Superfactorial
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Hyperfactorial, • Hyperfactorial introdu...
Note: on Wikipedia, it uses sf(n) for the Sloane and plouffe’s definition of the super factorial. But here it uses n$ for both mathworld.wolfr...
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For anyone wondering, the final results for 3$ is equal to 6 multiplied by itself:
3,048,038,843,157,366,051,156,451,562,587,960,914,683,843,070,009,432,577,720,371,743,970,260,248,988,253,388,734,087,776,516,378,135,275,070,793,705,483,810,168,992,174,078,819,457,229,916,982,909,272,972,117,802,705,538,546,043,924,392,880,249,338,669,607,593,617,565,773,503,681,777,562,894,648,411,193,991,668,219,111,024,140,760,301,610,404,587,920,245,316,338,496,104,651,309,961,792,720,642,767,736,988,127,034,593,657,473,200,942,901,910,918,843,867,439,542,131,034,903,455,333,688,218,414,524,168,236,538,840,952,909,280,093,318,279,899,396,079,306,730,083,204,209,917,234,906,371,378,011,735,099,583,017,148,436,711,446,474,063,192,400,558,782,588,545,007,076,563,578,617,663,198,436,110,485,321,365,453,724,515,867,573,206,330,463,738,004,176,650,289,141,225,099,730,033,358,933,544,597,672,427,393,530,251,774,965,047,211,695,506,312,946,306,762,294,998,433,468,262,930,173,135,236,327,015,973,465,126,381,014,128,785,885,984,902,934,363,536,428,840,380,015,651,386,113,570,744,138,512,750,613,093,972,490,684,498,513,211,851,441,804,268,854,063,729,284,096,776,385,783,954,252,916,394,679,043,883,583,422,253,317,285,565,365,659,687,584,910,629,328,675,433,251,717,593,584,447,460,445,269,780,478,280,767,340,984,528,217,318,932,238,565,096,396,250,980,632,747,766,225,123,954,084,888,838,021,720,715,193,156,305,805,969,620,729,856
Times.
Alex Paradise wow!!!! How did you do it? And this must be pinned!!!!!!
😇
Amazing
Programming
Wouldn't 3$ be 3! x 2! x 1! = 12 ?
Also 6^6^6^6^6^6 should be 10^(10^(10^(10^36305.31580191888)))
Or 6 multiplied by itself 10^(10^(10^36305.31580191888)) times, I don't get how multiplying 6 that many times (the small number you gave) will result in such a huge number such as 6^6^6^6^6^6.
n$ looks like an old code for a string variable to me. I am ancient and lost.
Yeah I thought the same :) Spent too much time coding in BASIC in my youth
You guys are lucky you had access to lowercase letters when you learned BASIC :)
Mário Brito you guys mean Visual Basic? If so, I did that too in high school long time ago
@@blackpenredpen how about basica from Microsoft, or worse TRS-80 BASIC from 1977. I also used a lot of QuickBasic
Just when i already started to learn programming in college lol
Me : How much does a diamond cost?
Owner: 2019$
Me: Wow what a deal!!😍😍
(A few seconds later)
Me: wait a minute!!!...😱😱
Indian dude counting rice grains on a chessboard: 😱
I think 3$ is enough to make me go 😱😱😱😱
Dollar signs are supposed to go before the value so you only have yourself to blame.
@@TheTdw2000 i think it's how the price is pronounced, the number first then "dollar"
blackpenredpen wait 3$ is equal to 6*2*1 why is that expensive
I need to ask my employer if I can have a "pay cut" to 3$. Mwahahahaaa!
alkankondo89 Hahahah!!
In a lot of countries (e.g. France), the currency symbol comes after the number, so this might backfire.
@@user-vn7ce5ig1z not in America 🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸🇺🇸
Kiwi yes it does, you just use it wrong
I like how maths always keeps surprising you... After so many years of learning, it's a subject that's still interesting. Thank you for another great video
Hyperfactorials look big but nevertheless grow a lot slower than TREE.
Head Librarian yeah it grows a heck of a lot slower than g. And g to tree is like a snail to the speed of light.
At least Sloane and plouffe's actually makes sense.
Pickover seems to have just gone "BIG NUMBERS OHHHHHHHHHHHHHHHHHHHHHHHH"
Darn I was hoping for h(n$).
@@DudeinatorMC n equals 1. Nah
H(n$) vs H(n)$❓
It's funny that 3$'s last digit is a 6
And the last 3 are 256 which is a power of 2
why
@@oboplays3413 6*6=36, which last digit is 6. Keep multiplying by 6 and you'll always get 6 as the last digit of your number!
@@pablofueros that's actually pretty interesting
You can actually calculate the last digit of any big number as 3^4556789. Using modular arithmetic
While it is true that there technically were two definitions of the superfactorial proprosed in 1990, the first definition, denoted by n!s, or denoted by sf(n), which is equal to n!·(n - 1)!···2!·1! is the definition most mathematicians use. Clover's definition is typically called the suprafactorial, or the Clover factorial, or confusingly, the tetrational factorial, which is the name for another operation already. Also, the notation n$ is reserved for Clover's definition, it is never used for the smaller definition. As I said, the smaller definition actually has proposed notations sf(n) and n!s, which is what mathematicians tend to use.
Ah!! Thanks.
@DejMar I'm completely aware that Clover did not invent the definition of the large superfactorial. That is completely irrelevant to my post, though, since I never made any claims about who invented it, I merely addressed the operator by its commonly used name. In fact, the operation being described can be concisely defined as simply n$ = n!^^n! for all n, since tetration is defined for all natural bases and heights.
Simon and Plouffe's definition is not a product of consecutive k^k terms, but rather consecutive k! terms. In other words, sf(n) is defined by sf(0) = 1 & sf(n + 1) = (n + 1)!·sf(n). The product of consecutive k^k terms is not the superfactorial, but the hyper factorial, denoted H(n), with the analytic continuation being the K function.
@DejMar Also, I know how repeated exponentiation and tetration works.
In here (Turkey) it is 1.30 AM and a new video from blackpenredpen. I clicked to video, and listen...
In here (Germany) it's 23:36. Good night 😊🌃
Also me
@@katarinakraus120 Thanks, same for you :)
@@Ali-mf4sl We are night people 🌕🌕🌕
In Brazil it is 08:06 pm
Isn’t the hyper factorial every term of the normal factorial to the tetration of 2?
Yup that's how it is defined
Is there a term that describes a single dimension of hyperoperation applied to factorials? As 3! Is 3x2x1 what would 3+2+1 be defined as?
@@decatessara5029 Well, you can use Gauss's addition formula for this, so you'd just end up with it being the same as a function f(x)=(x/2)(x+1)
Glass No, you misunderstood his question. What he is asking is if there exists an operator sequence H(n, •) such that, for example, H(1, 3) = 3 + 2 + 1, H(2, 3) = 3·2·1, H(3, 3) = 3^2^1, etc. In other words, the second variable determines the order of the hyperoperator that is applied to all the terms, and the first variable determines the upper bound of the terms, while the lower bound is 1. To my knowledge, no such sequence has been studies.
@@angelmendez-rivera351 Ah ok, his question makes more sense now
Let's put euro sign € for Pickover's factorial. Then my salary will seem a lot better:D
Viki V. Actually, the notation 3$ is alright. BPRP was wrong in stating that the other definition uses the notation 3$. In reality, its 3!s.
I'm trying to enjoy math for fun (I'm not good at it). You sir, are amazing at teaching, enjoying and helping me.
I can’t understand English.
However, I can understand this beauty.
I like mathematics very much!
I am not sure if I'll ever need that in my life, but it's very cool.
Those are some pretty fast-growing functions. I was expecting the hyperfactorial to be a sort of recursive definition, bearing the same relation to the superfactorial as the superfactorial does to the regular factorial.
And the nice thing about the hyperfactorial is the recursive definition works just fine for H(0).
0.5 super-factorial?
Try to find the value of this summations:
The sum of 1/k! with k=1 to infinity is ?
The sum of 1/k$ with k=1 to infinity is ?
The sum of 1/H(k) with k=1 to infinity is ?
The first one is easy: e, but there is a way to find the exact value of the other two ?
In the second I got approximately 1,58683
Thank you BrendaWork for sponsoring this video
Also, the dollar sign matters ;)
Pulling that work
The famous multiple question
The port and girl paradox
If you like to keep rolling the track
Brenda work slacks back from your pen
I like TREE(3), it's even bigger and raises even faster :)
TREE(Graham's number)$
@@JorgetePanete yeah, that's scary!
Amazing. This reminds me of numberphile's video on Graham's Number. they used like 4 arrows i believe.
But what is a super-hyper-ultra factorial?
Those 2 definitions for super factorial give different answers, or am I missing something? For example by def.1:
3$ = 3!*2!*1!=12
Эти определения от разных людей и ответы получатся разные
Yeah. So..
Yes, although the notation for the first definition is not 3$, it's 3!s. I'm not sure why BPRP presented it otherwise.
I like this blackpenredpen. I love your videos, all of it. I'm from Philippines and always watching your amazing tutorial. Please notice me.
Thank you! I appreciate your support.
2:24 but H(0) would give 0^0 which, although there are limits that approach 1, is undefined when trying to solve without limits. Why is it 1 in this case?
By definition, just like 0! is defined as 1
You could (and should) write H(n) using product notation: H(n) = product of k=1 to n over k^k. Then by definition H(0) would be an empty product, since starting at 1 you are already above the endpoint (namely 0). It is mathematically useful to define the empty product to be 1 and the empty sum to be 0, because the neutral element for multiplication is 1 (as in a*1 = a) and the neutral element for addition is 0 (a+0=a). You can think of the product sign as meaning: „start at the neutral element and multiply as many elements until you reach the upper bound.“ so if you already are above the upper bound, you do not multiply any elements, so your answer is the neutral element (1).
Edit: fixed a typo
just like √(0), it's 0, of course, but it has no limit from the left in the real numbers
Ace O'Spades Most mathematicians define 0^0 = 1, so saying it is undefined is stretching it. Regardless, though, H(0) is not defined as 0^0. It is defined as the empty product. The empty product is 1.
Angel Mendez-Rivera [citation needed]. I'm not aware of *any* mathematicians who define 0^0 as 1, it leads to such a mess.
Thanks man! My mind turns from REGULAR to SUPER HYPER after watching this video! Cool level: 2019$ 😎
Mak Vinci Hahahhaha
Btw, I really wonder they came up with all these. I can’t even calculate 3$ with the last definition lol
@@blackpenredpen Yes! Even wolframalhpa cannot handle 6^6^6^6^6^6, so crazy!
Those numbers have more digits that there are elementary particles in the entire observable universe.
I came up with superfactorial about 5 years ago (I was in 8th grade) and actually called it superfactorial too! Never really found much of a use for it so I scrapped the idea
It’s a very rough calculation but 3$ has a magnitude on the order of 10^10^10^36305. So needless to say, the super factorial grows very quickly
DragonFire17 yup, it’s some serious stuff!
I have seen the first kind of superfactorial befre but not the hyperfactorial or the double up arrow version.
How would i write down a normal factorial but with only even numbers? Like n*(n-2)*(n-4)*...*2
There is a !! for odd numbers. For even numbers you can just factor out the 2 so 6*4*2 =2^3(3*2*1)
So factorial for evens is just 2^n*n!
@@Green_Eclipse I haven't thought about that! Thanks :D
That’s why we put dollar symbol before the number to specify the cost
Amazing 😍. I know the superfactorial and hyperfactorial from the Google but I didn't see the relationship between them. This is wonderful ❤️
Is there a practical need for super or hyperfactorials? Normal factorials are used in combinations, permutations, taylor seriers, etc.
I love this channel for ridiculous maths I had to click
Wish I had a teacher like you! 💖
Simply awesome
Is there a name for n^(n-1)^(n-2)^...^3^2^1 ?
Henrix98 Yes. It is called the expofactorial, and it is sometimes denoted by n(!1), although there is no actual universally agreed upon notation.
I think we should write a hyperfactorial as n‽. It means
n$×(n-1)$×...3$×2$×1$. So:
3!=3×2×1=6
3$=3!×2!×1!=6×2×1=12
3‽=3$×2$×1$=12×2×1=24.
I want to use ¡, but it is reserved for inverted factorial. It means n÷(n-1)÷...3÷2÷1. So:
4¡=4÷3÷2÷1=2/3.
what are the uses for these operations?
I was expecting something like n^(n-1)^(n-2)^...^2^1 lmao
Matteuz lolll that will be so cool
Matteuz That already exists. It's called the "exponential factorial," or "expofactorial" for short. There is no agreed-upon notation, though, but the most common notation I have seen among googologists is n(!1), where n is the input and (!1) denotes expofactorial.
Angel Mendez-Rivera wow, that’s cool!
maybe notate it "nᵎ" or "n‽"
Wait. Isn’t hyper factioral smaller than super factorial for all values greater and equalt to 4?
For the hyper factorial, isn't 0^0 indeterminate? You mentioned H(0) = 1, but from the formula 0^0 can be many things, like e, pi, etc...?
What about a "flat" (an opposite of "super") factorial--(1/n!)[1/(n-1)!][1/(n-2)!]...(1/3!)(1/2!)1--or a
"hypo"-factorial--(1/n^n)[1/(n-1)^(n-1)][1/(n-2)^(n-2)]...(1/3^3)(1/2^2)1??
The second form of 3$ comes out to about 1.03144248E+28, or about 10.3 octillion, or we could just say about 10 thousand trillion trillion, or about 10 thousand trillion in the old British number system.
Wait a second is it supposed to evaulate left to right or right to left, because I could be wrong, if I was doing to operations in the wrong order.
I would think there should be a factorial like this. [ 6^5^4^2^1 ] I mean that seems to me like the next logical step from multiplying in a decreasing consecutive fashion to doing it exponentially
6^5^4^3^2^1 is equal to 6!1
Where is this applied? Or we don't do that here? Cuz factorials atleast have a use...
brushed my teeth, prepared for sleep, and then saw this notification... Read it as subfactorial and that's the reason I clicked on notif, hope I won't be disappointed, lmfao
Edit: video was dope... Now show application of those in real life please :)
aha, yes ..wanted to say, pls more video on combinatorics and subfactorial, yolo
Been there, done that czcams.com/video/dH7kt-xAlRA/video.html
Thank you!!
@@blackpenredpen yeah, I have watched, huge fan of those, thanks to You, that's why I want more :) Also edited my main comment, Liked this video, would like to see application in real life for every of those super and hyper factorials :)
Baba Froga The only applications are analysis. That's about it. Idk why people assume applications exist for everything. 90% of math doesn't have applications besides doing more math
@@angelmendez-rivera351 oh wouldn't agree, you can always see great application for everything... lol even if it is more math, it will lead to somewhere.... anyways "more math" is still real life... lmfao, not imaginary or whatsoever^^
Intuitively, superfactorial makes me think of something like:
n$ = n^(n-1)^(n-2)^(n-3}^...^3^2^1
Hey bprp!! I have a question for you. Please answer this _/\_
If n! has 17 zeros then what is the value of n?
( I know the thing about trailing zeros but this is about total number of zeros.)
casher: this orange is only $12$
costumer: What's $12$
casher: umm... 127313963299399416749559771247411200 Billion Dollars
This is very cool BPRP! Will you be continuing with more big functions/operations? I would like to see them 100% Maybe fast growing hierarchry? :)
The number of digits of the number of digits of the number of digits of the number of digits (4 times) of 3$ is 36,306. A very Big number
Isn't the first súper factorial the same as n!^1*(n-1)^2*(n-2)^3...*3^(n-2)*2^(n-1)*1?
There is also primorial #n - product of first n prime numbers.
Krutoy Gadget
Wow!!! I didn’t know about it. Thanks for letting me know and I will look into it
Actually, I'm fairly certain it is n#, not #n.
@@angelmendez-rivera351 Yes, in Wikipedia it is n#, but in one book it was written #n.
Very Interesting!
2 is very special. 2$ (second form) is the same as H(2) or 2² or 2×2 or 2+2. Daaaaamn
I was quite surprised at that simple relation, I thought that surely the superfactorial and hyperfactorial grow much faster than that.
Skallos They do grow very fast. They are just overshadowed by the suprafactorial, which is the second definition that BPRP gave for the superfactorial.
@@angelmendez-rivera351 After watching Numberfile's video on fast growing functions, my expectations for those functions were quite high. They do grow fast, but not as fast as I thought.
Skallos There are no studied functions that grow anywhere nearly as fast as Graham's function or the TREE function. The gap is humongous. It's the gap from 0 to ♾, and then some more.
What do you compute with super and hyper factorial?
I’ve only seen this term in Yugioh lol.
It is "easy" to expand the regular and super ! to the complex plane. The hyper! should be something like (Gamma(n+1))^(n+1)/n$ rigth?
gaier19 Yes
No
Why is the first definition different than the second?
I've seen all before! Are there any uses?
Are there any extensions to the rationals/reals/negative integers? How about derivatives?
Drew Michael Look up Barnes G function and K function
Solve integration of e^x×logx and sin^1/3 x
Would you use it in superpoker
Hey, Can you do a video about the Inverse function of the Factorial (for all numbers which the original function is defined at?)
Like the Inverse of the Gamma/Pi functions
holy moly , that second super factorial grow very quick, 6^6^6 already have 36306 digits
that should be called ultrafactorial instead
Lol!! superultrafactorial
David Franco,6^6^6=6^((7777-1)×6)
Both formuals for superfactorial are not the same
For example:
# using n$=n! (n-1)!(n-2)!...3!2!1! #
3$=3!2!1!=6*2*1=12
Using pickover's formula 3$=ridiculously huge number
Why is hyperfactorial of 0 equal to 1? At 2:22
Keep letting us know this type of new concept! Please....
what is d(n$)/dn and dH(n)/dn?
I propose naming the product n$ * H(n) as the "Ultra Factorial" of n.
Isn't n$ = (n!)^^(n!) = G(n+2)?
Where G(x) is the Barnes G-function which satisfies the functionial equation G(z+1)=Г(z)G(z)
Beethoviet Union TH
Yes I read about that too. I don’t understand the G function well enough yet so I didn’t mention that in the video. : )
Those definitions for n$ aren't equal, unless somehow 6^^6 = 12.
Overflow occurred in computation.
How much does it cost?
Oh, it's only 5$ USD
Uses of super and hyper factorial
To make CZcams videos : )
@@blackpenredpenyeah
can someone help me with this limit? lim n->infinity e^(sin(n! * pi * x)) , where x is in rational number.
What is the use of super and hyperfactorials?
"To make You tube videos", said this channel's owner somewhere in the comments
Is there any real application to these hyper-super-large number?
Wait, just to be sure, the two forms of the superfactorial do not have the same meaning right? Like they can't be equated right
AAK Music They do not.
This looks like a Numberphile content topic ;d
So when doing superfactorial do we use the definition define by sloune and plouffe or pickover?? Or the result will be the same ( which i dont think so ) ?? Thanksp
batu rock The definition by Sloune and Plouffe is the definition 99% of mathematicians who work with this thing use. So that's the definition to go with, and the notation is n!s, where s just means "super." The second definition of the super factorial, the n$ = n!^^n! is sometimes called the suprafactorial by said mathematicians, or just the Clover factorial.
All of these random CZcams videos with no relevant information for my life are what I live for
Hi, I am a math teacher as well. can we collaborate.
any app for it ???
But is there some convinient way to say: "/sum_{k=1}^{n} k"
"n*(n+1)/2",
"1+2+3+...+n",
"n+1 choose 2",
or most conveniently just "the nth triangular number"
@@philipphoehn3883 thank you :-)
So cool.
It's beautiful!
Imagine Graham's number after seeing the 3$ example...
So which definition of n$ is bigger? And are there analytic continuations of n$ like the Gamma function for n!?
Logan Kageorge
Definitely the one by Pickover
Yes, there are analytic continuations for the hyperfactorial and for sf(n) = n!·(n - 1)!···2!·1!. There is no known analytic continuation for x$ = x!^^x!, mainly because there is no known, undebatable analytic continuation for x^^x, although this in the research and it seems to exist, so far.
As math progresses, coding will be more integrated into it
Blackpenredpen, I love your channel so much and I wish I could meet you in real life. I have a question, would the absolute value of i be equal to 1? |i|=1? Is it possible to find the absolute value of i? If so, probably involves the complex number plane? Maybe since i = 0+i, then real = 0, imaginary = 1, would that be why |i|=1?
RudyTheMoose
Yes. Because that’s the distance of i from the origin on the complex plane.
And thank you for the nice comment.
The absolute value of any complex number z is defined by the equation |z|^2 = z·z*, where z* denotes the complex conjugate of z.
I call this the ¡ sign. Bassicly x¡=x^(x-1)^(x-2)^...^2^1.
H(0) shouldn't be undefined?
Ion Grădinaru Why should it be?
0^0 is undefined
Ion Grădinaru 1. No, it is not. Virtually every mathematician defines 0^0 = 1. This definition is basically required anyway, or else a lot of fields in mathematics fall apart. 2. H(0) is not identical to 0^0 in expression. H(n) is equal to the product of k^k with k running from 1 to n. So it would be impossible for 0^0 to show up: H(0) is simply the empty product, which is 1.
Lucky that it uses only the tetration and not the Nth operation
What's the point of the last super factorial, if it's even impossible to calculate 3$?
What's the point of Tree?
@@4snekwolfire813 just to promote MrBeast of course!
factorial gets superpower....... factorialman...... HAHA
6^6^6^6^6^6 is WAY LARGER than 6!•5!•4!•3!•2!•1!