Further optimizations are possible, e.g. you could special-case the smallest primes. With a trivial indexing change, you can have your sieve bits represent odd numbers only, and halve the memory requirement (or double the largest prime you can find with a fixed amount of memory). Subsequent primes (e.g. knocking out multiples of 3 so your bit vector only contains bits representing numbers of the form 6k±1) involve less trivial changes to the indexing logic with diminishing returns.

peter_retief3• 3 years ago

This is one of my favourite mysteries.

https://en.wikipedia.org/wiki/Goldbach%27s_conjecture

How is it that all even number are the sum of 2 primes and odd number the sum of 3 primes?

mik1998• 3 years ago

We don't know whether it's true.

qumpis• 3 years ago

Well, then it's a great mystery of up to 10^18: "T. Oliveira e Silva ran a distributed computer search that has verified the conjecture for n ≤ 4 × 10^18"

peter_retief3• 3 years ago

It is true but unproven.

misja1112• 3 years ago

The odd number sums follow from the fact about the even numbers.

peter_retief3• 3 years ago

Why though?

Poppys• 3 years ago

My browser did not enjoy that 40MB web page.

kfrzcode3• 3 years ago

It's because we live inside a spiral galaxy that's transitioning into a new galaxy type so we see through a spiral

tabtab• 3 years ago

Doesn't work in Australia.

anoncow• 3 years ago

Could there be a way to derive prime numbers based on an extrapolated graph which would be less resource intensive as compared to current prime identifying methods?

y42• 3 years ago

Shameless self-promotion, just because it seems to fit: I build an "animation engine" around primes and spirals, just in case you need some distraction or you are bored:

https://primes.nickyreinert.de/

Why do prime numbers make these spirals? (2019)

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