Sum of two adjacent numbers are primes

Sum of two adjacent numbers are primes 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

Show on this page the first 20 numbers and sum of two adjacent numbers which sum is prime.

C

Translation of: Wren

<lang c>#include <stdio.h>

1. define TRUE 1
2. define FALSE 0

int isPrime(int n) {

   if (n < 2) return FALSE;
   if (!(n%2)) return n == 2;
   if (!(n%3)) return n == 3;
   int 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 count = 0, n = 1;
   printf("The first 20 pairs of natural numbers whose sum is prime are:\n");
   while (count < 20) {
       if (isPrime(2*n + 1)) {
           printf("%2d + %2d = %2d\n", n, n + 1, 2*n + 1);
           count++;
       }
       n++;
   }
   return 0;

}</lang>

Output:

The first 20 pairs of natural numbers whose sum is prime are:
 1 +  2 =  3
 2 +  3 =  5
 3 +  4 =  7
 5 +  6 = 11
 6 +  7 = 13
 8 +  9 = 17
 9 + 10 = 19
11 + 12 = 23
14 + 15 = 29
15 + 16 = 31
18 + 19 = 37
20 + 21 = 41
21 + 22 = 43
23 + 24 = 47
26 + 27 = 53
29 + 30 = 59
30 + 31 = 61
33 + 34 = 67
35 + 36 = 71
36 + 37 = 73

Raku

<lang perl6>my @n-n1-triangular = map { $_, $_ + 1, $_ + ($_ + 1) }, ^Inf;

my @wanted = @n-n1-triangular.grep: *.[2].is-prime;

printf "%2d + %2d = %2d\n", |.list for @wanted.head(20);</lang>

Output:

 1 +  2 =  3
 2 +  3 =  5
 3 +  4 =  7
 5 +  6 = 11
 6 +  7 = 13
 8 +  9 = 17
 9 + 10 = 19
11 + 12 = 23
14 + 15 = 29
15 + 16 = 31
18 + 19 = 37
20 + 21 = 41
21 + 22 = 43
23 + 24 = 47
26 + 27 = 53
29 + 30 = 59
30 + 31 = 61
33 + 34 = 67
35 + 36 = 71
36 + 37 = 73

Ring

<lang ring> load "stdlibcore.ring" see "working..." + nl n = 0 num = 0

while true

    n++
    sum = 2*n+1
    if isprime(sum)
       num++
       if num < 21
         ? "n = " + n + "  sum= " + sum
       else
         exit
       ok
     ok

end

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

Output:

working...
n = 1  sum= 3
n = 2  sum= 5
n = 3  sum= 7
n = 5  sum= 11
n = 6  sum= 13
n = 8  sum= 17
n = 9  sum= 19
n = 11  sum= 23
n = 14  sum= 29
n = 15  sum= 31
n = 18  sum= 37
n = 20  sum= 41
n = 21  sum= 43
n = 23  sum= 47
n = 26  sum= 53
n = 29  sum= 59
n = 30  sum= 61
n = 33  sum= 67
n = 35  sum= 71
n = 36  sum= 73
done...

Wren

Library: Wren-math

Library: Wren-fmt

<lang ecmascript>import "./math" for Int import "./fmt" for Fmt

System.print("The first 20 pairs of natural numbers whose sum is prime are:") var count = 0 var n = 1 while (count < 20) {

   if (Int.isPrime(2*n + 1)) {
       Fmt.print("$2d + $2d = $2d", n, n + 1, 2*n + 1)
       count = count + 1
   }
   n = n + 1

}</lang>

Output:

The first 20 pairs of natural numbers whose sum is prime are:
 1 +  2 =  3
 2 +  3 =  5
 3 +  4 =  7
 5 +  6 = 11
 6 +  7 = 13
 8 +  9 = 17
 9 + 10 = 19
11 + 12 = 23
14 + 15 = 29
15 + 16 = 31
18 + 19 = 37
20 + 21 = 41
21 + 22 = 43
23 + 24 = 47
26 + 27 = 53
29 + 30 = 59
30 + 31 = 61
33 + 34 = 67
35 + 36 = 71
36 + 37 = 73

XPL0

Translation of: Ring

<lang XPL0> include xpllib; int N, Num, Sum; [Text(0, "Working...^M^J"); N:= 0; Num:= 0; loop

   [N:= N+1;
   Sum:= 2*N + 1;
   if IsPrime(Sum) then
       [Num:= Num+1;
       if Num < 21 then
         [Text(0,"N = "); IntOut(0,N); Text(0,"  Sum = "); IntOut(0,Sum); CrLf(0)]
       else
         quit
       ]
   ];

Text(0, "Done...^M^J"); ]</lang>

Output:

Working...
N = 1  Sum = 3
N = 2  Sum = 5
N = 3  Sum = 7
N = 5  Sum = 11
N = 6  Sum = 13
N = 8  Sum = 17
N = 9  Sum = 19
N = 11  Sum = 23
N = 14  Sum = 29
N = 15  Sum = 31
N = 18  Sum = 37
N = 20  Sum = 41
N = 21  Sum = 43
N = 23  Sum = 47
N = 26  Sum = 53
N = 29  Sum = 59
N = 30  Sum = 61
N = 33  Sum = 67
N = 35  Sum = 71
N = 36  Sum = 73
Done...