# Sum of all integers from 1 to n

I'm honestly surprised that this hasn't been done already. If you can find an existing thread, by all means mark this as a duplicate or let me know.

# Input

Your input is in the form of any positive integer greater than or equal to 1.

# Output

You must output the sum of all integers between and including 1 and the number input.

# Example

 In: 5
1+2+3+4+5 = 15
Out: 15


OEIS A000217 — Triangular numbers: a(n) = binomial(n+1,2) = n(n+1)/2 = 0 + 1 + 2 + ... + n.

Run the code snippet below to view a leaderboard for this question's answers. (Thanks to programmer5000 and steenbergh for suggesting this, and Martin Ender for creating it.)

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• Closely related – FryAmTheEggman Jul 18 '17 at 20:36
• @FryAmTheEggman Sorry - had a bit of a brain fart there. I see what you mean. – GarethPW Jul 18 '17 at 20:45
• @Aaron you got ninja'd by Husk, which was just posted with a 1 byte solution – Skidsdev Jul 18 '17 at 21:35
• I suggest a stack snippet. – programmer5000 Jul 19 '17 at 11:42
• – Jerry Jeremiah Jul 27 '17 at 12:20

# Kona - 8 bytes

{+/!x+1}


Explanation:

 +/      Add together
!      All numbers less than...
x+1   The input plus 1


### Alternative answer - 8 bytes:

{+/x,!x}

x      The input
,     Joined to
!     All the numbers less than...
x    The input

• If you dont need it to be a function you can do +/!1+ for 5 bytes. – streetster Aug 10 '17 at 14:12
• If a function is required, in most k's you can do {x+/!x} for 7 bytes. – coltim Dec 30 '20 at 20:13

# Emojicode, 50 bytes

🐖🔢➡️🚂🍇🍎➗✖🐕➕🐕1 2🍉


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# 4, 40 bytes

3.70060101002000120300026040230503045054


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# Braingolf, 1 byte

Q


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1-indexed (ie 7 returns 0-6 summed)

# Braingolf, 3 bytes

U&+


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U - range, &+ - sum.

# Cubically, 19 bytes

R3U1F1$:1/1+7*7/0%6  How it works: R3U1F1 Set the top face to 2$             Get the first input as a number
:1/1+7       Set the notepad to the input + 1
*7     Multiply the notepad by the input
/0   Divide the notepad by 2
%6 Output the notepad as a number


# MY, 4 bytes

Wow, MY is actually capable of something!

𝕫iΣ↵


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Explanation (hex/cp):

1A/𝕫 - push input as an integer
49/i - pop a; push [1 .. a]
53/Σ - pop a; push sum(a)
27/↵ - pop a; print(a) (with newline)


# Pyth, 11 9 bytes

VQ=+ZhN)Z


Explanation:

VQ       For N in range(0, Input)
=+ZhN)   Set Z to Z + 1 + N, suppress output and close function call
Z        Output Z


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• Beat you by 7 bytes – Tornado547 Dec 11 '17 at 16:38
• @Tornado547 I know, but that answers been taken. This is just a unique way of doing it. 2 bytes: sS. – Stan Strum Dec 11 '17 at 16:39

# ,,,, 6 bytes

::×+2÷


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Explanation:

::     Duplicates the input twice
×    Pops off top two values and muliples them
+   Adds the two values together. (Thus far it's basically n*n+n)
2÷ Divides by 2


D,f,@,Rb+


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ಠ_ಠ I forgot about functions. And then I forgot about the range command

## How it works

D,f,@,    - Create a monadic function called f (one argument)
R   - Generate a range from 1 to n
b+ - Reduce that range by addition (sum)


# Recursiva, 3 bytes

sBa


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# Recursiva, 5 bytes

H*a;a


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# Recursiva, 12 bytes

=a0:0!+a#~a$ Try it online! # Pyth - 2 Bytes sS  Explanation: sSQ Q added implicitly to resolve arity s sum of S integers from one to Q input  # SNOBOL4 (CSNOBOL4), 6240 38 bytes  N =INPUT OUTPUT =N * (N + 1) / 2 END  Try it online! # Alumin, 9 bytes jqdhcpfaf  Try it online! ## Explanation jqdhcpfaf j numeric input q p whlie TOS > 0 d duplicate TOS hc subtract 1 f f fold over... a ... addition  • @Riker shoot lol done – Conor O'Brien Dec 17 '17 at 21:06 # Kotlin, 13 bytes {it*(it+1)/2}  ## Beautified { it * (it + 1) / 2 }  ## Test var f: (Int) -> Unit = {it*(it+1)/2} fun main(args: Array<String>) { println(f(5)) }  ## TIO TryItOnline # brainfuck, 14 bytes ,[[>+<<.>-]>-]  Try it online! Takes input as character code, outputs as unary null bytes # Japt-x, 3 1 bytes ò  ## Explanation: ò Range [0...Input] -x Sum  Try it online! # Phooey, 10 bytes &.@+1*/2$i


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## Explanation

&.@+1*/2$i &. write input to the tape @ push same input to the stack +1 increment tape value * multiply tape value by popped stack value /2 divide it by 2$i    output as integer


# Whitespace, 71 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_integer][T   T   T   _Retrieve][S N
S _Duplicate][N
S S N
_Create_Label_LOOP][S N
T   _Swap][S S S T  N
_Push_1][T  S S T   _Subtract][S N
S _Duplicate][N
T   S S N
_If_0_Jump_to_Label_PRINT_AND_EXIT][S N
S _Duplicate][S T   S S T   S N
S N
N
_Jump_to_Label_LOOP][N
S S S N
_Create_Label_PRINT_AND_EXIT][S N
N
S T _Print_as_integer]


Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

Try it online (with raw spaces, tabs and new-lines only).

Example run: input = 3

Command   Explanation                     Stack        Heap   STDIN  STDOUT  STDERR

SSSN      Push 0                          [0]
SNS       Duplicate 0                     [0,0]
TNTT      Read STDIN as integer           [0]          {0:3}  3
TTT       Retrieve input from heap 0      [3]          {0:3}
SNS       Duplicate 3                     [3,3]        {0:3}
NSSN      Create Label_LOOP               [3,3]        {0:3}
SNT      Swap top two                    [3,3]        {0:3}
SSSTN    Push 1                          [3,3,1]      {0:3}
TSST     Subtract (3-1)                  [3,2]        {0:3}
SNS      Duplicate 2                     [3,2,2]      {0:3}
SNS      Duplicate 2                     [3,2,2]      {0:3}
STSSTSN  Copy (0-indexed) 2nd (3)        [3,2,2,3]    {0:3}

SNT      Swap top two                    [3,5,2]      {0:3}
SSSTN    Push 1                          [3,5,2,1]    {0:3}
TSST     Subtract (2-1)                  [3,5,1]      {0:3}
SNS      Duplicate 1                     [3,5,1,1]    {0:3}
SNS      Duplicate 1                     [3,5,1,1]    {0:3}
STSSTSN  Copy (0-indexed) 2nd (5)        [3,5,1,1,5]  {0:3}

SNT      Swap top two                    [3,5,6,1]    {0:3}
SSSTN    Push 1                          [3,5,6,1,1]  {0:3}
TSST     Subtract (1-1)                  [3,5,6,0]    {0:3}
SNS      Duplicate 0                     [3,5,6,0,0]  {0:3}
NSSSN     Create Label_EXIT               [3,5,6,0]    {0:3}
TNST     Print top (6) to STDOUT as int  [3,5]        {0:3}         6
error


Program stops with an error: No exit found. (Although I could add three trailing newlines NNN to get rid of that error.)

IEK+O@


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# @yBASIC, 10 bytes

?_*_+_>>!.


Input should be in _ (No input methods exist)

## Explanation

N(N+1) can be rewritten as N*N+N. Dividing by 2 would then require parentheses, but a left shift can be used instead. !. is the same as !0.0 (logical not of 0), which is 1.

# MathGolf, 2 bytes

╒Σ


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Pretty much exactly the sum (Σ) of the range from 1 to input (╒)

# Gol><>, 7 bytes

I:P*2,h


Courtesy of JoKing

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8 bytes

I::*+2,h


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# Go, 29 bytes

A function literal which uses the closed form, $$\ \frac n2(n+1) \$$.

func(n int)int{return-~n*n/2}


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# Symbolic Python, 14 bytes

_*=-~_*_/(_+_)


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This basically evaluates to $$\ \frac{ n^2 (n-1)}{2n} = \frac{ n (n-1)}{2}\$$, which is the usual form for triangular numbers.

# Shakespeare Programming Language, 154 bytes

S.Ajax,.Puck,.Act I:.Scene I:.[Enter Ajax and Puck]
Ajax:
Listen tothy.
You is the quotient betweenthe product ofyou the sum ofyou a cat a big cat.
Open heart


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Simple, just does n(n+1)/2.

# Charcoal, 5 bytes

ＩΣ…·Ｎ


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### Explanation:

 Σ        The sum of
…·Ｎ     the inclusive range from 0 to input
Ｉ          converted to a string


I tried to make it with the formula $$\\frac{x(x+1)}{2}\$$ but it took more characters.

## Keg, 2 bytes (SBCS)

Ï⅀


TIO

## Keg, 5 bytes (SBCS)

Ï∑+).


Explanation:

Ï#    Range from input to 0. The 0 will not affect the summation.
)# We have to complete the braces if we want to output as an integer.
.#Output as an integer


TIO

• That's available on the github interpreter and theoretically, it should work. – Lyxal Oct 13 '19 at 1:13
• -1 byte using the -hr flag – Lyxal Dec 6 '19 at 20:43

## Clojure - 19 bytes

#(apply +(range %))

# C (gcc), 19 bytes

#define f(n)n*-~n/2


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# C (gcc), 15 bytes

f(n){n*=-~n/2;}


-5 bytes thanks to ceilingcat!

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• @ceilingcat Huh. How does that work? – S.S. Anne Jan 10 '20 at 12:26
• @JL2210 It's abusing the way GCC compiles the code without optimizations. – NieDzejkob Jan 11 '20 at 14:37

# Labyrinth, 8 bytes

?:)*_2/!


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A linear program which halts by error at the bounced-off / (triggering division by zero).

?:)*  Push input, dup, increment, multiply (giving n*(n+1))
_2/   Divide by 2
!     Pop and print, bounce off the dead end
/     The top is (implicit) 0 so it triggers division by zero error