Sum of all integers from 1 to n

Question

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.

Leaderboard

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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Answer 1

Erlang 19 bytes

s(N) -> N*(N+1)/2.

Answer 2

NASM 32 bits, 719 bytes

I am learning Assembly and used this golf challenge as a context for learning. Probably not the best, but it works.

Note: This version can only take a digit as an input (1 byte character)

section .data
a: equ 1
section .bss
b: resb a
section .text
global _start
_start:
mov eax, 3
mov ebx, 0
mov ecx, b
mov edx, a
int 80H
cmp eax, -1
jz _exit_1
mov edi, eax
atoi:
mov eax, 0
mov ebx, 0
c:
movzx esi, byte [ecx]
cmp ebx, edi
je d
cmp esi, 48
jl _exit_1
cmp esi, 57
jg _exit_1
sub esi, 48
imul eax, 10
add eax, esi
inc ecx
inc ebx
jmp c
d:
mov ecx, eax
mov eax, 0
.count:
add eax, ecx
dec ecx
cmp ecx, 0
jne .count
print_uint32:
xor ebx, ebx
mov ecx, 10
push ecx
mov esi, esp
add esp, 4
.e:
inc ebx
xor edx, edx
div ecx
add edx, '0'
dec esi
mov [esi], dl
test eax, eax
jnz .e
mov edx, ebx
mov eax, 4
mov ebx, 1
mov ecx, esi
int 80h
_exit_0:
mov eax, 1
mov ebx, 0
int 80H
_exit_1:
mov eax, 1
mov ebx, 1
int 80H

Try it online (This is 32 bit assembly, it will not run on TIO's 64 bit assembly online compiler.)

You can find a "non factorized" and commented code here

Answer 3

PIC16F88x Machine Code, 5 words (70 bits)

A function that expects its input at address 0x70, and returns via register W.

70 bits can't be represented cleanly in hex, so here's the raw binary:

00 0001 0000 0000
00 0111 0111 0000
00 1011 1111 0000
10 1000 0000 0110
00 0000 0000 1000

This assumes the function is located at address 0x005. It would be the same length no matter where I put it, but I have to hardcode the GOTO destination.

In mpasm syntax:

    ORG 5         ; Linker directive: place the function at address 0x005
sum:
    CLRW          ; Zero the W register
sum_sub0:
    ADDWF 0x70,0  ; Add the value to W
    DECFSZ 0x70,1 ; Decrement the value, skipping the next instruction if it
                  ; reached zero
    GOTO sum_sub0 ; Loop
    RETURN        ; Return

The PIC16F88x have no multiply instruction, so I have to loop. It might be possible to do better if there's a clever bitwise way to avoid multiplication.

Answer 4

(,) i, 280 273 49 33 30 Chars or \$30\log_{256}(3)\approx\$ 5.94 Bytes

(,,((),(())(),,,(),,((()))()))

Massive golf (224 chars) by taking input in unary.
Another golf (16 Chars) by realizing that input length doesn’t have to be stored in a variable.
No longer taking input in unary!
TIO

Answer 5

Go, 29 bytes

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

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

Try it online!

Answer 6

bc, 9 bytes

n*(n+1)/2

Try it online!

Weird how some shell solutions call bc instead of just writing bc code.

Answer 7

Fortran (GFortran), 35 28 bytes

Try it online!

read*,n;print*,n*(n+1)/2;end

---

Old (but kinda fun) solution: Construct inline anonymous array, sum, and print.

read*,n;print*,sum([(j,j=1,n)]);end

Answer 8

CJam, 6 bytes

ri),:+

Try it online!

Explanation

ri    e# Read integer n
)     e# Add 1
,     e# Range from 0 to input argument minus 1
:+    e# Fold addition over array. Implicitly displa

Answer 9

BLua, 32 bytes

r=0;for i=1,n r=r+i end;return r

Try it out (Vanilla Lua version, 53 bytes)

I'm not very good at this

Answer 10

Swift, 13 bytes

Anonymous function:

{$0*($0+1)/2}

You can call it like this:

print({$0*($0+1)/2}(5))

Try it online!

Answer 11

Mathematica, 9 bytes

#(#+1)/2&

or, at 11 bytes,

Tr@Range@#&

Answer 12

AutoHotkey, 22 bytes

A(n){return n*(n+1)/2}

Defines a function that takes parameter n, and returns n*(n+1)/2, which is the nth triangle number, as shown in Leo's Husk answer.

Answer 13

Tcl, 27 bytes

proc T n {expr $n*($n+1)/2}

Try it online!

Answer 14

Casio Basic, 9 bytes

(n+1)n/2

8 bytes for the code, +1 to add n as parameter.

Answer 15

Factor, 12 bytes

[ iota sum ]

Input is given as an argument to this anonymous function (quotation).

Answer 16

Terrapin Logo, 16 bytes

OP (1+:N)*(:N/2)

Answer 17

Commentator, 22 bytes

//
;-} {-
 {-  -}<!-}!

Try it online!

Answer 18

GolfScript, 7 6 bytes

~.)2/*

Try it online!

Answer 19

Pip, 7 bytes

a*++a/2

Try it online!

Answer 20

><>, 20 15 bytes

:1-:?!v
n?=1l+<

@notatree saved me a couple bytes!

Try it online!

Answer 21

Carrot, 9 bytes

#^F+1/2*$

Explanation:

#  //Set the string stack to the input
^  //Convert to operations mode
F  //Change to float stack
+1 //Add one to the stack
/2 //Divide the stack by 2
*$ //Multiply the stack by the input
   //Implicitly output the result

Answer 22

Java 8 (39 37 bytes)

n->{System.out.println((n*(n+1))/2);}

saved 2 bytes thanks to @TheLethalCoder. I now realise that I could also omit print statements etc, but there already is a Java answer that I missed, which would end up being pretty much the same thing. Thus I will leave it as this :-)

(Is this the right way of scoring Java? I have no clue if this is valid, I saw it in another answer but this actually omits quite a bit of the code necessary to actually run it..)

try it online

Answer 23

Befunge, 9 bytes

&:1+*2/.@

Try it online!

Answer 24

Gaia, 2 bytes

+⊢

This is reduce ⊢ by addition +, which implicitly casts numbers to ranges beforehand.

You could also do ┅Σ (range ┅ and sum Σ), which is still 2 bytes.

Answer 25

Common Lisp, 26 bytes

(lambda(n)(/(+(* n n)n)2))

Answer 26

Clojure, 31 23 16 bytes

8 bytes saved thanks to @cliffroot

#(/(+(* % %)%)2)

Try it online!

Answer 27

Excel VBA, 14 Bytes

Anonymous VBE immediate window function that takes input from range [A1] and outputs to the VBE immediate window

?[A1*(A1+1)/2]

Answer 28

Java, 10 bytes

n->n*-~n/2

Admittedly, only slightly different from another Java answer, but uses two's complement trickery.

Answer 29

Ly, 26 7 bytes

n:1+*2/u

EDIT: Saved a bunch of bytes by using a far better algorithm.

Answer 30

Kona - 8 bytes

{+/!x+1}

Explanation:

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

Alternative answer - 8 bytes:

{+/x,!x}

 +/      Add together
   x      The input
    ,     Joined to
     !     All the numbers less than...
      x    The input