@@ChadTanker The subfactorial certainly has some (e.g. imagine n people meet and each of them brings a gift. The number of ways they can redistribute the gifts among them such that everyone gets exactly one gift and none gets his own gift is !n). It has some more serious applications as well (for instance in cryptography). The superfactorials grow insanely fast and I doubt they really have many application in reality tbh. It is easy to construct combinatoric problems that lead to them (however, I don't think those problems are likely to be relevant in practice) 😅
@@srevere7241 6^6^6^6^6^6 yeah, but I feel like that one is still more calculable than 4! Tetration 4! The real question is: Is 4! Tetration 4! bigger than Gram's Number or TREE(3)?
Two observations: 1. The double factorial is also known as the semifactorial, which I personally think makes more sense, since you are only multiplying half of the numbers less than or equal to n. 2. All this super-duper-mega-hyper factorial stuff reminds me of when we were kids, and got into an argument about things like whose car was faster, or whose daddy earned more money, like little boys often do. It usually went something like this: - A hundred. - Two hundred. - A thousand! - A thousand thousand! - Ten times more than you can say!!! (And no, that's not a triple factorial. It's just three exclamation marks.)
...but sometimes those who-can-name-the-biggest-number contests can end unexpectedly, as a colleague of mine once overheard his two boys compete: - One thousand. - Ten thousand. - One million. Elder brother now remembers that he has seen the infinity symbol, ∞, somewhere, and thinks he has a sure win: - Horizontal eight! But younger brother (who has no clue about infinity) is quick to respond: - Horizontal nine!
A factorial symbol is the exclamation mark, so it just depends on context. I bought a TV for my bedroom and won another 3!" Another 3! what? TVs or bedrooms? Did I win 3 and am just excited, or did I really win another 6?
I've set up my computer to run a quick python script. Hopefully, when every subatomic particle in the universe is a digit, I might be able to get you your answer. I'll set it up to read out from the milky way black hole..
For the primorial, ig 1# = 1 makes the most sense to me. Ways to arrive at this conclusion: 1: You also multiply by 1 even if it isn't a prime. 2: Since 2 is a prime number, (2-1)# must be 2#/2, which in this case is 1. 3: An empty multiplication is 1.
but also you can look at what the primes smaller than 2 are. That's the empty set. So you are multiplying the empty set over the empty set, witch is kinda undefined
@@owenbechtel That's how i thought of this. When you are suming things, but you don't have anyghing to sum, you get 0, that is the identity for sums. When you are multiplying things, but you don't have anything to multiply, you get 1, that is the identity for multiplications.
10:14 I calculated it, but UA-cam doesn't allow posting comments so large they physically create black holes in the server. I've submitted a bug report, when it's fixed I'll get back to you.
Usually the empty product is defined as 1 and empty sum as 0. So if the set of primes equal or lower than 1 is empty the product should be 1 by convention
@@Arthur0000100 oh no I didn't mean to call you out or something. Just a nice coincidence. It means that we're probably correct lol. You commented well before me as well anyways
It's not a convention, it's logic. The empty sum is 0 because whenever you add some number to it, it becomes that same number. So for all x, x+(empty sum)=x. That means empty sum=0. Similarly, (empty product).x=x, so empty product=1.
@@impwolf Maybe use google or Wikipedia before commenting? Pratchett was a comic fantasy writer, and the quote was a joke. He often used multiple exclamation marks himself.
The exponential factorial should use the euro (€) symbol. It's still a monetary symbol so it would remind us of the dollar symbol, and it symbolizes a E, just like 'exponential'.
It's a shame how most math students are never introduced to the double factorial and/or subfactorial during Calc 2. I feel that knowing these concepts would make comprehending series a little easier.
What level of math is this? I'd dare say most students are never exposed to any of this. I've done every math subject there is short of Masters level or PhD level math and I've never seen or heard of any of these... ever.
Yeah, these would just make writing out and defining a series super easy. Use of some of this notation would save a lot of hand cramps. But these don't really fall under a proper math class category. They are usedul tricks you pick up along the way. If ypu actually tried to use them you would probably have to provide a definition at the beginning to avoid confusion.
Wow this is pretty fascinating - I didn't know some these existed, and their uses are also interesting! Videos like yours inspire me to share my own maths content as well!
7:06 I don’t know the actual answer but I would guess 1#=1 for a reason similar to why 0!=1 We can define (n+1)# as =n# if n+1 isn’t prime and =(n+1) x n# if n+1 is prime 2 is prime and we know that 2#=2 so 2#=2=2 x 1# so 1#=1
@@rafiqhaq This is because Wolfram Alpha is using a different definition of n#. Wolfram Alpha defines n# not as the product of the prime numbers less than or equal to n, but as the product of the first few n prime numbers.
Some crazy stuff! And some not-so-crazy. I chuckled silently when you asked for calculator help with the power-tower, 24^(24^(24^(...^24)...)). I was picturing some poor cuss actually trying to work this out on a calculator. Even taking the log will only "reduce" the tower by 1 "level." And you didn't even crack a smile when you said that. Incidentally, I would say that 1# = 1, because it's a vacuous product - there are no primes ≤ 1. Fred
@@blackpenredpen Yes, that's good; and asking them to calculate this, would itself be instructive. Illustrates the power (pun intended!) of very large numbers. Fred
In the end my takeaway is: -the first 3 are useful notations -number 5 allows to write the biggest numbers with only few symbols -I don't see what 4 is good for but I have a feeling I could run into it naturally -I don't see what 6 is good for and have no idea when I'll ever need it -7 is bigger than 4
It's crazy that the number of derangements !n == the closest integer to n! / e. We looked at the formula for derangements on the first day of my combinatorics lecture because the formula was so cool.
pwr_twr(n!) x pwr_twr((n-1)!) x pwr_twr((n-2)!) x ... x pwr_twr(3!) x pwr_twr(2!) x pwr_twr(1!) = &(n) where pwr_twr = power tower of n, and &(n) = As I call it, Super Hyper Factorial.
Yikes, I thought I was in-the-know because I was familiar with the double factorial; I had no idea about the other factorial variants you showed. Very cool, thank you BPRP!!
24 tetrated to 2 = 1.3337357768502841244490814728438*10^33 tetration can either be written with double carets: 24^^2 or with the hyperoperation notation: 24[4]2 the "n" in x[n]y stands for the level of operation. x[1]y = x+y x[2]y = x*y x[3]y = x^y x[4]y = x^^y in greek, tetra means four. the carets come from knuth's up arrow notation.
I don't think that's correct. The example you've given should be written as p₅# = 2 x 3 x 5 x 7 x 11 = 2310 the product of the first 5 primes. If you write 5# that would be evaluated as 5# = the product of the primes ≤ 5 = 2 x 3 x 5 = 30 Note: p₅# = 12#
10:14 I'm pretty sure future Casio fx calculators will give the answer as 24^24^24.....^24. As the current ones are only limited in giving small answers like you enter 3/2 and press = button to see the answer 3/2.
I wish I saw this video before😅 I still remember when I was trying to solve an Olympiad combo problem and concluded that the answer was the multiplication of the odd numbers from 1 to 2n+1 Then I opened the solution and I was shocked when I saw the answer (2n+1)!! Only then to realise later "they are the same" 😂
7:05 by logic you need to define that if it’s less than or equal to 1. So you need to goes only +1 to the next p but in 0 (empty set) = 1 in a multiplicative way
He made a mistake when explaining what the primorial function does. He said that it multiples all the primes numbers which are less or equal than the number n but according to Wikipedia en.m.wikipedia.org/wiki/Primorial it multiples the first n prime numbers. So the answer to the question 1# is 2 because the first prime number is 2.
@@serbanhoban1517 Please, read the article carefully. Particularly, please, pay attention on this section en.m.wikipedia.org/wiki/Primorial#Definition_for_natural_numbers
@@serbanhoban1517 This article says that if p is a prime number, then p# is defined as the product of all primes from 2 to p. However, then it states that if you want to include any natural number n, the definition is another one, and it coincides with the definition given in this video. For example: If n = 5, then n# = 5# = 5 * 3 * 2 = 2 * 3 * 5 If n = 8, then n# = 8# = 7 * 5 * 3 * 2 Note that 8# = 7# For the cases n = 0 and n = 1, where there are any prime numbers ≤ n, it's defined that 0# = 1# = 1
I think I've seen all of these before. I play Four 4s a lot, so factorial extensions are key operations for me. Nice that he included both versions of the super factorial!
@@blackpenredpen pwr_twr(n!) x pwr_twr((n-1)!) x pwr_twr((n-2)!) x ... x pwr_twr(3!) x pwr_twr(2!) x pwr_twr(1!) = &(n) where pwr_twr = power tower of n, and &(n) = As I call it, Super Hyper Factorial.
@@sttlok btw there's an easier way to write down tetration, a^^b, there a is base (and each power) and b is the height. Thus, your example is simply 3^^5, and super factorial for n is n$ = (n!)^^(n!)
Here’s a question. Why does everyone solve factorial problems by multiplying integers from greatest to least. For example if a teacher teaches you how to solve for 4! they will likely tell you to multiply 4 by 3 by 2 by 1. Why not 1 by 2 by 3 by 4? You get the same result and it’s much more natural.
If you start calculating that n! with the biggest factor, n, then you'll follow a more standard factorial calculation procedure: you'll have to stop when the changing factor reached 1, a condition which is independent of the factorial you're calculating. If you start with 1, then you'll have to continuously compare the changing factor with n and hence keep remembering that value of n, and stop when that changing factor has become n & multiplied into the value you're calculating. I definitely prefer the first a.o. because it doesn't matter whether you don't multiply a value or multiply it with 1, and because I won't have to remember the value of n, the paper I'm writing on will do it for me.
I feel like the pickover one is impossible to apply. Like, you'd literally probably only be able to go to number 4, as with 5 you'd have 120 exponents which is just ridiculous
funfact: sf(n) * H(n) = (n!)^(n+1) it is very intuitive, but to prove it nicely you might have to use product of a product formula for switching indexes (if that's what it's called)
Not hard to prove at all. Just show that sf(n) = n^1 * (n-1)^2 * ... * 2^(n-1) * 1^n and the claim follows immediately sf(1) = 1 => claim trivially true for n=1. Now assume claim true for n. Then: sf(n+1) = (n+1)! * sf(n) [by definition of sf(n)] = (n+1) * n! * sf(n) = (n+1)! * n! * (n * (n-1)^2 * (n-2)^3 * ... * 1^n)) (Using the assumption) =(n+1)! * (n*(n-1)*...*1) * (n * (n-1)^2 * (n-2)^3 * ... * 1^n)) (Writing out the factorial) = (n+1)! * n * n * (n-1) * (n-1)^2 * ... 1 * 1^n (some rearranging) = (n+1)! * n!^2 * (n-1)^3 * ... * 2^n * 1^(n+1) (more rearranging) Thus, if claim is true for n, it is also true for n+1. qed.
@@arthur_p_dent didn't say it was hard I did it like this: (using P as a product, [...] is a step(?), (...) is a subject/base/whatever the hell it is called in english) P[1
I've seen the first three. I remember asking if there was a name for products of all primes up to a given number, and someone told me about primorial. I was (...and actually still am, kinda) messing around with prime generation, and so I had generalized the trick of ignoring even numbers (after 2), and was using what I found to be primorials for that (it's one of several projects that I've never finished, or quit, but just got distracted from.). To define primorial recursively, I'd say `n# = { isPrime(n) : n * (n-1)#, (n-1)# }` (or, just `n# = n^isPrime(n) * (n-1)#`), and we can start off with a base case of 2# = 2. But if we apply recursion to that anyway, 2# = 2 * 1#, but we know that 2# = 2, so 2 * 1# = 2 => 1# = 1... and 1# = 1 * 0# = 1 => 0# = 1. So we have 1, 1, 2, 6, 6, 30, etc.
7:07 We can find this answer with the same logic as why 0!=1. Let n be any integer and p be any prime. We can then define n# as p#/p, with p>n. Following this, we can see that 5#=7#/7=30, 4#=5#/5=6, 3#=5#/5=6, 2#=3#/3=2, and by consequence, 1#=2#/2=1. Following this same logic, not only can we deduce that 1#=1, but that, in fact, for any n
I learned about the super factorial right after this year’s Euclid Math Contest because one problem required a proof that involved the product of factorials
By trying to calculate the 4! tower (4$), I got back a memory error. However, I was much luckier calculating the 3! tower (3$): 3$=801905114177186421268233247183671872285611243790287670326429840266965276859090994232722804099071308208566642345342525473839197857922206826881247686613054597643639074114299814658910570299338387275018144418060451356204425587436618355894265899469206493496576567060902508216857234809659411883436856907262181406555792173257484458552977375606894392453200909034506894234184478236418421979962663479216120643800922939369420248674473362609602187661563551041157505739642033306712744000213561038789775549335115383195493100990320977797431849066454349854112351669394350351724119648421429675482501486302736500144621886523347992629826999974724330860189653089828532182794794248240477416274638167362282413526807854514320952096682617889397115584667137201322422937457729214489407907405518444344340089061930346769872400573045001311080100230425970533942745847972064970363330555794582550644070075448682407064391762605241178885977478172470439245614352782718873090563810918058676016196022517960964002392982148152622058158104958518830487349863461522737045419079805176828913337987237167998461268815906214056666240308532663321889986375962262141989078341225419274892934633471601337630145021177561682163361588301146273292029772181095793682371661321565671179250200873481397054591452273317157196303425228704984654767851075710532634534940796785677558890950799401875263511992661902169258890278086716291023843497372147231848593552275703330179333395157137953888601584226588131426100524625525615311244683340215525755193173697123985498932994880224661923242660863038692352636818818091446575100518750311622740988660944192795623802082203241025300988864720691114284336174884722725160551906710564699824148484730470707902578930619626494023221095499047958286617225276486876179287677463797214957475199592111410409161111024724320181524607190511675442364059199832339531178389332438871670894278123643702026198922090184989766828514386825218944751917133528352820304932965893847129193929732262192111912880919222840357641983028044015106742642713134002917504796175868158080020653346101062376128143166925008124162624778493310053821947745097837762493928482536937358487491224793636348213860230948090092608071270697036421316013417589210684049327427491895567716870540159334726003182535675968082210912512117117036411988561552555424135025992192431252311247070107037564320408519913415791972361428643569407291782230769633403762980911951260235335468415654697223881790965348650156255150470465709634202169556242801373930782315697735699489821418879261442079714412155375949060050935369523298480393127780154774697206538820578852481294171389639340821243198793285107034663451816584313178509573270340714717653972268811979935455568659825920079977104240044757023571324964943766412817014787831726000431239296277568149403379174685366513529096824121631549336050517240784764044158530092410468898790882906726991168235676755052595083949405892993514487989629327303507999701858400364951812663411243218524311814960565403396906101566037518454582866326674740652656967374738643546913572072027015270654024870872914125274032777679768834616330289620042855458464404935752253141307743949799679373788177021131263060724194551523232678825949835712984835004658258078967038721817894573819554326478723879110512134676175579870238496958283594595247111635504199858696576767040558179086446871276735764539552108394244368401906598270272523213985019325867597404117299522896174182781347656228133260501669599573840643828131130837868317552037425215982186057658406291543623646877113038178380490129752610988187060310837787799219303381539699528293723206372177059719935531506073859021197524406579643039883039728628836461474751067864431977032358675848360773708387211420116787599737621317224241346875009176863639530452676627730931378159457365569487241901935734071637648678771531953675914311001534496147038332750307708867979198279698026903039770263012642154401276299002427289117685602673262358039948743624480371236137632544504304823818957992107773203870105130812284336956828027729321903579499814164578180299915045407689667530374597860119037107839602699845102433609954824008871263055281424268092422912559273889700924995226448267306343535545322900135542162984089368300143981387952516535890373585769044768270079232745085310534780379433679641764412570375902770137404074177820073270088260988742823688892707845709507869126201853287365775198969687579436875786108977542040269149258582213880806730504418248217557255761673402533058045211820437282641288015597565632574887136806808091337017274509640585947630061378243713693613162003445998800513844020356593674967439236032719297765887804559453426094291753338337320872533167029618779345490908355556740326053560776376448793273729369475913183616635968036303958961312252848799884953039291437629677310491001983631561495387558374254249597009726836978531354929462178177642763033790164067445673502415866746505721852575827258860644876762985518399443861444129789611155823260748613960983738802730799807870324833863673572794179621716686213597175126065963043765314408250036111188043650982973774434447477841745166609106376305766597815630308332278922332012868449774553692733717992022275716188668002733820424048869010692647287753683032329124547512690629495028349649028761229072342231520826626527689967862367744521152658974319063649327835030970627742864238920810668385925185216817124523427167003892110153204070727224612710173873389921936290442205620640819677053163599111244195701659784290628033387794423384897379043640715550904349542341988051448696644729119321923974170788984946987136512729765351867471308995876186529082842949528120694579172451660355612447630749890773691802401321948599241617171873740187460875541452669196018430458379320978910452677708740121149389289049260368909671797571587872574361576403325458450829959641703568470576948819313050657979060435743564740553565911085870118497098825973672356583186516354715506718750007325734787689281138147193205163931032061943134231140199543095420684425751639787908398865190601747112700042196582032481766506799648617686643106868998527331337192639617847034473260672095881810378587492712587519328256 If I'm correct, that should be 3!↑↑3!↑↑3!↑↑3!↑↑3!↑↑3!.
Calculating 3$ should have given you a memory error as well. 3$ is equal to 6^6^6^6^6^6, which, according to the Googoloy Wiki, is greater than 10^10^10^10^36305.
6:59 1# will just be an empty product, which is very often chosen to be 1 by convention. (just like how the empty product a^0 is 1 by convention) Empty sums are picked as 0 by convention. So if we'd define, say n€ as the SUM of all primes
People say this is a "convention", but it is not actually a convention by any sensible meaning of the word. Consider a monoid (M, °, e) where ° is just an arbitrary associative binary operation (could be addition, functional composition, multiplication, anything), and e is the identity element of this operation. Because e exists, and because ° is associative (by the definition of a monoid), ° can be uniquely extended to a function on n-tuples for arbitrary n, and due to the associativity of ° and the properties of such an extension, this function necessarily evaluates the 0-tuple to e. This is not a matter of notational convention, it is a theorem about the operations on a monoid.
If 1 is prime, then the unique prime factorization of positive numbers larger than 1 isn't true anymore, which is a property you like to have, especially in higher level maths it turns out to be a useful property. For example , if 1 is prime, then 2=2 but also 2*1 but also 2*1*1 etc. So it's not unique
@@helloitsme7553 and that breaks what? You're argument is the same as all of the others, a semantics game. You lose nothing by making 1 prime, but you lose a lot of functionality by excluding it. This video literally contains an example of this.
@@Harkmagic you do lose something, it is unique prime factorization! It is extremely useful in fields like abstract algebra and numbertheory for example. what functionality do you lose by excluding it? It's not a prime by definition: a prime is a positive number divisible by exactly two positive numbers
You don’t need to define 1 as a prime. Just define 1# = 1 and you’re done! No need to change the existing definition of prime numbers just to include 1 in the calculations for #. But like others said above, the concept of unique prime factorization of positive integers above 1 is extremely, extremely useful.
@@Harkmagic Even though 1 is not prime, 1# = 1 is still true, because the product of the empty tuple is 1. *I never liked any of the arguments for 1 not being prime.* You may not like the arguments, but unless you can syntactically deconstruct those arguments and demonstrate that they are invalid, you not liking the arguments has 0 implications. *Somebody needs to show me what breaks if 1 is prime.* I am not sure what you are referring to. 1 not being a prime number is not a matter of "breaking" mathematics. It is a matter of definition. The prime numbers have to satisfy a definition in order to be called "prime numbers". 1 does not satisfy that definition in the same way that composite numbers do not satisfy it. So 1 is not a prime number. It is that simple. *and that breaks what?* It breaks the fact that the integers form what is called a "unique factorization domain". Also, having 1 be a prime number is inconsistent with itself: that would make 1 the only prime number, since every number is divisible by 1. *Your argument is the same as all others, a semantics game.* This is an incredibly moot point, since literally EVERYTHING is semantics. You do know that, in order to have a conversation, a set of agreed-upon definitions that are completely arbitrary and not practically supported have to be established, right? Language is built on definitions, and language is everything in the world, not just in mathematics. Your complaint is the equivalent to complaining that we define the English word "house" to refer to a specific type of building that shelters living being, rather than defining it instead to refer to, say, a kind of food. This complaint is a non-argument. The term "prime number" is defined in the way that it is defined, whether you like such a definition, or not. The natural number 1 does not satisfy this definition, so it is not a prime number. If you have an issue with 1 not being a prime number, then what you really have an issue with is the definition of "prime number" as a whole. So the onus is on you to explain, what about the current definition of "prime number" is problematic? Because as I understand it, a better definition for the phrase "prime number" could not exist, and it just so happens that 1 does not satisfy this definition. It causes exactly 0 problems. *You lose nothing by making 1 prime, but you lose a lot of functionality by excluding it.* No, this is just false. There is no functionality lost from defining 1 to be a prime number, and there is much to be lost from arbitrarily changing the definition of "prime number" to include the number 1 arbitrarily, with no other changes, as such a definition would be mostly useless and meaningless, as such a label would no identify a set of numbers that satisfy any particularly important property warranting such a label to begin with. Explain: what number-theoretic or algebraic property is sufficiently important that is satisfied by 1 as well as the prime numbers, and no other numbers? *This video literally contains an example of it.* I literally prefaced my comment by explaining how said "example" is not an example at all.
Figure this out while playing around with Gamma function :) Here's the formulas for double factorials : (2x)!! = 2^x ×Gamma(X+1) (2x-1)!! = 2^(1-x) ×[Gamma(2x)/Gamma(x)] Anyways, thanks for the useful video!
10:16 wolfram alpha couldnt calculate it so i decided to resort to a funny method: i got 10^(1.8408x10^33) edit: yay i got a heart Edit 2: I have recalculated it, turns out I got 10^10^10…(repeat the 10, 20 more times)^795
My maths profesor always told us about the person that invented the subfactorial, or the left factorial because he is Serbian. The name of the mathematician is Đuro Kurepa ( Ђуро Курепа ), but never took the time to explain what it actually does. I finally remembered by myself and found a video about it, thanks.
4:34 you didn't state _why_ it's always a whole number, but it's pretty clear just by looking at it. The reason is n! by definition will be contain as a factor any and all factorials below n!. So 1!, 2!, 3!, . . . n! all divide n! cleanly. Very neat property of the factorial.
I am a simple man. Too dumb for maths. But I LOVE your content and your style of teaching/presenting, so I watch your videos anyways. Oh, and I have subscribed as well. Thank you for your work!
Intersting thing to point out: Sloane's super factorial and the hyper factorial are very similar in pi notation! Let's take the example of sf(4) and H(4). sf(4) = 4!*3!*2!*1! = (4^1)*(3^2)*(2^3)*(1^4) H(4) = (4^4)*(3^3)*(2^2)*(1^1) and more generally, sf(n) is the product from k=1 to n of: k^(n-k+1) Whereas H(n) is the product from k=1 to n of: k^k (I'll try to write it in pi notation like this Π(index; upper bound; expression) ) sf(n) = Π(k=1; n; k^(n-k+1)) H(n) = Π(k=1; n; k^(k)) Neat!
Me: grows 1 inch up after the last meeting with my grandmother My grandmother: adds a dollar sign and wonders who's that giant guy she has never seen before
This video made me excited for math! I can see these operations being useful, which they must be because they exist, but still. I could see ME using them, which is awesome.
so for sf(n) = pi(k!) we end up with a sort of triangle 1 * 1 * 2 * 1 * 2 * 3 * ... 1 * 2 *...* n which if we look at vertically equals 1^n * 2^(n-1) * ... * n^1 so in fact we can also write sf(n) as pi(k=1, n, k^(n-k+1))
Why are you asking us to do 24^^24 on a calculator? You know it goes from top to bottom right? 24^24~1.334e33 24^(1.334e33)=10^(log(24)*1.334e33)~10^(1.841e33) 24^(10^(1.841e33))=10^(log(24)*10^(1.841e33))=10^(10^(1.841e33+log(log(24))=10^(10^(1.841e33+.1400))=... and that is just the first 4 24's. Each 24 after that will give a new power of 10 in the base. Needless to say, this is way to big for a calculator as even by the third 24, the number will have 1.841e33 digits which is WAY too big for the 100 digit calculator limit. P.S. 24^^24 ~ 10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(1.841e33+.1400))+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)
I’m now letting my iPad calculator calculate the power tower ²⁴24 as you asked but spitting out the digits my iPad is developing into a black hole. Cool, I always wanted to see a singula
7:03 Yeah it's 1 because 1 is the neutral element of multiplication and the set below the Big-Pi-Operator is empty since there is no prime lower or equal 1
For the primorial, we can find a condition for n# based on (n-1)#. If n is composite, then n# = (n - 1)#. It will just be the same as the previous number, and this goes back to the previous prime number. If n is prime, then n# = n * (n - 1)#. You have to multiply your previous result by the now new prime. So if we look at 1#, we can get there by taking 2#. 2 is a prime, so 2# = 2 * 1#. We also no that 2# = 2, as it is the first prime number and therefore it is just itself. So now we get 2# = 2 = 2 * 1#. So 2 * 1# = 2, or 1# = 2/2 = 1. Therefore 1# should logically be 1. Now we also can do the same for 0#: 1 is not a prime, so we simply get that 1# = (1 - 1)# = 0#. So 0# = 1 as well.
Link to my IG notes instagram.com/p/CSMu_IJnzik/?
does any of those have any normal application? XD
Heheh that's all 4 in every examples.XD.But actually I like it.
ua-cam.com/video/tMnG7h1J61s/v-deo.html Ghostbusters..3
@@ChadTanker The subfactorial certainly has some (e.g. imagine n people meet and each of them brings a gift. The number of ways they can redistribute the gifts among them such that everyone gets exactly one gift and none gets his own gift is !n). It has some more serious applications as well (for instance in cryptography). The superfactorials grow insanely fast and I doubt they really have many application in reality tbh. It is easy to construct combinatoric problems that lead to them (however, I don't think those problems are likely to be relevant in practice) 😅
Now make a video on relation between them
If that is the Hyper factorial, the Pickover factorial should be named UltraMegaBlaster factorial instead of merely Super.
Oh yeah 😄👍
This reminds me of Celeste speedruning
Or just Ultra Factorial
even inputting 3 into the pickover factorial will get you a number indescribable.
@@srevere7241 6^6^6^6^6^6 yeah, but I feel like that one is still more calculable than 4! Tetration 4! The real question is: Is 4! Tetration 4! bigger than Gram's Number or TREE(3)?
Two observations:
1. The double factorial is also known as the semifactorial, which I personally think makes more sense, since you are only multiplying half of the numbers less than or equal to n.
2. All this super-duper-mega-hyper factorial stuff reminds me of when we were kids, and got into an argument about things like whose car was faster, or whose daddy earned more money, like little boys often do. It usually went something like this:
- A hundred.
- Two hundred.
- A thousand!
- A thousand thousand!
- Ten times more than you can say!!!
(And no, that's not a triple factorial. It's just three exclamation marks.)
...but sometimes those who-can-name-the-biggest-number contests can end unexpectedly, as a colleague of mine once overheard his two boys compete:
- One thousand.
- Ten thousand.
- One million.
Elder brother now remembers that he has seen the infinity symbol, ∞, somewhere, and thinks he has a sure win:
- Horizontal eight!
But younger brother (who has no clue about infinity) is quick to respond:
- Horizontal nine!
A factorial symbol is the exclamation mark, so it just depends on context. I bought a TV for my bedroom and won another 3!" Another 3! what? TVs or bedrooms? Did I win 3 and am just excited, or did I really win another 6?
@@X22GJP I do hope you won 6 quotation marks, and not 3, because 3 would be unbalanced and would certainly cause a syntax error.
There'e even more, look up hyperfactorial array notation in googology wiki
Woah a thousand thousand factorial, that’s a lot
My TI-nspire sadly passed away calculating the 24 Power Tower... Rest in Pieces
rip
I am sorry to hear that….
I've set up my computer to run a quick python script. Hopefully, when every subatomic particle in the universe is a digit, I might be able to get you your answer. I'll set it up to read out from the milky way black hole..
That's not really a big loss 😉
@@camrouxbg Leave. NOW! 😤 😤
The real question is: "How do you seamlessly switch between pens?!"
there is a video of him showing how to switch the pens
Hacks
Link?
@@lorenzohsu5133 ua-cam.com/video/-HQrpaveZJo/v-deo.html
that is the whole point of his channel name
For the primorial, ig 1# = 1 makes the most sense to me.
Ways to arrive at this conclusion:
1: You also multiply by 1 even if it isn't a prime.
2: Since 2 is a prime number, (2-1)# must be 2#/2, which in this case is 1.
3: An empty multiplication is 1.
but also you can look at what the primes smaller than 2 are. That's the empty set. So you are multiplying the empty set over the empty set, witch is kinda undefined
@@emmata98
The empty product is 1, as the original comment noted
I agree
Same ans
@@owenbechtel That's how i thought of this. When you are suming things, but you don't have anyghing to sum, you get 0, that is the identity for sums. When you are multiplying things, but you don't have anything to multiply, you get 1, that is the identity for multiplications.
"Five times three times one. You can do that by yourself."
*Finally* he gets to a level of mathematics I can do!
😂😂😂😂😂😂
😂😂😂
😂😂
10:14 I calculated it, but UA-cam doesn't allow posting comments so large they physically create black holes in the server. I've submitted a bug report, when it's fixed I'll get back to you.
😂
Post it in pastebin then give us the link
@@yat_ii Got another error:
Domain error: value cannot fit into the universe.
@@MagnusSkiptonLLC show an approximation in standard form then
@@yat_ii doesnt work either, the exponent creates another black hole
Usually the empty product is defined as 1 and empty sum as 0. So if the set of primes equal or lower than 1 is empty the product should be 1 by convention
I said exactly the same thing
@@skylardeslypere9909 sorry. Didn't see. Great to know you're on board
@@Arthur0000100 oh no I didn't mean to call you out or something. Just a nice coincidence. It means that we're probably correct lol.
You commented well before me as well anyways
Agreed. Otherwise, it makes no sense to say 0! = 1 either.
It's not a convention, it's logic. The empty sum is 0 because whenever you add some number to it, it becomes that same number. So for all x, x+(empty sum)=x. That means empty sum=0.
Similarly, (empty product).x=x, so empty product=1.
"Multiple exclamation marks are a sure sign of a diseased mind."
Sir Terry Pratchett
I agree!!!
@@fgvcosmic6752 !!!!!!!!!!!!!!!
True!!!!!!!!!!
idk who that is but he sounds like a mega incel based on that quote
@@impwolf Maybe use google or Wikipedia before commenting? Pratchett was a comic fantasy writer, and the quote was a joke. He often used multiple exclamation marks himself.
6:46 There's no primes less than or equal to one. Therefore, the solution is the product of the empty set, which is 1: the multiplicative identity.
Correct.
Then why is (-1)! Not defined as 1?
@@luffnis What?
@@anshumanagrawal346 yes same Argument. It would be an empty set
@@luffnis I don't follow?
The exponential factorial should use the euro (€) symbol. It's still a monetary symbol so it would remind us of the dollar symbol, and it symbolizes a E, just like 'exponential'.
That's a great idea
It's a shame how most math students are never introduced to the double factorial and/or subfactorial during Calc 2. I feel that knowing these concepts would make comprehending series a little easier.
What level of math is this? I'd dare say most students are never exposed to any of this. I've done every math subject there is short of Masters level or PhD level math and I've never seen or heard of any of these... ever.
@@taekwondotime I'm guessing it's some journals or papers or certain professors. idk for sure though.
Yeah, these would just make writing out and defining a series super easy. Use of some of this notation would save a lot of hand cramps.
But these don't really fall under a proper math class category. They are usedul tricks you pick up along the way. If ypu actually tried to use them you would probably have to provide a definition at the beginning to avoid confusion.
@@Harkmagic I'm willing to bet these are all recently invented mathematical notations. I doubt any of these existed ~40 years ago.
Gregory Rolfe Why not use the product notation?
Wow this is pretty fascinating - I didn't know some these existed, and their uses are also interesting! Videos like yours inspire me to share my own maths content as well!
You didn't share any maths content here
@@X22GJP No, I share my maths content on my channel.
hello ali khan, #1 UA-camr
7:06
I don’t know the actual answer but I would guess 1#=1 for a reason similar to why 0!=1
We can define (n+1)# as =n# if n+1 isn’t prime and =(n+1) x n# if n+1 is prime
2 is prime and we know that 2#=2 so 2#=2=2 x 1# so 1#=1
Product of nothing = 1
Exactly.
WolframAlpha has 1# = 2
@@rafiqhaq This is because Wolfram Alpha is using a different definition of n#. Wolfram Alpha defines n# not as the product of the prime numbers less than or equal to n, but as the product of the first few n prime numbers.
You are absolutely correct!
1# is called an Empty Product. The value of an empty product is 1.
It's funny how the Hyper factorial gives way smaller numbers then the Super factorials (Pickover)
What new factorial will you define next?
fig factorial
n★
@@LostArcadeMachine cool! and what does it mean?
@@KyleTheFolf Come up with yourself, don't know what could that be 😅
!N! is N! but multiplied by 2π and exponentiated by N!!
Some crazy stuff! And some not-so-crazy.
I chuckled silently when you asked for calculator help with the power-tower, 24^(24^(24^(...^24)...)).
I was picturing some poor cuss actually trying to work this out on a calculator. Even taking the log will only "reduce" the tower by 1 "level."
And you didn't even crack a smile when you said that.
Incidentally, I would say that 1# = 1, because it's a vacuous product - there are no primes ≤ 1.
Fred
😆 I guess that phrase came pretty naturally for me since I often ask my students to calculate certain things for me during class. Hahaha
@@blackpenredpen Yes, that's good; and asking them to calculate this, would itself be instructive.
Illustrates the power (pun intended!) of very large numbers.
Fred
My Casio gave an instant answer "Math ERROR".
@@koharaisevo3666 Your calculator is correct. In the early days of computers, this was called, "floating point overflow."
Fred
Using this superfactorial shouldn't be hard to write down something bigger than graham's number.
In the end my takeaway is:
-the first 3 are useful notations
-number 5 allows to write the biggest numbers with only few symbols
-I don't see what 4 is good for but I have a feeling I could run into it naturally
-I don't see what 6 is good for and have no idea when I'll ever need it
-7 is bigger than 4
It's crazy that the number of derangements !n == the closest integer to n! / e. We looked at the formula for derangements on the first day of my combinatorics lecture because the formula was so cool.
Totally crazy. Almost deranged.
Amazing!
I suppose we have to show that the truncation error from the infinite sum is at most 0.5 in absolute value?
@@Grassmpl Perhaps, but this falls out for free when you consider derangements in Sn for n > 2.
@@lego312 how does counting alone justify proximity to the transcendental number n!/e?
Do mathematicians secretly hate humanity
I don't think so hmm. But they may hate each other.
Cause they are not equal.
Jk
I am surprised no one has come up with a Super Hyper Factorial
pwr_twr(n!) x pwr_twr((n-1)!) x pwr_twr((n-2)!) x ... x pwr_twr(3!) x pwr_twr(2!) x pwr_twr(1!) = &(n)
where pwr_twr = power tower of n, and &(n) = As I call it, Super Hyper Factorial.
@@mysticdragonex815 you should write a paper, and you'll go down in mathematical history 👍
Yikes, I thought I was in-the-know because I was familiar with the double factorial; I had no idea about the other factorial variants you showed. Very cool, thank you BPRP!!
Still waiting for the five-star-super-deluxe-premium-factorial.
Sounds like something you can order from a restaurant.
If That's true Rip person that solves that
"And only for $11.99, you can buy this pack of -horse armor- factorials to add to your -game- calculations"
Le Giraffe:
Calculates all difficult factorials and leaves us the easiest(24 power tower) to solve
24 tetrated to 2 = 1.3337357768502841244490814728438*10^33
tetration can either be written with double carets: 24^^2
or with the hyperoperation notation: 24[4]2
the "n" in x[n]y stands for the level of operation.
x[1]y = x+y
x[2]y = x*y
x[3]y = x^y
x[4]y = x^^y
in greek, tetra means four.
the carets come from knuth's up arrow notation.
12:47 for print screen
CORRECTION: The primorial n# does NOT multiply all the primes
I don't think that's correct.
The example you've given should be written as p₅# = 2 x 3 x 5 x 7 x 11 = 2310 the product of the first 5 primes.
If you write 5# that would be evaluated as 5# = the product of the primes ≤ 5 = 2 x 3 x 5 = 30
Note: p₅# = 12#
10:14 I'm pretty sure future Casio fx calculators will give the answer as 24^24^24.....^24. As the current ones are only limited in giving small answers like you enter 3/2 and press = button to see the answer 3/2.
I love him for how reluctantly he called it a hashtag and not a pound sign.
Just call it hash symbol. Pound sign could also mean £.
I wish I saw this video before😅
I still remember when I was trying to solve an Olympiad combo problem and concluded that the answer was the multiplication of the odd numbers from 1 to 2n+1
Then I opened the solution and I was shocked when I saw the answer (2n+1)!!
Only then to realise later "they are the same" 😂
I remember watching your videos about the subfactorial, double, super and hyper factorials. Thank you for always giving me new information!
I was always curious about those ever since I met the subfactorial on another video - thanks a lot for feeding mine and probably others' curiosities!
7:05 by logic you need to define that if it’s less than or equal to 1. So you need to goes only +1 to the next p but in 0 (empty set) = 1 in a multiplicative way
1# according to your definition is an empty product (there is no p
Agree
ua-cam.com/video/fh7T5qlIVtA/v-deo.html Cinderella.3
He made a mistake when explaining what the primorial function does. He said that it multiples all the primes numbers which are less or equal than the number n but according to Wikipedia en.m.wikipedia.org/wiki/Primorial it multiples the first n prime numbers.
So the answer to the question 1# is 2 because the first prime number is 2.
@@serbanhoban1517 Please, read the article carefully. Particularly, please, pay attention on this section en.m.wikipedia.org/wiki/Primorial#Definition_for_natural_numbers
@@serbanhoban1517 This article says that if p is a prime number, then p# is defined as the product of all primes from 2 to p.
However, then it states that if you want to include any natural number n, the definition is another one, and it coincides with the definition given in this video.
For example:
If n = 5, then n# = 5# = 5 * 3 * 2 = 2 * 3 * 5
If n = 8, then n# = 8# = 7 * 5 * 3 * 2
Note that 8# = 7#
For the cases n = 0 and n = 1, where there are any prime numbers ≤ n, it's defined that 0# = 1# = 1
This is why I love this channel ,I got surprised.
Glad to hear 😃
I think I've seen all of these before. I play Four 4s a lot, so factorial extensions are key operations for me. Nice that he included both versions of the super factorial!
10:11 My calculator handed me a letter of resignation.
8:50 man you're making me laugh throughout the video 😂
😆😆
@@blackpenredpen
pwr_twr(n!) x pwr_twr((n-1)!) x pwr_twr((n-2)!) x ... x pwr_twr(3!) x pwr_twr(2!) x pwr_twr(1!) = &(n)
where pwr_twr = power tower of n, and &(n) = As I call it, Super Hyper Factorial.
10:16 my calculator says "Timed out. Value may be infinite or undefined."
Yeah pretty sure it was just a joke, we can't even calculate stuff like 3^3^3^3^3.
@@sttlok btw there's an easier way to write down tetration, a^^b, there a is base (and each power) and b is the height. Thus, your example is simply 3^^5, and super factorial for n is n$ = (n!)^^(n!)
@@МОЩЬ32СТВОЛОВ yeah I know, I am not used to the “^^” notation, but I am to writing down the exponent at the left.
5:12 "Don't be too crazy"
5:15 Puts factorial on n's head
@2:51 The way he stopped while saying, "yeah" and then "I am not gonna do it" ! Bro are you reading my mind?!! xD
Here’s a question. Why does everyone solve factorial problems by multiplying integers from greatest to least. For example if a teacher teaches you how to solve for 4! they will likely tell you to multiply 4 by 3 by 2 by 1. Why not 1 by 2 by 3 by 4? You get the same result and it’s much more natural.
I guess most people don't have a preference but if you have the latter as preference, go for it
@@helloitsme7553 hi now i see u on this channel!
Because 4!=4.3! and we write 4x instead of x.4
If you start calculating that n! with the biggest factor, n, then you'll follow a more standard factorial calculation procedure: you'll have to stop when the changing factor reached 1, a condition which is independent of the factorial you're calculating.
If you start with 1, then you'll have to continuously compare the changing factor with n and hence keep remembering that value of n, and stop when that changing factor has become n & multiplied into the value you're calculating.
I definitely prefer the first a.o. because it doesn't matter whether you don't multiply a value or multiply it with 1, and because I won't have to remember the value of n, the paper I'm writing on will do it for me.
@@Apollorion ok.
I feel like the pickover one is impossible to apply. Like, you'd literally probably only be able to go to number 4, as with 5 you'd have 120 exponents which is just ridiculous
I didn't knew how math can be interesting and fun before, thank you for teaching me these new factorials.
There is also the subrecursive factorial:
srf(n)=n * product(k
funfact:
sf(n) * H(n) = (n!)^(n+1)
it is very intuitive, but to prove it nicely you might have to use product of a product formula for switching indexes (if that's what it's called)
It could be a nice thing to try to prove in Coq (or Metamath, but I didn't manage to have any success with Metamath yet)
Not hard to prove at all. Just show that sf(n) = n^1 * (n-1)^2 * ... * 2^(n-1) * 1^n and the claim follows immediately
sf(1) = 1 => claim trivially true for n=1.
Now assume claim true for n. Then:
sf(n+1) = (n+1)! * sf(n) [by definition of sf(n)]
= (n+1) * n! * sf(n)
= (n+1)! * n! * (n * (n-1)^2 * (n-2)^3 * ... * 1^n)) (Using the assumption)
=(n+1)! * (n*(n-1)*...*1) * (n * (n-1)^2 * (n-2)^3 * ... * 1^n)) (Writing out the factorial)
= (n+1)! * n * n * (n-1) * (n-1)^2 * ... 1 * 1^n (some rearranging)
= (n+1)! * n!^2 * (n-1)^3 * ... * 2^n * 1^(n+1) (more rearranging)
Thus, if claim is true for n, it is also true for n+1. qed.
@@arthur_p_dent didn't say it was hard
I did it like this:
(using P as a product, [...] is a step(?), (...) is a subject/base/whatever the hell it is called in english)
P[1
I like how you ask everyone to try 4 super factorial on the calculators when you know that they won't be able to display the answer
whenever I see a new function, I try to graph it on desmos. I'd be very interested to see a video on how you would try to graph these
7:10 1#=undefined
10:21 none of my calculators can handle more that ²24. 😭
For the subfactorial, you can also calculate it by taking the floor of (n!/e +½).
Did you know that lim n goes to infinity of !n/n! is 1/e? That's one of my favorite results in all of math!
Its really cool, and its pretty easy to prove.
Hi...............
Will u make a video on the proof of it?
That mean that the series at 5:56 when n→infinity is 1/e
@@aashsyed1277 u can prove it very easily simply by using the taylor series expansion of e^x, and simply pulg x= -1
- Hey can I borrow some money?
- Sure how much?
- 4$
- ...
I worked out the Pickover super factorial for 24. It's exactly equal to ERR.
What is ERR?
@@hasan_issa It's short for "error".
Just compute it mod p for a bunch of primes p. Then use the Chinese remainder theorem to narrow down some options.
scream the number to make it bigger, but scream too loud and you’ll scare it
I've seen the first three. I remember asking if there was a name for products of all primes up to a given number, and someone told me about primorial. I was (...and actually still am, kinda) messing around with prime generation, and so I had generalized the trick of ignoring even numbers (after 2), and was using what I found to be primorials for that (it's one of several projects that I've never finished, or quit, but just got distracted from.). To define primorial recursively, I'd say `n# = { isPrime(n) : n * (n-1)#, (n-1)# }` (or, just `n# = n^isPrime(n) * (n-1)#`), and we can start off with a base case of 2# = 2. But if we apply recursion to that anyway, 2# = 2 * 1#, but we know that 2# = 2, so 2 * 1# = 2 => 1# = 1... and 1# = 1 * 0# = 1 => 0# = 1. So we have 1, 1, 2, 6, 6, 30, etc.
this is a great video. thank you for making a video explaining all of these in one place :)
If there's a superfactorial and a hyperfactorial, does that imply the existence of the maxfactorial? (Pokemon games reference)
Fullrestorial?
7:07 We can find this answer with the same logic as why 0!=1. Let n be any integer and p be any prime. We can then define n# as p#/p, with p>n. Following this, we can see that 5#=7#/7=30, 4#=5#/5=6, 3#=5#/5=6, 2#=3#/3=2, and by consequence, 1#=2#/2=1. Following this same logic, not only can we deduce that 1#=1, but that, in fact, for any n
Hello bprp, small doubt.....
Is subfactorial the same as the number of dearrangements??
Yes
@@blackpenredpen Thank you.....
love the kobe shout out big respect
Aqui no Brasil o subfatorial é conhecido como permutação caótica.
I learned about the super factorial right after this year’s Euclid Math Contest because one problem required a proof that involved the product of factorials
By trying to calculate the 4! tower (4$), I got back a memory error. However, I was much luckier calculating the 3! tower (3$):
3$=80190511417718642126823324718367187228561124379028767032642984026696527685909099423272280409907130820856664234534252547383919785792220682688124768661305459764363907411429981465891057029933838727501814441806045135620442558743661835589426589946920649349657656706090250821685723480965941188343685690726218140655579217325748445855297737560689439245320090903450689423418447823641842197996266347921612064380092293936942024867447336260960218766156355104115750573964203330671274400021356103878977554933511538319549310099032097779743184906645434985411235166939435035172411964842142967548250148630273650014462188652334799262982699997472433086018965308982853218279479424824047741627463816736228241352680785451432095209668261788939711558466713720132242293745772921448940790740551844434434008906193034676987240057304500131108010023042597053394274584797206497036333055579458255064407007544868240706439176260524117888597747817247043924561435278271887309056381091805867601619602251796096400239298214815262205815810495851883048734986346152273704541907980517682891333798723716799846126881590621405666624030853266332188998637596226214198907834122541927489293463347160133763014502117756168216336158830114627329202977218109579368237166132156567117925020087348139705459145227331715719630342522870498465476785107571053263453494079678567755889095079940187526351199266190216925889027808671629102384349737214723184859355227570333017933339515713795388860158422658813142610052462552561531124468334021552575519317369712398549893299488022466192324266086303869235263681881809144657510051875031162274098866094419279562380208220324102530098886472069111428433617488472272516055190671056469982414848473047070790257893061962649402322109549904795828661722527648687617928767746379721495747519959211141040916111102472432018152460719051167544236405919983233953117838933243887167089427812364370202619892209018498976682851438682521894475191713352835282030493296589384712919392973226219211191288091922284035764198302804401510674264271313400291750479617586815808002065334610106237612814316692500812416262477849331005382194774509783776249392848253693735848749122479363634821386023094809009260807127069703642131601341758921068404932742749189556771687054015933472600318253567596808221091251211711703641198856155255542413502599219243125231124707010703756432040851991341579197236142864356940729178223076963340376298091195126023533546841565469722388179096534865015625515047046570963420216955624280137393078231569773569948982141887926144207971441215537594906005093536952329848039312778015477469720653882057885248129417138963934082124319879328510703466345181658431317850957327034071471765397226881197993545556865982592007997710424004475702357132496494376641281701478783172600043123929627756814940337917468536651352909682412163154933605051724078476404415853009241046889879088290672699116823567675505259508394940589299351448798962932730350799970185840036495181266341124321852431181496056540339690610156603751845458286632667474065265696737473864354691357207202701527065402487087291412527403277767976883461633028962004285545846440493575225314130774394979967937378817702113126306072419455152323267882594983571298483500465825807896703872181789457381955432647872387911051213467617557987023849695828359459524711163550419985869657676704055817908644687127673576453955210839424436840190659827027252321398501932586759740411729952289617418278134765622813326050166959957384064382813113083786831755203742521598218605765840629154362364687711303817838049012975261098818706031083778779921930338153969952829372320637217705971993553150607385902119752440657964303988303972862883646147475106786443197703235867584836077370838721142011678759973762131722424134687500917686363953045267662773093137815945736556948724190193573407163764867877153195367591431100153449614703833275030770886797919827969802690303977026301264215440127629900242728911768560267326235803994874362448037123613763254450430482381895799210777320387010513081228433695682802772932190357949981416457818029991504540768966753037459786011903710783960269984510243360995482400887126305528142426809242291255927388970092499522644826730634353554532290013554216298408936830014398138795251653589037358576904476827007923274508531053478037943367964176441257037590277013740407417782007327008826098874282368889270784570950786912620185328736577519896968757943687578610897754204026914925858221388080673050441824821755725576167340253305804521182043728264128801559756563257488713680680809133701727450964058594763006137824371369361316200344599880051384402035659367496743923603271929776588780455945342609429175333833732087253316702961877934549090835555674032605356077637644879327372936947591318361663596803630395896131225284879988495303929143762967731049100198363156149538755837425424959700972683697853135492946217817764276303379016406744567350241586674650572185257582725886064487676298551839944386144412978961115582326074861396098373880273079980787032483386367357279417962171668621359717512606596304376531440825003611118804365098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If I'm correct, that should be 3!↑↑3!↑↑3!↑↑3!↑↑3!↑↑3!.
😮
I knew the last digit must be 6 xD btw I think it's too small. Did you evaluate this left to right?
Calculating 3$ should have given you a memory error as well. 3$ is equal to 6^6^6^6^6^6, which, according to the Googoloy Wiki, is greater than 10^10^10^10^36305.
6:59
1# will just be an empty product, which is very often chosen to be 1 by convention. (just like how the empty product a^0 is 1 by convention)
Empty sums are picked as 0 by convention.
So if we'd define, say n€ as the SUM of all primes
People say this is a "convention", but it is not actually a convention by any sensible meaning of the word. Consider a monoid (M, °, e) where ° is just an arbitrary associative binary operation (could be addition, functional composition, multiplication, anything), and e is the identity element of this operation. Because e exists, and because ° is associative (by the definition of a monoid), ° can be uniquely extended to a function on n-tuples for arbitrary n, and due to the associativity of ° and the properties of such an extension, this function necessarily evaluates the 0-tuple to e. This is not a matter of notational convention, it is a theorem about the operations on a monoid.
I never liked any of the arguments for 1 not being prime. Somebody needs to show me what breaks if 1 is prime.
As such 1#=1.
If 1 is prime, then the unique prime factorization of positive numbers larger than 1 isn't true anymore, which is a property you like to have, especially in higher level maths it turns out to be a useful property.
For example , if 1 is prime, then 2=2 but also 2*1 but also 2*1*1 etc. So it's not unique
@@helloitsme7553 and that breaks what?
You're argument is the same as all of the others, a semantics game. You lose nothing by making 1 prime, but you lose a lot of functionality by excluding it. This video literally contains an example of this.
@@Harkmagic you do lose something, it is unique prime factorization! It is extremely useful in fields like abstract algebra and numbertheory for example. what functionality do you lose by excluding it?
It's not a prime by definition: a prime is a positive number divisible by exactly two positive numbers
You don’t need to define 1 as a prime. Just define 1# = 1 and you’re done! No need to change the existing definition of prime numbers just to include 1 in the calculations for #.
But like others said above, the concept of unique prime factorization of positive integers above 1 is extremely, extremely useful.
@@Harkmagic Even though 1 is not prime, 1# = 1 is still true, because the product of the empty tuple is 1.
*I never liked any of the arguments for 1 not being prime.*
You may not like the arguments, but unless you can syntactically deconstruct those arguments and demonstrate that they are invalid, you not liking the arguments has 0 implications.
*Somebody needs to show me what breaks if 1 is prime.*
I am not sure what you are referring to. 1 not being a prime number is not a matter of "breaking" mathematics. It is a matter of definition. The prime numbers have to satisfy a definition in order to be called "prime numbers". 1 does not satisfy that definition in the same way that composite numbers do not satisfy it. So 1 is not a prime number. It is that simple.
*and that breaks what?*
It breaks the fact that the integers form what is called a "unique factorization domain". Also, having 1 be a prime number is inconsistent with itself: that would make 1 the only prime number, since every number is divisible by 1.
*Your argument is the same as all others, a semantics game.*
This is an incredibly moot point, since literally EVERYTHING is semantics. You do know that, in order to have a conversation, a set of agreed-upon definitions that are completely arbitrary and not practically supported have to be established, right? Language is built on definitions, and language is everything in the world, not just in mathematics. Your complaint is the equivalent to complaining that we define the English word "house" to refer to a specific type of building that shelters living being, rather than defining it instead to refer to, say, a kind of food. This complaint is a non-argument. The term "prime number" is defined in the way that it is defined, whether you like such a definition, or not. The natural number 1 does not satisfy this definition, so it is not a prime number. If you have an issue with 1 not being a prime number, then what you really have an issue with is the definition of "prime number" as a whole. So the onus is on you to explain, what about the current definition of "prime number" is problematic? Because as I understand it, a better definition for the phrase "prime number" could not exist, and it just so happens that 1 does not satisfy this definition. It causes exactly 0 problems.
*You lose nothing by making 1 prime, but you lose a lot of functionality by excluding it.*
No, this is just false. There is no functionality lost from defining 1 to be a prime number, and there is much to be lost from arbitrarily changing the definition of "prime number" to include the number 1 arbitrarily, with no other changes, as such a definition would be mostly useless and meaningless, as such a label would no identify a set of numbers that satisfy any particularly important property warranting such a label to begin with. Explain: what number-theoretic or algebraic property is sufficiently important that is satisfied by 1 as well as the prime numbers, and no other numbers?
*This video literally contains an example of it.*
I literally prefaced my comment by explaining how said "example" is not an example at all.
Primorials are actually the first n primes. for example:
5#=2*3*5*7*11
You should cover the rising factorial, it’s used in the hypergeometric function
I love how much it takes me to notice the pokeball, it gets me in every video, I'm so focused that I just dont notice
Amazing! Ironically I Heard them when I was 4th grade but I had no Idea Of the applications. Thank you BPRP!
Figure this out while playing around with Gamma function :)
Here's the formulas for double factorials :
(2x)!! = 2^x ×Gamma(X+1)
(2x-1)!! = 2^(1-x) ×[Gamma(2x)/Gamma(x)]
Anyways, thanks for the useful video!
10:16 wolfram alpha couldnt calculate it so i decided to resort to a funny method:
i got
10^(1.8408x10^33)
edit: yay i got a heart
Edit 2: I have recalculated it, turns out I got 10^10^10…(repeat the 10, 20 more times)^795
😮
@@blackpenredpen big number
My maths profesor always told us about the person that invented the subfactorial, or the left factorial because he is Serbian. The name of the mathematician is Đuro Kurepa ( Ђуро Курепа ), but never took the time to explain what it actually does. I finally remembered by myself and found a video about it, thanks.
Great video! I didn't know most of these!
You: "Here are 7 less common factorials that you probably didn't know..."
Me: "What's a factorial?"
5:10 "No. Don't be too crazy" 😂
Person at the bank: here is 10$
Me: 10$ what?
Wow, I'm actually reading a Ken Wilber book and I was about to search something related to his work and this video popped up
Excellent presentation!!
I'm convinced mathematicians are just trolling at this point
I confirm
pickover's superfactorial is pretty crazy huh
They're really crazy factorials😱
4:34 you didn't state _why_ it's always a whole number, but it's pretty clear just by looking at it. The reason is n! by definition will be contain as a factor any and all factorials below n!. So 1!, 2!, 3!, . . . n! all divide n! cleanly. Very neat property of the factorial.
I am a simple man. Too dumb for maths. But I LOVE your content and your style of teaching/presenting, so I watch your videos anyways. Oh, and I have subscribed as well. Thank you for your work!
Thank you, Dániel!
Intersting thing to point out:
Sloane's super factorial and the hyper factorial are very similar in pi notation!
Let's take the example of sf(4) and H(4).
sf(4) = 4!*3!*2!*1! = (4^1)*(3^2)*(2^3)*(1^4)
H(4) = (4^4)*(3^3)*(2^2)*(1^1)
and more generally, sf(n) is the product from k=1 to n of: k^(n-k+1)
Whereas H(n) is the product from k=1 to n of: k^k
(I'll try to write it in pi notation like this Π(index; upper bound; expression) )
sf(n) = Π(k=1; n; k^(n-k+1))
H(n) = Π(k=1; n; k^(k))
Neat!
Me: grows 1 inch up after the last meeting with my grandmother
My grandmother: adds a dollar sign and wonders who's that giant guy she has never seen before
"Woah" Wikipedia:"This is Worthless!"
This video made me excited for math! I can see these operations being useful, which they must be because they exist, but still. I could see ME using them, which is awesome.
so for sf(n) = pi(k!)
we end up with a sort of triangle
1 *
1 * 2 *
1 * 2 * 3 *
...
1 * 2 *...* n
which if we look at vertically equals
1^n * 2^(n-1) * ... * n^1
so in fact we can also write sf(n) as
pi(k=1, n, k^(n-k+1))
I have seen most of them! Pretty fascinating! Thanks for inspiring me so much!
I don't actually know about the triple factorial
Why are you asking us to do 24^^24 on a calculator? You know it goes from top to bottom right?
24^24~1.334e33
24^(1.334e33)=10^(log(24)*1.334e33)~10^(1.841e33)
24^(10^(1.841e33))=10^(log(24)*10^(1.841e33))=10^(10^(1.841e33+log(log(24))=10^(10^(1.841e33+.1400))=...
and that is just the first 4 24's. Each 24 after that will give a new power of 10 in the base. Needless to say, this is way to big for a calculator as even by the third 24, the number will have 1.841e33 digits which is WAY too big for the 100 digit calculator limit.
P.S. 24^^24 ~
10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(1.841e33+.1400))+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)+.14)
I remember when this channel was at 500K subs.
5:04 Is it equal to 3?
I’m now letting my iPad calculator calculate the power tower ²⁴24 as you asked but spitting out the digits my iPad is developing into a black hole. Cool, I always wanted to see a singula
7:03 Yeah it's 1 because 1 is the neutral element of multiplication and the set below the Big-Pi-Operator is empty since there is no prime lower or equal 1
7:05
Assuming the definition you've given is precise, it is undefined.
There are no primes smaller than or equal to 1 to find the product of.
Today I have learned something new which that the most some of them I have not know before,thank you sir
Thank you for this list. This is an interesting set of operations.
For the primorial, we can find a condition for n# based on (n-1)#. If n is composite, then n# = (n - 1)#. It will just be the same as the previous number, and this goes back to the previous prime number. If n is prime, then n# = n * (n - 1)#. You have to multiply your previous result by the now new prime. So if we look at 1#, we can get there by taking 2#. 2 is a prime, so 2# = 2 * 1#. We also no that 2# = 2, as it is the first prime number and therefore it is just itself. So now we get 2# = 2 = 2 * 1#. So 2 * 1# = 2, or 1# = 2/2 = 1. Therefore 1# should logically be 1. Now we also can do the same for 0#: 1 is not a prime, so we simply get that 1# = (1 - 1)# = 0#. So 0# = 1 as well.