Awesome Prime Number Constant (Mills' Constant) - Numberphile

Have you ever heard of Mills' Constant? Featuring James Grime. bit.ly/grimevideos
More links & stuff in full description below ↓↓↓
Several people have pointed out the n=4 prime is 2521008887 (we missed an 8)
More on prime numbers: bit.ly/primevids
This video features Dr James Grime - / jamesgrime
The Mills Proof is at: bit.ly/MillsProof
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[@singingbanana]

Hey! (This is James *waves*)

[@7infernalphoenix]

singingbanana hi James!

[@TabooGroundhog]

*🌊*

[@ommoprakashmalviya9071]

आप हिंदी में नहीं बोल पाव हिंदी में दो ना नंबर हिंदी में इंडियन इंडियन

[@ommoprakashmalviya9071]

आई लव यू

[@ommoprakashmalviya9071]

🤝

[@icisne7315]

Do a video on " what if the Riemann hypothesis is wrong"

[@НикитаЛубин]

Phonzo Cisne heck that’s going to be a great watch

[@General12th]

Yes please! +++

[@SuperYtc1]

* whispers * It's probably true..

[@AlexandrKovalenko]

I agree. That would be the best video ever. _EVEN IF_ Riemann hypotesis is true

[@luker.6967]

That would be a long video.

[@2Cerealbox]

"The rockstars of math are..." In any other field they would have named people, but he names numbers.

[@noamtashma2859]

oh, there are human rockstars too in math too.

[@jonathanschossig1276]

Ryan N In chemistry, you usually name elements.

[@megatrix500]

So the rockstars would be... maybe Hydrogen and the noble gases?

[@MuzikBike]

no wouldn't it be carbon?

[@MrInsdor]

Muzik Bike Carbon for organic chemistry definitely, along with Hydrogen

[@Ocklepod]

"Wanna know what my conclusion is? That number is awesome!" "I know! I know!" I didn't know I there were people like me.

[@helloitsme7553]

niklas schüller same

[@numberphile]

For those wanting to see Mills' proof, Dr Grime sent me a link and I have put it on the video description!

[@usageunit]

Clicked over expecting some very long, in depth thing. Guys, it's only a single page long. You can print it on one side of one sheet of paper.

[@logiclrd]

I think I see what's going on. Basically, Mills proved that a factor 3 in the exponent is enough of a difference that you can always fine-tune θ to find a prime p2 without the *previous* value going out of the range [p1, p1+1). Tiny changes to θ only affect primes past a particular n, see, so once you've locked onto a prime for one n, you then move to smaller changes to find n+1's prime, and so on. This proves that θ exists; finding θ then means tuning it more and more finely to find the primes.

[@rosiefay7283]

You've nailed it. It's claimed that Mills's constant generates primes. The point is that this claim is bogus. You need to generate some primes in a different way in the process of fine-tuning the constant, and as you correctly say, this is possible.

[@wideawake3080]

"I KNOW! I-I I completely agree!" Made me happy

[@MrEnkelmagnus]

What if i disprove the Riemann Hypothesis ? I'd be the most hated man in maths ever. I wouldn't even get the Fields medal. They just couldn't give it to me. "You ? You destroyed everything."

[@ykl1277]

+Enkel Magnus get a fast computer, use the calculated value for mill's constant and check to see if the successive numbers are actually primes. If you find 1 that isn't a prime, then there is a proof that RH is false. The proof will be in the form: If RH is true, then mill's constant is x. (James in the video said that they are calculating it on the assumption that RH is true) x is not a mill's constant. (a number in the form MC^3^n is not a prime) Therefore, RH is not true.

[@ykl1277]

You KNOW that you are working with very precise numbers here. You would never use a simple 'double' primitive type or anything like that. You can be extra careful and calculate extra digits of Mills Constant and be sure that it is not an issue. The bigger issue is that you can't find a counterexample and that proves nothing (although evidence that RH is true, it is not a proof).

[@coopergates9680]

+YK L Can the RH be used to calculate constants for other formats, such as floor(k^(n^3)) being prime or floor(k^(2^n)) prime? I expect such ones exist.

[@irene4733]

Actually, it's such a famous hypothesis in maths (so many other branches on maths have rested on this unproved hypothesis for millenniums) that proving or disproving it would make you famous. MyThey would definitely still give you the medal. Mathematicians are concerned with the pursuit of truth, not what they want something to be.

[@theresamay4280]

+Irene Pretty sure he was joking.

[@gtziavelis]

in the video there was a typo with Mills' Constant to the power of 81. he said it was 252,100,887 but it is actually 2,521,008,887.

[@kirbyurner]

Glad you mentioned this, as I was about to point out that 252100887 = 3 * 84033629.

[@patrickwienhoft7987]

3:15 I love his reaction :DD

[@guepardiez]

It's even funnier in slow motion (speed 0.5). :)

[@dfess]

+Guepardo Guepárdez They sound like little kids! So cute!

[@DaffyDaffyDaffy33322]

+Patrick Wienhöft A true mathematician there

[@fergusmaclachlan1404]

Actually the whole video is hilarious at 0.5 speed.

[@David_Last_Name]

I would love to see a video explaining both how this guy figured this formula out, how he calculated theta, and what the proof is that this will always work. Because this number is AWESOME!

[@stargazer7644]

They did that in this video. You start with a prime, and work backwards. They told you that nobody knows the constant, that you have to calculate it from known primes.

[@hakarraji5723]

@@stargazer7644 yes but how do you know which primes to choose? I think it obvious that the user meant that

[@stargazer7644]

@@hakarraji5723 You use the ones that appear to fit the progression. You have to calculate the primes using other methods, as mentioned in the video. I suspect how to figure out the original formula would be apparent from an explanation of the proof. There's a deeper level here than just plugging exponents into a constant.

[@hakarraji5723]

@@stargazer7644 i think there is a misunderstanding:) You cant just calculate the number without knowing the primes and you cant just know the primes without knowing the number. So the user wanted to know where it all began. either there is another way to calculate those primes or another way to calcucate the number.

[@topilinkala1594]

@@hakarraji5723 Start with an approximation. Calculate the power and if it's not a prime adjust the number until it is the nearest prime. Rinse & repeat. The amount the exponent increases guarantees that the fidly bits you add & adjust in the end won't affect the smaller exponents. If you listen James carefully from the start he implies that there are more such numbers than just the Mill's constant. Probably is. You can try to find the k that guarantees that k^(4^n) gives you primes with integers 1 to infinity.

[@javipdr19]

James Grimes is my favorite of all. He always seems excited to share what he knows!

[@ppancho188]

We want more videos with James Grime!

[@chasduff8186]

Go on @singingbanana

[@mcmuffincakes]

I love how excited he gets when Brady says "That number's awesome!" His face at 3:19 is priceless

[@ZebraF4CE]

3:18

[@mikethompson5839]

I know!! It's genuinely adorable how excited he gets :)

[@charlottebikech]

5:21

[@rakesh3136]

well it is numberphile, ya know ;)

[@Aviationlover-belugaxl]

Charlotte Cole - Hossain u

[@itze_]

lost

[@modestorosado1338]

You gotta love James' mischievous grin. It's a shame he rarely features in Numberphile's videos nowadays. By the way, I'm not saying the rest of the people making the videos are boring or don't bring interesting content to the channel, but James has this unique way of explaining things, that drives your interest even if you're not into math that much.

[@Martmists]

From wikipedia: "Currently, the largest known Mills prime (under the Riemann hypothesis) is (((((((((2^3+3)^3+30)^3+6)^3+80)^3+12)^3+450)^3+894)^3+3636)^3+70756)^3+97220, which is 20,562 digits long."

[@coopergates9680]

+Mart Mists That's what I was wondering, if you could cube one Mills prime and add something to get the next, but those differences seem pretty random.

[@Martmists]

you mean like cubing the same number over and over again until it is a palindrome? might make a script on that :) sounds like a new type of numbers is coming... gates's numbers :P

[@coopergates9680]

+Mart Mists 11^3 = 1331 is already a palindrome, my point was that you already wrote the derivation of a Mills prime and showed that you have to add a bit to the cube each time and the amount you have to add isn't too predictable. Never mind palindromes anyway, write 'em in another base and they aren't. What are the differences between the cube of one Mills prime and the next, from what you wrote? 3, 30, 6, 80, 12, 450, 894, 3636, 70756, 97220, ..... 894 is a multiple of the prime 149 and 97220 is a multiple of the prime 4861. Mills's constant sounds like a transcendental irrational number.

[@coopergates9680]

+Mart Mists It sounds like such a constant does exist for k^(2^n).

[@cyndie26]

0:31 I've also seen "math.floor".

[@coopergates9680]

+cyndie26 It's the floor function, yea.

[@muhammadmoazzam4817]

can we just appreciate how beautiful this equation looks

[@luciuscaeciliuslucundus3647]

Its beauty is in its simplicity.

[@marasmusine]

I love the way James says "we don't know" in these videos.

[@QuasarRiceMints]

"Dear Excellent Translator, Your translation in Portuguese (Portugal) (Português (Portugal)) for the video "Awesome Prime Number Constant - Numberphile" has been approved! It should appear on KZbin very soon at: *[this video]* Thank you very much for your support!" :D *.* ♥

[@blahasuk7398]

A fun side note is that this is fundamental to most computer algorithms. We force data into data-structures, that allows for accessing, deletion and insertion in a logarithmic bound. So what would have taken ages to search for linearly, is now lightning fast. Instead of doing a million computations, you do ~19.

[@evanfortunato2382]

I was gonna ask if Ø was transcendental, but we don't even know if it's irrational. Guess I gotta start working up a proof

[@angelmendez-rivera351]

Grass Observator You cannot be sure of it without a proof.

[@jandor6595]

*Any math problem:* *exists *Riemann Hypothesis, pi or e randomly appearing from nowhere:* bonjour

[@DestinyQx]

if there are an infinite number of primes.. then theta must be irrational.. otherwise.. if theta were rational.. then theta would terminate eventually.. not allowing you to generate further primes as guaranteed by the proof that Mill's formula works for n ≥ 1

[@lythd]

No. 3.3333 never terminates but is rational. Although it is probably irrational you can't prove it without knowing it's formula.

[@nosuchthing8]

So it would seem

[@3dward67]

I love James' enthusiasm towards Mathematics, he makes these videos addictive.

[@bplabs]

Dear Dr. Grime: I remember hearing, back when I was in high school, of some sort of prize (à la the x-prize) for the first and/or best formula for finding primes; do you know anything about this? Your friend and fan, Brad

[@Luffy_wastaken]

3:15 James was expecting some serious conclusion but became overjoyed with Brady's comment haha

[@Quasar2456]

6:04 turn on the captions "it's related to crimes" lol

[@scottdebrestian9875]

Reading the closed captions: "Riemann's hypothesis is a very important hypothesis in mathematics that hasn't been proven yet, that is related to crimes and how they are distributed." Wow, the Riemann hypothesis really does appear in all sorts of apparently unrelated phenomena, doesn't it?

[@alial3802]

One day i will find a general formula to predict all primes.

[@offchan]

I like this channel because it has English subtitles for those who aren't English native speaker like me! So I can understand them more easily.

[@jchenergy]

Well, I think that your Theta actually is not a constant. It is a number that is being constructed according to the quantities of primes that you want to represent by the algorithm. As n increases, more decimal you need to include in theta, and that additional decimals should be such that they have not effect on the previuos calculation (and rounding) for n-1. The construction of theta is a simple mechanic, only a bit boring. Conclusion:Theta in your algorithm is not a constant, in the sense thar are e, pi , 2 , i or others.

[@StevenR0se]

No, but, there's been proven to be such a constant. As in, the theorem goes, 'there exists some constant theta such that floor(theta^(3^n)) is prime. And that's been proven. The value of that constant is unimportant for theoretical purposes.

[@jamma246]

Your comment is about as useful as saying that irrationals don't exist because you can't write them down. In other words, your comment is ridiculous.

[@zackyezek3760]

Actually, e is such an algorithmic constant too! "e" is constructed by defining an infinite power series whose derivatives are itself, I.e. an invariant under the differential operator. Since differentiation amounts to the limit of a computation- the delta of f(x) divided by the delta in x- it is effectively defined by an algorithm. You compute e and pi both to arbitrary accuracy by truncating some infinite sum or product to a finite # of terms, This constant is far less useful or widely known because it offers no new insight or data into the primes- it skips most of them and provides no new primality test. It would be interesting to see if a lot of results like the paper defining this constant still held up if the Riemann Hypothesis was falsified. I guess it would depend on whether the disproof was merely an explicit counter example or if there was more of a theorem behind it.

[@orangegold1]

That "rounding down" notation is called the "floor function" (alternatively rounding up is called the ceiling function and is the same symbols but flipped so the horizontal lines are on top... Rounding the normal way is just denoted like this [3.4] (or sometimes double brackets [[3.4]])

[@leexyz6398]

I guess this implies there are infinitely many primes, since n has no upper bound.

[@tidorith]

It's very easy to prove there are infinitely many primes. Imagine there were a finite number of primes. If there were, you could multiply them all together and add one to get a new number that is greater than all primes. If a number is greater than all primes, it can't be a prime number itself. But the new number would not be divisible by any prime, so it must *be* a prime number. Because we've reached a contradiction, we know our original assumption (finite primes) is false. And so there are infinitely many primes.

[@ryanofarrell186]

Another proof (using the zeta function) is this: Zeta(1) is the harmonic series, which diverges to infinity. However, zeta values greater than or equal to 1 can be written as a product of primes. The only way to multiply finite numbers to get infinity is to multiply numbers. Therefore, there are an infinite number of primes.

[@torresfan1143]

interesting work

[@nikoyochum6974]

having infinitely many primes is one of the most basic proofs in number theory

[@Tiqerboy]

Actually that new number could be composed of primes that weren't on the list used to generate it. Either way, it results in other primes that weren't on your finite list, and therefore the number of primes must go on forever. I consider it an 'inefficient' proof because the number of primes it generates is actually very small compared to the actual number of primes out there. For example 2*3*5*7 + 1 = 211. The proof generated just one extra prime on the number line to 211 yet we know there are a lot more than that. So even these 'proof generating primes' is a small subset to the total number of primes out there, it's still an infinite list.

[@johnnybikesalot]

FUN FACT: 252,100,887 is NOT in fact a prime. 3 * 84,033,629 = 252,100,887. But Mills Constant still works, he just forgot an 8. (Mills Constant)^81 = 2,521,008,887 (which is prime)

[@KasabianFan44]

I think it's non-transcendental but irrational.

[@prae197]

Well, statistically speaking, it's probably transcendental, as are most real numbers.

[@coopergates9680]

+KasabianFan44 Using a huge Mills prime to approximate theta is (massive prime) ^ (super tiny power). These approximations are algebraic irrationals, but the limit should be transcendental.

[@scarletice]

agreed, I just love how excited he gets. you can really tell how much he loves this stuff.

[@NeosimianSapiens]

5:47 "I'm not so impressed by it" ... this was my gut reaction the moment I saw him put the dots at the end of the constant. I became even less impressed as I saw that it produced far fewer primes than it had digits. Still, it was fun to watch this video.

[@NashvilleMonkey1000]

This is a very interesting case of algorithmic hashing! The amount of information stored in the decimal encoding of the constant itself can completely account for the information that each answer contains, because there will be a prime that "lines up" and adjusts the trajectory. If this is true, then there are other constants that can be encoded to represent other types of information, which is effectively hashing.

[@WilliametcCook]

How exactly would you prove that this works?

[@diegosanchez894]

William1234567890123 Cook there are many ways to test for primes.

[@ostheer]

But testing the formula's output doesn't prove that it holds in general

[@Jim-cr9ut]

Cheeki Breeki Nice profile pic

[@karapuzo1]

The way I see it the primes are encoded in the digits of theta. Think about, you calculate additional digits of theta using primes you found from another source and fitting the additional digits of theta in such a way as to produce these primes. If your process of calculating the additional digits is rigorous that is the proof that it will always produce primes. It's a circular argument and not really impressive. I have a constant alpha which constitutes a picture of infinitely many cats, you just find a picture of a cat and encode it as additional digits of alpha, there you got a constant with infinitely many pictures of cats.

[@laxrulz7]

I wonder why this wouldn't be true for all integers (not just 3) then.

[@anticorncob6]

If you question everything you won't get anywhere. I questioned the fact that if two parallel lines are cut by a transversal, the corresponding angles are congruent. "What if they're just super close?", I asked myself. I developed a formal logical system with a set of sixteen axioms (axioms like "If P is true and Q is true, then the conditional statement P -> Q is locally true) and turned two column proofs into three column proofs and didn't allow myself to do so much.

[@ham8utube]

It is really interesting but for n=4, the value 252100887 is divisible by 3. It might be because Theta is irrational and it needs its precise value to the hundreds of its digits to be raised to 3^n and then the floor function applied, for the resultant number to be prime.

[@Scy]

Actually it's 3 8's at the end. He only wrote 2... 2 521 008 887

[@randomasdf97]

They wrote in the upper right corner in very small letters that it's actually 2 521 008 887. I didn't notice those letters at first. I believe they should make the text slightly bigger to avoid downvotes.

[@DarkYuan]

When referring to "real world" circles, in which the physical world we live in applies constraints to how perfect a true circle can be, your calculation of pi could work out to have different degrees of accuracy depending on how perfect your circle is; usually proportional to how big you can make it.

[@PrivateEyeYiYi]

What if n is a fractional value? n = 3.4 n = 5.25 Could this somehow be a way fill in all the "in between" primes?

[@hanifhasan9320]

PrivateEyeYiYi that was what i was thinking

[@hyperghoul]

PrivateEyeYiYi it has to be integer

[@ewiem4351]

adi paramartha Care to explain why?

[@hyperghoul]

Its on the paper

[@amc8038]

Deboogs But that would give you a fractional power.

[@meretruant]

CORRECTION: 252100887 is not prime. 2521008887 however, is. (sorry for the pedantry but I just wrote a program to check for primes and couldn't help but run 252100887 through it)

[@genius11433]

Question: If we are only now coming up with a way to calculate Theta, then how did MIlls originally come up with this number? Did he get some kind of revelation from heaven or something?

[@OnamKingtheKing]

Mills only proved that there exist such a number, he did not calculate it

[@stargazer7644]

They showed you in the video how he came up with as much of it as he came up with. He started with known primes.

[@mookiewana]

Okay, after a bit more research and help from a friend, my original query has been addressed. RH is not assumed to prove the existence of Mills' Constant. It's assumed in order to compute Mills' Constant to greater precision. Mills uses a result which states that there's always a prime number between N^3 and (N+1)^3. Basically, the fact that Mills' Constant has to exist comes from that result.

[@MrGammaGoblin]

I wish you would cover where did that Theta constant come from in first place.

[@nikoyochum6974]

by using different values of n and finding primes close there and retroactively fitting a curve

[@angelmendez-rivera351]

That is not something you can discuss in a video. These proofs are very complicated.

[@Sharaton]

I'm not familiar with this problem, but I'd guess it goes something like this: Prove that there is a prime between n^3 and (n+1)^3 for all n. Let {p_n} be a sequence of primes where p_1 = 2 and p_{n+1} is a prime between p_n^3 and (p_n + 1)^3. Now find the number by taking the limit of the 3^n-roots of the sequence.

[@abcdefzhij]

So we know that there are other numbers with this same property. My question is, are there other numbers that always give primes for (theta^(x^n)) for all n, where x isn't 3?

[@remuladgryta]

I was wondering this too! The proof is fairly short (linked in the description, only one page) and seems reasonably straightforward to understand even without deep knowledge of mathematics if you give it enough time, though I'll admit I still don't get it after reading it over a few times. I can't see a step where x=3 is required, but since the paper doesn't claim the general case, I'm guessing there's a reason I'm not seeing.

[@Jimpozcan]

If it's true for the case where _x_ is 3, it clearly must be true for the case where _x_ is any positive integer power of 3.

[@abcdefzhij]

jimpozcaner True.

[@eragontherider123]

remuladgryta well, it seems that they calculate it by taking the inverse function and seeing if it matches up. So if you wanted to, you could program/excel a spreadsheet where you take a list of primes and apply that function to many constant values and see if any strings start showing up.

[@S4MJ4M]

The last sequence of videos on prime numbers are simply fantastic, I always kind of assumed that prime numbers are random and unpredictable in distribution. Keep up the great work Brady!

[@mixcocam]

If the Riemann Hypothesis is true and we can calculate huge primes easier than we can now, is this going to affect cryptography? If so, how is it going to affect it?

[@DaffyDaffyDaffy33322]

+Rodrigo Camacho It probably won't. The mills primes get exponentially spaced out as the numbers involved get bigger. There will only be a handful of mills primes in the 2^2048 range (which is what we're using currently). If mills primes are used in cryptography, and it's easy to calculate them, then someone guessing someone else's key will be pretty easy. Because of this, we're probably just going to stick to the usual methods of calculating large primes.

[@joshuajurgensmeier4534]

+Rodrigo Camacho P vs NP (another millennial problem) on the other hand...

[@zwz.zdenek]

I have a hunch that theta is transcendental. That's because it contains an infinite number of primes encoded in it and everything that we've ever managed to use for describing primes were infinite series that couldn't be written down otherwise.

[@monk_marius4338]

There are an uncountable amount of transcendental numbers (I've proved it today, consider the set of all polynomials with rational coefficients and prove that this set is countable) so theta probably is transcendental. Numbers like gamma(Euler's constant) and most values of the Riemann zeta function probably are too.

[@coopergates9680]

+icemaster523 So this means that the probability of a random irrational number being transcendental is >99.9999% but

[@danobot12]

Hi, Im looking for a beautiful mathematical proof or conjecture or solution to a problem that would fit on an A3 piece of paper. I'm renovating my room, love maths and I want to frame the proof. Any suggestions for a proof that looks beautiful, is super elegant and links seemingly unrelated areas of mathematics? (last one is that surprise factor some proofs have.)

Pretty cool idea :)

[@danobot12]

Moisés Prado Yeah, i wanna have something with Integrals. Its so annoying I've seen lots of great proofs but I cant remember what they're called or how they go.

[@pivotman64]

danobot Perhaps use a proof of Euler's identity. If you must have integrals, you could use the comparison of Gabriel's horn's surface area to its volume. My personal favorite "math thing" is the derivation of the quadratic formula from standard form.

[@KryalSDA]

this seems provable easily enough by induction? i'm too lazy to do it though

[@monk_marius4338]

Proof of infinitude of primes: Let N be composite. Then N(N+1) has more prime factors. Q.E.D This is probably the best proof I've ever seen.

[@Cuix]

Him getting all excited over a cool number completely made my night.

[@simemetti8733]

so this prove that there's an infinite amount of prime numbers

[@ToastyOs]

Gamer placE yeah true

[@simemetti8733]

cul

[@panescudumitru]

Gamer placE Except that it was already proven more than 1000 years ago.

[@diegorojaslaluz962]

panescudumitru about 2300 years ago

[@ykl1277]

+Diego Rojas La Luz 2300 is still more than 1000. The statement by panescudumitru is not wrong.

[@ThePharphis]

I'm not even remotely interested in being a video journalist and yet I envy how great his job is.

[@heloswelo6309]

What if n=0?

[@luciuscaeciliuslucundus3647]

If n=0, then the mill's prime would be theta (mill's constant) ^3^0. Anything to the power of zero is one because anything divided by itself is one. Therefore, the mill's prime would be 1.306...^1 which equals 1.306... . If you round this down, as it says to do in this video, you get one and one isn't a prime. Therefore, if n=0, it wouldn't be a mill's prime because it's not a prime.

[@essboarder23]

Gosh this is really awesome. Amazing how people can come up with this stuff

[@Taqu3]

If theta supposed to give infinity of primes how on earth can it be rational ?

[@anarcho.pacifist]

+Taqu3 Well, it's not that simple. For example: 1.5^3^n also goes to infinity as n gets larger, but nevertheless, 1.5 is rational.

[@DaffyDaffyDaffy33322]

+Daniel Șuteu I think the argument is that since primes are distributed more or less randomly, having a constant that generates them all should contain infinite information. A constant that contains infinite information would not only be irrational, but transcendental as well.

[@massimilianotron7880]

+DaffyDaffyDaffy33322 Well, it doesn't generate them all, just some of them.

[@ishwar8119]

Yes but it would give an infinite SUBSET of them

[@akanegally]

It will contain infinite information if you ASSUME that primes are distributed randomly.

[@gricka31]

Dr Grime's face when Brady said the constant was awesome. Inspired.

[@branthebrave]

Isn't the name of rounding down "floor"ing?

[@FourTwentyMagic]

+Brandon Boyer yeah, it's called the floor of a number

[@chrisandtrenton5808]

+Brandon Boyer Same thing as a greatest integer function

[@4snekwolfire813]

@@chrisandtrenton5808 no, floor is end rounding. greatest unteger is intermediate rounding

[@LittlePeng9]

If we always choose smallest P_n+1 after P_n then yes, we are converging to single value. You may find interesting that existence of theta variation can be proven from Legrendre's conjecture, if you know proof of Mills' result you will probably figure out this correspondence yourself.

[@katzen3314]

How is theta calculated then?

[@katzen3314]

Commented before I watched the whole video, that seems like the main reason we cant use it to find primes.

[@johannschiel6734]

Right, but you have to know theta to find these numbers (Pn) effectively... Damn ^^

[@andrewkepert923]

Recursively - P_{n+1} is always the next prime after P_n^3 and Θ=lim P_n^{3^{-n}}. Something on the density of primes (Bertrand's postulate or something sharper?) is needed to get a handle on the growth of P_n^{3^{-n}} so that the rounding down isn't ruined for earlier values of n. Essentially you need to ensure that the intervals [ P_n^{3^{-n}} , (1+P_n)^{3^{-n}} ] are nested.

[@andrewkepert923]

Bertrand's postulate doesn't cut it, I don't think. It needs P_n^3 < P_{n+1} < P_n^3 + 3P_n^2 + 3P_n, so we need a version that says for any x, there is a prime between x^3 and (x+1)^3. Very close to Legendre's conjecture! I think it's true, but can't find a quick reference. If Legendre's conjecture is true, then there is a number Θ such that ⌊Θ^{2^n}⌋ is always prime.

[@andrewkepert923]

Ah - the wikipedia page for Mill's Constant has what I just worked out, but the fact that it works for P_1=2 relies on the Riemann hypothesis. So yes, a small result on the density of primes is required.

[@nakamakai5553]

Amazed. Every day I think I know "a bit" about maths, Brady comes along with something like this. I don't know the tiniest slice of the whole thing. If we have a field of study that is itself infinite, it its probably mathematics

[@binky2819]

If we don't know how to get to this constant, how did we even figure it out in the first place?

[@SayNOtoGreens]

It helps if you watch the video to the end, you know. He named TWO different ways there...

[@thisisrtsthree9992]

+SayNOtoGreens gg

[@robin-vt1qj]

just test it

[@snbeast9545]

Experimentation in boredom.

[@RKBock]

rather simple: many mathematical proofs, most actually, don't use numbers directly. this one was probably shown by showing that there are constants that have that property. finding such a number can then be done numerically. for example starting with cube root(2), and then adjusting it numerically, by iteration using higher prime numbers.

[@JacobManson]

From what I understand, theta is a number that is calculated from the formula given in reverse, with theta being the testable variable as you check the result for primality. And because it is a cube root, you can be sure that the numbers that come out will be odd. It might could be the case that any odd integer exponent of a variation of theta can do the same thing. You could test this theory for a couple of digits of n, but because the exponent will be 5, it will grow extremely fast.

[@lolmaker1984]

Make a video on the Riemann Hypothesis.

[@Neuroprophet]

damn, i don't know anything about Prime Numbers but Dr. James Grime makes me learn about it further.

[@unecomedy13]

I think its irrational.

[@MichaelFrancisRay]

Thanks for the vid Brady. Great job. Im always happy and greatful for my weekly dose of numberphile. What's more is that its a viddy with dr James Grime, arguably the most genuine quirky and interesting person on the channel.

[@magnus264]

How Mill found this number?

[@robin-vt1qj]

test

[@Ocklepod]

Mill: "Which number is awesome..?" . . . "Let's take 1.306...!!!!"

[@ophello]

For fun. He basically realized that since the value grows so rapidly, he only has to make sure it passes through a few known primes in the beginning, then tweak the value to guarantee that it passes through other primes as n goes up. It's not really that impressive of a feat because it's just basically creating a number that satisfies your arbitrary rule.

[@BeckGregorFL]

I've always been interested in maths. But since I watch Numberphile, especially the videos with James Grime, I absolutely love maths. He has so much passion! You can really feel the awesomeness of numbers in his voice. Keep on the great work!

[@prathameshsawant5574]

thetha=1.3... put n=0 you get thetha which 1.3.., round it you get 1. that proves 1 is prime.

[@FrostMonolith]

n is a natural number.

[@MadocComadrin]

If one was considered prime, then we could allow n=0. n > 0 is explicit stated to not generate 1.

[@MikeJones-ue7ux]

prathamesh sawant XD what a beautiful and flawless piece if logic

[@megatrix500]

Y U tryin 2 break math?

[@CaseyShontz]

prathamesh sawant 1 is prime but it doesn’t behave like other primes so people leave it off the list now because they don’t want to have to keep saying “primes except 1”

[@aaronleperspicace1704]

What i learned from this video: 1) that number is awesome 2) as n gets bigger, not only do the resulting values of theta^(3^n) get bigger but they also get closer and closer to the integer that is a prime. Eg: He had to round 2.22 down to 2 at the beginning, therefore subtracting 0.22 but with 1361, it was 1361.000 something. 3) Riemann's hypothesis is probably true

[@multicelled]

Yes. If n were any number, then any integer could be constructed with floor(theta^n). So n is limited to integers

[@cooperstimson]

Mill's constant and Foundation in one video? I'm sold.

[@beeble2003]

We knew that every circle of diameter N has circumference pi x N but, until 1737, we didn't know that pi is irrational. Knowing some property of a number doesn't tell you if it's rational or not.

[@The_Conspiracy_Analyst]

Simon Plouffe has a new (well, 2022) paper out revisiting Mills and Wright. He produces some new formulae, and discusses the order of growth of the Mills constant functions, and creates some new constants and prime generating functions with slower growth rates. You should really check it out! I believe he set some records for the longest prime generating polynomial in the paper. The paper is titled "A set of Formula for Primes" and is on the Arxiv. By the way, if you aren't familiar with Simon Plouffe, he's one of the co creators of the Bailey-Borwein-Plouffe formula (BBP formula) , which is is an explicit formula for the digits of Pi.

[@stevenvh17]

No, I think it's in engineering we say that. Like that quote from that professor: "The summation theoretically goes to infinity, but in practice infinity is five." Mathematicians won't be happy with the close-enough number. Already for n=3 the result isn't a prime...

[@beeble2003]

The 3 is related to the fact that, if N is big enough, there's always a prime between N^3 and (N+1)^3. I'm not sure if that holds for other exponents.

[@beeble2003]

"Mod 1" means "the remainder when you divide by 1". If that means anything at all (in mathematics, modular arithmetic applies only to the integers), it returns the fractional part, not the integer part.

[@jimi02468]

Prime numbers are interesting because they are not completely chaotic but not completely orderly either. They have both chaos and patterns.

[@tabamal]

Thanks.. Yes, if we accept the Legendre's Conjecture, then there is always at least one prime number between two consecutive square integers.. Then the formula: P_n = theta^(2^n) would yield a unique theta. The associated algorithm is then: P_(n+1)= smallest prime larger than (P_n)^2. These P_n will grow less rapidly than Mills' primes: a lower bound is given by: P_n > 2^(2^(n-1)) A lower bound on the Mills' primes is: P_n > 2^(3^(n-1)). Giving: P_20 > 10^(349,875,564)

[@beeble2003]

N has to be a whole number because, if you let it be any number, then theta^(3^N) can take any positive value, such as 8, which isn't prime. N normally means a natural number (1, 2, 3, ...).

[@mookiewana]

Agreed. If RH is assumed, he probably should have stated that from the start of the video. Either that, or when the listener asks if Mills' constant is proven to only generate primes, he should have said "Yes, but only if we assume Riemann's Hypothesis, which states..."

[@beeble2003]

Mills proved that there are numbers with this property, without needing to calculate an example. In a way, it's like saying that we know that the (sqrt(2))^2 = 2, even though we don't know all the decimal digits of sqrt(2).

[@laboheme8]

That should't matter, all that matters is that primes go to infinitely high numbers, and that this always creates primes.

[@caiheang]

03:18 - 03:20 my response when someone voiced an unpopular opinion that I never dare say myself

[@sonarbangla8711]

Talking about primes, it is discovered lately that all elements of the periodic table fits perfectly in permissible slots of primes, implying divine design.

[@beeble2003]

2:30 "You get gaps: it's not every consecutive prime." More precisely, each prime you get is the first prime bigger than the cube of the previous one. So 2^3=8 and the next prime is 11; 11^3=1331 and the next prime bigger than that is 1361; 1361^3=2521008881 and the next prime after that is 2521008887. And so on.

[@beeble2003]

Mills proved that at least one value for theta exists and that that value is the limit of a convergent series. Mills's constant was later defined to be the least value for theta that works; that's not necessarily the same constant that Mills's proof would imply.

[@nosuchthing8]

Doesn't the mills number need to be irrational?

[@eragontherider123]

I suspect it is as well. My reasoning. Since the Mills primes need increasingly high amounts of accuracy in order to calculate it, and both the set n, all positive integers, and p, all primes, are also infinitely large, Mill’s constant would necessitate an infinite length. There is one thing I would need to actually prove however. It is conceivable that the number of digits you’d need for each consecutive number might decrease over time thus converging to a rational value. However, I don’t think that’s likely but I can’t PROVE it.

[@nosuchthing8]

@@eragontherider123 thanks for the answer!

[@MissionFailed2]

As I recall (hopefully correctly), in a nutshell, you have a huge number that is a multiple of two prime numbers and the difficulty in cracking the encryption it is that you don't know what those prime numbers are. I might be wrong about the algorithm, but calculating large prime numbers would enable you to crack most of today's computer systems.

[@kevinestabrook6100]

are there more constants (infinite constants?) that produce primes in similar ways?

[@beeble2003]

theta^(3^n) is actually never a whole number so the rounding is essential. However, I think the construction used by Mills guarantees that the rounding becomes less and less significant. That is, as n gets bigger, the answer before rounding gets closer and closer to being a whole number. You can see this to an extent in the video: for n=1, the rounding cuts off about 10%; for n=2, it's about 1%; for n=3, it's at most one part in a million.

[@Toksyuryel]

It's pretty easy to deduce that the set of primes is infinite. If whole numbers are infinite (which they are because you can always add one), then at least one of the sets of composite numbers and prime numbers must be infinite. Composite numbers being infinite requires that primes also be infinite, and the other option is just that primes are infinite. Therefore primes are infinite. Primes being infinite also implies infinite composite numbers, so the two sets are pretty tightly linked.

[@mrmimeisfunny]

if it has a limit then there is a finite number of primes and if you multiply them together and subtract/add 1 you get either a prime or a number with a prime factor bigger than the "biggest prime" so it makes it impossible for an end, so it is infinite

[@alaapsarkar]

thanks, your channel has created such a big interest about numbers in me!