Find prime numbers of the form n*n*n+2

Find prime numbers of the form n*n*n+2 is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Task

Find prime numbers of form   n³+2,   where 0 < n < 200

C

Translation of: Wren

<lang c>#include <stdio.h>

1. include <stdbool.h>
2. include <locale.h>

bool isPrime(int n) {

   int d;
   if (n < 2)  return false;
   if (!(n%2)) return n == 2;
   if (!(n%3)) return n == 3;
   d = 5;
   while (d*d <= n) {
       if (!(n%d)) return false;
       d += 2;
       if (!(n%d)) return false;
       d += 4;
   }
   return true;

}

int main() {

   int n, p;
   const int limit = 200;
   setlocale(LC_ALL, "");
   for (n = 1; n < limit; ++n) {
       p = n*n*n + 2;
       if (isPrime(p)) {
           printf("n = %3d => n³ + 2 = %'9d\n", n, p);
       }
   }
   return 0;

}</lang>

Output:

n =   1 => n³ + 2 =         3
n =   3 => n³ + 2 =        29
n =   5 => n³ + 2 =       127
n =  29 => n³ + 2 =    24,391
n =  45 => n³ + 2 =    91,127
n =  63 => n³ + 2 =   250,049
n =  65 => n³ + 2 =   274,627
n =  69 => n³ + 2 =   328,511
n =  71 => n³ + 2 =   357,913
n =  83 => n³ + 2 =   571,789
n = 105 => n³ + 2 = 1,157,627
n = 113 => n³ + 2 = 1,442,899
n = 123 => n³ + 2 = 1,860,869
n = 129 => n³ + 2 = 2,146,691
n = 143 => n³ + 2 = 2,924,209
n = 153 => n³ + 2 = 3,581,579
n = 171 => n³ + 2 = 5,000,213
n = 173 => n³ + 2 = 5,177,719
n = 189 => n³ + 2 = 6,751,271

Factor

Using the parity optimization from the Wren entry:

Works with: Factor version 0.99 2021-02-05

<lang factor>USING: formatting kernel math math.functions math.primes math.ranges sequences tools.memory.private ;

1 199 2 <range> [

   dup 3 ^ 2 + dup prime?
   [ commas "n = %3d => n³ + 2 = %9s\n" printf ] [ 2drop ] if

] each</lang> Or, using local variables:

Translation of: Wren

Works with: Factor version 0.99 2021-02-05

<lang factor>USING: formatting kernel math math.primes math.ranges sequences tools.memory.private ;

[let

   199 :> limit
   1 limit 2 <range> [| n |
       n n n * * 2 + :> p
       p prime?
       [ n p commas "n = %3d => n³ + 2 = %9s\n" printf ] when
   ] each

]</lang>

Output:

n =   1 => n³ + 2 =         3
n =   3 => n³ + 2 =        29
n =   5 => n³ + 2 =       127
n =  29 => n³ + 2 =    24,391
n =  45 => n³ + 2 =    91,127
n =  63 => n³ + 2 =   250,049
n =  65 => n³ + 2 =   274,627
n =  69 => n³ + 2 =   328,511
n =  71 => n³ + 2 =   357,913
n =  83 => n³ + 2 =   571,789
n = 105 => n³ + 2 = 1,157,627
n = 113 => n³ + 2 = 1,442,899
n = 123 => n³ + 2 = 1,860,869
n = 129 => n³ + 2 = 2,146,691
n = 143 => n³ + 2 = 2,924,209
n = 153 => n³ + 2 = 3,581,579
n = 171 => n³ + 2 = 5,000,213
n = 173 => n³ + 2 = 5,177,719
n = 189 => n³ + 2 = 6,751,271

Go

<lang go>package main

import "fmt"

func isPrime(n int) bool {

   switch {
   case n < 2:
       return false
   case n%2 == 0:
       return n == 2
   case n%3 == 0:
       return n == 3
   default:
       d := 5
       for d*d <= n {
           if n%d == 0 {
               return false
           }
           d += 2
           if n%d == 0 {
               return false
           }
           d += 4
       }
       return true
   }

}

func commatize(n int) string {

   s := fmt.Sprintf("%d", n)
   if n < 0 {
       s = s[1:]
   }
   le := len(s)
   for i := le - 3; i >= 1; i -= 3 {
       s = s[0:i] + "," + s[i:]
   }
   if n >= 0 {
       return s
   }
   return "-" + s

}

func main() {

   const limit = 200
   for n := 1; n < limit; n++ {
       p := n*n*n + 2
       if isPrime(p) {
           fmt.Printf("n = %3d => n³ + 2 = %9s\n", n, commatize(p))
       }
   }

}</lang>

Output:

n =   1 => n³ + 2 =         3
n =   3 => n³ + 2 =        29
n =   5 => n³ + 2 =       127
n =  29 => n³ + 2 =    24,391
n =  45 => n³ + 2 =    91,127
n =  63 => n³ + 2 =   250,049
n =  65 => n³ + 2 =   274,627
n =  69 => n³ + 2 =   328,511
n =  71 => n³ + 2 =   357,913
n =  83 => n³ + 2 =   571,789
n = 105 => n³ + 2 = 1,157,627
n = 113 => n³ + 2 = 1,442,899
n = 123 => n³ + 2 = 1,860,869
n = 129 => n³ + 2 = 2,146,691
n = 143 => n³ + 2 = 2,924,209
n = 153 => n³ + 2 = 3,581,579
n = 171 => n³ + 2 = 5,000,213
n = 173 => n³ + 2 = 5,177,719
n = 189 => n³ + 2 = 6,751,271

Ring

<lang ring> load "stdlib.ring"

see "working..." + nl

for n = 1 to 200 step 2

   pr = pow(n,3)+2
   if isprime(pr)
      see "n = " + n + " => n³+2 = " + pr + nl
   ok

next

see "done..." + nl </lang>

Output:

working...
n = 1 => n³+2 = 3
n = 3 => n³+2 = 29
n = 5 => n³+2 = 127
n = 29 => n³+2 = 24391
n = 45 => n³+2 = 91127
n = 63 => n³+2 = 250049
n = 65 => n³+2 = 274627
n = 69 => n³+2 = 328511
n = 71 => n³+2 = 357913
n = 83 => n³+2 = 571789
n = 105 => n³+2 = 1157627
n = 113 => n³+2 = 1442899
n = 123 => n³+2 = 1860869
n = 129 => n³+2 = 2146691
n = 143 => n³+2 = 2924209
n = 153 => n³+2 = 3581579
n = 171 => n³+2 = 5000213
n = 173 => n³+2 = 5177719
n = 189 => n³+2 = 6751271
done...

Wren

Library: Wren-math

Library: Wren-trait

Library: Wren-fmt

If n is even then n³ + 2 is also even, so we only need to examine odd values of n here. <lang ecmascript>import "/math" for Int import "/trait" for Stepped import "/fmt" for Fmt

var limit = 200 for (n in Stepped.new(1...limit, 2)) {

   var p = n*n*n + 2
   if (Int.isPrime(p)) Fmt.print("n = $3d => n³ + 2 = $,9d", n, p)

}</lang>

Output:

n =   1 => n³ + 2 =         3
n =   3 => n³ + 2 =        29
n =   5 => n³ + 2 =       127
n =  29 => n³ + 2 =    24,391
n =  45 => n³ + 2 =    91,127
n =  63 => n³ + 2 =   250,049
n =  65 => n³ + 2 =   274,627
n =  69 => n³ + 2 =   328,511
n =  71 => n³ + 2 =   357,913
n =  83 => n³ + 2 =   571,789
n = 105 => n³ + 2 = 1,157,627
n = 113 => n³ + 2 = 1,442,899
n = 123 => n³ + 2 = 1,860,869
n = 129 => n³ + 2 = 2,146,691
n = 143 => n³ + 2 = 2,924,209
n = 153 => n³ + 2 = 3,581,579
n = 171 => n³ + 2 = 5,000,213
n = 173 => n³ + 2 = 5,177,719
n = 189 => n³ + 2 = 6,751,271