[unreality]

if you want to participate, send me a message by 6 pm Eastern Standard Time (UTC -5) tomorrow (wednesday may 26).

Title your message something along the lines of "LPI contest" or whatever.

The body of the message can be no longer than 100 characters. A character if you don't know is a single instance of keyboard input, such as a number, letter, space, pound symbol, etc. These are the only characters you can use in your message:

0123456789.,( )+-*/^!%\

^ = exponent (raise the thing directly to the left to the power of the thing directly to the right; ie, 4*5^6*7 is equal to 28 * (5^6). If you wanted the 4*5 unit to be raised, you would need to surround it in parentheses, that's how the order of operations works).

! = factorial (0! = 1 and n! = n * (n-1)!, only valid for positive integers)

\ = integer division, ie, divide and truncate. The goal is for your number to be a positive integer; it will be invalid otherwise

% = modulus. The remainder function. Not sure why you would want to use this but there you go.

Anyway, remember, you only have 100 characters to express your number. You can't use pre established mathematical constants like Graham's number or whatever.

Good luck!

P.s. I may need help with the mathematics figuring out what is the biggest, if it comes to that, so also tell me in your PM if you're good at math haha

[Molly Mae]

Correct me if I'm wrong.

9!^9!^9! would be larger than ((9^9)^9)

In the first example, you continue to increase the exponents vice the bases. Further, it uses fewer characters than the second. I'll go through and do the math in an attempt to see if the below is largest...

9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!^9!!!

EDIT: I guess I'm wrong. A little research shows that I am trying to do the same thing just in a different (and incorrect) method.

Edited June 1, 2010 by Molly Mae

[Guest]

The question is... is 9!! or 9^9 more.

And the answer is that 9!! is A LOT more.

so unreality's trumps mine (and I think mine was the largest of the exponent based approaches)

Yeah, I'm thinking that unreality's is the largest possible number with 100 characters, but I'm not completely certain. I was messing around with 3s on my TI-83 Plus (must seem pretty old and archaic in today's terms ) and while I got it to overflow pretty quickly, it was definitely the case that for the same number of characters, the factorial was growing faster, since it could have at least two factorials for every exponent.

3!! = 720

3^3 = 27

3!!! = <more than 1700 digits, I used a Java applet to calculate that >

3^33 = 5559060566555523

I'm not sure if you took it to the extreme, if the exponential might overtake it:

9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9!

or:

(9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9^9)!

I'm not sure which is larger...

The real question is does factorial or tetration grow faster and I'm not sure of the answer to that question, though the fact that unreality can increase the factorial at twice the rate (two !! for every ^9) might make up the difference, even if tetration did increase faster.

Hmm, I'm also getting a little confused on PEMDAS rules. I would interpret

2^2^2^2 as 2^(2^(2^2)) = 65536, but my calculator treats it as ((2^2)^2)^2 = 256

Argh, the article writer is correct, we really don't think about large numbers. I've been completely wrong the whole time. We think about multiplication all the time 2*2*2 = (2*2)*2, but rarely do we think about the rules as applied to exponentiation, where following the same logic, 2^2^2 = (2^2)^2. It also doesn't help that exponentiation is not commutative like multiplication is. But thinking about it I realize that the calculator's rules for exponents is in keeping with it's rules for multiplying, so I was thinking about it the wrong way. If you want to do tetration (what I was trying to do with my submission), you do need the parentheses, though glycerine could have made his number much larger by omitting them since they would have been the same result without and he could have done more iterations without them.

9^99^9^9 = ((9^99)^9)^9 < 9^99^9^9^9^9

but tetration (which should be much larger) would have been for the same number of characters:

9^(99^(9^9)) > ((9^99)^9)^9

It should be larger than the same number of digits sans parentheses too:

9^(99^(9^9)) ?> 9^99^9^9^9^9

So I think that maybe the largest possible competitor to 9!...! would be along the lines of:

9^(9^(9^...9)...) out to 100 characters. That forces tetration of 9s which I think is the best bet at overtaking factorial, though I admittedly haven't done the calculations...

[Glycereine]

Yeah i realized as soon as I started analyzing these that I put my parethesis backwards. The way I have them they aren't even necessary, but I meant to put them the other way around lol.

I have a hard time analyzing harvey or dawh's in comparison with unreality.

I am working on it though :-p.

And yeah at least initially factorals win. Double checking after that.

[unreality]

9^(9^(9^...9)...) out to 100 characters. That forces tetration of 9s which I think is the best bet at overtaking factorial, though I admittedly haven't done the calculations...

the only way I can think of "calculating" is by repeatedly doing logarithms until one is obviously bigger but I'm not sure how well that approach would work for some of them

[plainglazed]

Yeah i realized as soon as I started analyzing these that I put my parethesis backwards. The way I have them they aren't even necessary, but I meant to put them the other way around lol.

I have a hard time analyzing harvey or dawh's in comparison with unreality.

I am working on it though :-p.

And yeah at least initially factorals win. Double checking after that.

yeah, me too, except didn't even realize the mistake until dawh pointed it out above. you're right about my extra unneeded parens as well. D'oh! does seem unreality's is greatest but as to largest possible, would think you could substitute one ^9 for a !! somewhere and get a greater number yet. As to where within the 100 digits is an interesting problem. and if one exponent makes it bigger, would another ^9 and a set of parens be greater yet? and then... yikes, big numbers, small tired brain. has been fun to think about...

[unreality]

yeah, me too, except didn't even realize the mistake until dawh pointed it out above. you're right about my extra unneeded parens as well. D'oh! does seem unreality's is greatest but as to largest possible, would think you could substitute one ^9 for a !! somewhere and get a greater number yet. As to where within the 100 digits is an interesting problem. and if one exponent makes it bigger, would another ^9 and a set of parens be greater yet? and then... yikes, big numbers, small tired brain. has been fun to think about...

No, raising N to the 9th power is guaranteed to be less than N! if N is sufficiently big (actually I just did some tests, starting with N=14 and higher)

[plainglazed]

well, reread your OP again and am wrong yet again. what I was thinking would require a set of parentheses as well, 9!!!!!!!!!!!!!...^(9!!!!!!!!!...). was then expanding that concept to be kinda like Harvey45's entry subbing 9!!!!!!!! for his 99999999!s. so you would loose four !s for each exponent. hmmmm...

[Glycereine]

The thing to keep in mind is that the number at which factorals overtake exponents is low enough that exponents are never worth using. 14 is the first time a factoral overtakes. 9! is way larger than 14 and 9^ is nothing. so 9^9 is severally lacking compared to 9!!.

Now, the only way exponents could win would be with them back loaded. 9^(9^(9^.....))) However I can't prove that even this would overtake factorals. However if it does it would always overtake them...

I don't believe that a mix of ! and ^ is going to be the solution (with the exception of using one or the other to fill a final spot to get to 100 characters without waste (as would be the case in mine if done correctly and with 9! instead of 99 to get the 100th character.)

anyways long story short.... 9^x is lower than x! starting with x = 22. So even with the exponents on the back end factorals win.

Unrealities is the highest number that can be achieved with 100 characters.

[unreality]

The thing to keep in mind is that the number at which factorals overtake exponents is low enough that exponents are never worth using. 14 is the first time a factoral overtakes. 9! is way larger than 14 and 9^ is nothing. so 9^9 is severally lacking compared to 9!!.

Now, the only way exponents could win would be with them back loaded. 9^(9^(9^.....))) However I can't prove that even this would overtake factorals. However if it does it would always overtake them...

I don't believe that a mix of ! and ^ is going to be the solution (with the exception of using one or the other to fill a final spot to get to 100 characters without waste (as would be the case in mine if done correctly and with 9! instead of 99 to get the 100th character.)

anyways long story short.... 9^x is lower than x! starting with x = 22. So even with the exponents on the back end factorals win.

Unrealities is the highest number that can be achieved with 100 characters.

I appreciate the vote of confidence but I think first we need to prove that the back-loaded method doesn't work too because that has the potential to get realllllly big.

[Glycereine]

The problem with the backload method is that its still multiplying each number by 9. So even with backload, the x! is higher than 9^x after 21. x being the permutations that have already happened. If x is the same size in these two cases the factorial still wins (after 21). Since the starting number 9! is far more than 21, 9^x will never catch up to x! even if they used the same number of characters. Since backloading exponents actually uses 4 characters it falls even further behind.

so 9^(9^9) is less than 9!!! (we already proved that 9!! is more than 9^9 (same characters)). Since both 9!! and 9^9 are over 21, 9^(9^9) is less than (9^9)! Also 9!! is greater than 9^9, so we wouldnt use that anyways. 9!!! is greater than 9^9!! and still uses one less character.

[unreality]

ok so pretty much, it seems that I won Does anyone dispute this though? Does anyone think they can trump that?

The only thing I could think to do would be to replace the last two factorials with a ^9 but that would just test for a comparison of N!! versus N^9.

By testing with google (putting n!! - n^9 starting with n=1 and going up until the subtracting is positive), at n=4 it goes from -18thousand or so to wayyy positive. And certainly the number before all that is 4 or bigger

So yeah I really can't think of anything that can trump a 9 followed by 99 factorials

Here's an interesting challenge though: what is the last digit of said number? In other words, take the modulo by 10 of the number

[Izzy]

*nevermind* Haha.

Edited June 5, 2010 by Izzy

[Guest]

Anyone can name the largest positive integer. It is "infinity".

Who can construct the largest number from the characters shown

using no more than 100 characters? Hmmm....

I'd go for 9 followed by ^^^^^^^^^^ up to 98 total carets followed by a 9.

Being a non algebraic exponential expression parenthesis should not be required.

[Guest]

ok so pretty much, it seems that I won Does anyone dispute this though? Does anyone think they can trump that?

The only thing I could think to do would be to replace the last two factorials with a ^9 but that would just test for a comparison of N!! versus N^9.

By testing with google (putting n!! - n^9 starting with n=1 and going up until the subtracting is positive), at n=4 it goes from -18thousand or so to wayyy positive. And certainly the number before all that is 4 or bigger

So yeah I really can't think of anything that can trump a 9 followed by 99 factorials

Here's an interesting challenge though: what is the last digit of said number? In other words, take the modulo by 10 of the number

I'm not a total math whiz and I don't know of any readily-available tools that can calculate numbers on the scale that we're discussing, but I do have a nagging feeling that we may be missing something somewhere down the line. The factorials are multiplying by larger and larger numbers, but at the same time, tetrating the nines is multiplying by nine an increasingly large number of times.

I was trying to calculate 9^(9^9) using a program I wrote to deal with arbitrarily large numbers in Java. Unfortunately, I achieved that effect by storing it as a string and multiplying digit by digit, so it's monstrously slow and I eventually gave up since 9^9 = 387,420,489, meaning that the next iteration will be multiplying 9 almost 390 million times. And that's just the third iteration. The one with 100 characters has 25 iterations, plus a factorial to fill in the gap.

I haven't done the equivalent with factorials, but I'm sure that it will grow extremely quickly as well. I haven't seen anyone provide a mathematical proof that the factorial number is an upper bound of the tetrated version, so as we go out to 100 characters, I'm not completely convinced that the tetration won't be able to overtake the factorial.

My number theory is a little rusty and I was never very good at Big O calculations. And usually when people study Big O examples, the purpose is to find more efficient algorithms, not to see which is larger.

I think that the factorials are probably larger, but I don't think that we can just ignore tetration as a possible contender.

[unreality]

I'm not a total math whiz and I don't know of any readily-available tools that can calculate numbers on the scale that we're discussing, but I do have a nagging feeling that we may be missing something somewhere down the line. The factorials are multiplying by larger and larger numbers, but at the same time, tetrating the nines is multiplying by nine an increasingly large number of times.

I was trying to calculate 9^(9^9) using a program I wrote to deal with arbitrarily large numbers in Java. Unfortunately, I achieved that effect by storing it as a string and multiplying digit by digit, so it's monstrously slow and I eventually gave up since 9^9 = 387,420,489, meaning that the next iteration will be multiplying 9 almost 390 million times. And that's just the third iteration. The one with 100 characters has 25 iterations, plus a factorial to fill in the gap.

I haven't done the equivalent with factorials, but I'm sure that it will grow extremely quickly as well. I haven't seen anyone provide a mathematical proof that the factorial number is an upper bound of the tetrated version, so as we go out to 100 characters, I'm not completely convinced that the tetration won't be able to overtake the factorial.

My number theory is a little rusty and I was never very good at Big O calculations. And usually when people study Big O examples, the purpose is to find more efficient algorithms, not to see which is larger.

I think that the factorials are probably larger, but I don't think that we can just ignore tetration as a possible contender.

note that the following upper/lower bounds are kind of obvious and can be shrunk inward more but here they are to start with:

The upper bound of a factorial n! is n^n and the lower bound is n.

the upper bound of a doubled-up factorial n!! is (n^n)^(n^n) and its lower bound is n!

To prove more concisely though that n!! > n^n we need to figure out WHERE roughly n! falls in between n and n^n.

Here's a little chart

n   n!     n^n        n! / n^n

1   1      1          1.0000

2   2      4          .50000

3   6      27         .22222

4   24     256        .09375

5   120    3125       .03840

6   720    46656      .01543

7   5040   823543     .00612

8   40320  6777216    .00240

9   362880 387420489  .00093

10                    .00036

11                    .00014

12                    .00005

13                    .00002

so n^n is monstrous compared to n!, get roughly twice as big as n! is getting per new step. If that's true: d (n! / n^n) / dn = - (n! / n^n) / 2 d (n! / n^n) / (n! / n^n) = -dn/2 ln( n! / n^n) = -n/2 + c ln(n!) - ln(n^n) = -n/2 + c ln(n!) = n*ln(n) - n/2 + c Testing that:

n    ln(n!)     n*(ln(n) - 1/2)

1    0          1/2

2    .693       .386

3    

I had to stop because of dinner time, I don't think this will work because the 1/2 was more of an eyeball approximation from the column, but anyway I'm thinking it's possible to prove that n!! > n^n, I guess there really isn't an elegant method, but after dinner I'll make a chart comparing n!! to n^n

[Glycereine]

I am trying to think of how to prove it but I'm certain that factorials beats the tetration of 9^n.

hmm I messed with this post for 30 minutes and ended up deleting all my math but I still stand by my above statement.

I was thinking I had proven it already but it seems circular (alot of proofs are )

[unreality]

also I don't think anyone answered the riddle of what the last digit of 9!!!!!!!!etc...!!!! is, but the answer is 0

[Glycereine]

also I don't think anyone answered the riddle of what the last digit of 9!!!!!!!!etc...!!!! is, but the answer is 0

Oops, I completely missed it! but yeah as soon as you have a number over 10 you have a 0 in the one's place after the factorial.

[Guest]

i have to agree the 9 followed by the 99 factorials is the biggest.

while all the others are quite big,

9^9^9 = 9^387420489

10^x = 9^387420489

log 10^x = log 9^387420489

x = 387420489*log 9

x = 369693099

9^9^9 aprox. = 10^369693099

9!!!! is phenomenally huge;

9! = 362880

362880! aprox = (181440^362880)/ln 362880

10^x = 181440^362880 /ln 362880

x *ln 10 = ln 181440^362880 /ln 362880

x *ln 10 *ln 362880 = 362880*ln 181440

x*29.477 = 4393977

x = 149065

10^149065 ! is monstrous, not even gonna attempt to calculate it, let alone adding another !.

[Guest]

The multifactorial is much smaller than the factorial. For simplicity in the following table, instead of displaying

! a number of times, I shall display !(k) where k is the number of ! following the integer.

9!(1) = 362880 [6 digits]

9!(2) = 945 [3 digits]

9!(3) = 162 [3 digits]

9!(4) = 45 [2 digits]

...

9!(9) = 10 [2 digits]

9!(9+)= 9 [1 digit]

If one is trying to create large numbers, one need note, for example, 9!!! is not equal to ((9!)!)!. For the large

number, the factorials should be nested --> ((9!)!)!.

9! = 362880 [6 digits]

(9!)! = ??...00 [1859934 digits]

...

Another method of creating a large number is the use of exponentiation.

9^9 = 387420489 [9 digits]

Unlike the factorial, for creating larger numbers with the finite characters, nesting should not be used. Without nesting, evaluation occurs at the highest level first, of which is where we want our largest number to be.

Compare:

(9^9)^9 = 150094635296999121 [18 digits]

9^(9^9) = 9^9^9 = 196627050475.....9 [78 digits]

Another type of exponentiation is the use of tetration, a power tower. The ASCII character representation for tetration is to use the double caret(^^):

9^^9 = 9^9^9^9^9^9^9^9^9 = ?????????????????...? [?? digits]

Now the mix of factorials and exponentiation can create huge numbers as well. The question arises whether 9!^9! is greater than 9^^9!. In general, the much larger number is the number with more levels of exponents if the highest level of exponent is the same: 1.1^(1.1^(1.1^1000)) > 1000^(1000^1000).

If this holds true with the super-large numbers, then 9^^9! > 9!^9! Thus, with with the given character set, I believe the largest number representable might be:

9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^9^^99!

For those who are interested in reading about large numbers:

http://www.mrob.com/pub/math/largenum.html

Edited July 8, 2010 by Dej Mar

[unreality]

for example, 9!!! is not equal to ((9!)!)!.

what do you mean? When we've been saying 9!!!!!!!!!!, we mean 9! then THAT factorialized, then that, etc, etc, etc. From this base:

9!! = (9!)!

I understand one could make up other uses of "!!", like skipping every other number or whatnot. But I think everyone on the thread assumed that

3!!! would be:

((3!)!)!

aka the factorial of 720, whatever that is

[Guest]

I do realize that many of the posters on this thread had assumed 9!!! to mean ((9!)!)!, but that had been an incorrect assumption. Without nesting in brackets, 9!!! should be interpreted as a triple factorial of 9 and not as the factorial of the factorial of the factorial of 9.

The value of the novemnonagintuple factorial of 9 is simply 9. So, though many assumed it might have been the largest submitted number, it was the smallest number submitted.

[Guest]

Reading about Large Numbers, I learned the double caret (^^) used in tetration may be extended in the representation of higher hyper operators. Thus, following this extension, the largest positive integer that may be represented, which uses the hyper100 operator, is:

9^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^9

The above number may also be represented as a 9 followed by another 9 separated by a superscripted circled 100. This value is smaller than one boogol (a 10 followed by another 10 separated by a superscripted circled 100),

named by Johnathan Bowers who invented the BEAF (Bowers Exploding Array Function), a combined notation of the Array Notation and Extended Array Notation used to represent huge numbers. [Note, the superscripted circled number will be 2 greater than the number of carets).

[Guest]

As I made an error in my earlier post for the value of 9!(9), ie., 9!!!!!!!!!, and to explain what I proffered for the largest number in my last post.

In reference to multifactorials, I shall use <f> to represent f factorials. For exampl e, n<3> = n!!!.

The multifactorial, for which the factorial is a subset, can be defined by: n<f> = n*(n-1f)*(n-2f)*(n-3f)*...*2. Therefore, for the simple factorial, f would be equal to 1 and is defined by: n! = n*(n-1)*(n-2)*(n-3)*...*2.

Operation notation w/base# = 9 value digits

factorial -------------------- n! | 9! 362880 6

double factorial ------------- n!! | 9!! 945 3

triple factorial ------------- n!!! | 9!!! 162 3

quadruple factorial ---------- n!!!! | 9!!!! 45 2

quintuple factorial ---------- n!!!!! | 9!!!!! 36 2

sextuple factorial ----------- n!!!!!! | 9!!!!!! 27 2

septuple factorial ----------- n!!!!!!! | 9!!!!!!! 18 2

octuple factorial ------------ n!!!!!!!! | 9!!!!!!!! 9 1

nontuple factorial ----------- n!!!!!!!!! | 9!!!!!!!!! 9 1

... ... | ... ... ...

novemnonagintuple factorial -- n<99> | 9<99> 9 1

... and so forth.

Simple exponentiation is subset of iterated exponentiation. Tetration and above are sometimes referred to as hyperoperations. Each hyperoperation is defined recursively in terms of the previous one. Tetration, for example,

is defined in terms of exponetiation: 5^^9 = 5^5^5^5^5^5^5^5^5. Exponentiation itself is defined as iterated multiplication: 5^9 = 5*5*5*5*5*5*5*5*5. In reference to iterated exponentiation, I shall use <x> to represent x carets, the ASCII notation for the hyperoperation of iterated exponetiation.

w/base# = 9

(hyper)operation notation |& exponent = 9 value digits

exponentiation --------------- n^p | 9^9 387420489 9

tetration (hyper4) ----------- n^^p | 9^^9 ? ?

pentation (hyper5) ----------- n^^^p | 9^^^9 ? ?

hexation (hyper6) ------------ n^^^^p | 9^^^^9 ? ?

heptation (hyper7) ----------- n^^^^^p | 9^^^^^9 ? ?

... ... | ... ... ...

hecatontation (hyper100) ----- n<98>p | 9<98>9 ? ?

...

The wiki article on the factorial's rate of growth - http://en.wikipedia.org/wiki/Factorial#Rate_of_growth -

(not an ultimate authority, but nonetheless a good reference), states that "As n grows, the factorial n! becomes larger than all polynomials and exponential functions (but slower than double exponential functions) in n."

As a double exponential function is a power tower of only two levels, if the exponents of each level are the same for the same base, a tetrated number will be greater. Thus, in general, iterated exponentiation is larger than factorialization. Of course, there are hyperfunctions for the factorial -- the hyperfactorial and superfactorials. Each of which can produce "infinity scrapers" as well. Yet, the characters for the notation for these [H(n) and n$] are not among those presented in the contest.

Edited July 10, 2010 by Dej Mar

[unreality]

nice I didn't know it was acceptable to put multiple carats next to each other for other hyper operators (for example google won't accept 3^^2 in its calculator ) But more seriously, nice job!

Here's a question though. If we did "proper notation" for 9!!!!!!...etc and wrapped it in parentheses:

9!)!)!)!)!)!)!)..

with an appropriate number in the beginning, do you think that still beats the other submissions (though not 9<98>9 i'm sure)

1 + x + 2x = 100

1 + 3x = 100

x = 33

If i did that right, it means the proper version of my original submission would be 9 factorialized 33 times (instead of 99).

Wait! We don't need outer parentheses!

So we can save two more digits on the 33rd factorial

9!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!

that's 9 followed by 32 "!)" units and a "!". Hence 1 + 32*2 + 1 = 66

plus the 32 more opening parentheses that need to go in front = 98

((((((((((((((((((((((((((((((((9!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!

the above is 98 characters and represents a 9 factorialized 33 times. But we still have 2 extra characterse

It would be a waste to package another set of parentheses because that would use up what we have left and not change anything. We can't get any more factorial signs out of this. So the only thing I can think to do is:

((((((((((((((((((((((((((((((((999!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!)!

of course 9 to 999 is no substitute for another factorial, but it's not possible with all the parentheses. So the largest number creatable with just factorials is 999 factorialized 33 times

edit: a better option (but it goes down the slippery slope of the exponents and might as well say "change it all to exponents") but it is to do 9^9 instead of 999

because 9^9 = 387 420 489

but since the next char is "!", does that mean 9^9!.. is that 9 raised to the power of 9! or 9^9 factorialized?

Edited July 10, 2010 by unreality