Calculate the sum of the first n prime numbers

Question

I'm surprised that this challenge isn't already here, as it's so obvious. (Or I'm surprised I couldn't find it and anybody will mark it as a duplicate.)

Task

Given a non-negative integer \$n\$, calculate the sum of the first \$n\$ primes and output it.

Example #1

For \$n = 5\$, the first five primes are:

• 2
• 3
• 5
• 7
• 11

The sum of these numbers is \$2 + 3 + 5 + 7 + 11 = 28\$, so the program has to output \$28\$.

Example #2

For \$n = 0\$, the "first zero" primes are none. And the sum of no numbers is - of course - \$0\$.

Rules

• You may use built-ins, e.g., to check if a number is prime.
• This is code-golf, so the lowest number of bytes in each language wins!

Answer 1

6502 machine code routine, 75 bytes

A0 01 84 FD 88 84 FE C4 02 F0 32 E6 FD A0 00 A5 FD C9 04 90 1F 85 64 B1 FB 85
65 A9 00 A2 08 06 64 2A C5 65 90 02 E5 65 CA D0 F4 C9 00 F0 DC C8 C4 FE D0 DB
A5 FD A4 FE 91 FB C8 D0 C8 A9 00 18 A8 C4 FE F0 05 71 FB C8 D0 F7 60

Expects a pointer to some temporary storage in $fb/$fc and the number of primes to sum up in $2. Returns the sum in A (the accu register).

Never did some prime checks in 6502 machine code, so here it finally comes ;)

Note this starts giving wrong results for inputs >= 14. This is because of overflow, the code works with the "natural" number range of the 8bit platform which is 0 - 255 for unsigned.

Commented disassembly

; function to sum the first n primes
;
; input:
;   $fb/$fc: pointer to a buffer for temporary storage of primes
;   $2:      number of primes to sum (n)
; output:
;   A:       sum of the first n primes
; clobbers:
;   $fd:     current number under primality test
;   $fe:     number of primes currently found
;   $64:     temporary numerator for modulo check
;   $65:     temporary divisor for modulo check
;   X, Y
 .primesum:
A0 01       LDY #$01            ; init variable for ...
84 FD       STY $FD             ; next prime number to test
88          DEY                 ; init number of found primes
 .mainloop:
84 FE       STY $FE             ; store current number of found primes
C4 02       CPY $02             ; compare with requested number
F0 32       BEQ .sum            ; enough primes -> calculate their sum
 .mainnext:
E6 FD       INC $FD             ; check next prime number
A0 00       LDY #$00            ; start check against first prime number
 .primecheckloop:
A5 FD       LDA $FD             ; load current number to check
C9 04       CMP #$04            ; smaller than 4?
90 1F       BCC .isprime        ; is a prime (shortcut to get list started)
85 64       STA $64             ; store to temp as numerator
B1 FB       LDA ($FB),Y         ; load from prime number table
85 65       STA $65             ; store to temp as divisor
A9 00       LDA #$00            ; init modulo to 0
A2 08       LDX #$08            ; iterate over 8 bits
 .bitloop:
06 64       ASL $64             ; shift left numerator
2A          ROL A               ; shift carry into modulo
C5 65       CMP $65             ; compare with divisor
90 02       BCC .bitnext        ; smaller -> to next bit
E5 65       SBC $65             ; otherwise subtract divisor
 .bitnext:
CA          DEX                 ; next bit
D0 F4       BNE .bitloop
C9 00       CMP #$00            ; compare modulo with 0
F0 DC       BEQ .mainnext       ; equal? -> no prime number
C8          INY                 ; next index in prime number table
C4 FE       CPY $FE             ; checked against all prime numbers?
D0 DB       BNE .primecheckloop ; no -> check next
 .isprime:
A5 FD       LDA $FD             ; prime found
A4 FE       LDY $FE             ; then store in table
91 FB       STA ($FB),Y
C8          INY                 ; increment number of primes found
D0 C8       BNE .mainloop       ; and repeat whole process
 .sum:
A9 00       LDA #$00            ; initialize sum to 0
18          CLC
A8          TAY                 ; start adding table from position 0
 .sumloop:
C4 FE       CPY $FE             ; whole table added?
F0 05       BEQ .done           ; yes -> we're done
71 FB       ADC ($FB),Y         ; add current entry
C8          INY                 ; increment index
D0 F7       BNE .sumloop        ; and repeat
 .done:
60          RTS

---

Example C64 assembler program using the routine:

Online demo

Code in ca65 syntax:

.import primesum   ; link with routine above

.segment "BHDR" ; BASIC header
                .word   $0801           ; load address
                .word   $080b           ; pointer next BASIC line
                .word   2018            ; line number
                .byte   $9e             ; BASIC token "SYS"
                .byte   "2061",$0,$0,$0 ; 2061 ($080d) and terminating 0 bytes

.bss
linebuf:        .res    4               ; maximum length of a valid unsigned
                                        ; 8-bit number input
convbuf:        .res    3               ; 3 BCD digits for unsigned 8-bit
                                        ; number conversion
primebuf:       .res    $100            ; buffer for primesum function

.data
prompt:         .byte   "> ", $0
errmsg:         .byte   "Error parsing number, try again.", $d, $0

.code
                lda     #$17            ; set upper/lower mode
                sta     $d018

input:
                lda     #<prompt        ; display prompt
                ldy     #>prompt
                jsr     $ab1e

                lda     #<linebuf       ; read string into buffer
                ldy     #>linebuf
                ldx     #4
                jsr     readline

                lda     linebuf         ; empty line?
                beq     input           ; try again

                lda     #<linebuf       ; convert input to int8
                ldy     #>linebuf
                jsr     touint8
                bcc     numok           ; successful -> start processing
                lda     #<errmsg        ; else show error message and repeat
                ldy     #>errmsg
                jsr     $ab1e
                bcs     input

numok:          
                sta     $2
                lda     #<primebuf
                sta     $fb
                lda     #>primebuf
                sta     $fc
                jsr     primesum        ; call function to sum primes
                tax                     ; and ...
                lda     #$0             ; 
                jmp     $bdcd           ; .. print result

; read a line of input from keyboard, terminate it with 0
; expects pointer to input buffer in A/Y, buffer length in X
.proc readline
                dex
                stx     $fb
                sta     $fc
                sty     $fd
                ldy     #$0
                sty     $cc             ; enable cursor blinking
                sty     $fe             ; temporary for loop variable
getkey:         jsr     $f142           ; get character from keyboard
                beq     getkey
                sta     $2              ; save to temporary
                and     #$7f
                cmp     #$20            ; check for control character
                bcs     checkout        ; no -> check buffer size
                cmp     #$d             ; was it enter/return?
                beq     prepout         ; -> normal flow
                cmp     #$14            ; was it backspace/delete?
                bne     getkey          ; if not, get next char
                lda     $fe             ; check current index
                beq     getkey          ; zero -> backspace not possible
                bne     prepout         ; skip checking buffer size for bs
checkout:       lda     $fe             ; buffer index
                cmp     $fb             ; check against buffer size
                beq     getkey          ; if it would overflow, loop again
prepout:        sei                     ; no interrupts
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and clear 
                and     #$7f            ;   cursor in
                sta     ($d1),y         ;   current row
output:         lda     $2              ; load character
                jsr     $e716           ;   and output
                ldx     $cf             ; check cursor phase
                beq     store           ; invisible -> to store
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and show
                ora     #$80            ;   cursor in
                sta     ($d1),y         ;   current row
                lda     $2              ; load character
store:          cli                     ; enable interrupts
                cmp     #$14            ; was it backspace/delete?
                beq     backspace       ; to backspace handling code
                cmp     #$d             ; was it enter/return?
                beq     done            ; then we're done.
                ldy     $fe             ; load buffer index
                sta     ($fc),y         ; store character in buffer
                iny                     ; advance buffer index
                sty     $fe
                bne     getkey          ; not zero -> ok
done:           lda     #$0             ; terminate string in buffer with zero
                ldy     $fe             ; get buffer index
                sta     ($fc),y         ; store terminator in buffer
                sei                     ; no interrupts
                ldy     $d3             ; get current screen column
                lda     ($d1),y         ; and clear 
                and     #$7f            ;   cursor in
                sta     ($d1),y         ;   current row
                inc     $cc             ; disable cursor blinking
                cli                     ; enable interrupts
                rts                     ; return
backspace:      dec     $fe             ; decrement buffer index
                bcs     getkey          ; and get next key
.endproc

; parse / convert uint8 number using a BCD representation and double-dabble
.proc touint8
                sta     $fb
                sty     $fc
                ldy     #$0
                sty     convbuf
                sty     convbuf+1
                sty     convbuf+2
scanloop:       lda     ($fb),y
                beq     copy
                iny
                cmp     #$20
                beq     scanloop
                cmp     #$30
                bcc     error
                cmp     #$3a
                bcs     error
                bcc     scanloop
error:          sec
                rts
copy:           dey
                bmi     error
                ldx     #$2
copyloop:       lda     ($fb),y
                cmp     #$30
                bcc     copynext
                cmp     #$3a
                bcs     copynext
                sec
                sbc     #$30
                sta     convbuf,x
                dex
copynext:       dey
                bpl     copyloop
                lda     #$0
                sta     $fb
                ldx     #$8
loop:           lsr     convbuf
                lda     convbuf+1
                bcc     skipbit1
                ora     #$10
skipbit1:       lsr     a
                sta     convbuf+1
                lda     convbuf+2
                bcc     skipbit2
                ora     #$10
skipbit2:       lsr     a
                sta     convbuf+2
                ror     $fb
                dex
                beq     done
                lda     convbuf
                cmp     #$8
                bmi     nosub1
                sbc     #$3
                sta     convbuf
nosub1:         lda     convbuf+1
                cmp     #$8
                bmi     nosub2
                sbc     #$3
                sta     convbuf+1
nosub2:         lda     convbuf+2
                cmp     #$8
                bmi     loop
                sbc     #$3
                sta     convbuf+2
                bcs     loop
done:           lda     $fb
                clc
                rts
.endproc

Answer 2

Python 2, 49 bytes

f=lambda n,t=1,p=1:n and p%t*t+f(n-p%t,t+1,p*t*t)

Uses Wilson's theorem, (as introduced to the site by xnor, I believe here)

Try it online!

The function f is recursive, with an initial input of n and a tail when n reaches zero, yielding that zero (due to the logical and); n is decremented whenever t, a test number which increments with every call to f, is prime. The prime test is then whether \$(n-1)!\ \equiv\ -1 \pmod n\$ for which we keep a track of a square of the factorial in p.

Answer 3

Jelly, 4 bytes

My first Jelly program

RÆNS

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Answer 4

05AB1E, 3 bytes

ÅpO

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Explanation:

Åp     # List of the first N primes (N being the implicit input)
       #  i.e. 5 → [2,3,5,7,11]
  O    # Sum of that list
       #  i.e. [2,3,5,7,11] → 28

Answer 5

Java 8, 89 bytes

n->{int r=0,i=2,t,x;for(;n>0;r+=t>1?t+0*n--:0)for(t=i++,x=2;x<t;t=t%x++<1?0:t);return r;}

Try it online.

Explanation:

n->{               // Method with integer as both parameter and return-type
  int r=0,         //  Result-sum, starting at 0
      i=2,         //  Prime-integer, starting at 2
      t,x;         //  Temp integers
  for(;n>0         //  Loop as long as `n` is still larger than 0
      ;            //    After every iteration:
       r+=t>1?     //     If `t` is larger than 1 (which means `t` is a prime):
           t       //      Increase `r` by `t`
           +0*n--  //      And decrease `n` by 1
          :        //     Else:
           0)      //      Both `r` and `n` remain the same
    for(t=i++,     //   Set `t` to the current `i`, and increase `i` by 1 afterwards
        x=2;       //   Set `x` to 2
        x<t;       //   Loop as long as `x` is still smaller than `t`
      t=t%x++<1?   //    If `t` is divisible by `x`:
         0         //     Set `t` to 0
        :          //    Else:
         t);       //     `t` remains the same
                   //   If `t` is still the same after this loop, it means it's a prime
  return r;}       //  Return the result-sum

Answer 6

Brachylog, 8 7 bytes

~lṗᵐ≠≜+

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Saved 1 byte thanks to @sundar.

Explanation

~l        Create a list of length input
  ṗᵐ      Each element of the list must be prime
    ≠     All elements must be distinct
     ≜    Find values that match those constraints
      +   Sum

Answer 7

Perl 6, 31 bytes

{sum grep(&is-prime,2..*)[^$_]}

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The is-prime built-in is unfortunately long.

Answer 8

J, 8 bytes

1#.p:@i.

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Answer 9

Husk, 4 bytes

Σ↑İp

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Generates the İnfinite list of prime numbers, and computes the sum of the first \$N\$ (Σ↑)

Answer 10

Attache, 10 bytes

Sum@Primes

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ho hum

Answer 11

Retina, 41 bytes

K`_
"$+"{`$
$%"_
)/¶(__+)\1+$/+`$
_
^_

_

Try it online! I wanted to keep adding 1 until I had found n primes but I couldn't work out how to do that in Retina so I resorted to a nested loop. Explanation:

K`_

Start with 1.

"$+"{`

Loop n times.

$
$%"_

Make a copy of the previous value, and increment it.

)/¶(__+)\1+$/+`$
_

Keep incrementing it while it's composite. (The ) closes the outer loop.)

^_

Delete the original 1.

_

Sum and convert to decimal.

Answer 12

MATL, 4 bytes

:Yqs

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Explanation:

       % Implicit input: 5
:      % Range: [1, 2, 3, 4, 5]
 Yq    % The n'th primes: [2, 3, 5, 7, 11]
   s   % Sum: 28

Answer 13

PHP, 66 bytes

using my own prime function again ...

for(;$k<$argn;$i-1||$s+=$n+!++$k)for($i=++$n;--$i&&$n%$i;);echo$s;

Run as pipe with -nr or try it online.

breakdown

for(;$k<$argn;      # while counter < argument
    $i-1||              # 3. if divisor is 1 (e.g. $n is prime)
        $s+=$n              # add $n to sum
        +!++$k              # and increment counter
    )
    for($i=++$n;        # 1. increment $n
        --$i&&$n%$i;);  # 2. find largest divisor of $n smaller than $n:
echo$s;             # print sum

Answer 14

Haskell, 48 bytes

sum.(`take`[p|p<-[2..],all((>0).mod p)[2..p-1]])

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Note: \p-> all((>0).mod p)[2..p-1] is not a valid prime check, since it is True for \$0,1\$ as well. But we can work around that by beginning with \$2\$, so in this case it suffices.

Answer 15

C (gcc), 70 bytes

f(n,i,j,s){s=0;for(i=2;n;i++)for(j=2;j/i?s+=i,n--,0:i%j++;);return s;}

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Answer 16

Japt -x, 11 bytes

;@_j}a°X}hA

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Saved several bytes thanks to a new language feature.

Explanation:

;@      }hA    :Get the first n numbers in the sequence:
     a         : Get the smallest number
      °X       : Which is greater than the previous result
  _j}          : And is prime
               :Implicitly output the sum

Answer 17

C, C++, D : 147 142 bytes

int p(int a){if(a<4)return 1;for(int i=2;i<a;++i)if(!(a%i))return 0;return 1;}int f(int n){int c=0,v=1;while(n)if(p(++v)){c+=v;--n;}return c;}

5 bytes optimization for C and C++ :

-2 bytes thanks to Zacharý

#define R return
int p(int a){if(a<4)R 1;for(int i=2;i<a;++i)if(!(a%i))R 0;R 1;}int f(int n){int c=0,v=1;while(n)if(p(++v))c+=v,--n;R c;}

p tests if a number is a prime, f sums the n first numbers

Code used to test :

C/C++ :

for (int i = 0; i < 10; ++i)
    printf("%d => %d\n", i, f(i));

D Optimized answer by Zacharý, 133 131 bytes

D has a golfy template system

T p(T)(T a){if(a<4)return 1;for(T i=2;i<a;)if(!(a%i++))return 0;return 1;}T f(T)(T n){T c,v=1;while(n)if(p(++v))c+=v,--n;return c;}

Answer 18

JavaScript (ES6), 55 bytes

f=(k,n=2)=>k&&(g=d=>n%--d?g(d):d<2&&k--&&n)(n)+f(k,n+1)

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Answer 19

Stax, 6 bytes

ê☺Γ☼èY

Run and debug it

Explanation:

r{|6m|+ Unpacked program, implicit input
r       0-based range
 {  m   Map:
  |6      n-th prime (0-based)
     |+ Sum
        Implicit output

Answer 20

APL (Dyalog Unicode), 7 + 9 = 16 bytes

+/pco∘⍳

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9 additional bytes to import the pco (and every other) Dfn: ⎕CY'dfns'

How:

+/pco∘⍳
      ⍳ ⍝ Generate the range from 1 to the argument
     ∘  ⍝ Compose
  pco   ⍝ P-colon (p:); without a left argument, it generates the first <right_arg> primes.
+/      ⍝ Sum

Answer 21

Ruby, 22 + 7 = 29 bytes

Run with ruby -rprime (+7)

->n{Prime.take(n).sum}

Answer 22

Pari/GP, 20 bytes

n->vecsum(primes(n))

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Answer 23

JAEL, 5 bytes

#&kȦ

Explanation (generated automatically):

./jael --explain '#&kȦ'
ORIGINAL CODE:  #&kȦ

EXPANDING EXPLANATION:
Ȧ => .a!

EXPANDED CODE:  #&k.a!,

#     ,                 repeat (p1) times:
 &                              push number of iterations of this loop
  k                             push nth prime
   .                            push the value under the tape head
    a                           push p1 + p2
     !                          write p1 to the tape head
       ␄                print machine state

Answer 24

Python 2, 63 59 56 51 bytes

f=lambda n:n and prime(n)+f(n-1)
from sympy import*

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---

Saved:

• -5 bytes, thanks to Jonathan Allan

---

Without libs:

Python 2, 83 bytes

n,s=input(),0
x=2
while n:
 if all(x%i for i in range(2,x)):n-=1;s+=x
 x+=1
print s

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Answer 25

Pyke, 4 bytes

~p>s

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~p   -   prime_numbers
  >  -  first_n(^, input)
   s - sum(^)

Answer 26

CJam, 21 bytes

0{{T1+:Tmp!}gT}li*]:+

Explanation:
0{{T1+:Tmp!}gT}li*]:+ Original code

 {            }li*    Repeat n times
  {        }          Block
   T                  Get variable T | stack: T
    1+                Add one | Stack: T+1 
      :T              Store in variable T | Stack: T+1
        mp!           Is T not prime?     | Stack: !(isprime(T))
            g         Do while condition at top of stack is true, pop condition
             T        Push T onto the stack | Stack: Primes so far
0                 ]   Make everything on stack into an array, starting with 0 (to not throw error if n = 0)| Stack: array with 0 and all primes up to n
                   :+ Add everything in array

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Answer 27

F#, 111 bytes

let f n=Seq.initInfinite id|>Seq.filter(fun p->p>1&&Seq.exists(fun x->p%x=0){2..p-1}|>not)|>Seq.take n|>Seq.sum

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Seq.initInfinite creates an infinitely-long sequence with a generator function which takes, as a parameter, the item index. In this case the generator function is just the identity function id.

Seq.filter selects only the numbers created by the infinite sequence that are prime.

Seq.take takes the first n elements in that sequence.

And finally, Seq.sum sums them up.

Answer 28

cQuents, 3 bytes

;pz

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Explanation

;    sum of first n terms for input n
 pz  each term is the next prime after the previous term

Answer 29

MY, 4 bytes

⎕ṀΣ↵

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Still regretting no implicit input/output in this garbage language, would've been two bytes otherwise.

• ⎕ = input
• Ṁ = 1st ... nth prime inclusive
• Σ = sum
• ↵ = output

Answer 30

APL(NARS), 27 chars, 54 bytes

{⍵=0:0⋄+/{⍵=1:2⋄¯2π⍵-1}¨⍳⍵}

{¯2π⍵} here would return the n prime different from 2. So {⍵=1:2⋄¯2π⍵-1} would return the n prime 2 in count in it...

Answer 31

Japt -mx, 8 bytes

T=_j}a°T

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Answer 32

Reticular, 52 40 bytes

indQ2j;o_1-2~d:^=[d@P~1-]:^`*[+]:^`1+*o;

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Explanation

Fun fact: Reticular does not count 2 as a prime number, so the instruction @P which gives the \$n\$-th prime in reality gives the \$(n+1)\$-th prime and due to this we have to add the first prime 2 manually.

in           # Read input and convert to int
dQ2j;o_      # Check if input is 0. If so, output and exit
1-2~d:^=     # Subtract 1 from input and save it as  `^`
[d@P~1-]     # Duplicate the top of the stack (call it k)
               and push the k-th prime. Finally swap the two top items
               in the stack and subtract 1.
               Stack before: [k]
               Stack after: [k-1, k-th prime]
:^`*         # Repeat the above a `^` number of times. 
               Stack before: [n]
               Stack after: [0, 3, 5, ...,  n-th prime, 2]
[+]:^`1+*    # Add the two top items in the stack a
               total of (`^`+1) number of times
o;           # Output the sum and exit.

Answer 33

Pyth, 5 bytes

s.fP_

Try it online here.

s.fP_ZQ   Implicit: Q=eval(input())
          Trailing ZQ inferred
 .f   Q   Starting at Z=1, return the first Q numbers where...
   P_Z    ... Z is prime
s         Sum the resulting list, implicit print