Find prime n such that reversed n is also prime: Difference between revisions

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Task

Find prime n for 0 < n < 500 which are also primes when the (decimal) digits are reversed.
Nearly Emirp Primes, which don't count palindromatic primes like 101.

11l

F is_prime(a)
   I a == 2
      R 1B
   I a < 2 | a % 2 == 0
      R 0B
   L(i) (3 .. Int(sqrt(a))).step(2)
      I a % i == 0
         R 0B
   R 1B

L(n) 1..499
   I is_prime(n) & is_prime(Int(reversed(String(n))))
      print(n, end' ‘ ’)

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Action!

Library: Action! Sieve of Eratosthenes

INCLUDE "H6:SIEVE.ACT"

BYTE FUNC IsPrimeAndItsReverse(INT i BYTE ARRAY primes)
  INT rev
  BYTE d

  IF primes(i)=0 THEN
    RETURN (0)
  FI
  rev=0
  WHILE i#0
  DO
    d=i MOD 10
    rev==*10
    rev==+d
    i==/10
  OD
RETURN (primes(rev))

PROC Main()
  DEFINE MAX="999"
  BYTE ARRAY primes(MAX+1)
  INT i,count=[0]

  Put(125) PutE() ;clear the screen
  Sieve(primes,MAX+1)
  FOR i=2 TO 499
  DO
    IF IsPrimeAndItsReverse(i,primes) THEN
      PrintI(i) Put(32)
      count==+1
    FI
  OD
  PrintF("%E%EThere are %I primes",count)
RETURN

Output:

Screenshot from Atari 8-bit computer

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

There are 34 primes

Ada

with Ada.Text_Io;

procedure Reverse_Prime is

   type Number is new Long_Integer range 0 .. Long_Integer'Last;
   package Number_Io is new Ada.Text_Io.Integer_Io (Number);

   function Is_Prime (A : Number) return Boolean is
      D : Number;
   begin
      if A < 2       then return False; end if;
      if A in 2 .. 3 then return True;  end if;
      if A mod 2 = 0 then return False; end if;
      if A mod 3 = 0 then return False; end if;
      D := 5;
      while D * D <= A loop
         if A mod D = 0 then
            return False;
         end if;
         D := D + 2;
         if A mod D = 0 then
            return False;
         end if;
         D := D + 4;
      end loop;
      return True;
   end Is_Prime;

   function Reverse_Num (N : Number) return Number is
      N2  : Number := N;
      Res : Number := 0;
   begin
      while N2 /= 0 loop
         Res := 10 * Res;
         Res := Res  + (N2 mod 10);
         N2  := N2 / 10;
      end loop;
      return Res;
   end Reverse_Num;

   use Ada.Text_Io;
   Count : Natural := 0;
begin
   for N in Number range 1 .. 499 loop
      if Is_Prime (N) and then Is_Prime (Reverse_Num (N)) then
         Count := Count + 1;
         Number_Io.Put (N, Width => 3); Put ("  ");
         if Count mod 8 = 0 then
            New_Line;
         end if;
      end if;
   end loop;
   New_Line;
   Put_Line (Count'Image & " primes.");
end Reverse_Prime;

Output:

  2    3    5    7   11   13   17   31
 37   71   73   79   97  101  107  113
131  149  151  157  167  179  181  191
199  311  313  337  347  353  359  373
383  389
 34 primes.

ALGOL 68

BEGIN # find primes whose reversed digits are also prime         #
    INT max number = 500;   # largest prime we will reverse      #
    INT max prime  = 1 000; # enough primes to handle reversing  #
                            # max number                         #
    [ 1 : max prime ]BOOL p; FOR n TO UPB p DO p[ n ] := ODD n OD;
    FOR i FROM 3 BY 2 TO ENTIER sqrt( UPB p ) DO
        IF p[ i ] THEN
            FOR s FROM i * i BY i + i TO UPB p DO p[ s ] := FALSE OD
        FI
    OD;
    p[ 1 ] := FALSE; p[ 2 ] := TRUE;
    OP FMT = ( INT n )STRING: whole( n, 0 );
    INT count := 0;
    FOR n TO max number DO
        IF p[ n ] THEN
            INT r := n MOD 10;
            INT v := n;
            WHILE ( v OVERAB 10 ) > 0 DO
                r *:= 10 +:= v MOD 10
            OD;
            IF p[ r ] THEN 
                print( ( " ", whole( n, -3 ) ) );
                IF ( count +:= 1 ) MOD 10 = 0 THEN print( ( newline ) ) FI
            FI
        FI
    OD;
    print( ( newline, "Found ", whole( count, 0 )
           , " reversable primes up to ", whole( max number, 0 )
           )
         )
END

Output:

   2   3   5   7  11  13  17  31  37  71
  73  79  97 101 107 113 131 149 151 157
 167 179 181 191 199 311 313 337 347 353
 359 373 383 389
Found 34 reversable primes up to 500

ALGOL W

begin % find some primes whose digits reversed is also prime %
    % sets p( 1 :: n ) to a sieve of primes up to n %
    procedure Eratosthenes ( logical array p( * ) ; integer value n ) ;
    begin
        p( 1 ) := false; p( 2 ) := true;
        for i := 3 step 2 until n do p( i ) := true;
        for i := 4 step 2 until n do p( i ) := false;
        for i := 3 step 2 until truncate( sqrt( n ) ) do begin
            integer ii; ii := i + i;
            if p( i ) then for pr := i * i step ii until n do p( pr ) := false
        end for_i ;
    end Eratosthenes ;
    integer MAX_NUMBER, maxPrime;
    MAX_NUMBER := 500;
    % approximate the largest prime we need to consider ( 10 ^ number of digits in MAX_NUMBER ) %
    begin
        integer v;
        v        := MAX_NUMBER;
        maxPrime := 1;
        while v > 0 do begin
            v := v div 10;
            maxPrime := maxPrime * 10
        end while_v_gt_0
    end;
    begin
        logical array prime( 1 :: maxPrime);
        integer       pCount;
        % sieve the primes to maxPrime %
        Eratosthenes( prime, maxPrime );
        % find the primes that are reversable %
        pCount := 0;
        for i := 1 until MAX_NUMBER - 1 do begin
            if prime( i ) then begin
                integer pReversed, v;
                v         := i;
                pReversed := 0;
                while v > 0 do begin
                    pReversed := ( pReversed * 10 ) + v rem 10;
                    v         := v div 10
                end while_v_gt_0 ;
                if prime( pReversed ) then begin
                    writeon( i_w := 4, s_w := 0, " ", i );
                    pCount := pCount + 1;
                    if pCount rem 20 = 0 then write()
                end if_prime_pReversed
            end if_prime_i
        end for_i ;
        write( i_w := 1, s_w := 0, "Found ", pCount, " reversable primes below ", MAX_NUMBER )
    end
end.

Output:

    2    3    5    7   11   13   17   31   37   71   73   79   97  101  107  113  131  149  151  157
  167  179  181  191  199  311  313  337  347  353  359  373  383  389
Found 34 reversable primes below 500

Arturo

print 
    select 1..499 'x ->
        and? [prime? x][prime? to :integer reverse to :string x]

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

AWK

# syntax: GAWK -f FIND_PRIME_N_FOR_THAT_REVERSED_N_IS_ALSO_PRIME.AWK
BEGIN {
    start = 1
    stop = 500
    for (i=start; i<=stop; i++) {
      if (is_prime(i) && is_prime(revstr(i,length(i)))) {
        printf("%3d%1s",i,++count%10?"":"\n")
      }
    }
    printf("\nReversible primes %d-%d: %d\n",start,stop,count)
    exit(0)
}
function is_prime(x,  i) {
    if (x <= 1) {
      return(0)
    }
    for (i=2; i<=int(sqrt(x)); i++) {
      if (x % i == 0) {
        return(0)
      }
    }
    return(1)
}
function revstr(str,start) {
    if (start == 0) {
      return("")
    }
    return( substr(str,start,1) revstr(str,start-1) )
}

Output:

  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389
Reversible primes 1-500: 34

BASIC

10 DEFINT A-Z: MP=999: MX=500
15 MP=10^FIX(LOG(MX)/LOG(10)+1)
20 DIM C(MP): C(0)=-1: C(1)=-1
30 FOR P=2 TO SQR(MP)
40 FOR C=P+P TO MP STEP P: C(C)=-1: NEXT
50 NEXT
60 FOR N=1 TO MX: IF C(N) THEN 100
70 R=0: V=N
80 IF V>0 THEN R=10*R+V MOD 10: V=V\10: GOTO 80
90 IF NOT C(R) THEN PRINT N,
100 NEXT

Output:

 2             3             5             7             11
 13            17            31            37            71
 73            79            97            101           107
 113           131           149           151           157
 167           179           181           191           199
 311           313           337           347           353
 359           373           383           389

bc

define t(n) {
    auto i, p
    if (n == 2) return(1)
    for (i = 0; n % (p = a[i]) != 0; ++i) if (p * p > n) return(1)
    return(0)
}

define r(n) {
    auto m
    for (m = 0; n > 9; n /= 10) m = (n % 10 + m) * 10
    return(m + n)
}

for (n = 2; n != 500; ++n) if (t(n) != 0) a[o++] = n

for (i = 0; i != o; ++i) if (t(r(n = a[i])) != 0) n

BCPL

get "libhdr"

let sieve(prime, n) be
$(  0!prime := false
    1!prime := false
    for i = 2 to n do i!prime := true
    for p = 2 to n/2 if p!prime
    $(  let c = p*2
        while c <= n
        $(  c!prime := false
            c := c+p
        $)
    $)
$)

let reverse(n) = valof
$(  let r=0
    while n>0
    $(  r := 10*r + n rem 10
        n := n/10
    $)
    resultis r
$)

let start() be
$(  let prime = vec 999
    sieve(prime, 999)
    for i = 2 to 500 
        if i!prime & reverse(i)!prime do
            writef("%N ",i)
    wrch('*N')
$)

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

C

#include <stdbool.h>
#include <stdio.h>

bool is_prime(unsigned int n) {
    if (n < 2)
        return false;
    if (n % 2 == 0)
        return n == 2;
    if (n % 3 == 0)
        return n == 3;
    for (unsigned int p = 5; p * p <= n; p += 4) {
        if (n % p == 0)
            return false;
        p += 2;
        if (n % p == 0)
            return false;
    }
    return true;
}

unsigned int reverse(unsigned int n) {
    unsigned int rev = 0;
    for (; n > 0; n /= 10)
        rev = rev * 10 + n % 10;
    return rev;
}

int main() {
    unsigned int count = 0;
    for (unsigned int n = 1; n < 500; ++n) {
        if (is_prime(n) && is_prime(reverse(n)))
            printf("%3u%c", n, ++count % 10 == 0 ? '\n' : ' ');
    }
    printf("\nCount = %u\n", count);
    return 0;
}

Output:

  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389 
Count = 34

CLU

sieve = proc (max: int) returns (array[bool])
    prime: array[bool] := array[bool]$fill(0,max+1,true)
    prime[0] := false
    prime[1] := false
    for p: int in int$from_to(2,max/2) do
        if ~prime[p] then continue end
        for c: int in int$from_to_by(p*p,max,p) do
            prime[c] := false
        end
    end
    return(prime)
end sieve

reverse = proc (n: int) returns (int)
    r: int := 0
    while n>0 do
        r := 10*r + n//10
        n := n/10
    end
    return(r)
end reverse

start_up = proc ()
    po: stream := stream$primary_output()
    prime: array[bool] := sieve(999)
    
    for i: int in int$from_to(2, 500) do
        if prime[i] cand prime[reverse(i)] then
            stream$puts(po, int$unparse(i) || " ")
        end
    end
end start_up

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Cowgol

include "cowgol.coh";

const PRIME_LIMIT := 999;
const LIMIT := 500;

var prime: uint8[PRIME_LIMIT + 1];
typedef P is @indexof prime;
sub sieve() is
    prime[0] := 0;
    prime[1] := 0;
    MemSet(&prime[2], 0xFF, @bytesof prime-2);
    var p: P := 2;
    while p*p <= PRIME_LIMIT loop
        if prime[p] != 0 then 
            var c := p*p;
            while c <= PRIME_LIMIT loop
                prime[c] := 0;
                c := c + p;
            end loop;
        end if;
        p := p + 1;
    end loop;
end sub;

sub reverse(n: P): (r: P) is
    r := 0;
    while n>0 loop
        r := 10*r + n%10;
        n := n/10;
    end loop;
end sub;

sieve();
var n: P := 1;
while n <= LIMIT loop
    if prime[n] != 0 and prime[reverse(n)] != 0 then
        print_i32(n as uint32);
        print_char(' ');
    end if;
    n := n + 1;
end loop;
print_nl();

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Delphi

Library: System.SysUtils

Library: PrimTrial

Translation of: Ring

program Find_prime_n_for_that_reversed_n_is_also_prime;

{$APPTYPE CONSOLE}

uses
  System.SysUtils,
  PrimTrial;

function Reverse(s: string): string;
var
  i: Integer;
begin
  Result := '';
  if Length(s) < 2 then
    exit(s);
  for i := Length(s) downto 1 do
    Result := Result + s[i];
end;

begin
  writeln('working...'#10);
  var row := 0;
  var count := 0;
  var limit := 500;

  for var n := 1 to limit - 1 do
  begin
    if not isPrime(n) then
      Continue;

    var val := n.ToString;
    var valr := reverse(val);
    var nr := valr.ToInteger;

    if not isPrime(nr) then
      Continue;

    write(n: 4);

    inc(row);
    inc(count);
    if row mod 10 = 0 then
      writeln;
  end;
  writeln(#10#10, 'found ', count, ' primes');
  Writeln('done...');
  readln;
end.

Output:

working...

   2   3   5   7  11  13  17  31  37  71
  73  79  97 101 107 113 131 149 151 157
 167 179 181 191 199 311 313 337 347 353
 359 373 383 389

found 34 primes
done...

EasyLang

fastfunc reverse n .
   while n > 0
      r = r * 10 + n mod 10
      n = n div 10
   .
   return r
.
fastfunc isprim num .
   i = 2
   while i <= sqrt num
      if num mod i = 0
         return 0
      .
      i += 1
   .
   return 1
.
fastfunc nextprim prim .
   repeat
      prim += 1
      until isprim prim = 1
   .
   return prim
.
prim = 2
while prim < 500
   if isprim reverse prim = 1
      write prim & " "
   .
   prim = nextprim prim
.

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

F#

This task uses Extensible Prime Generator (F#)

// Reversible Primes. Nigel Galloway: March 22nd., 2021
let emirp2=let rec fN g=function |0->g |n->fN(g*10+n%10)(n/10) in primes32()|>Seq.filter(fN 0>>isPrime)
emirp2|>Seq.takeWhile((>)500)|>Seq.iter(printf "%d "); printfn ""

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Factor

Works with: Factor version 0.99 2021-02-05

USING: formatting grouping io kernel math math.primes sequences ;

: reverse-digits ( 123 -- 321 )
    0 swap [ 10 /mod rot 10 * + swap ] until-zero ;

499 primes-upto [ reverse-digits prime? ] filter
dup length "Found %d reverse primes < 500.\n\n" printf
10 group [ [ "%4d" printf ] each nl ] each nl

Output:

Found 34 reverse primes < 500.

   2   3   5   7  11  13  17  31  37  71
  73  79  97 101 107 113 131 149 151 157
 167 179 181 191 199 311 313 337 347 353
 359 373 383 389

Forth

Works with: Gforth

: prime? ( n -- ? ) here + c@ 0= ;
: not-prime! ( n -- ) here + 1 swap c! ;

: prime-sieve ( n -- )
  here over erase
  0 not-prime!
  1 not-prime!
  2
  begin
    2dup dup * >
  while
    dup prime? if
      2dup dup * do
        i not-prime!
      dup +loop
    then
    1+
  repeat
  2drop ;

: reverse ( n -- n )
  0 swap
  begin
    dup 0 >
  while
    10 /mod swap rot 10 * + swap    
  repeat drop ;

: main
  1000 prime-sieve
  0
  500 1 do
    i prime? if i reverse prime? if
      1 +
      i 3 .r
      dup 10 mod 0= if cr else space then
    then then
  loop
  cr ." Count: " . cr ;

main
bye

Output:

  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389 
Count: 34

FreeBASIC

Use one of the primality testing algorithms as an include as I can't be bothered putting these in all the time.

#include "isprime.bas"

function isbackprime( byval n as integer ) as boolean
    if not isprime(n) then return false
    dim as integer m = 0
    while n
        m *= 10
        m += n mod 10
        n \= 10
    wend
    return isprime(m)
end function

for n as uinteger = 2 to 499
    if isbackprime(n) then print n;" ";
next n
print

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Frink

select[primes[2,500], {|n| isPrime[parseInt[join["", reverse[integerDigits[n]]]]]}]

Output:

[2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389]

Go

package main

import "fmt"

func sieve(limit int) []bool {
    limit++
    // True denotes composite, false denotes prime.
    c := make([]bool, limit) // all false by default
    c[0] = true
    c[1] = true
    for i := 4; i < limit; i += 2 {
        c[i] = true
    }
    p := 3 // Start from 3.
    for {
        p2 := p * p
        if p2 >= limit {
            break
        }
        for i := p2; i < limit; i += 2 * p {
            c[i] = true
        }
        for {
            p += 2
            if !c[p] {
                break
            }
        }
    }
    return c
}

func reversed(n int) int {
    rev := 0
    for n > 0 {
        rev = rev*10 + n%10
        n /= 10
    }
    return rev
}

func main() {
    c := sieve(999)
    reversedPrimes := []int{2}
    for i := 3; i < 500; i += 2 {
        if !c[i] && !c[reversed(i)] {
            reversedPrimes = append(reversedPrimes, i)
        }
    }
    fmt.Println("Primes under 500 which are also primes when the digits are reversed:")
    for i, p := range reversedPrimes {
        fmt.Printf("%5d", p)
        if (i+1) % 10 == 0 {
            fmt.Println()
        }
    }
    fmt.Printf("\n\n%d such primes found.\n", len(reversedPrimes))
}

Output:

Primes under 500 which are also primes when the digits are reversed:
    2    3    5    7   11   13   17   31   37   71
   73   79   97  101  107  113  131  149  151  157
  167  179  181  191  199  311  313  337  347  353
  359  373  383  389

34 such primes found.

Haskell

import Data.List (intercalate, transpose)
import Data.List.Split (chunksOf)
import Data.Numbers.Primes (isPrime, primes)
import Text.Printf (printf)

------------------------ PREDICATE -----------------------

p :: Int -> Bool
p n = isPrime (read (reverse $ show n) :: Int)

--------------------------- TEST -------------------------
main :: IO ()
main =
  mapM_
    putStrLn
    [ "Reversible primes below 500:",
      (table " " . chunksOf 10 . fmap show) $
        takeWhile (< 500) (filter p primes)
    ]

------------------------ FORMATTING ----------------------

table :: String -> [[String]] -> String
table gap rows =
  let widths =
        maximum . fmap length
          <$> transpose rows
   in unlines $
        fmap
          ( intercalate gap
              . zipWith
                ( printf
                    . flip intercalate ["%", "s"]
                    . show
                )
                widths
          )
          rows

Output:

Reversible primes below 500:
  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389

J

   (#~ 1 p: |.&.":"0) i.&.(p:inv) 500
2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

jq

Works with: jq

Works with gojq, the Go implementation of jq

Using the definition of is_prime at Erdős-primes#jq:

# Generate a stream of reversible primes.
# If . is null the stream is unbounded;
# otherwise only integers less than . are considered.

def reversible_primes:
  def r: tostring|explode|reverse|implode|tonumber;
  (if . == null then infinite else . end) as $n
  | 2, (range(3; $n; 2) | select(is_prime and (r|is_prime)));

"Primes under 500 which are also primes when the digits are reversed:",
 (500 | reversible_primes)

Output:

Primes under 500 which are also primes when the digits are reversed:
2
3
5
7
11
13
17
31
37
71
73
79
97
101
107
113
131
149
151
157
167
179
181
191
199
311
313
337
347
353
359
373
383
389

Julia

using Primes

let
    pmask, pcount = primesmask(1, 994), 0
    isreversibleprime(n) = pmask[n] && pmask[evalpoly(10, reverse(digits(n)))]

    println("Reversible primes between 0 and 500:")
    for n in 1:499
        if isreversibleprime(n)
            pcount += 1
            print(rpad(n, 4), pcount % 17 == 0 ? "\n" : "")
        end
    end
    println("Total found: $pcount")
end

Output:

Reversible primes between 0 and 500:
2   3   5   7   11  13  17  31  37  71  73  79  97  101 107 113 131 
149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389
Total found: 34

Mathematica/Wolfram Language

Select[Range[499], PrimeQ[#] \[And] PrimeQ[IntegerReverse[#]] &]

Output:

{2,3,5,7,11,13,17,31,37,71,73,79,97,101,107,113,131,149,151,157,167,179,181,191,199,311,313,337,347,353,359,373,383,389}

Lua

do  -- find primes whose reversed digits are also prime
    local function isPrime( p )
        if p <= 1 or p % 2 == 0 then
            return p == 2
        else
            local prime = true
            local i     = 3
            local rootP = math.floor( math.sqrt( p ) )
            while i <= rootP and prime do
                prime = p % i ~= 0
                i     = i + 2
            end
            return prime
        end
    end
    local function reverseDigits( n )
        return tonumber( string.reverse( tostring( n ) ) )
    end
    local count = 0
    for n = 1,500 do
        if isPrime( n ) then
            if isPrime( reverseDigits( n ) ) then
                count = count + 1
                io.write( string.format( "%3d", n ), count % 10 == 0 and "\n" or " " )
            end
        end
    end
    io.write( "\nFound ", count, " reversable primes up to 500" )
end

Output:

  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389
Found 34 reversable primes up to 500

MAD

            NORMAL MODE IS INTEGER
            BOOLEAN PRIME
            DIMENSION PRIME(1000)
            VECTOR VALUES FMT = $I3*$
            
            PRIME(0)=0B
            PRIME(1)=0B
            THROUGH INIT, FOR P=2, 1, P.GE.1000
INIT        PRIME(P)=1B

            THROUGH SIEVE, FOR P=2, 1, P.GE.32
            THROUGH SIEVE, FOR C=P*P, P, C.GE.1000
SIEVE       PRIME(C)=0B
            
            THROUGH TEST, FOR N=2, 1, N.GE.500
            WHENEVER .NOT. PRIME(N), TRANSFER TO TEST
            NN=N
            R=0
RVRSE       WHENEVER NN.G.0
                NXT=NN/10
                R=R*10+NN-NXT*10
                NN=NXT
                TRANSFER TO RVRSE
            END OF CONDITIONAL
            WHENEVER .NOT. PRIME(R), TRANSFER TO TEST
            PRINT FORMAT FMT,N
TEST        CONTINUE
            END OF PROGRAM

Output:

Modula-2

MODULE ReversePrime;
FROM InOut IMPORT WriteCard, WriteLn;

CONST 
    Primes = 1000;
    Max = 500;  

VAR prime: ARRAY [1..Primes] OF BOOLEAN;
    n, col: CARDINAL;

PROCEDURE reverse(n: CARDINAL): CARDINAL;
VAR r: CARDINAL;
BEGIN
    r := 0;
    WHILE n > 0 DO
        r := r*10 + n MOD 10;
        n := n DIV 10;
    END;
    RETURN r;
END reverse;

PROCEDURE Sieve;
VAR i, j: CARDINAL;
BEGIN
    prime[1] := FALSE;
    FOR i := 2 TO Primes DO
        prime[i] := TRUE;
    END;
    FOR i := 2 TO Primes DIV 2 DO
        j := i*2;
        WHILE j <= Primes DO
            prime[j] := FALSE;
            j := j + i;
        END;
    END;
END Sieve;

BEGIN
    Sieve();
    col := 0;
    FOR n := 2 TO Max DO
        IF prime[n] AND prime[reverse(n)] THEN
            WriteCard(n,5);
            col := col + 1;
            IF col MOD 8 = 0 THEN
                WriteLn();
            END;
        END;
    END;
    WriteLn();
END ReversePrime.

Output:

    2    3    5    7   11   13   17   31
   37   71   73   79   97  101  107  113
  131  149  151  157  167  179  181  191
  199  311  313  337  347  353  359  373
  383  389

Nim

import math, strutils

const
  N1 = 499    # Limit for the primes.
  N2 = 999    # Limit for the reverses of primes.

# Sieve of Erathosthenes.
var composite: array[2..N2, bool]     # Default is false.
for p in 2..sqrt(N2.toFloat).int:
  if not composite[p]:
    for k in countup(p * p, N2, p):
      composite[k] = true

template isPrime(n: int): bool = not composite[n]

func reversed(n: int): int =
  var n = n
  while n != 0:
    result = 10 * result + n mod 10
    n = n div 10

var result: seq[int]
for n in 2..N1:
  if n.isPrime and reversed(n).isPrime:
    result.add n

for i, n in result:
  stdout.write ($n).align(3)
  stdout.write if (i + 1) mod 10 == 0: '\n' else: ' '
echo()

Output:

  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389

OCaml

Using the function seq_primes from Extensible prime generator#OCaml:

let int_reverse =
  let rec loop m n =
    if n < 10 then m + n else loop ((m + n mod 10) * 10) (n / 10)
  in loop 0

let is_prime n =
  let not_divisible x = n mod x <> 0 in
  seq_primes |> Seq.take_while (fun x -> x * x <= n) |> Seq.for_all not_divisible

let () =
  seq_primes |> Seq.filter (fun p -> is_prime (int_reverse p))
  |> Seq.take_while ((>) 500) |> Seq.iter (Printf.printf " %u") |> print_newline

Output:

 2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Pascal

Free Pascal

Type Tboolarr = array Of boolean;

Const MAX = 1000;
Function SieveOfEratosthenes(limit: Integer): Tboolarr;
Var 
  sieve: Tboolarr;
  i, j: Integer;
Begin
  SetLength(sieve, limit + 1);
  sieve[2] := True;
  For i := 3 To limit Do
    sieve[i] := (i Mod 2 <> 0);
  For i := 3 To Trunc(Sqrt(limit)) Do
    Begin
      If sieve[i] Then
        Begin
          j := i * i;
          While j <= limit Do
            Begin
              sieve[j] := False;
              Inc(j, 2 * i);
            End;
        End;
    End;
  SieveOfEratosthenes := sieve;
End;

Function ReverseNumber(number: Integer): Integer;
Var 
  reversed: Integer;
Begin
  reversed := 0;
  While number <> 0 Do
    Begin
      reversed := reversed * 10 + (number Mod 10);
      number := number Div 10;
    End;
  ReverseNumber := reversed;
End;

Var 
  primes: Tboolarr;
  i: Integer;
Begin
  primes := SieveOfEratosthenes(MAX);
  For i := 2 To 500 Do
    Begin
      If primes[i] And Primes[ReverseNumber(i)] Then
        Write(i,' ');
    End;
End.

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Perl

Library: ntheory

use strict;
use warnings;
use List::Util 'max';
use ntheory 'is_prime';

sub pp {
    my $format = ('%' . (my $cw = 1+length max @_) . 'd') x @_;
    my $width  = ".{@{[$cw * int 60/$cw]}}";
    (sprintf($format, @_)) =~ s/($width)/$1\n/gr;
}

my($limit, @rp) = 500;
is_prime($_) and is_prime(reverse $_) and push @rp, $_ for 1..$limit;
print @rp . " reversible primes < $limit:\n" . pp(@rp);

Output:

34 reversible primes < 500:
   2   3   5   7  11  13  17  31  37  71  73  79  97 101 107
 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353
 359 373 383 389

Phix

function rp(integer p) return is_prime(to_integer(reverse(sprint(p)))) end function
procedure test(sequence args)
    {integer n, string fmt} = args
    sequence res = apply(true,sprintf,{{"%3d"},filter(get_primes_le(n),rp)})
    string r = sprintf(fmt,{join_by(res,1,ceil(length(res)/2)," ")})
    printf(1,"%,d reverse primes < %,d found%s\n",{length(res),n,r})
end procedure
papply({{500,":\n%s"},{1000,":\n%s"},{10000,"."},{10_000_000,"."}},test)

Output:

34 reverse primes < 500 found:
  2   3   5   7  11  13  17  31  37  71  73  79  97 101 107 113 131
149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

56 reverse primes < 1,000 found:
  2   3   5   7  11  13  17  31  37  71  73  79  97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337
347 353 359 373 383 389 701 709 727 733 739 743 751 757 761 769 787 797 907 919 929 937 941 953 967 971 983 991

260 reverse primes < 10,000 found.
82,439 reverse primes < 10,000,000 found.

Python

#!/usr/bin/python

def isPrime(n):
    for i in range(2, int(n**0.5) + 1):
        if n % i == 0:
            return False        
    return True

def isBackPrime(n):
    if not isPrime(n):
        return False
    m = 0
    while n:
        m *= 10
        m += n % 10
        n //= 10
    return isPrime(m)

if __name__ == '__main__':
    for n in range(2, 499):
        if isBackPrime(n):
            print(n, end=' ');

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

PILOT

C :n=1
*number
C :p=n 
U :*prime
J (pr=0):*next
U :*rev
C :p=r
U :*prime
T (pr):#n
*next
C :n=n+1 
J (n<500):*number
E :

*rev
C :r=0
*digit
C :z=p/10
  :r=r*10+(p-z*10)
  :p=z
J (z):*digit
E :

*prime
C :pr=0
E (p=1):
C :pr=1
E (p<4):
C :pr=0
E (p=2*(p/2)):
C :i=3
*div
E (p=i*(p/i)):
C :i=i+2
J (i<p/2):*div
C :pr=1
E :

Output:

PL/I

reversePrimes: procedure options(main);
    declare prime(1:999) bit;
    
    sieve: procedure;
        declare (i,j,hi) fixed;
        hi = hbound(prime,1);
        prime(1) = '0'b;
        do i=2 to hi; prime(i) = '1'b; end;
        do i=2 to sqrt(hi);
            do j=i*i to hi by i; prime(j) = '0'b; end;
        end;
    end sieve;
    
    reverse: procedure(nn) returns(fixed);
        declare (n, nn, r) fixed;
        r=0;
        do n=nn repeat(n/10) while (n>0);
            r = 10*r + mod(n,10);
        end;
        return(r);
    end reverse;
    
    declare (n, found) fixed;
    call sieve;
    found = 0;
    do n=1 to 499;
        if prime(n) & prime(reverse(n)) then do;
            put edit(n) (F(4));
            found = found + 1;
            if mod(found,20) = 0 then put skip;
        end;
    end;
    
    put skip list('Reverse primes found:',found);
end reversePrimes;

Output:

   2   3   5   7  11  13  17  31  37  71  73  79  97 101 107 113 131 149 151 157
 167 179 181 191 199 311 313 337 347 353 359 373 383 389
Reverse primes found:        34

PL/M

Works with: 8080 PL/M Compiler

... under CP/M (or an emulator)

100H: /* FIND PRIMES THAT ARE STILL PRIME WHEN THEIR DIGITS ARE REVERSED     */

   /* CP/M BDOS SYSTEM CALL                                                  */
   BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
   /* I/O ROUTINES                                                           */
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
   PR$NL:     PROCEDURE;   CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
   PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH  */
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
      V = N;
      W = LAST( N$STR );
      N$STR( W ) = '$';
      N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;

   /* RETURNS TRUE IF N IS PRIME, FALSE OTHERWISE, USES TRIAL DIVISION      */
   IS$PRIME: PROCEDURE( N )BYTE;
      DECLARE N ADDRESS;
      DECLARE PRIME BYTE;
      IF      N < 3       THEN PRIME = N = 2;
      ELSE IF N MOD 3 = 0 THEN PRIME = N = 3;
      ELSE IF N MOD 2 = 0 THEN PRIME = 0;
      ELSE DO;
         DECLARE ( F, F2, TO$NEXT ) ADDRESS;
         PRIME   = 1;
         F       = 5;
         F2      = 25;
         TO$NEXT = 24;            /* NOTE: ( 2N + 1 )^2 - ( 2N - 1 )^2 = 8N */
         DO WHILE F2 <= N AND PRIME;
            PRIME   = N MOD F <> 0;
            F       = F + 2;
            F2      = F2 + TO$NEXT;
            TO$NEXT = TO$NEXT + 8;
         END;
      END;
      RETURN PRIME;
   END IS$PRIME;

   REVERSE: PROCEDURE( N )ADDRESS;  /* RETURNS THE REVERSED DIGITS OF N */
      DECLARE N ADDRESS;
      DECLARE ( R, V ) ADDRESS;
      V = N;
      R = V MOD 10;
      DO WHILE( ( V := V / 10 ) > 0 );
         R = ( R * 10 ) + ( V MOD 10 );
      END;
      RETURN R;
   END REVERSE ;

   /* FIND THE NUMBERS UP TO 500                                             */

   DECLARE ( I, COUNT ) ADDRESS;

   COUNT = 0;
   DO I = 1 TO 500;
      IF IS$PRIME( I ) THEN DO;
         IF IS$PRIME( REVERSE( I ) ) THEN DO;
            IF I <  10 THEN CALL PR$CHAR( ' ' );
            IF I < 100 THEN CALL PR$CHAR( ' ' );
            CALL PR$NUMBER( I );
            IF ( COUNT := COUNT + 1 )  MOD 20 = 0 THEN CALL PR$NL;
                                                  ELSE CALL PR$CHAR( ' ' );
         END;
      END;
   END;

EOF

Output:

  2   3   5   7  11  13  17  31  37  71  73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353 359 373 383 389

Quackery

eratosthenes and isprime are defined at Sieve of Eratosthenes#Quackery.

  1000 eratosthenes

  [ number$ reverse $->n drop ] is revnum     ( n --> n ) 

  [ dup isprime iff 
      [ revnum isprime ]
    else [ drop false ] ]       is isrevprime ( n --> b )

  [] [] 500 times
    [ i^ isrevprime if
      [ i^ join ] ]
  witheach [ number$ nested join ]
  60 wrap$

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151
157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

Raku

unit sub MAIN ($limit = 500);
say "{+$_} reversible primes < $limit:\n{$_».fmt("%" ~ $limit.chars ~ "d").batch(10).join("\n")}",
    with ^$limit .grep: { .is-prime and .flip.is-prime }

Output:

34 reversible primes < 500:
  2   3   5   7  11  13  17  31  37  71
 73  79  97 101 107 113 131 149 151 157
167 179 181 191 199 311 313 337 347 353
359 373 383 389

REXX

/*REXX program counts/displays the number of reversed primes under a specified number N.*/
parse arg n cols .                               /*get optional number of primes to find*/
if    n=='' |    n==","  then    n= 500          /*Not specified?   Then assume default.*/
if cols=='' | cols==","  then cols=  10          /* "      "          "     "       "   */
call genP copies(9, length(n) )                  /*generate all primes under  N.        */
w= 10                                            /*width of a number in any column.     */
if cols>0  then say ' index │'center(" reversed primes that are  < "  n,  1 + cols*(w+1) )
if cols>0  then say '───────┼'center(""                           ,  1 + cols*(w+1),  '─')
Rprimes= 0;                idx= 1                /*initialize # of additive primes & idx*/
$=                                               /*a list of additive primes  (so far). */
       do j=2  until j>=n; if \!.j  then iterate /*Is  J  not a prime? No, then skip it.*/
       _= reverse(j);      if \!._  then iterate /*is sum of J's digs a prime? No, skip.*/
       Rprimes= Rprimes + 1                      /*bump the count of additive primes.   */
       if cols<1             then iterate        /*Build the list  (to be shown later)? */
       $= $  right( commas(j), w)                /*add the additive prime to the $ list.*/
       if Rprimes//cols\==0  then iterate        /*have we populated a line of output?  */
       say center(idx, 7)'│'  substr($, 2);  $=  /*display what we have so far  (cols). */
       idx= idx + cols                           /*bump the  index  count for the output*/
       end   /*j*/

if $\==''  then say center(idx, 7)"│"  substr($, 2)  /*possible display residual output.*/
say
say 'found '      commas(Rprimes)       " reversed primes  < "       commas(n)
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: parse arg h;   @.=.; @.1=2;  @.2=3;  @.3=5;  @.4=7;  @.5=11;  @.6=13;  @.7=17;  #= 7
      w= length(h);  !.=0; !.2=1;  !.3=1;  !.5=1;  !.7=1;  !.11=1;  !.13=1;  !.17=1
            do j=@.7+2  by 2  while j<h          /*continue on with the next odd prime. */
            parse var  j  ''  -1  _              /*obtain the last digit of the  J  var.*/
            if _      ==5  then iterate          /*is this integer a multiple of five?  */
            if j // 3 ==0  then iterate          /* "   "     "    "     "     " three? */
                                                 /* [↓]  divide by the primes.   ___    */
                  do k=4  to #  while  k*k<=j    /*divide  J  by other primes ≤ √ J     */
                  if j//@.k == 0  then iterate j /*÷ by prev. prime?  ¬prime     ___    */
                  end   /*k*/                    /* [↑]   only divide up to     √ J     */
            #= # + 1;          @.#= j;  !.j= 1   /*bump prime count; assign prime & flag*/
            end   /*j*/
     return

output when using the default inputs:

 index │                                        reversed primes that are  <  500
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │          2          3          5          7         11         13         17         31         37         71
  11   │         73         79         97        101        107        113        131        149        151        157
  21   │        167        179        181        191        199        311        313        337        347        353
  31   │        359        373        383        389

found  34  reversed primes  <  500

output when using the inputs: 10000 0

found  260  reversed primes  <  10,000

Ring

load "stdlib.ring"

see "working..." + nl

row = 0
num = 0
limit = 500

for n = 1 to limit
    strm = ""
    strn = string(n)
    for m = len(strn) to 1 step -1
        strm = strm + strn[m]
    next
    strnum = number(strm)
    if isprime(n) and isprime(strnum)
       num = num + 1
       row = row + 1
       see "" + n + " "
       if row%10 = 0
          see nl
       ok
     ok       
next

see nl + "found " + num + " primes" + nl
see "done..." + nl

Output:

working...
2 3 5 7 11 13 17 31 37 71 
73 79 97 101 107 113 131 149 151 157 
167 179 181 191 199 311 313 337 347 353 
359 373 383 389 
found 34 primes
done...

RPL

This program uses the words RVSTR and BPRIM?, respectively made to revert a string and to test primality by trial division

≪ { } 1 499 FOR n
     IF n R→B 'BPRIM? THEN 
        IF n →STR RVSTR STR→ R→B BPRIM? THEN n + END 
     END 
  NEXT 
≫ 'TASK' STO

Straightforward approach

Numbers are here reversed without any string conversion.

Works with: HP version 49

« { }
   1 499 FOR n
      IF n ISPRIME? THEN
         n 0
         WHILE OVER REPEAT
            SWAP 10 IDIV2 ROT 10 * + 
         END NIP
         IF ISPRIME? THEN n + END
      END
   NEXT
» 'TASK' STO

Output:

1: { 2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389 }

Ruby

p Prime.each(500).select{|pr| pr.digits.join.to_i.prime? }

Output:

[2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389]

Rust

use prime_tools ;

fn myreverse( n : u32 ) -> u32 {
   let forward : String = n.to_string( ) ;
   let numberstring = &forward[..] ;
   let mut reversed : String = String::new( ) ;
   for c in numberstring.chars( ).rev( ) {
      reversed.push( c ) ;
   }
   *&reversed[..].parse::<u32>( ).unwrap( )
}

fn main() {
   let mut reversible_primes : Vec<u32> = Vec::new( ) ;
   for num in 2..=500 {
      if prime_tools::is_u32_prime( num ) && prime_tools::is_u32_prime( 
               myreverse( num ))  {
            reversible_primes.push( num ) ;
      }
   }
   println!("{:?}" , reversible_primes ) ;
}

Output:

[2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389]

Seed7

$ include "seed7_05.s7i";

const func boolean: isPrime (in integer: number) is func
  result
    var boolean: prime is FALSE;
  local
    var integer: upTo is 0;
    var integer: testNum is 3;
  begin
    if number = 2 then
      prime := TRUE;
    elsif odd(number) and number > 2 then
      upTo := sqrt(number);
      while number rem testNum <> 0 and testNum <= upTo do
        testNum +:= 2;
      end while;
      prime := testNum > upTo;
    end if;
  end func;

const func integer: revDigits (in var integer: number) is func
  result
    var integer: revNum is 0;
  begin
    while number > 0 do
      revNum *:= 10;
      revNum +:= number rem 10;
      number := number div 10;
    end while;
  end func;
  
const func boolean: isRevPrime (in integer: number) is
  return isPrime(number) and isPrime(revDigits(number));

const proc: main is func
  local
    var integer: number is 0;
    var integer: count is 0;
  begin
    for number range 1 to 499 do
      if isRevPrime(number) then
        write(number <& " ");
        incr(count);
      end if;
    end for;
    writeln;
    writeln("Found " <& count <& " reverse primes < 500.");
  end func;

Output:

2 3 5 7 11 13 17 31 37 71 73 79 97 101 107 113 131 149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389
Found 34 reverse primes < 500.

Sidef

say primes(500).grep { .reverse.is_prime }

Output:

[2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, 97, 101, 107, 113, 131, 149, 151, 157, 167, 179, 181, 191, 199, 311, 313, 337, 347, 353, 359, 373, 383, 389]

Wren

Library: Wren-math

Library: Wren-fmt

import "./math" for Int
import "./fmt" for Fmt

var reversed = Fn.new { |n|
    var rev = 0
    while (n > 0) {
        rev = rev * 10 + n % 10
        n = (n/10).floor
    }
    return rev
}

var primes = Int.primeSieve(499)
var reversedPrimes = []
for (p in primes) {
    if (Int.isPrime(reversed.call(p))) reversedPrimes.add(p)
}
System.print("Primes under 500 which are also primes when the digits are reversed:")
Fmt.tprint("$3d", reversedPrimes, 17)
System.print("\n%(reversedPrimes.count) such primes found.")

Output:

Primes under 500 which are also primes when the digits are reversed:
  2   3   5   7  11  13  17  31  37  71  73  79  97 101 107 113 131
149 151 157 167 179 181 191 199 311 313 337 347 353 359 373 383 389

34 such primes found.

XPL0

func IsPrime(N);        \Return 'true' if N is a prime number
int  N, I;
[if N <= 1 then return false;
for I:= 2 to sqrt(N) do
    if rem(N/I) = 0 then return false;
return true;
];

func Reverse(N);        \Return the reverse of the digits in N
int  N, M;
[M:= 0;
while N do
    [N:= N/10;
    M:= M*10 + rem(0);
    ];
return M;
];

int Count, N;
[Count:= 0;
for N:= 1 to 499 do
    [if IsPrime(N) & IsPrime(Reverse(N)) then
        [IntOut(0, N);
        Count:= Count+1;
        if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
        ]
    ];
CrLf(0);
IntOut(0, Count);
Text(0, " reversible primes found.");
]

Output:

2       3       5       7       11      13      17      31      37      71
73      79      97      101     107     113     131     149     151     157
167     179     181     191     199     311     313     337     347     353
359     373     383     389     
34 reversible primes found.