明らかに、コードゴルフは、最小限のコードを最大限に活用することです。実際の出力が本当に重要なのは誰ですか？

一方で、私たちが持っていた最高の入力から出力への挑戦比率を、これがためのコールで最も有限と決定論与えられた符号長で出力。皮肉なことに、この課題はcode-golfではありません。

ルール：

1. 3つの独立したスニペットを作成します（完全なプログラム/機能ではありません）。

2. スニペットは同じ言語である必要があります。

3. スコアは、出力されたバイトの総数です。

4. 出力は、結果、STDOUTなどの形式になります。

5. スニペットによりエラーが発生することはありません。

6. スニペットは、さまざまな形式の出力を引き起こす可能性があります。

7. 末尾の改行文字はカウントされません。

8. 最初のスニペットは1バイト、または少なくとも1バイトの出力を生成する最小長でなければなりません。

9. 2番目のスニペットは、それより1バイト長くする必要があります。

10. 3番目のスニペットは、最初のスニペットよりも2バイト長くする必要があります。

gs2、412 + 5.37 * 10 ⁹⁰² + 10 ^{10 903.1}バイト

1. f 押す 1\n2\nFizz\n4\nBuzz\n...\nFizzBuzz412-byte文字列としてします。

2. fô すべての順列を出力するため、 412! * 412文字を出力します。

3. fôôその412！要素リストのすべての順列を出力し412 * (412!)!ます。各要素は412文字なので、バイトです。

編集：物事を見通しに入れるために、これは少なくとも

10^{1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000}

これまでのところ、他のすべての答えを小さくしました。

Pyth、26 + 1140850688 +（> 4.37×10 ^20201781）

3番目のプログラムの出力の正確な長さを計算できるかどうかはわかりません。境界線のみを与えることができます。との間に何かを出力4.37 × 10^20201781し1.25 × 10^20201790ます。

G
yG
yyG

これは印刷します：

abcdefghijklmnopqrstuvwxyz
['', 'a', 'b', ..., 'abcdefghijklmnopqrstuvwxyz']
[[], [''], ['a'], ['b'], ['c'], ..., ['', 'a', 'b', ..., 'abcdefghijklmnopqrstuvwxyz']]

最初の文字はアルファベットを出力し、2番目の文字はアルファベットのすべてのサブセットを出力し、3番目の文字はアルファベットのサブセットのサブセット（長さのリスト）を出力し2^(2^26) ~= 1.09 × 10^20201781ます。

明らかに、この大きなリストを計算して出力できるコンピューターはありません。

CJam、17 + 34 + 72987060245299200000 = 72987060245299200051バイトの出力

比較を簡単にするために、これは約7.3 * 10 ¹⁹です。

P
PP
Ke!

プリント：

3.141592653589793
3.1415926535897933.141592653589793
012345678910111213141516171819012345678910111213141516171918012...

さて、最後のものはのすべての順列で構成されています [0 1 2 ... 19]、数字が一緒に押しつぶされたています。私はそれを試すことをお勧めしません...（として試してみてください4e!味を得るために。）

ここでテストします：プログラム1、プログラム2、プログラム3の健全なバージョン。

Jelly、1.2×10 ²⁵⁶⁸バイトの出力

ȷ
ȷ*
ȷ*!

計算し1000年、1000年^1000年と1000年^1000年！。

オンラインで試す：最初のプログラム | セカンドプログラム | 3番目のプログラム（変更）

バイトカウントの目的ȷで、現在のバージョンのJellyでバイト0xa0としてエンコードできます。

使い方

Jellyでは、ȷ数値リテラル内でPython e（科学表記法）として使用できます。たとえば、30000を3ȷ4返します。ゼリーの科学的表記での係数のデフォルト値1とする指数のデフォルト3は、そう、そしてすべて同じ番号を返します。ȷ1ȷ31000

ȷ      Return 1000.

ȷ      Return 1000.
       Parse the remaining code as a program with input 1000.
 *     Hook; compute 1000 ** 1000.

ȷ      Return 1000.
       Parse the remaining code as a program with input 1000.
  !    Return 1000!.
 *     Fork; compute 1000 ** 1000!.

六角形、1 + 3 + 6 = 10バイトの出力

まあ...あまり印象的なスコアではありませんが、少なくとも私はそれが最適だと断言できます。1バイトでは、何かを出力して終了することは不可能なので、2バイトから始めます。

!@

展開されたコードは

 ! @
. . .
 . .

これはバイトを出力して終了します。

3バイトのコードの場合、3バイトの出力を印刷できます。例えば：

o!@

または展開：

 o !
@ . .
 . .

印刷し111ます。からdまでの小文字はすべてz機能し、その文字コードを出力します。これらは、3バイトのコードで3バイトを印刷する唯一の23の方法です。

最後に、4バイトの場合、6バイトを印刷する169の方法があります。それらはどれも、単純な解決策よりも面白い（奇妙な制御フローを除く）ことは何もしないので、以下を提示します。

o!!@

展開：

 o !
! @ .
 . .

ご推察通り。印刷する111111ます。

これらが最適であることをどのように知ることができますか？真実のマシンカタログ用に書いたブルートフォーサーを採用しましたし、7000サイクルで最大の有限出力を探すよう（7000サイクルで動作するが、後で終了する4バイトのビジービーバーを書くことはできないと思います）。

真剣に、2025409バイト

1バイト：

N

（11,756バイトの出力を生成）

2バイト：

Nñ

153,717バイトの出力を生成します

3バイト：

9!!

1,859,936バイトの出力を生成します

真剣に「すべてのサブセット」や「すべての組み合わせ」などの機能をまだ備えていないため、これに関するスコアは比較的低くなっています。

Python 3、1 + 22 + 23 = 56

9
id
abs

出力

9
<built-in function id>
<built-in function abs>

9を印刷してから、との定義を印刷idしabsます。

ラビリンス, 1 + 2 + 4 = 7 bytes

言語に最適であることを証明したため、主に投稿している別の低スコア。

Hexagonyと同様に、Labyrinthは1バイトで印刷および終了することができないため、2バイトで開始します。

!@

Prints a zero and terminates.

3バイトの場合、単純なソリューションに勝るものはありません。

!!@

これは、終了する前に2バイトを出力します。印刷のようないくつかの他のオプションがある-1と(!@か、~!@または,!@。ただし、ソースコードのローテーションを使用する非常に優れたソリューションが1つあります。

!>@

これはゼロを出力し、ソースをにシフトします@!>。その時点で、行き止まりにぶつかり、向きを変え、実行します!、途中で再びから終了します。

4バイトの場合、4文字を印刷する唯一の方法は上記のトリックを使用することであるため、もう少し楽しいです。

!!>@

2つのゼロを出力し、シフトして@!!>、さらに2つのゼロを出力します。

これらのすべての場合、\または.でバイトを印刷することもできますが、これらは常に1バイトだけ!を印刷しますが、少なくとも1つ、場合によっては複数のバイトを印刷します。

Bash、1726バイト

（今修正しました。投票を検討してください。）

1バイト："

出力：

>

307バイト： id

出力：

uid=501(geokavel) gid=20(staff) groups=20(staff),701(com.apple.sharepoint.group.1),12(everyone),61(localaccounts),79(_appserverusr),80(admin),81(_appserveradm),98(_lpadmin),33(_appstore),100(_lpoperator),204(_developer),395(com.apple.access_ftp),398(com.apple.access_screensharing),399(com.apple.access_ssh)

1418バイト： zip（STDOUTに出力）

Copyright（c）1990-2008 Info-ZIP-ソフトウェアライセンスに「zip "-L"」と入力します。
Zip 3.0（2008年7月5日）。使用法：
zip [-options] [-b path] [-t mmddyyyy] [-n suffixes] [zipfile list] [-xi list]
  デフォルトのアクションは、リストからzipfileエントリを追加または置換することです。
  特別な名前を含めることができます-標準入力を圧縮します。
  zipfileとlistを省略すると、zipはstdinをstdoutに圧縮します。
  -f freshen：変更されたファイルのみ-u update：変更されたファイルまたは新しいファイルのみ
  -d zipfileのエントリを削除する-m zipfileに移動する（OSファイルを削除する）
  -rディレクトリへの再帰-jジャンク（記録しない）ディレクトリ名
  -0ストアのみ-l LFからCR LFへの変換（-ll CR LFからLF）
  -1より速く圧縮-9よりよく圧縮
  -q quiet operation -v詳細な操作/バージョン情報の出力
  -c 1行のコメントを追加-z zipファイルのコメントを追加
  -@ stdinから名前を読み取ります-o zipファイルを最新のエントリと同じくらい古いものにします
  -x次の名前を除外-i次の名前のみを含める
  -F zipファイルを修正します（-FFはさらに試行します）-Dディレクトリエントリを追加しません
  -A自己解凍exeの調整-Jジャンクzipファイルプレフィックス（unzipsfx）
  -T zipファイルの整合性をテストします
  -yシンボリックリンクを参照ファイルの代わりにリンクとして保存します
  -e暗号化-nこれらのサフィックスを圧縮しません
  -h2さらにヘルプを表示

MATL, 313

The current version of the language (3.1.0) is used, which is earlier than this challenge.

1. Code (predefined literal: produces number 2, which is implicitly printed):

   H
   

   Output (1 byte):

   2
   

2. Code (produces number pi, which is implicitly printed with 15 decimals):

   YP
   

   Output (17 bytes):

   3.141592653589793
   

3. Code (numbers from 1 to 99, which are printed by default with spaces in between):

    99:
   

   Output (295 bytes):

   1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99
   

Processing, 39 bytes

Deterministic

1 byte:

print(X);

Outputs 0.

9 bytes:

print(PI);

Outputs 3.1415927

29 bytes:

print(P3D);

Outputs processing.opengl.PGraphics3D

Non-deterministic, >= 129 bytes

>= 32 bytes:

print(g);

Outputs processing.awt.PGraphicsJava2D@ + [mem-address]

>= 32 bytes:

print( g);

Outputs processing.awt.PGraphicsJava2D@ + [mem-address]

>= 65 bytes: (Thank you to @anOKsquirrel for this suggestion.)

print(g,g);

Outputs

processing.awt.PGraphicsJava2D@ + [mem-address]
processing.awt.PGraphicsJava2D@ + [mem-address]

JavaScript, 1 + 3 + 18 = 18 22

Not a very interesting answer but probably the best JavaScript is capable of.

alert(1)
alert(.1)
alert(1/9)

Added 4 score thanks to @UndefinedFunction!

Outputs as text:

1
0.1
0.1111111111111111

Befunge, 2 + 4 + 6 = 12

.@
..@
...@

Any snippet shorter than length 2 either cannot output, or cannot terminate its output.

In Befunge, . outputs the top value of the stack as an integer, followed by a space. A space is not a newline, so it is included in the count. Additionally, the stack is "infinitely" filled up with 0's, so the programs output (respectively):

0 
0 0 
0 0 0 

SmileBASIC, 1+4+10= 15 bytes

Program 1:

The shortest way to print something is with ? (PRINT) and a single character. This can be either a number or a variable name, and it doesn't matter since they're all the same length.

?1
1

Program 2:

Now we have access to a few more things. The longest expression which can be made would be one of the constants #Y, #L, or #R, which have values 128, 256, and 512, respectively. However, instead of that, I use a comma so that (in this case) 3 extra spaces are printed.

?1,
1   

Program 3:

With 3 characters, you can write E-notation numbers:

?1E9
1000000000

HQ9+, 71304

9

Prints the 11,884-character lyrics of "99 bottles of beer"

99

Prints "99 bottles of beer" twice

999

Prints "99 bottles of beer" three times

Japt -Q, Outputs 1.0123378918474279e+150 bytes

The full number is

1,012,337,891,847,427,807,734,770,805,740,683,255,348,979,141,331,502,541,182,800,555,980,960,810,784,280,906,237,433,006,787,771,597,919,201,659,212,694,207,520,340,705,280,000,000,000,000,000,000,568

bytes.

# 1

M

Outputs

{"P":3.141592653589793,"Q":1.618033988749895,"T":6.283185307179586}

For 67 bytes. (Credit to Shaggy)

# 2

LÆ

Outputs

[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null]

which is 501 bytes.

(Credit to @Shaggy)

# 3

;Eá

Outputs all permutations of the 95 printable ASCII characters in the format ["...","...","..."...], which is

$1,012,337,891,847,427,807,734,770,805,740,683,255,348,979,141,331,502,541,182,800,555,980,960,810,784,280,906,237,433,006,787,771,597,919,201,659,212,694,207,520,340,705,280,000,000,000,000,000,000,001$ bytes of output.

You can get an infinite amount of output bytes if you use the -F flag in Japt. What it does is that if the last expression of the program evaluates to false, it outputs the value specified in the flag instead. So I guess the score for Japt-F"Insert Super Long String Here" is infinity.

Malbolge, 1 + 2 + 3 = 6 bytes

(=aN
(&a`M
(=a`_L

Try it online: first, second, third

Outputs:

r
ll
xxx

Brute forced. Assumes \0 is not a valid output character

With \0:

cP
cbO
cbaN

Outputs:

\0
\0\0
\0\0\0

scg, 1 + 27 + 188 = 216

First one:

1

Just prints 1, as the stack is outputted at the end of program.

Second:

.d

Prints debug info, which should look like this:

Stack: []
Array Markers: []

Third:

99r

adds 99 to stack, then uses range function. Outputs 01234567891011.... (this is one of those times I wish I implemented the factorial function. I haven't)

Marbelous 1 + 1 + 2 = 4 bytes of output

Marbelous is hamstrung here by having two-byte instructions. Pointless comments or unnecessary whitespace are the only ways to get an odd byte count.

print "A" and terminate:

`A

print "B" and terminate, with an empty EOL comment

`B#

print "CD" and terminate:

`C`D

Mathematica, 6 + 461 + 763 = 1230 1225 618 163 bytes of output

%  (* -> Out[0] *)

?D  (* ->

D[f, x] gives the partial derivative ∂ f/∂ x. 
                                              n        n
D[f, {x, n}] gives the multiple derivative ∂  f/∂ x . 
D[f, x, y, …] differentiates f successively with respect to x, y, ….
D[f, {{x , x , …}}] for a scalar f
        1   2
     gives the vector derivative (∂ f/∂ x , ∂ f/∂ x , …). 
                                             1             2
D[f, {array}] gives a tensor derivative.

*)

?Do (* ->

Do[expr, {i   }] evaluates expr i    times. 
           max                   max
Do[expr, {i, i   }] evaluates expr with the variable i successively taking on the values 1 through i    (in steps of 1). 
              max                                                                                   max
Do[expr, {i, i   , i   }] starts with i = i   . 
              min   max                    min
Do[expr, {i, i   , i   , di}] uses steps di. 
              min   max
Do[expr, {i, {i , i , …}}] uses the successive values i , i , ….
               1   2                                     1   2
Do[expr, {i, i   , i   }, {j, j   , j   }, …] evaluates expr looping over different values of j, etc. for each i. 
              min   max        min   max

*)

Currently, the last two use Information to get documentation about the symbols, which can output many bytes. Note that this was run on the 10.1 command-line MathKernel.

Javascript, 72 bytes

This works in the Mozilla JSShell Javascript command line interpreter.

1 byte: 1

Outputs 1

35 bytes: gc

Outputs

function gc() {
    [native code]
}

36 bytes: run

Outputs

function run() {
    [native code]
}

Octave, 2818417 bytes

e

14 bytes for ans = 2.7183\n

pi

14 bytes for ans = 3.1416\n

doc

Display the entire documentation. 2818389 bytes, counted with dd Try it online! because evalc didn't work.

SmileBASIC 4, 1 + 13 + 15 = 29 bytes

This is going to be similar to 12Me21's SmileBASIC 3 answer, with a couple adjustments.

1

As before, the shortest amount of code to produce some output is 2 bytes: ? (PRINT) and some single-byte expression. The consensus is that PRINT does not produce a newline when it advances to the next line, due to the way the text screen works. So this results in one byte of output.

?1
1

2

With 3 bytes, we can do something different. SB4 introduces INSPECT, aliased as ??, which prints info about a single value. If we give it an empty string, for example, this can produce much more output than SB3 could. This gets us 13 bytes.

??"
STRING: (0)""

3

We have 4 bytes to work with, so we have to decide what we should do to maximize our output. Going with ?? is a safe bet; we only have 2 bytes to use on our expression, but the additional output of INSPECT is basically free. So I use it to print a label string. This is 15 bytes.

??@A
STRING: (2)"@A"

The total is 29 bytes.

Microscript II, 23+47+71=141 bytes

1: C

The stringification of continuations is not strictly defined by the specs, but in the reference implementation this, run on its own, yields a 23 byte string.

<Continuation @t=\d\d\dus> (\d represents a digit, which digits varies).

On my computer, at least, this does, in fact, always take between about 180 and about 400 microseconds to run.

The first use I've ever actually had for this instruction.

2: CP

47 bytes of output- the output from the first one twice with a newline in between.

3: CPP

Fairly straightforward. 71 bytes of output- the output from the first one three times with newlines in between.

PowerShell, ~4300 bytes

Approximate output length, given the system that it's run on. All the snippets below are deterministic, in that if given the same initial state of the computer will output the same text, just that in practice the output could change from execution to execution.

Length 1, 107 bytes

?

This is an alias for Where-Object. It will output a user prompt asking for additional information:

cmdlet Where-Object at command pipeline position 1
Supply values for the following parameters:
Property: 

Length 2, 113 bytes

rp

This is an alias for Remove-ItemProperty. It will output a user prompt asking for additional information:

cmdlet Remove-ItemProperty at command pipeline position 1
Supply values for the following parameters:
Path[0]: 

Just barely longer than the length 1 snippet.

Length 3, ~4100 bytes

gps

This is an alias for Get-Process which will output a formatted table of all running processes on the system:

Handles  NPM(K)    PM(K)      WS(K) VM(M)   CPU(s)     Id ProcessName                                                                                          
-------  ------    -----      ----- -----   ------     -- -----------                                                                                
     85       8     1232       4452    46     0.14    544 armsvc                                                                                               
    151      10     6800       9888    39     0.58   5116 audiodg                                                                                              
    480      25     6060      17200   124     0.84   4536 AuthManSvr            
...

Javascript, 312 + 318 + 624 = 1254 bytes of output

$
$$
$+$

The two functions $ and $$ are available in all major browsers' consoles, as shortcuts for document.querySelector and document.querySelectorAll respectively. Different browsers have native code coerced to strings somewhat differently from each other, and IE uses plain JS in each function resulting in much longer representation.

For the byte count, I'm taking the length of the string representation of each rather than the sometimes-modified console display, so the total bytes are, for each of the following browsers:

• Chrome: 56 + 57 + 112 = 225 bytes
• IE: 312 + 318 + 624 = 1254
• Firefox: 33 + 33 + 66 = 132

(I'm considering the IE result to be the "official" count because it's the longest.)

For non-console browser environments, the largest outputs come from the following:

1
!1 // Or '{}' if being fed to a function instead of beginning a statement
Map

Results length by browser:

• Chrome: 1 + 5 (or 15 if {} is usable) + 32 = 38 (or 48) bytes
• IE: 1 + 5 (or 15 if {} is usable) + 38 = 44 (or 54)
• Firefox: 1 + 5 (or 15 if {} is usable) + 36 = 42 (or 52)

These two sets of input produce the largest output possible in all of these browsers and consoles. To prove this, let's check all alternatives:

• Existing variables: We can find all native available variables with maximum three characters with Object.getOwnPropertyNames(window).filter(x=>x.length<=3), and then map them to their string outputs to determine the larger ones. (Note that in some browsers such as Firefox, certain special console variables cannot be accessed this way as they're not a property of the window.)
• JS has a limited number of types of literals that can be created with so few characters:
  • Per the spec, decimal numbers must be a decimal digit (1234567890) optionally followed by . and more decimal digits and/or an exponent part, or be a . followed by one or more decimal digits and optionally an exponent part. Other kinds of numbers must be either 0o, 0x, or 0b (or uppercase forms), followed by one or more digits. For our purposes, we can deduce the following:
    • There are only ten single-character numbers, which are the integers 0 through 9. The only two-character numbers are integers 0-99 (0-9 by adding a decimal after), and the numbers 0.1 through 0.9 by omitting the initial zero. The only three-character numbers are 0-999, including 0x0-0xf, 0o0-0o7, 0b0, and 0b1, the exponential 1e1 through 9e9 (10 characters), and the numbers 0.1-9.9 and .01 through .99. While not technically a number literal, NaN can also be referenced in three characters.
  • Strings, arrays, and regexps, each of which can take at max only a single character of content. An array could be filled with $, a comma, or a single-digit number. Arrays with only one element are coerced to strings as the element itself. Empty arrays become empty strings.
  • Depending on environment, it may or may not be possible to create simple objects with three characters. Normally, {} alone at the beginning of a script would be treated as an enclosure rather than creating an object. eval({}) returns undefined, eval({$}) returns the $ function. There are insufficient characters to surround the {} in ().
• There are no keywords that are short enough to use. The keywords if, in, do, new, for, try, var, and let would all require a minimum of two other characters to use, exceeding the limit.
• Available unary operators include ~, +, -, !, ++, and --. The two-character operators can only be used with a single character variable, of which there is only one ($), which yields NaN. The other four operators can be used with any one- or two- character value, of which there are:
  • Several variables. (On some consoles $, $_, $0, $1, $2, $3, $4, $, $$, $x ). When used with these operators, the results are limited to -1, true, false, and NaN.
  • 109 numbers. Results: Integers -100 through 99, -0.9 through -0.1, true, false.
  • Empty strings. Results: -1, 0 (-0 becomes 0 on toString), true.
  • The results of any of the one-character values above associated with a one-character unary operator. New results: None.
• Usable binary operators (which must be a single character to have room left for both operands) are +, -, *, /, %, <, >, &, |, ^. They can only be used with a single-character value on each side. Options for values include $ and integers 0-9. Results of all combinations of these include Infinity, some numbers and binary values mentioned above, and numerous fractions which are coerced to strings of 19 characters or less (1/7 is 19 characters, unlike 1/9 suggested above which is only 18), and the text representation of $ preceded or followed by a single-digit integer or itself.
• Finally, all remaining operators and expressions: The member operator . requires an existing variable and a identifier referring to a property. All uses of this here result in undefined. Surrounding a value in ( ) returns the value, as does assigning it with =. Using () or `` to call a value as a function results in undefined or errors with all available values.

Adding all this up, there are a grand total of 1651 possible outputs when using a Chrome console. The longest outputs for one, two, and three characters are from $, $$, and $+$ respectively.

dc, 2+5+18=25 bytes

1: Ff yields (Try it online!):

15

2: Fdf yields (Try it online!):

15
15

3: Fd^f yields (Try it online!)

437893890380859375

None of which are particularly interesting, but dc isn't really great for spitting out piles of output. I do like that each answer builds on the previous. Anyway, F is just the number 15; f prints the entire stack; d duplicates top-of-stack; ^ raises next-to-top-of-stack to the power of top-of-stack (in this case, 15^15). I don't believe this can be topped in dc.

Ruby, 3+14+28 = 45 bytes

Why did I do this.

p

Prints nil.

$>

Prints #<IO:<STDOUT>>.

dup

Prints something along the lines of #<Object:0x0000000003610988>.

irb

Launches an instance of Interactive Ruby. Upon exiting, the returned object is #<IRB::Irb: @context=#<IRB::Context:0x0000000003643040>, @signal_status=:IN_EVAL, @scanner=#<RubyLex:0x00000000038900a0>> for 121, but since it requires you to press ^D or something to actually exit the irb instance, I wasn't sure if it'd actually count as a solution in "3 bytes" so I'm not actually including it in the score unless it gets an OK.

Perl 6, 53 (17 + 18 + 18) bytes

e
-e
e*e

1. e outputs $e$ to 15 decimal places
2. -e outputs $-e$ to 15 decimal places
3. e*e outputs $e^2$, but oddly, it gives one more decimal place

Runic Enchantments, 4,000,000 bytes of output

The first program is:

a@

Takes 2 bytes to: push a value to the stack, print a value from the stack, and terminate. In this case it prints 10 (though any integer value from 0 to 16 are also just as valid)

For 3 bytes:

aY@

Prints 10000, Again, a could be 1 through 16 inclusive (in order to generate more output than the original program, 0 is potentially valid under other operators) and there aren't a whole lot of operators that take a single input and produce any output, much less longer output. XCYZ:E are the only real options. aY@ is just the one that results in the most output.

P$!;

According to language specification, this runs infinitely. However as the interpreter has a built in "ok, that's enough" maximum execution limit, this is the most output achievable in 4 characters (and TIO cuts off execution after ~130,000 bytes for exceeding 128kib) and as the interpreter defines the language, this works. And while I have raised that threshold once before (from 10k steps to 1 million), I don't plan on messing with it any time soon.

Bigger?

If I invoke the three assumptions I made here, then sure.

`AA@

Which works out to Ack(65,Ack(65,64)), which aren't terribly large values to go shoving into the Ackerman function initially--certainly smaller than the 255 in the older post--but its ok, we can call Ack twice in 4 instructions.

And only god knows what it'll print.

_{Note: the A instruction has since then been made the Math meta-instruction, which consumes 3 objects on the stack: a char for what instruction to perform and then two inputs, x and y. As such this program doesn't actually do anything, both because neither A nor @ map to a math function and because two subsequent calls results in a stack underflow.}