Numbers by Position

Question

Challenge

Print the numbers:

In that order.

I/O

Takes no input. The numbers can have any delimiters desired (or none). That includes lists, cell arrays, .jpeg, etc.... Example outputs:

122333444455555666666777777788888888999999999

[1,22,333,4444,55555,666666,7777777,88888888,999999999]

etc....

Code Example

This is an un-golfed example that may perhaps act as algorithm guide (or maybe not):

Turing Machine Code, 535 bytes

0 * 1 r L
L * _ r 2
2 * 2 r a
a * 2 r M
M * _ r 3
3 * 3 r b
b * 3 r c
c * 3 r N
N * _ r 4
4 * 4 r d
d * 4 r e
e * 4 r f
f * 4 r O
O * _ r 5
5 * 5 r g
g * 5 r h
h * 5 r i 
i * 5 r j
j * 5 r P
P * _ r 6
6 * 6 r k
k * 6 r l
l * 6 r m
m * 6 r n
n * 6 r o
o * 6 r Q
Q * _ r 7
7 * 7 r p
p * 7 r q
q * 7 r r
r * 7 r s
s * 7 r t
t * 7 r u
u * 7 r R
R * _ r 8
8 * 8 r v
v * 8 r w
w * 8 r x
x * 8 r y
y * 8 r z
z * 8 r A
A * 8 r B
B * 8 r S
S * _ r 9
9 * 9 r C
C * 9 r D
D * 9 r E
E * 9 r F
F * 9 r G
G * 9 r H
H * 9 r I
I * 9 r J
J * 9 r halt

Try it online!

This prints out the numbers with a space delimiter:

1 22 333 4444 55555 666666 7777777 88888888 999999999

Challenge Type

code-golf, so shortest answer in bytes (by language) wins.

Based on a submission in the sandbox.

@AdHocGarfHunter, No. Good catch. Edit: Actually, I think '0' should be acceptable. — ouflak, Commented Mar 17, 2020 at 17:17
Could you verify that they "strange delimiters" version of this answer, is valid? It definitely seems cheaty. — Wheat Wizard, Commented Mar 17, 2020 at 17:32
Honestly I think it's a clever 'outside-of-the-box' solution. I'd upvote, but I'm out of votes until tomorrow. — ouflak, Commented Mar 17, 2020 at 17:35
@ouflak Thanks for the algorithm guide! How did you know I always write my prototypes with Turing Machines :p — AviFS, Commented Mar 19, 2020 at 1:01

Community · Accepted Answer · 2020-06-17 09:04:33Z

31

Google Sheets, 35 bytes

=ArrayFormula(Rept(Row(1:9),Row(1:9

Sheets will automatically add three trailing parentheses when you exit the cell. Output is one line per row.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 17, 2020 at 12:34

Engineer Toast

9,5291 gold badge23 silver badges53 bronze badges

\$\begingroup\$ I've run out of votes. \$\endgroup\$
– ouflak
Commented Mar 17, 2020 at 12:37
\$\begingroup\$ @ouflak Darn... \$\endgroup\$
– S.S. Anne
Commented Mar 17, 2020 at 13:18

Add a comment |

Dorian · Accepted Answer · 2020-03-17 08:58:18Z

16

05AB1E, 3 bytes

9L×

Try it online!

9L    Build a list from 1 to 9 {1, 2, 3, 4, 5, 6, 7, 8, 9}
×     copy each number that many times

answered Mar 17, 2020 at 8:58

Dorian

3,04514 silver badges12 bronze badges

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Mitchell Spector · Accepted Answer · 2020-03-19 09:11:15Z

16

Bash + Core utilities, 27, 25 bytes

seq -f8d%f*7-v1+2/n 45|dc

Try it online!

Changed seq formatting from %0.f to %f for a 2-byte savings.

Modified to print on one line, with no delimiters, instead of having a newline after each number, just because I like that better. Same number of bytes.

This uses the formula $$\left\lfloor\frac{\big\lfloor\sqrt{8n-7}\big\rfloor+1}2\right\rfloor$$

for the \$n^{th}\$ digit, where \$n\$ goes from 1 to 45.

edited Mar 19, 2020 at 9:11

answered Mar 17, 2020 at 7:47

Mitchell Spector

7,9631 gold badge17 silver badges35 bronze badges

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Chas Brown · Accepted Answer · 2020-03-17 07:24:02Z

11

Python 2, 28 bytes

i=1;exec"print`i`*i;i+=1;"*9

Try it online!

answered Mar 17, 2020 at 7:24

Chas Brown

9,7571 gold badge13 silver badges39 bronze badges

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Robin Ryder · Accepted Answer · 2020-03-17 08:18:44Z

10

R, 18 bytes

a=1:9;(10^a-1)/9*a

Try it online!

Use the formula \$\frac{10^n-1}{9}\times n\$ for the \$n\$th number.

answered Mar 17, 2020 at 8:18

Robin Ryder

15.6k2 gold badges24 silver badges70 bronze badges

\$\begingroup\$ Hmm, I may be out of date but isn't the usual ruling that it needs cat() or equivalent to print the output? \$\endgroup\$
– user2390246
Commented Mar 17, 2020 at 16:44
1

\$\begingroup\$ @user2390246 yeah, the consensus was changed...a while ago? Late 2017 by my quick search of meta. This is probably what you're aware of but taking that approach with this meta post on compiler flags basically means all R answers are just "R invoked with source(...,echo=TRUE)" \$\endgroup\$
– Giuseppe
Commented Mar 17, 2020 at 18:44
\$\begingroup\$ @Giuseppe thanks for the clarification, I've not been around for a while! Seems like the trend for cat()-type answers ended up being quite a short phase then. \$\endgroup\$
– user2390246
Commented Mar 17, 2020 at 19:33
\$\begingroup\$ Beat you with a string-only approach! :-) \$\endgroup\$
– Giuseppe
Commented Mar 18, 2020 at 19:07
\$\begingroup\$ @Giuseppe Ooh, nice! \$\endgroup\$
– Robin Ryder
Commented Mar 18, 2020 at 19:16

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Adám · Accepted Answer · 2020-03-17 07:21:32Z

9

APL (dzaima/APL), 6 bytes

Full program, requiring ⎕IO←0.

⍋⍛⌿⍨⎕D

Try it online!

⎕D on the string "0123456789",

⍛⌿⍨ replicate the characters by

⍋ their grade (0, 1, 2, …, 9)

answered Mar 17, 2020 at 7:21

Adám

31.1k3 gold badges124 silver badges287 bronze badges

\$\begingroup\$ Isn't \⍨⍳9 at 4 bytes sufficient? I've added it as my own answer, though it doesn't seem to be gaining any traction. Would love feedback! \$\endgroup\$
– AviFS
Commented Mar 19, 2020 at 0:58
\$\begingroup\$ @AviF.S. It wasn't clear that this was a permitted format, but now that OP has clarified, you have my upvote. \$\endgroup\$
– Adám
Commented Mar 19, 2020 at 6:52

Add a comment |

snail_ · Accepted Answer · 2020-03-17 07:51:49Z

9

Kotlin, 30 bytes

{(1..9).map{"$it".repeat(it)}}

Try it online!

answered Mar 17, 2020 at 7:51

snail_

2,67012 silver badges16 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

APL (Dyalog Unicode), 4 bytes

\⍨⍳9

Try it online!

How it works

⍳9        ⍝ Integers 1..9
   ⍨      ⍝ Duplicate argument on each side
     \    ⍝ Replicate each element *n* times

Examples

Index Generator:    ⍳5          =  1 2 3 4 5
Expand:             2 3 \ 1 4   =  1 1 4 4 4
Commute:            +⍨4         =  4 + 4 = 8

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 18, 2020 at 5:24

AviFS

2,17116 silver badges26 bronze badges

\$\begingroup\$ Feedback: I'd use /⍨⍳9 as it is simpler. / is really "replicate" while \ is the more complex "expand". \$\endgroup\$
– Adám
Commented Mar 19, 2020 at 6:53

Add a comment |

andytech · Accepted Answer · 2020-03-17 08:41:50Z

6

Perl 5, 18 bytes

map{say$_ x$_}1..9

Try it online!

answered Mar 17, 2020 at 8:41

andytech

3491 silver badge4 bronze badges

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Dorian · Accepted Answer · 2020-03-17 09:11:40Z

6

brainfuck, 56 bytes

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

Try it online!

+++++++[>+++++++<-]     49 (ASCII "1")
+++++++++[              do 9 times
  <+                    add 1 to output counter
  [                     do that many times
    >>.<<               print character
    -<+>                move value of output counter to temp
  ]
  <[>+<-]               move value of temp back to output counter
  >>>+                  increment character
  <-                    decrement loop counter
]

answered Mar 17, 2020 at 9:11

Dorian

3,04514 silver badges12 bronze badges

1

\$\begingroup\$ 50 bytes \$\endgroup\$
– Jo King ♦
Commented Mar 17, 2020 at 9:22
2

\$\begingroup\$ Thank you. Your code is so different from mine, you should post it as a separate answer. \$\endgroup\$
– Dorian
Commented Mar 17, 2020 at 13:27

Add a comment |

Neil · Accepted Answer · 2020-03-17 10:55:27Z

6

Retina, 12 bytes


9*

$.`*$.`

Try it online! Outputs a leading _ to each number, which appears to be acceptable (would cost 2 bytes to fix if not). Explanation:

9*

Insert 9 _s.


$.`*$.`

Around each _, insert its position repeated appropriately.

answered Mar 17, 2020 at 10:55

Neil

177k12 gold badges74 silver badges281 bronze badges

\$\begingroup\$ The leading _ is absolutely acceptable. \$\endgroup\$
– ouflak
Commented Mar 17, 2020 at 11:31

Add a comment |

ovs · Accepted Answer · 2020-03-17 13:58:43Z

6

Haskell, 30 29 bytes

(<$)<*>g<$>g '9'
g c=['1'..c]

Try it online!

edited Mar 17, 2020 at 13:58

answered Mar 17, 2020 at 13:53

ovs

60.6k3 gold badges47 silver badges161 bronze badges

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Bubbler · Accepted Answer · 2020-03-17 07:20:35Z

5

APL (Dyalog Unicode), 7 bytes^SBCS

⎕D/⍨⍳10

Try it online!

Uses ⎕IO←0.

How it works

⎕D/⍨⍳10
⎕D       ⍝ The string '0123456789'
  /⍨     ⍝ Replicate each of them the following times...
    ⍳10  ⍝ 0..9

answered Mar 17, 2020 at 7:20

Bubbler

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Wheat Wizard · Accepted Answer · 2020-03-18 01:17:29Z

Brain-Flak, 90 82 bytes

([(()()())({}){}]){((({})()<([{}]((((()()()){}){}){}){})>)<{({}()<(({}))>)}{}>)}{}

Try it online!

Explanation:

Compare this with the output of JoKing's autogolfer

Brain-Flak, 142 bytes

(((((((((((((((((((((((((((((((((((((((((((((((((()()()){}){}){}){}())()))())))()))))())))))()))))))())))))))()))))))))()))))))))){({}<>)<>}<>

Try it online!

Strange delimiters, 78 bytes

([(()()())({}){}]){((({})()<([{}]((((()()()){}){}){}){})>)<{({}()<(({}))>)}>)}

Try it online!

If we decide to play around with our delimiters a bit, we can shave off 4 bytes. This version outputs the correct stuff but with two leading null bytes and null bytes between the chunks:

This is a tiny bit cheaty but it meets the specs of the challenge.

And for posterity here is the old super cheaty version that has been made obsolete by my golfs.

Calga · Accepted Answer · 2022-09-08 13:33:58Z

5

Husk, 5 bytes

d´Ṙḣ9

Try it online!

Output: 122333444455555666666777777788888888999999999.

answered Sep 8, 2022 at 13:33

Calga

1011 silver badge4 bronze badges

\$\begingroup\$ Welcome to Code Golf! \$\endgroup\$
– rydwolf
Commented Sep 8, 2022 at 13:34
\$\begingroup\$ Thank you a lot! :) \$\endgroup\$
– Calga
Commented Sep 9, 2022 at 11:57

Add a comment |

Nick Kennedy · Accepted Answer · 2020-03-17 09:56:41Z

4

Jelly, 5 bytes

9ẋ`€Ḍ

Try it online!

A niladic link returning a list of integers. If a program printing the numbers is preferred, subsitute Y for Ḍ.

Explanation

9     | Literal 9
 ẋ`€  | Repeat each that many times
    Ḍ | Convert from decimal digits to integer

answered Mar 17, 2020 at 9:56

Nick Kennedy

21.2k3 gold badges17 silver badges43 bronze badges

1

\$\begingroup\$ You can drop the trailing Ḍ, based on the comments below the challenge description between xnor and OP. \$\endgroup\$
– Kevin Cruijssen
Commented Mar 18, 2020 at 15:37

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Jo King · Accepted Answer · 2020-03-17 11:57:21Z

4

Perl 6, 13 bytes

{1..9 Zx^9+1}

Try it online!

Anonymous code block that returns a list of strings by zip string multiplying the range 1 to 9 with itself.

answered Mar 17, 2020 at 11:57

Jo King♦

47.5k5 gold badges125 silver badges186 bronze badges

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Giuseppe · Accepted Answer · 2020-03-17 23:37:25Z

4

R, 15 bytes

strrep(1:9,1:9)

Try it online!

answered Mar 17, 2020 at 23:37

Giuseppe

28.8k3 gold badges31 silver badges105 bronze badges

1

\$\begingroup\$ Just rep(1:9,1:9) is shorter (and gives exactly the output of the first example)? \$\endgroup\$
– Dominic van Essen
Commented Jun 24, 2020 at 22:50

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Galen Ivanov · Accepted Answer · 2020-03-25 19:26:57Z

4

J, 18 bytes

echo u:(#48+])i.10

Try it online!

K (oK), 11 10 bytes

-1 byte thanks to ngn!

,/${x}#!10

Try it online!

edited Mar 25, 2020 at 19:26

answered Mar 17, 2020 at 8:12

Galen Ivanov

21.2k3 gold badges24 silver badges61 bronze badges

1

\$\begingroup\$ ,/${x}#!10 - using filter ({ }#) for replication \$\endgroup\$
– ngn
Commented Mar 25, 2020 at 19:23
\$\begingroup\$ @ngn Thanks, of course! \$\endgroup\$
– Galen Ivanov
Commented Mar 25, 2020 at 19:28
1

\$\begingroup\$ I think echo#~1+i.10 is ok too since any delimiter is allowed. \$\endgroup\$
– Jonah
Commented Jan 3, 2022 at 4:50

Add a comment |

dingledooper · Accepted Answer · 2020-04-22 16:18:19Z

4

C (gcc), 46 bytes

s;main(n){n>9||main(puts(memset(&s,n+48,n)));}

Try it online!

edited Apr 22, 2020 at 16:18

answered Mar 17, 2020 at 16:42

dingledooper

22.9k1 gold badge36 silver badges125 bronze badges

\$\begingroup\$ can't be called twice \$\endgroup\$
– l4m2
Commented Apr 22, 2020 at 16:13
\$\begingroup\$ Thanks for catching that :) \$\endgroup\$
– dingledooper
Commented Apr 22, 2020 at 16:18

Add a comment |

640KB · Accepted Answer · 2020-09-25 14:15:52Z

x86-16 machine code, IBM PC DOS, 18 16 bytes

Binary:

00000000: b839 0ab2 09b1 2dcd 1048 2aca 4a75 f8c3  .9....-..H*.Ju..

Listing:

B8 0A39     MOV  AX, 0A39H              ; AH = 0AH, AL = '9'
B2 0A       MOV  DL, 10                 ; DL as counter value
B1 2D       MOV  CL, 1+2+3+4+5+6+7+8+9  ; start digit repeat 45 times
        NLOOP:
CD 10       INT  10H                    ; call BIOS - write digit * CX times
48          DEC  AX                     ; decrement ASCII digit
4A          DEC  DX                     ; decrement counter value
2A CA       SUB  CL, DL                 ; reduce digit repeat value by counter
75 F8       JNZ  NLOOP                  ; loop until 0 
C3          RET                         ; return to DOS

Try it online!

Explanation:

This uses the PC BIOS API's INT 10H / 0AH function to write the ASCII char in AL to the screen CX number of times. However, this function does not update the cursor position to the end of the output -- it just stays where it started. In other words, the next call simply overstrikes existing characters writing over them. Making a BIOS call to advance the cursor is expensive byte-wise.

Since going forward isn't going to work, we go backwards starting from '9'. It writes '9' 45 times, then '8' 36 times, '7' 28 times, etc -- each time starting from the first column overwriting like so:

999999999999999999999999999999999999999999999
888888888888888888888888888888888888999999999
777777777777777777777777777788888888999999999
666666666666666666666777777788888888999999999
555555555555555666666777777788888888999999999
444444444455555666666777777788888888999999999
333333444455555666666777777788888888999999999
222333444455555666666777777788888888999999999
122333444455555666666777777788888888999999999

Output:

caird coinheringaahin g · Accepted Answer · 2020-10-08 19:38:22Z

4

Java 11 (JDK), 60 59 bytes

Not sure if thats the shortest approach but couldn`t make it shorter even without System.out.print. Output is without delimiters.

-1 byte thanks to Kevin Cruijssen

v->{for(int i=0;i++<9;System.out.print((i+"").repeat(i)));}

Try it online!

edited Oct 8, 2020 at 19:38

caird coinheringaahin g♦

50.3k11 gold badges128 silver badges354 bronze badges

answered Mar 17, 2020 at 14:33

greinet

1014 bronze badges

1

\$\begingroup\$ You can save 1 byte changing the ()-> to v->, by taking an empty unused (Void null) argument, which is allowed when it's completely unused. \$\endgroup\$
– Kevin Cruijssen
Commented Mar 18, 2020 at 10:04
\$\begingroup\$ @Kevin Cruijssen Thanks for the suggestion, didn't know about that. I now use an Object as parameter and pass null, is this the intended way or can you actually do it with Void? \$\endgroup\$
– greinet
Commented Mar 18, 2020 at 10:26
\$\begingroup\$ Object null is fine as well. As long as the argument isn't used in any way, not even for static calls, it's ok. That's why I personally use Void, since it has no available static calls anyway. :) \$\endgroup\$
– Kevin Cruijssen
Commented Mar 18, 2020 at 10:27
\$\begingroup\$ Well thats awkward, could've sworn that it just didn't compile with type Void, changed it again and now it works, thanks. \$\endgroup\$
– greinet
Commented Mar 18, 2020 at 10:32

Add a comment |

pxeger · Accepted Answer · 2022-01-01 07:04:45Z

4

Haskell, 21 bytes

[c<$[1..c]|c<-[1..9]]

Try it online!

A list of lists of numbers.

23 bytes

replicate<*>id=<<[1..9]

Try it online!

A flat list of numbers.

edited Jan 1, 2022 at 7:04

pxeger

24.6k4 gold badges53 silver badges141 bronze badges

answered Mar 17, 2020 at 17:54

xnor

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\$\begingroup\$ I think that the delimiter has to be in between differing digits (e.g. 1 and 2 as opposed to 8 and another 8). \$\endgroup\$
– Wheat Wizard ♦
Commented Mar 17, 2020 at 18:06
\$\begingroup\$ @AdHocGarfHunter Looking again at the challenge, I'm not clear on this so I asked. I think it would be unfortunate if the flat list in the second version were invalid too. \$\endgroup\$
– xnor
Commented Mar 17, 2020 at 18:14

Add a comment |

amelies · Accepted Answer · 2022-09-11 08:18:20Z

4

Julia, 22 bytes

1:9 .|>i->show("$i"^i)

Attempt This Online!

prints

"1""22""333""4444""55555""666666""7777777""88888888""999999999"

Relying on the repl for printing output we can get to 17 bytes

~i="$i"^i;.~(1:9)

edited Sep 11, 2022 at 8:18

answered Sep 9, 2022 at 15:06

amelies

6814 silver badges7 bronze badges

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Seggan · Accepted Answer · 2022-09-15 16:58:18Z

4

Fig, \$6\log_{256}(96)\approx\$ 4.939 bytes

Mcd'*_

Try it online!

Lyxal tried to outgolf me in my own language again.

Mcd'*_ # Full program
 cd    # Digits
M  '   # Map each digit string
    *  # Repeat the digit
     _ # By the number form of it

edited Sep 15, 2022 at 16:58

answered Sep 8, 2022 at 18:50

Seggan

6,64223 silver badges50 bronze badges

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emanresu A · Accepted Answer · 2021-02-24 09:19:11Z

3

Deadfish~, 194 bytes

{{i}ddddd}dc{d}{d}iiic{i}{i}ddcc{d}{d}iic{i}{i}dccc{d}{d}ic{i}{i}cccc{d}{d}c{i}{i}iccccc{d}{d}dc{i}{i}iicccccc{d}{d}ddc{i}{i}iiiccccccc{d}{d}dddc{i}{i}iiiicccccccc{d}{d}ddddc{i}{i}iiiiiccccccccc

Try it online!

answered Feb 24, 2021 at 9:19

emanresu A

44.2k5 gold badges100 silver badges240 bronze badges

\$\begingroup\$ Understandable. \$\endgroup\$
– lyxal
Commented Jun 12, 2021 at 11:38
\$\begingroup\$ @lyxal ???? (filler) \$\endgroup\$
– Wasif
Commented Jun 12, 2021 at 11:39
\$\begingroup\$ @wasif the reason it was accepted \$\endgroup\$
– lyxal
Commented Jun 12, 2021 at 11:40
1

\$\begingroup\$ 159 bytes, and 128 separating by zeros. \$\endgroup\$
– naffetS
Commented Sep 8, 2022 at 23:53
1

\$\begingroup\$ Actually, 61 since delimiters aren't even required. \$\endgroup\$
– naffetS
Commented Sep 8, 2022 at 23:55

| Show 1 more comment

mazzy · Accepted Answer · 2020-03-17 07:33:12Z

3

PowerShell, 15 bytes

1..9|%{"$_"*$_}

Try it online!

answered Mar 17, 2020 at 7:33

mazzy

7,1652 gold badges12 silver badges22 bronze badges

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Galen Ivanov · Accepted Answer · 2020-03-17 07:47:12Z

3

Icon, 27 bytes

write(1(i:=1to 9,1to i))&\z

Try it online!

answered Mar 17, 2020 at 7:47

Galen Ivanov

21.2k3 gold badges24 silver badges61 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

3

T-SQL, 69 bytes

SELECT top 9replicate(1+number,1+number)FROM spt_values WHERE'p'=type

Try it online

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Mar 17, 2020 at 8:47

t-clausen.dk

3,76412 silver badges16 bronze badges

\$\begingroup\$ Nice code. Sometimes I wonder how well T-SQL would do in these challenges if we had a built-in number table or some equivalent to PostgreSQL's generate-series() function. \$\endgroup\$
– BradC
Commented May 26, 2020 at 15:48
1

\$\begingroup\$ @BradC normally SQL will not win in these golfing questions even with improved features. When we use a command like replicate(a,b), some of the other languages can solve problem with fewer characters than the 14 characters(minimum) it takes to call that function. The best we can do it solve the problems using a language which usually isn't optimal, and hope some people are impressed enough to up-vote those solutions. \$\endgroup\$
– t-clausen.dk
Commented May 27, 2020 at 11:29

Add a comment |

Neil · Accepted Answer · 2020-03-17 10:48:45Z

3

Charcoal, 5 bytes

⭆χ⭆ιι

Try it online! Link is to verbose version of code. Outputs without separators. The first StringMap could be changed into a for statement for the same byte count. Explanation:

 χ       Predefined variable 10
⭆        Map over implicit range and join
   ι     Current index
  ⭆      Map over implicit range and join
    ι    Outer index
         Implicitly print

answered Mar 17, 2020 at 10:48

Neil

177k12 gold badges74 silver badges281 bronze badges

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Numbers by Position

Turing Machine Code, 535 bytes

122 Answers 122

Google Sheets, 35 bytes

05AB1E, 3 bytes

Bash + Core utilities, 27, 25 bytes

Python 2, 28 bytes

R, 18 bytes

APL (dzaima/APL), 6 bytes

Kotlin, 30 bytes

APL (Dyalog Unicode), 4 bytes

How it works

Examples

Perl 5, 18 bytes

brainfuck, 56 bytes

Retina, 12 bytes

Haskell, 30 29 bytes

APL (Dyalog Unicode), 7 bytesSBCS

How it works

Brain-Flak, 90 82 bytes

Explanation:

Brain-Flak, 142 bytes

Strange delimiters, 78 bytes

Husk, 5 bytes

Jelly, 5 bytes

Explanation

Perl 6, 13 bytes

R, 15 bytes

J, 18 bytes

K (oK), 11 10 bytes

C (gcc), 46 bytes

x86-16 machine code, IBM PC DOS, 18 16 bytes

Java 11 (JDK), 60 59 bytes

Haskell, 21 bytes

Julia, 22 bytes

Fig, \$6\log_{256}(96)\approx\$ 4.939 bytes

Deadfish~, 194 bytes

PowerShell, 15 bytes

Icon, 27 bytes

T-SQL, 69 bytes

Charcoal, 5 bytes

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