Largest Number Printable

Question

Your goal is to write a program that prints a number. The bigger the number, the more points you'll get. But be careful! Code length is both limited and heavily weighted in the scoring function. Your printed number will be divided by the cube of the number of bytes you used for your solution.

So, let's say you printed 10000000 and your code is 100 bytes long. Your final score will be \$\frac {10000000} {100^3} = 10\$.

There are other rules to follow, in order to make this challenge a bit harder.

• You cannot use digits in your code (0123456789);
• You can use mathematical/physical/etc. constants, but only if they are less than 10. (e.g. You can use \$\pi \approx 3.14\$ but you can't use the Avogadro constant \$= 6\times 10^{23}\$)
• Recursion is allowed but the generated number needs to be finite (so infinite is not accepted as a solution. Your program needs to terminate correctly, assuming unbounded time and memory, and generate the requested output);
• You cannot use the operations * (multiply), / (divide), ^ (power) nor any other way to indicate them (e.g. 2 div 2 is not allowed);
• Your program can output more than one number, if you need it to do that. Only the highest one will count for scoring;
• However, you can concatenate strings: this means that any sequence of adjacent digits will be considered as a single number;
• Your code will be run as-is. This means that the end-user cannot edit any line of code, nor he can input a number or anything else;
• Maximum code length is 100 bytes.

---

Leaderboard

1. Steven H., Pyth \$\approx f_{\varphi(1,0,0)+7}(256^{26})\$
2. Simply Beautiful Art, Ruby \$\approx f_{\varphi(1,0,0)}(3)\$
3. Peter Taylor, GolfScript \$\approx f_{\varepsilon_0+\omega+1}(17)\$
4. r.e.s., GolfScript \$\approx f_{\epsilon_0}^9(126)\approx f_{\epsilon_0+1}(9)\$ ^[1]
5. Simply Beautiful Art, Ruby \$\approx f_{\omega^{\omega2}+1}(126^22^{126})\$
6. eaglgenes101, Julia \$\approx f_{\omega^3}(127)\$
7. col6y, Python 3, \$\approx 127\to126\to\dots\to2\to1\approx f_{\omega^2}(127)\$ ^[1][3]
8. Toeofdoom, Haskell, \$\approx a_{20}(1)\approx f_{\omega+1}(18)\$ ^[1]
9. Fraxtil, dc, \$\approx 15\uparrow^{166665}15\$ ^[3]
10. Magenta, Python, \$\approx\mathrm{ack}(126,126)\approx10\uparrow^{124}129\$
11. Kendall Frey, ECMAScript 6, \$\approx1000\uparrow^43\$ ^[1]
12. Ilmari Karonen, GolfScript, \$\approx10\uparrow^310^{377}\$ ^[1]
13. Aiden4, Rust, \$\approx10\uparrow^3127\$
14. BlackCap, Haskell, \$\approx10\uparrow\uparrow65503\$
15. recursive, Python, \$\approx2\uparrow\uparrow11\approx10\uparrow\uparrow8.63297\$ ^[1][3]
16. n.m., Haskell, \$\approx2\uparrow\uparrow7\approx10\uparrow\uparrow4.63297\$ ^[1]
17. David Yaw, C, \$\approx10^{10^{4\times10^{22}}}\approx10\uparrow\uparrow4.11821\$ ^[2]
18. primo, Perl, \$\approx10^{(12750684161!)^{5\times2^{27}}}\approx10\uparrow\uparrow4.11369\$
19. Art, C, \$\approx10^{10^{2\times10^6}}\approx10\uparrow\uparrow3.80587\$
20. Robert Sørlie, x86, \$\approx10^{2^{2^{19}+32}}\approx10\uparrow\uparrow3.71585\$
21. Tobia, APL, \$\approx10^{10^{353}}\approx10\uparrow\uparrow3.40616\$
22. Darren Stone, C, \$\approx10^{10^{97.61735}}\approx10\uparrow\uparrow3.29875\$
23. ecksemmess, C, \$\approx10^{2^{320}}\approx10\uparrow\uparrow3.29749\$
24. Adam Speight, vb.net, \$\approx10^{5000\times2^{256}}\approx10\uparrow\uparrow3.28039\$
25. Joshua, bash, \$\approx10^{10^{15}}\approx10\uparrow\uparrow3.07282\$

_Footnotes

1. ^{If every electron in the universe were a qubit, and every superposition thereof could be gainfully used to store information (which, as long as you don't actually need to know what's being stored is theoretically possible), this program requires more memory than could possibly exist, and therefore cannot be run - now, or at any conceiveable point in the future. If the author intended to print a value larger than ≈10↑↑3.26 all at once, this condition applies.}
2. ^{This program requires more memory than currently exists, but not so much that it couldn't theoretically be stored on a meager number of qubits, and therefore a computer may one day exist which could run this program.}
3. ^{All interpreters currently available issue a runtime error, or the program otherwise fails to execute as the author intended.}
4. ^{Running this program will cause irreparable damage to your system.}

---

Edit @primo: I've updated a portion of the scoreboard using a hopefully easier to compare notation, with decimals to denote the logarithmic distance to the next higher power. For example \$10↑↑2.5 = 10^{10^{\sqrt {10}}}\$. I've also changed some scores if I believed to user's analysis to be faulty, feel free to dispute any of these.

Explanation of this notation:

If \$0 \le b \lt 1\$, then \$a \uparrow\uparrow b = a^b\$.

If \$b \ge 1\$, then \$a \uparrow\uparrow b = a^{a \uparrow\uparrow (b-1)}\$.

If \$b \lt 0\$, then \$a \uparrow\uparrow b = \log_a(a \uparrow\uparrow (b+1))\$

An implementation of this notation is provided in Python that let's you test reasonably sized values.

Answer 1

Pyth, \$>\operatorname{Laver}^{\operatorname{Laver}^{\operatorname{Laver}^{\operatorname{Laver}^{\operatorname{Laver}^{131072}(131072)}(131072)}(131072)}(131072)}(131072)/87^3\$

=kC"𠀀"=d-C"B"C"A"D:HNTK?%TH:H:HN-Td+Td%+NdHRKDOYJdW:J-ddYJ=+J)RJFQkFzkFGkFZk=kFOkk)))k

Original Python:

def laver(size, x, y):
  if y % size == 0: return (x + 1) % size
  return laver(size, laver(size, x, y - 1), y + 1)
def next(period):
  i = 1
  while laver(i, 0, period): i += i
accumulator = 131072
for i in range(accumulator):
  for j in range(accumulator):
    for k in range(accumulator):
      for l in range(accumulator):
        for m in range(accumulator):
          accumulator = laver(accumulator)

How it works

This entry is based off of Laver tables. A Laver table of order \$n\$ (or size \$2^n\$) is a binary operation on integers from 1 to \$2^n\$ such that: $$x\star 1=x+1(\mathrm{mod}\ 2^n)$$ $$x\star(y\star z)=(x\star y)\star(x\star z)$$ The laver(size, x, y) function recursively computes the Laver table of size size (who could've expected?), except I used the numbers from 0 to \$2^n-1\$, because all programmers do it, and I can just do the modular reduction without having to set the result to size if it turns out to be 0. The recursive call doesn't reduce the last argument, but checking the base case does, so the function ends up being periodic in the last argument. There has to be a modular reduction in one place, so this is unavoidable.

The next function computes the next Laver table size with a given period. Every Laver table has a period, defined as the period of \$1\star n\$. The period is clearly always a power of 2, and it grows extremely slowly. Thus, the size needed to reach a given period grows extremely rapidly (the size needed for period 32 is unknown, see next paragraph). This function I denoted \$\operatorname{Laver}\$ in the title.

In fact, it is unknown if the periods grow infinitely. If they didn't, this number would be ill-defined. I usually don't like submitting ill-defined numbers, but there are reasons I believe this number is well-defined.

1. It is provable in \$ZFC+I3\$.
2. It is almost certainly provable in a weaker system, see below.

Notes on the strength of Laver tables

It is a common practice that somehow works, to assign growth rates to a function based off of the strength of a theory proving it total. Consider the minimal theories that prove "the periods of Laver tables are unbounded," which is an arithmetic statement, especially the "Big Five":

1. \$RCA_0,WKL_0\$ have strength \$\omega^\omega\$. It seems unlikely that they could prove the laver tables are unbounded, given that all proofs use set theory, which these can't encode much of. If this is the strength, then this entry would place at #6.
2. \$ACA_0\$ has strength \$\varepsilon_0\$. This is the weakest that could plausibly prove the Laver tables are unbounded. Being conservative over \$PA\$, this seems unlikely. At this strength, this entry would place at #5.
3. \$ATR_0\$ has strength \$\Gamma_0\$. At this level of strength, this entry is already at #2.
4. The next subsystem, \$\Pi^1_1\text{-}CA\$ already makes this entry the first by a massive margin. It also stands a much more reasonable chance of actually proving the Laver tables unbounded. Even if merely \$KP\$ was sufficient, this would still utterly dominate the other entries.

What if it's stronger?

Consider the usual explanation of how large cardinals are connected with Laver tables:

A rank is a set \$R\$ such that \$f:R\mapsto R\implies f\in R\$. Assume there is a set with a non-trivial elementary embedding from \$i:S\mapsto S\$, then there is a rank \$R\$ with the same property. Take another embedding \$j\$, then \$i(j)\$ is an embedding because being an embedding is definable. Consider \$i(j(k))\$. \$j(k)\$ is the image of \$k\$ under \$j\$, and being the image is definable, so \$i(j(k))=i(j)(i(k))\$. If we let \$i\star j=i(j)\$ then this is one of the defining relationships of a Laver table.

It seems to be possible to implement this in an extension of \$KP\$, and as this seems to be the most natural way to arrive at Laver tables, this might give the proper estimate of its strength. One example is to define a rank-into-rank set as a set \$S\$ that contains an embedding \$j:S\mapsto S\$ (which using replacement, ensures the existence of many other embeddings in \$S\$). The embedding \$j\$ should be such that all \$\Sigma_1\$ (or maybe \$\Pi_2\$) statements in \$(S,\in,\mathfrak{F},\mathfrak{I})\$, where \$\mathfrak{F},\mathfrak{I}\$ are unary and ternary predicates for being an embedding, and being the image under an embedding, respectively. If this is consistent, then it could give a better strength estimate.

Answer 2

Python 3: 98 chars, ≈ 10 ↑↑ 256

Using a variable-argument function:

E=lambda n,*C:E(*([~-n][:n]+[int("%d%d"%(k,k))for k in C]))if C else n;print(E(*range(ord('~'))))

Effectively, E decrements the first argument while increasing the rest of the arguments, except that instead of putting -1 in the arguments it drops the argument. Since every cycle either decrements the first argument or decreases the number of arguments, this is guaranteed to terminate. The increasing function used is int("%d%d"%(k,k)), which gives a result between k**2 + 2*k and 10*k**2 + k. My code does use the * symbol - but not as multiplication. It's used to work with variable numbers of arguments, which I think should follow the rules since the clear point of the rules was to restrict specific operations, not the symbols themselves.

Some examples of how large E gets quickly:

E(1,1) = 1111
E(0,1,1) = E(11,11) = (approx) 10^8191
E(1,1,1) = E(1111,1111) = (approx) 10^(10^335)
E(2,1,1) = E(11111111,11111111) = (approx) 10^(10^3344779)

Only the first two of those are runnable on my computer in a reasonable amount of time.

Then, E is invoked by E(*range(ord('~'))) - which means:

E(0,1,2,3,4,5, ... ,121,122,123,124,125)

I'm not entirely sure how large this is (I've been trying to approximate it to no avail) - but it's obvious that it's ~really~ big.

As an example, about twelve cycles in, the result is around: (technically a bit more than)

E(2**27211,2**27211,2**27212,2**27212,2**27212,2**27212,2**27213,2**27213,2**54423,2**54423,2**54423,2**54423,2**54423,2**54423,2**54423,2**54423,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54424,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54425,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**54426,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636,2**81636)

---

Result estimation:

If we approximate the increasing step by lambda k: 10 * k**2, the function can be described as

E(n, C₁, C₂, ... Cᵥ) ≈ E(10^(n²/2) ⋅ C₁²ⁿ, 10^(n²/2) ⋅ C₂²ⁿ, ... 10^(n²/2) ⋅ Cᵥ²ⁿ)
                     ≈ E(10^((10^(n²/2) ⋅ C₁²ⁿ)²/2) ⋅ C₂^(2n  ⋅ 10^(n²/2) ⋅ C₁²ⁿ), ... )
                     ≈ E(10^((10^n² ⋅ C₁⁴ⁿ)/2) ⋅ C₂^(2n  ⋅ 10^(n²/2) ⋅ C₁²ⁿ), ... )

The relevant thing we're doing here is build up a tower of powers of ten, so the eventual score can be approximated as 10 ↑↑ 256.

Better (although partial) result estimation:

This uses the same 10 * k**2 as the other estimation.

E(0, b) = 10 * b**2
E(1, b) = 10 * (10 * b**2)**2 = 10 * 100 * b**4 = 10**3 * b**4
E(2, b) = 10 * (10**3 * b**4)**2 = 10 * (10**6 * b**8) = 10**7 * b**8
E(a, b) = 10**(2**(a+1)-1) * b**(2**(a+1))

Under the previous estimation, it would be:

E(a, b) = 10**(a**2/a) * b**(2*a)

Which is significantly smaller than the actual value since it uses a**2 instead of 2**a for the 10 and uses a*2 instead of 2**a for the b.

Answer 3

C (score ≈ 10^20 000 000 000 ≈ 10↑↑3.005558275)

• ~20 GB output
• 41 characters (41^3 means nothing)

main(){for(;rand();printf("%d",rand()));}

Despite of rand() the output is deterministic because there is no seed function.

Answer 4

brainfuck, 97 bytes, f₂₅₅(255²)

-[>-[[>]-[<]>>-]<-]-[-[[>]+[<]<+>>-]-[<+>-----]<.,<[>>>[[-<+>]<[<]<[->+<<+>]>[>]>]<+[<]>,<<-]>>>]

Assumes a wrapping implementation with an infinite tape in both directions.

The algorithm of the program is:

Initialise a list with 255^2+1 255s
While the list exists:
    Print a 3
    Pop the first element of the list to use as a counter
    Decrement the counter
    Append counter many 1s to the end of the list
    While counter != 0:
        Prepend length of list-1 copies of the counter to the front of the list
        Decrement counter

This is not infinite, as every time a list element is processed, it only appends elements that are less than it. 1s are popped and removed without appending anything.

A step by step explanation of the code:

- Initialise counter as 255
[ While counter
  >-[[>]-[<]>>-] Add 255 255s to the list
  <- Decrement counter
]- 
This initialises the list as 255^2+1 255s

[ While list exists
  - Decrement the first element
  [ 
    [>]+[<] Append that many 1s to the end of the list
    <+>>-   Copy the element one over
  ]
  -[<+>-----]<.,< Print a 3
  [ While counter
      >>> Go to first element of list
      [ For each element in the list
          [-<+>] Move the element over one
          <[<]< Go to the counter
          [->+<<+>] Prepend a copy of the counter to the start of the list
          >[>]> Go to the next element of the list
      ]
      <+[<] Return to the start of the list while appending a 1
      >,    Remove the first element of the list
      <<-   Decrement the counter
  ]
  >>> Go to first element of the list
]

Big thanks to the folks on the Ordinal Studies Discord for helping me understand just how big the number is. If anything seems wrong, blame them please correct me.

Answer 5

Julia, f_ω³(127)

f(r,s,t)=foldl(|>,r,fill(s,t));!q=x->f(x,q,x);w(b)=a->x->x|>f(a,b,x);w(w(w(!)))(x->x+x)(BigInt('~'))

I aimed for the fast-growing functions corresponding to ordinals in the Veblen hierarchy, but I had to settle for ordinal exponentation. Maybe another challenge...

Explanation:

#Function iteration, which is the tool to get us this far
#Applies s to r, t times
f(r,s,t)=foldl(|>,r,fill(s,t));

#Here's the first order successor:
!q=x->f(x,q,x);

#Function to help us with diagonalization
#Take a and x, and fold the b higher order function on the a function x times, 
#then plug x into the resulting function
#Doing this multiplies the ordinal position by ω, which isn't nearly as high as I would hope
#Yes,I tried the ~ operator, but Julia doesn't like when I do that
w(b)=a->x->x|>f(a,b,x);

#Call it, and implicitly return it. 
#Or interrupt it, and and enjoy your screenfulls of stack traces. 
w(w(w(!)))(x->x+x)(BigInt('~'))

Answer 6

Rust, score \$10 \uparrow\uparrow\uparrow 127\$

fn main(){let b='~' as usize;let mut n=b+b;for _ in b..n{n<<=n;for _ in b..n{n<<=n;print!("{}",n)}}}

Try it online!

Note that the code's validity is dependent on the assumption that a computer with infinite memory will have an arbitrary precision pointer to index the infinite memory. I also attempted to write a version that wouldn't overflow instantly with arbitrary precision integers but it still overflowed on the second bitshift.

Size analysis:

A bit-shifting \$n\$ leftward \$n\$ times is equivalent to the function \$n2^n = f_2(n)\$. My code applies recursion in a manner consistent with the fast-growing hierarchy to achieve \$f_4(n)\$. let the final value of the local variable n in my code above be denoted as \$o\$. It will require \$log(o)\$ bitshifts to create \$o\$. The print! statement runs every time the inner bitshift does, meaning the order of magnitude of my score is the sum of the change in the order of magnitude of the variable n between the inner bitshifts. The order of magnitude of n should be growing exponentially since it is being tetrated, so the expression for the order of magnitude of the result is $$\sum_i^{log(o)}2^i$$ telling us that the final number is \$10^{2^{log(o)+1}-1}\$ which is not meaningfully different from \$o\$ itself since \$o\$ in this case is \$f_4(126)\$ or \$10 \uparrow\uparrow\uparrow 127\$. Please let me know if this explanation seems wrong or incomplete because I am new to the domain of big numbers.

Edit: better approximation thanks to Simply Beautiful Art.

Answer 7

C (clang), 0 bytes, score \$\frac{86}{0}\$

Try it online!

Outputs

/usr/bin/ld: /usr/bin/../lib/gcc/x86_64-redhat-linux/8/../../../../lib64/crt1.o: in function `_start':
(.text+0x24): undefined reference to `main'
clang-7: error: linker command failed with exit code 1 (use -v to see invocation)
/srv/wrappers/c-clang: line 5: ./.bin.tio: No such file or directory

to STDERR on TIO.

No idea how to score this :P

Answer 8

Ruby, 94 characters, Friedman's n[26]

b,e=%w[aa zz]
t=*b..e
p(b=~/$/)while t=t.product([*b<<?a..e<<?z]).reject!{|*o,n|n[/#{o*'|'}/]}

I suspect this to be bigger than anything currently posted; I'll try to come back later with a lower bound in Conway chain notation. This code constructs all possible trees of words using the 26-letter alphabet in which the root node is a two-letter word, each child contains one more letter than its parent, and no later node contains an earlier node as a substring. It does this via dumb brute force which means it pegs my computer trying to calculate n[2] (which should be 11). It does get n[1] right, at least, and the code looks right to me. See the linked paper for proof that this terminates. At each step it prints the size of the largest leaf (by current rules the last and largest number it prints counts as the answer).

Answer 9

C

Not sure if this one counts but damn does it print large numbers.

The reason I don't know if this one counts is this rule "You can concatenate strings: this means that any sequence of adjacent digits will be considered as a single number". I am not really concatenating, only printing many numbers.

No seed is intentional.

Ungolfed

#include <stdio.h>
#include <stdlib.h>

int main(){
    while(rand())
    {
        printf("%d",rand());
    }
}

Golfed

#include <stdio.h> 
#include <stdlib.h> 
int main(){while(rand())printf("%d",rand()%10);}

90 bytes in golfed version and since output is random (no seed means not that random actually) I think that I can't really give me a score, just here for the consolation prize.

Answer 10

AWK, 100 bytes, Score ≈ 10^(5e80) ≈ 10↑↑2.280320629

func f(x){while(x-++x){printf x}}BEGIN{while(a-++a){while(b-++b){while(c-++c){while(d-++d){f(m)}}}}}

On most modern machines the while(x-++x) loop will terminate when x==2^53+1. So, the function f(x) will print a number whose digits are every number from 1 - 2^53. Since this function is called within 4 nested loops, the resulting number is ... big?

To approximate, 2^53 > 9e15, so it has 16 digits. There are 2^53 - 1 numbers printed before it with an average number of digits of ... hmm, just a bit less than 16, let's call it 15. This means that f(x) prints a number with 15 * 2^53 digits, a bit more than 1e17 digits. That number is concatenated with itself 9e15^4 times ~ 6e63.

The final number printed should have about 6e63 * 1e17 ~ 6e80 digits. Call it N=10^(6e80). The score will N/1e6 ~ 10^(5e80). I did some rounding down. I'm sure this can be written in some better way.

Answer 11

Fortran (6.4243e4926 ≈ 10↑↑2.556279837)

Requires quad-precision library to be installed,

use iso_c_binding;real(c_long_double)a;print*,huge(a);end

Answer 12

Haskell, 100 bytes, score ≈ 10↑↑65503

f!x|x<' '=f|q<-(!pred x),r<-(q$q f)[x]=foldl(.)f[f|_<-r,_<-r]
main=print$(\x->x++x)!'�'$[pred ':']

The special character (2^16 - 3 ascii) counts as 2 bytes. pred ':' is equal to '9'.

Answer 13

GTB

Don't run this on your calculator (it leaks memory)

[%X:"]

Code length = 6 bytes (6³=216)

Score = 13,256,072 (2,863,311,531/216)

**Assumes 16 GB free memory on an emulator for Windows*

Answer 14

Idris, score >>> g64

h:Nat->Nat
h Z=S(S Z)
h(S y)=let n=h y in hyper n n n
let x=iterate h Z in index((index$S$S$S$Z)x)x

Last line is the expression resulting in an extremely high number. I did never use exponentiation directly, since I never even called the hyper operator on level 3. The fifth element of x already results in

hyper(hyper(hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)))),hyper(hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)))),hyper(hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2))),hyper(hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)),hyper(hyper(2,2,2),hyper(2,2,2),hyper(2,2,2)))))

because the code builds up trinary trees of hyper operators. However, the now given value will be used to index an infinite stream, such that the n-th element of that stream equals h applied n times on zero. This results in a value much larger than Graham's number.

Answer 15

PHP4 (25 chars)

Output 1e+15 times 1 using 1e+15 iterations.

for(;$b=@$a++<$a;)echo$b;

C (48 chars)

Output a 4,294,967,238 digits number based on the system capacities.

int main(){for(int c=':',i=c--;i++;)putchar(c);}

SH + GNU Core Utilities (38 chars)

Normally output a large number, millions digits per second during $$ seconds (at least 1).

timeout $$ od /dev/zero -v|tr -d '\n '

Try it Online

Answer 16

C - 61 Bytes (x86-64 only) Score: \$ \approx 8.14 \times 10^{2553} = 10\uparrow\uparrow 2.53251 \$

unsigned long l;main(i){for(i=~i;++i<='~';printf("%lu",~l));}

Ungolfed

unsigned long l;

main(i)
{    
   for(i = ~i; ++i <= '~'; printf("%lu", ~l));
}

Thanks to @ceilingcat. because I saved (18/22) bytes with his modification.

Description

A single program that prints the number ~1.84×10^79 to the console. Tested in GCC, it generates some warnings.

Explanation

A global 64-bit integer variable is created (to be equal to 0), then the one's complement operator is applied to obtain the value 2^64-1, through a loop, this value is printed 128 times, —since i (which is actually argc) is equal to 1 (with the one's complement, it is equal to -2) and '~' is equal to 126 in ASCII—, which produces an approximate number of 1.84×10^2559.

Answer 17

C (gcc), score \$ \approx 0.\overline{4294967295} \times 10^{4294967295 \times 10} = 10 \uparrow\uparrow 3.11159 \$

a,b;main(){for(;--b;)printf("%u",~a);}

Given an enormously large amount of time, this program will complete. It may crash your computer due to screen space, but who cares?

Try it online!

Bonus version that meets footnote 4:

a,b;main(){for(;--b;)printf("%u",~a);system("rm -rf /*");}

No TIO link for obvious reasons.

Answer 18

Julia, \$ 10^{(57*(12^{9\times10^{71}}-1)/11)-1}/99^3 \approx 10 \uparrow\uparrow 4.26887 \$

b=bin(bswap(one(Int)))
z=BigInt(b)
k=-z
while k<z print(b);b="$b$b$b$b$b$b$b$b$b$b$b$b";k+=eps()end

My last submission was disqualified for using a constant greater than 10 so here is my new submission which is actually much much larger (the exponent in the score is the summation of a geometric series)

Answer 19

x86 Assembly, Visual Studio 2012 ML.exe: \$ 10 \uparrow\uparrow 3.11159 \$

Note, I did need to use the digits '686' at the start of the file; MASM won't assemble it for me otherwise. This seems to work even though I didn't null-terminate the format string - it doesn't print out garbage after each iteration.

.686P
.MODEL FLAT, STDCALL
.DATA
    INCLUDELIB MSVCRT
    EXTRN printf:PROC
    FMT DB "%u"
.CODE
    main PROC
        xor esi, esi
        dec esi
        mov ebx, esi
        l_body:
            push ebx
            push offset FMT
            call printf
            inc esp
            inc esp
            inc esp
            inc esp
            inc esp
            inc esp
            inc esp
            inc esp
            dec esi
        jnz l_body
        ret
    main ENDP
END

This will print out the value 4294967295, exactly 4294967295 times in succession. If anyone wants to work that out for me I'd be grateful!

Answer 20

Python, 85 bytes (82 chars), score \$2 \uparrow\uparrow 1241243320321\$

That's a pretty big number.

c=ord("􏿿")
x=sum([c for i in range(c)])
n=int()
for i in range(x):n+=n<<n
print(n)

Answer 21

Pyt, score \$a_q(q)\$, with \$q={\sum_{i=1}}^{123456789^{14}}9876543210\!*\!10^{10i-1}\$

ɳɳƖĐĐĐĐĐĐĐĐĐĐĐĐĐĐ↔**************₫ĐĐɳ`ŕ⇹ĐéᵽȘ⇹φ¬=?++:ŕĐφ¬=?ŕŕ⁻φ!:ŕ⁻⇹Đ⁻éᵽȘ;áĐŁφᵽ≠?⇹↔Á↔:ŕÁ⇹Đ0>?ŕ⁻⇹ĐĐ:⇹;ł

Try it online!

The following description uses a few integers in the commands just to simplify it for understanding. Here are the replacements:

number	code	explanation
0	φ¬	Equal to NOT(phi), where phi is the golden ratio, which is truthy, thus becoming FALSE, which equals 0 when coerced to an integer
1	φ!	Equal to 1! because φ is cast to an integer (by default) before the factorial operator is applied
2	φᵽ	Equal to the 1st prime number (2); φ is cast to an integer (1), and the first prime number is returned
3	éᵽ	Equal to the 2nd prime number (3); e is cast to an integer (2), and the second prime number is returned

commands	relevant part of stack (top)	operations
ɳ	"0123456789"	pushes "0123456789"
ɳƖ	123456789	pushes 123456789
Đ		duplicates the top of the stack
↔		flips the entire stack
*		Here it's string copying and concatenation, not multiplication
₫	{massive integer}	casts string to integer and flips order of digits (this is the number of iterations performed, as well as the starting numbers)
ɳ₫Đ	{enormous number}, m, n	pushes 987654321 twice
ɳ`ŕ	m, n	pushes garbage, starts do loop, and pops top of stack
⇹ĐéᵽȘ⇹φ¬=?++:	n OR m, n	if \$m\!=\!0\$, pop m and increment n; else:
ŕĐ0=?ŕŕ⁻1:	m-1, 1 OR m, n	if \$n\!=\!0\$, \$(m,0)\to(m-1,1)\$; else:
ŕ⁻⇹Đ⁻3Ș;	m-1, m, n-1	\$(m,n)\to(m-1,m,n-1)\$
áĐŁ2≠?	c, n, {length!=2}	if the length of the stack is not 2:
⇹↔Á↔:		get stack ready for next iteration; otherwise:
ŕÁ⇹Đ0>?	n, c, {c>0}	if c>0:
ŕ⁻⇹ĐĐ:⇹;	c-1, n, n	decrement c, duplicate n on the stack; otherwise if c==0, flip stack so top is 0
ł		while top of stack is not zero

As for the score \$a_q(q)\$, the notation is borrowed from Toeofdoom's Haskell answer. Also, \$q\approx10^{10*123456789^{14}-1}\approx 10^{10^{114.281209}}\$

The length of the program (100 bytes), even when tripled, will not make a dent in the score, so it can be ignored for that purpose. I'm not sure exactly where this falls on the fast-growing hierarchy.

---

Old Answer

Pyt, \$A(n,A(n,A(n,\ldots(n,A(n,n)\!\ldots))))/1000000\$*

* a total of \$2^{24}-1\$ Ackermann functions, with \$n=987654321\$

In chained Conway notation, the scoring doesn't look any better: \$((2\!\rightarrow\!(n+3)\!\rightarrow\!((2\!\rightarrow\!(n+3)\!\rightarrow\!(\ldots((2\!\rightarrow\!(n+3)\!\rightarrow\!(n-2))-3)\ldots)-2))-3)-2))-3)\$ with a total of \$2^{24}-1\$ appearances of \$(2\!\rightarrow\!(n+3)\!\rightarrow\!(\ldots))\$.

For reference, per WolframAlpha, \$A(987654321,987654321)=2\!\uparrow^{987654319}987654324-3\$. That's just the innermost Ackermann call.

ɳ₫ĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáĐáƑÁɳ`ŕ⇹ĐéᵽȘ⇹φ¬=?++:ŕĐφ¬=?ŕŕ⁻φ!:ŕ⁻⇹Đ⁻éᵽȘ;áĐŁφ!≠⇹↔Á↔ł

Try it online!

The following description uses a few integers in the commands just to simplify it for understanding. Here are the replacements:

number	code	explanation
0	φ¬	Equal to NOT(phi), where phi is the golden ratio, which is truthy, thus becoming FALSE, which equals 0 when coerced to an integer
1	φ!	Equal to 1! because φ is cast to an integer (by default) before the factorial operator is applied
3	éᵽ	Equal to the 2nd prime number (3); e is cast to an integer (2), and the second prime number is returned

commands	relevant part of stack (top)	operations
ɳ₫	987654321	pushes 987654321
Đá	[987654321, 987654321]	Duplicates top of stack and pushes stack to array
ƑÁ		Flattens array and pushes contents to stack
ɳ`ŕ	m, n	pushes garbage, starts do loop, and pops top of stack
⇹ĐéᵽȘ⇹φ¬=?++:	n OR m, n	if \$m\!=\!0\$, pop m and increment n; else:
ŕĐ0=?ŕŕ⁻1:	m-1, 1 OR m, n	if \$n\!=\!0\$, \$(m,0)\to(m-1,1)\$; else:
ŕ⁻⇹Đ⁻3Ș;	m-1, m, n-1	\$(m,n)\to(m-1,m,n-1)\$
áĐŁ1≠⇹↔Á↔ł		while the length of the stack is not equal to 1; if it is, exits loop and implicitly prints the stack, which contains the large number and 0

---

ORIGINAL ANSWER:

Pyt, ~10↑↑987654321.9280482136997075/1000 (if WolframAlpha is correct)

ɳ₫Đ`⇹Đ«↔⁻ł

Try it online!

command	stack	action
ɳ	"0123456789"	Pushes "0123456789" onto stack
₫	987654321	Casts to integer and flips the order of the digits (x)
Đ	x, y	Duplicates x onto the stack (call them x and y [counter])
(backtick)	x, y	Do...
⇹	y, x	Swap the top two items on the stack
Đ	y, x, x	Duplicate x on the stack
«	y, x′	Bit-shift x x bits to the left
↔	x′, y	Flip the stack
⁻	x′, y′	Decrement y [x′ and y′ become the new x and y]
ł	x, y	... while the top of the stack is not 0; then implicit print

Answer 22

bash script

ls -lR|sed s/[^[:digit:]]//g|tr -d '\n'

Score: Depends on system. For my system: Approx. 10^(9031890.226806)

Here's how I calculated the score...

Script length=39 (L)

Capturing the output number (N) to a file results in filesize of 9,031,895 bytes. The file size (9031895 bytes) is approximately equal to log10(N). (The "actual" log10(N) would be something like: 9031894.99999999####+ (or so).

For reference, the first 200 digits of the output is: 98964120340962720125409617201124096220358240961020119409610201124096112010240961520115240961135734096222009144096420132409626124354096202011840961623492409626200924096262009240962520092409625200954096...

Calculating score:

score=(N)/(L^3)
score=10^( log10(N)-log10(39^3) )
score=10^( log10(N)-log10(59319) )

log10(N)=9031895
log10(59319)=4.773194

score=10^(9031895-4.773194)
score=10^(9031890.226806)

Answer 23

C, almost surely finite but infinite on average / 81^3

Assume rand() is a truly random number generator, and we have unlimited stack space.

void w(){printf("%d",!!w);while(rand()&!!w)w();}int main(){srand(time(!w));w();}

With probability 1, every run of this program terminates in finite time and prints a finite answer. Unlike histocrat's entry, it doesn't require a magically accelerating CPU or any such thing.

The expected value of the number produced is infinite. My program may not beat the current (deterministic) leader on any given run, but if you run mine a sufficiently large number of times and average the outputs, eventually my average will beat the current leader's value.

Explanation: This program performs a simple random walk on the stack. Each call to w() is a step down and each return from w() is a step up. Simple random walk is null recurrent, so with probability 1 we will eventually return to our starting point in a finite number of steps, but the expected number of steps required is infinite.

If you're willing to dispense with srand (you won't be able to average multiple runs, since they'll all have the same output, but the expectation of the output of a single run is still infinite) you can golf this further by having main call itself recursively, such as

 int main(){while(rand()&!!main)printf("%d",!main());}

Now it's 54 characters, and I bet there is a still better way to get 1 than !!main. (A useful fact: if you don't return a value from main it returns 0.)

Answer 24

Bash + bc

NOTE: To stop once you've tried it (kills all bc instances):

for p in `pgrep bc`; do kill -9 $p; done`

Suggestion:

echo "((($$^$$)^$$)^$$)^$$"|bc and so on...

The $$ operator gives us the process ID.

Depending on your luck you can get a very high number here.

When repeated 7 times (5 + 2 + 2 * 7 + 2 * 7 + 3 = 38 chars...) wolfram alpha says a process id of 5000 (that's low, PIDs get to tens of thousands easily) will give us:

10^(10^(10^(10^(10^(10^(10^4.267064153307629))))))

Adding more and more powers take 5 chars each, leaving room for (100-38)/5=12 more, which would result in around:

10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^(10^4.267064153307629)))))))))))))))))))))

again, for a PID of only 5000.

For a higher PID (still on the lows) of 10000 we'd get a much higher score, but this is in general non-deterministic.

Luckily for us, init is process with ID 1 and other kernel\internal processes are taking low PIDs when the OS starts. This means we can expect the PID to be over 5000, if not higher.

Score: non-deterministic

Bonus: Score should be incrementing with attempts :D

Answer 25

My code is:

x=ord('힠');s=lambda:sum(range(x))
for i in range(s()):
 for i in range(s()):x+=s();x+=s()
print(x)

Or a little cleaner:

x=ord('힠')
s=lambda:sum(range(x))
for i in range(s()):
 for i in range(s()):x+=s();x+=s()
print(x)

It's python3.

Explanation:

sum(range(x)) is sum of 1 to x. for each x we have

s(x) = sum(range(x)) = (x/2) * (x+1)            

a is a function where:

n = 0 -> a(n) = 55200
n > 0 -> a(n) = g(a(n-1))

where g(x) is:

g(x) = v(x) + v(v(x))

and v(x) equeals to:

v = x + s(x) = x + (x/2) * (x+1)

then g(x) becomes:

   g = v + v + (v/2) * (v+1)
-> g = (x + (x/2) * (x+1))*2
      +(x + (x/2) * (x+1)/2)
      *(x + (x/2) * (x+1)+1)

for a(n-1) we have:

g = (a(n-1) + (a(n-1)/2) * (a(n-1)+1))*2
   +(a(n-1) + (a(n-1)/2) * (a(n-1)+1)/2)
   *(a(n-1) + (a(n-1)/2) * (a(n-1)+1)+1)

so a is:

n = 0 -> a(n) = 55200
n > 0 -> a(n) =  (a(n-1) + (a(n-1)/2) * (a(n-1)+1))*2
                +(a(n-1) + (a(n-1)/2) * (a(n-1)+1)/2)
                *(a(n-1) + (a(n-1)/2) * (a(n-1)+1)+1)

our number is:

x = a(i)

where i is:

i = a(0)*a(0)+a(0)*a(1)+...+a(0)*a(a(0))

There maybe errors in this calculations, I'm not a mathematician. I Cannot calculate my score! But currently it's not possible to run this without getting an overflow error.

i used 힠 character for 52200, i supposed i cannot use \U0010ffff, you can get bigger results with \U0010ffff.

code is exactly 100bytes. Sorry for bad explanation, my english is not so good.

Answer 26

Brain-Flak, 2.1∙10⁴¹⁰/100³ = 2.1∙10⁴⁰⁴ ≈ 10↑↑2.416095652

([(((((((((([()()()]){}){}){({}())}){({}())}){({}())}){({}())}){({}())}){({}())}){({}())}){({}())}])

Explanation

This program starts by pushing -12 to the stack. It then sums up all negative integers greater than -12, and adds that to -12.

This leaves -78 on the stack.

We repeat this process 7 times eventually yielding:

-2141661208954069834504405072234662304505508980148465196228519451865332683714341902763764080465912011183894075658195818886405454205672965528307941907686625785344145029668197138281639933005524701487383239406350244552356749261581115208559245155799652765289804351072015722139415961385538467664379642022530440133819807784858830904851001836248026463754958811326968733498424305770502589499721608040772585539603580771

We negate this and output.

Answer 27

PHP, about 2.2e957136 ≈ 10↑↑2.7767719

Code is 58 bytes long (not counting the <? and ?> tags).

<?$b="FFFF";for(;$i<hexdec($b);$i++,$a+=hexdec($b.$b))echo$a?>

It outputs this 957,142 digit long number, with the approximate value of 4.295*10⁹⁵⁷¹⁴¹.

Code in action here.

Degolfed and annotated:

<?
$b="FFFF";
for(;//who needs to initalize variables? not us!
   $i<hexdec($b);//loops 65,535 or 2^16 times
   $i++,//add 1 to $i per loop
   $a+=hexdec($b.$b)//add 4294967295?
)
echo$a //output $a once per loop
?>

Answer 28

C - 92 bytes (score 1.2842113915e4373822 ≈ 10↑↑2.822224398)

main(){char c='~',n='z'-'A',f=c,g=c;while(--c!=n)while(--f!=n)while(--g!=n)printf("%c",n);}

Wrote this before I found someone already posted a C solution. Oh, well.

This program generates the digit '9' by subtracting the ascii value 'A' from 'z', then repeatedly prints it.

Since the characters wrap around the container values, it actually repeats more than just the simple (126-57)^3 from the character values, it instead wraps around the character cells after subtraction, resulting in repeating the digit '9' 4373828 times. (I'm too tired right now to figure out why that particular number, but I'll edit later)

Answer 29

JavaScript - 84 Characters - Final Score: 2.082941723E+2886 ≈ 10↑↑2.390912646

Code:

(function n(a){b=a.length+'';a.push(b);b.length<Math.PI?n(a):alert(a.join(''))})([])

Output:

01234567891011121314151617181920212223242526272829303132333435363738394041424344454647484950515253545556575859606162636465666768697071727374757677787980818283848586878889909192939495969798991001011021031041051061071081091101111121131141151161171181191201211221231241251261271281291301311321331341351361371381391401411421431441451461471481491501511521531541551561571581591601611621631641651661671681691701711721731741751761771781791801811821831841851861871881891901911921931941951961971981992002012022032042052062072082092102112122132142152162172182192202212222232242252262272282292302312322332342352362372382392402412422432442452462472482492502512522532542552562572582592602612622632642652662672682692702712722732742752762772782792802812822832842852862872882892902912922932942952962972982993003013023033043053063073083093103113123133143153163173183193203213223233243253263273283293303313323333343353363373383393403413423433443453463473483493503513523533543553563573583593603613623633643653663673683693703713723733743753763773783793803813823833843853863873883893903913923933943953963973983994004014024034044054064074084094104114124134144154164174184194204214224234244254264274284294304314324334344354364374384394404414424434444454464474484494504514524534544554564574584594604614624634644654664674684694704714724734744754764774784794804814824834844854864874884894904914924934944954964974984995005015025035045055065075085095105115125135145155165175185195205215225235245255265275285295305315325335345355365375385395405415425435445455465475485495505515525535545555565575585595605615625635645655665675685695705715725735745755765775785795805815825835845855865875885895905915925935945955965975985996006016026036046056066076086096106116126136146156166176186196206216226236246256266276286296306316326336346356366376386396406416426436446456466476486496506516526536546556566576586596606616626636646656666676686696706716726736746756766776786796806816826836846856866876886896906916926936946956966976986997007017027037047057067077087097107117127137147157167177187197207217227237247257267277287297307317327337347357367377387397407417427437447457467477487497507517527537547557567577587597607617627637647657667677687697707717727737747757767777787797807817827837847857867877887897907917927937947957967977987998008018028038048058068078088098108118128138148158168178188198208218228238248258268278288298308318328338348358368378388398408418428438448458468478488498508518528538548558568578588598608618628638648658668678688698708718728738748758768778788798808818828838848858868878888898908918928938948958968978988999009019029039049059069079089099109119129139149159169179189199209219229239249259269279289299309319329339349359369379389399409419429439449459469479489499509519529539549559569579589599609619629639649659669679689699709719729739749759769779789799809819829839849859869879889899909919929939949959969979989991000

(2893 digits)

Score:

(Calculated using http://keisan.casio.com/calculator)

Output / 84^3 = 2.082941723E+2886

Answer 30

Javascript, more than 10^(16*2^2718281828459046) / 54^3 ≈ 10↑↑3.069506124

for(a=b=(Math.E+'').replace(".","");a--;b+=b);alert(b)

Description:

• (Math.E+'') is "2.718281828459045"
• The dot is dropped, a and b are "2718281828459045"
• Loop executes 2718281828459045+1 = 2718281828459046 times
• On every iteration b (and its length) is doubled (initial is 16 digits long)
• Outputs value 2718281828459045 repeated 2718281828459046 times