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.)


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\$.


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

33 Answers 33


Japt -mx, 8 bytes


Try it


Pyth, 5 bytes


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

Reticular, 52 40 bytes


Try it online!


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.

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