# Circular primes

Circular primes
You are encouraged to solve this task according to the task description, using any language you may know.
Definitions

A circular prime is a prime number with the property that the number generated at each intermediate step when cyclically permuting its (base 10) digits will also be prime.

For example: 1193 is a circular prime, since 1931, 9311 and 3119 are all also prime.

Note that a number which is a cyclic permutation of a smaller circular prime is not considered to be itself a circular prime. So 13 is a circular prime, but 31 is not.

A repunit (denoted by R) is a number whose base 10 representation contains only the digit 1.

For example: R(2) = 11 and R(5) = 11111 are repunits.

• Find the first 19 circular primes.

• If your language has access to arbitrary precision integer arithmetic, given that they are all repunits, find the next 4 circular primes.

• (Stretch) Determine which of the following repunits are probably circular primes: R(5003), R(9887), R(15073), R(25031), R(35317) and R(49081). The larger ones may take a long time to process so just do as many as you reasonably can.

## ALGOL 68

```BEGIN # find circular primes - primes where all cyclic permutations  #
# of the digits are also prime                                 #
# genertes a sieve of circular primes, only the first            #
# permutation of each prime is flagged as TRUE                   #
OP   CIRCULARPRIMESIEVE = ( INT n )[]BOOL:
BEGIN
[ 0 : n ]BOOL prime;
prime[ 0 ] := prime[ 1 ] := FALSE;
prime[ 2 ] := TRUE;
FOR i FROM 3 BY 2 TO UPB prime DO prime[ i ] := TRUE  OD;
FOR i FROM 4 BY 2 TO UPB prime DO prime[ i ] := FALSE OD;
FOR i FROM 3 BY 2 TO ENTIER sqrt( UPB prime ) DO
IF prime[ i ] THEN
FOR s FROM i * i BY i + i TO UPB prime DO prime[ s ] := FALSE OD
FI
OD;
INT first digit multiplier := 10;
INT max with multiplier    := 99;
# the 1 digit primes are non-curcular, so start at 10    #
FOR i FROM 10 TO UPB prime DO
IF i > max with multiplier THEN
# starting a new power of ten                    #
first digit multiplier *:= 10;
max with multiplier    *:= 10 +:= 9
FI;
IF prime[ i ] THEN
# have a prime #
# cycically permute the number until we get back #
# to the original - flag all the permutations    #
# except the original as non-prime               #
INT permutation := i;
WHILE permutation :=   ( permutation OVER 10 )
+ ( ( permutation MOD  10 ) * first digit multiplier )
;
permutation /= i
DO
IF NOT prime[ permutation ] THEN
# the permutation is not prime           #
prime[ i ] := FALSE
ELIF permutation > i THEN
# haven't permutated e.g. 101 to 11      #
IF NOT prime[ permutation ] THEN
# i is not a circular prime          #
prime[ i ] := FALSE
FI;
prime[ permutation ] := FALSE
FI
OD
FI
OD;
prime
END # CIRCULARPRIMESIEVE # ;
# construct a sieve of circular primes up to 999 999              #
# only the first permutation is included                          #
[]BOOL prime = CIRCULARPRIMESIEVE 999 999;
# print the first 19 circular primes #
INT c count := 0;
print( ( "First 19 circular primes: " ) );
FOR i WHILE c count < 19 DO
IF prime[ i ] THEN
print( ( " ", whole( i, 0 ) ) );
c count +:= 1
FI
OD;
print( ( newline ) )
END```
Output:
```First 19 circular primes:  2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## ALGOL W

```begin % find circular primes - primes where all cyclic permutations  %
% of the digits are 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 ;
% find circular primes in p in the range lo to hi, if they are circular, flag the %
% permutations as non-prime so we do not consider them again                      %
% non-circular primes are also flageed as non-prime                               %
% lo must be a power of ten and hi must be at most ( lo * 10 ) - 1                %
procedure keepCircular ( logical array p ( * ); integer value lo, hi ) ;
for n := lo until hi do begin
if p( n ) then begin
% have a prime %
integer       c, pCount;
logical       isCircular;
integer array permutations ( 1 :: 10 );
c          := n;
isCircular := true;
pCount     := 0;
% cyclically permute c until we get back to p or find a non-prime value for c %
while begin
integer first, rest;
first      := c div lo;
rest       := c rem lo;
c          := ( rest * 10 ) + first;
isCircular := p( c );
c not = n and isCircular
end do begin
pCount := pCount + 1;
permutations( pCount ) := c
end while_have_another_prime_permutation ;
if not isCircular
then p( n ) := false
else begin
% have a circular prime - flag the permutations as non-prime %
for i := 1 until pCount do p( permutations( i ) ) := false
end if_not_isCircular__
end if_p_n
end keepCircular ;
integer       cCount;
% sieve the primes up to 999999 %
logical array p ( 1 ::   999999 );
Eratosthenes( p,         999999 );
% remove non-circular primes from the sieve %
% the single digit primes are all circular so we start at 10 %
keepCircular( p,     10,     99 );
keepCircular( p,    100,    999 );
keepCircular( p,   1000,   9999 );
keepCircular( p,  10000,  99999 );
keepCircular( p, 100000, 200000 );
% print the first 19 circular primes %
cCount := 0;
write( "First 19 circular primes: " );
for i := 1 until 200000 do begin
if p( i ) then begin
writeon( i_w := 1, s_w := 1, i );
cCount := cCount + 1;
if cCount = 19 then goto end_circular
end if_p_i
end for_i ;
end_circular:
end.```
Output:
```First 19 circular primes: 2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## Arturo

```perms: function [n][
str: repeat to :string n 2
result: new []
lim: dec size digits n
loop 0..lim 'd ->
'result ++ slice str d lim+d

]

circulars: new []

circular?: function [x][
if not? prime? x -> return false

loop perms x 'y [
if not? prime? y -> return false
if contains? circulars y -> return false
]

'circulars ++ x

return true
]

i: 2
found: 0
while [found < 19][
if circular? i [
print i
found: found + 1
]
i: i + 1
]
```
Output:
```2
3
5
7
11
13
17
37
79
113
197
199
337
1193
3779
11939
19937
193939
199933```

## AWK

```# syntax: GAWK -f CIRCULAR_PRIMES.AWK
BEGIN {
p = 2
printf("first 19 circular primes:")
for (count=0; count<19; p++) {
if (is_circular_prime(p)) {
printf(" %d",p)
count++
}
}
printf("\n")
exit(0)
}
function cycle(n,  m,p) { # E.G. if n = 1234 returns 2341
m = n
p = 1
while (m >= 10) {
p *= 10
m /= 10
}
return int(m+10*(n%p))
}
function is_circular_prime(p,  p2) {
if (!is_prime(p)) {
return(0)
}
p2 = cycle(p)
while (p2 != p) {
if (p2 < p || !is_prime(p2)) {
return(0)
}
p2 = cycle(p2)
}
return(1)
}
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)
}
```
Output:
```first 19 circular primes: 2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## C

Library: GMP
```#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <gmp.h>

bool is_prime(uint32_t n) {
if (n == 2)
return true;
if (n < 2 || n % 2 == 0)
return false;
for (uint32_t p = 3; p * p <= n; p += 2) {
if (n % p == 0)
return false;
}
return true;
}

// e.g. returns 2341 if n = 1234
uint32_t cycle(uint32_t n) {
uint32_t m = n, p = 1;
while (m >= 10) {
p *= 10;
m /= 10;
}
return m + 10 * (n % p);
}

bool is_circular_prime(uint32_t p) {
if (!is_prime(p))
return false;
uint32_t p2 = cycle(p);
while (p2 != p) {
if (p2 < p || !is_prime(p2))
return false;
p2 = cycle(p2);
}
return true;
}

void test_repunit(uint32_t digits) {
char* str = malloc(digits + 1);
if (str == 0) {
fprintf(stderr, "Out of memory\n");
exit(1);
}
memset(str, '1', digits);
str[digits] = 0;
mpz_t bignum;
mpz_init_set_str(bignum, str, 10);
free(str);
if (mpz_probab_prime_p(bignum, 10))
printf("R(%u) is probably prime.\n", digits);
else
printf("R(%u) is not prime.\n", digits);
mpz_clear(bignum);
}

int main() {
uint32_t p = 2;
printf("First 19 circular primes:\n");
for (int count = 0; count < 19; ++p) {
if (is_circular_prime(p)) {
if (count > 0)
printf(", ");
printf("%u", p);
++count;
}
}
printf("\n");
printf("Next 4 circular primes:\n");
uint32_t repunit = 1, digits = 1;
for (; repunit < p; ++digits)
repunit = 10 * repunit + 1;
mpz_t bignum;
mpz_init_set_ui(bignum, repunit);
for (int count = 0; count < 4; ) {
if (mpz_probab_prime_p(bignum, 15)) {
if (count > 0)
printf(", ");
printf("R(%u)", digits);
++count;
}
++digits;
mpz_mul_ui(bignum, bignum, 10);
}
mpz_clear(bignum);
printf("\n");
test_repunit(5003);
test_repunit(9887);
test_repunit(15073);
test_repunit(25031);
test_repunit(35317);
test_repunit(49081);
return 0;
}
```
Output:

With GMP 6.2.0, execution time on my system is about 13 minutes (3.2 GHz Quad-Core Intel Core i5, macOS 10.15.4).

```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19), R(23), R(317), R(1031)
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
R(35317) is not prime.
R(49081) is probably prime.
```

## C++

Library: GMP
```#include <cstdint>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <gmpxx.h>

typedef mpz_class integer;

bool is_prime(const integer& n, int reps = 50) {
return mpz_probab_prime_p(n.get_mpz_t(), reps);
}

std::string to_string(const integer& n) {
std::ostringstream out;
out << n;
return out.str();
}

bool is_circular_prime(const integer& p) {
if (!is_prime(p))
return false;
std::string str(to_string(p));
for (size_t i = 0, n = str.size(); i + 1 < n; ++i) {
std::rotate(str.begin(), str.begin() + 1, str.end());
integer p2(str, 10);
if (p2 < p || !is_prime(p2))
return false;
}
return true;
}

integer next_repunit(const integer& n) {
integer p = 1;
while (p < n)
p = 10 * p + 1;
return p;
}

integer repunit(int digits) {
std::string str(digits, '1');
integer p(str);
return p;
}

void test_repunit(int digits) {
if (is_prime(repunit(digits), 10))
std::cout << "R(" << digits << ") is probably prime\n";
else
std::cout << "R(" << digits << ") is not prime\n";
}

int main() {
integer p = 2;
std::cout << "First 19 circular primes:\n";
for (int count = 0; count < 19; ++p) {
if (is_circular_prime(p)) {
if (count > 0)
std::cout << ", ";
std::cout << p;
++count;
}
}
std::cout << '\n';
std::cout << "Next 4 circular primes:\n";
p = next_repunit(p);
std::string str(to_string(p));
int digits = str.size();
for (int count = 0; count < 4; ) {
if (is_prime(p, 15)) {
if (count > 0)
std::cout << ", ";
std::cout << "R(" << digits << ")";
++count;
}
p = repunit(++digits);
}
std::cout << '\n';
test_repunit(5003);
test_repunit(9887);
test_repunit(15073);
test_repunit(25031);
test_repunit(35317);
test_repunit(49081);
return 0;
}
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19), R(23), R(317), R(1031)
R(5003) is not prime
R(9887) is not prime
R(15073) is not prime
R(25031) is not prime
R(35317) is not prime
R(49081) is probably prime
```

## D

Translation of: C
```import std.bigint;
import std.stdio;

immutable PRIMES = [
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,
211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293,
307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997
];

bool isPrime(BigInt n) {
if (n < 2) {
return false;
}

foreach (p; PRIMES) {
if (n == p) {
return true;
}
if (n % p == 0) {
return false;
}
if (p * p > n) {
return true;
}
}

for (auto m = BigInt(PRIMES[\$ - 1]); m * m <= n ; m += 2) {
if (n % m == 0) {
return false;
}
}

return true;
}

// e.g. returns 2341 if n = 1234
BigInt cycle(BigInt n) {
BigInt m = n;
BigInt p = 1;
while (m >= 10) {
p *= 10;
m /= 10;
}
return m + 10 * (n % p);
}

bool isCircularPrime(BigInt p) {
if (!isPrime(p)) {
return false;
}
for (auto p2 = cycle(p); p2 != p; p2 = cycle(p2)) {
if (p2 < p || !isPrime(p2)) {
return false;
}
}
return true;
}

BigInt repUnit(int len) {
BigInt n = 0;
while (len > 0) {
n = 10 * n + 1;
len--;
}
return n;
}

void main() {
writeln("First 19 circular primes:");
int count = 0;
foreach (p; PRIMES) {
if (isCircularPrime(BigInt(p))) {
if (count > 0) {
write(", ");
}
write(p);
count++;
}
}
for (auto p = BigInt(PRIMES[\$ - 1]) + 2; count < 19; p += 2) {
if (isCircularPrime(BigInt(p))) {
if (count > 0) {
write(", ");
}
write(p);
count++;
}
}
writeln;
}
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933```

## F#

This task uses rUnitP and Extensible Prime Generator (F#)

```// Circular primes - Nigel Galloway: September 13th., 2021
let fG n g=let rec fG y=if y=g then true else if y>g && isPrime y then fG(10*(y%n)+y/n) else false in fG(10*(g%n)+g/n)
let rec fN g l=seq{let g=[for n in g do for g in [1;3;7;9] do let g=n*10+g in yield g] in yield! g|>List.filter(fun n->isPrime n && fG l n); yield! fN g (l*10)}
let circP()=seq{yield! [2;3;5;7]; yield! fN [1;3;7;9] 10}
circP()|> Seq.take 19 |>Seq.iter(printf "%d "); printfn ""
printf "The first 5 repunit primes are "; rUnitP(10)|>Seq.take 5|>Seq.iter(fun n->printf \$"R(%d{n}) "); printfn ""
```
Output:
```2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
The first 5 repunit primes are R(2) R(19) R(23) R(317) R(1031)
```

## Factor

Unfortunately Factor's miller-rabin test or bignums aren't quite up to the task of finding the next four circular prime repunits in a reasonable time. It takes ~90 seconds to check R(7)-R(1031).

Works with: Factor version 0.99 2020-03-02
```USING: combinators.short-circuit formatting io kernel lists
lists.lazy math math.combinatorics math.functions math.parser
math.primes sequences sequences.extras ;

! Create an ordered infinite lazy list of circular prime
! "candidates" -- the numbers 2, 3, 5 followed by numbers
! composed of only the digits 1, 3, 7, and 9.

: candidates ( -- list )
L{ "2" "3" "5" "7" } 2 lfrom
[ "1379" swap selections >list ] lmap-lazy lconcat lappend ;

: circular-prime? ( str -- ? )
all-rotations {
[ [ infimum ] [ first = ] bi ]
[ [ string>number prime? ] all? ]
} 1&& ;

: circular-primes ( -- list )
candidates [ circular-prime? ] lfilter ;

: prime-repunits ( -- list )
7 lfrom [ 10^ 1 - 9 / prime? ] lfilter ;

"The first 19 circular primes are:" print
19 circular-primes ltake [ write bl ] leach nl nl

"The next 4 circular primes, in repunit format, are:" print
4 prime-repunits ltake [ "R(%d) " printf ] leach nl
```
Output:
```The first 19 circular primes are:
2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933

The next 4 circular primes, in repunit format, are:
R(19) R(23) R(317) R(1031)
```

## Forth

Forth only supports native sized integers, so we only implement the first part of the task.

```create 235-wheel 6 c, 4 c, 2 c, 4 c, 2 c, 4 c, 6 c, 2 c,
does> swap 7 and + c@ ;

0 1 2constant init-235    \ roll 235 wheel at position 1
: next-235   over 235-wheel + swap 1+ swap ;

\ check that n is prime excepting multiples of 2, 3, 5.
: sq  dup * ;
: wheel-prime? ( n -- f )
>r init-235 begin
next-235
dup sq r@ >    if  rdrop 2drop true  exit  then
r@ over mod 0= if  rdrop 2drop false exit  then
again ;

: prime? ( n -- f )
dup 2 < if drop false exit then
dup 2 mod 0= if 2 = exit then
dup 3 mod 0= if 3 = exit then
dup 5 mod 0= if 5 = exit then
wheel-prime? ;

: log10^ ( n -- 10^[log n], log n )
dup 0<= abort" log10^: argument error."
1 0 rot
begin dup 9 > while
>r  swap 10 *  swap 1+  r> 10 /
repeat drop ;

: log10  ( n -- n )  log10^ nip ;

: rotate ( n -- n )
dup log10^ drop /mod swap 10 * + ;

: prime-rotation? ( p0 p -- f )
tuck <= swap prime? and ;

: circular? ( n -- f )  \ assume n is not a multiple of 2, 3, 5
dup wheel-prime? invert
if  drop false exit
then dup >r true
over log10 0 ?do
swap rotate j over prime-rotation? rot and
loop nip rdrop ;

: .primes
2 . 3 . 5 .
16 init-235  \ -- count, [n1 n2] as 2,3,5 wheel
begin
next-235 dup circular?
if dup . rot 1- -rot
then
third 0= until 2drop drop ;

." The first 19 circular primes are:" cr .primes cr
bye
```
Output:
```The first 19 circular primes are:
2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## FreeBASIC

```#define floor(x) ((x*2.0-0.5)Shr 1)

Function isPrime(Byval p As Integer) As Boolean
If p < 2 Then Return False
If p Mod 2 = 0 Then Return p = 2
If p Mod 3 = 0 Then Return p = 3
Dim As Integer d = 5
While d * d <= p
If p Mod d = 0 Then Return False Else d += 2
If p Mod d = 0 Then Return False Else d += 4
Wend
Return True
End Function

Function isCircularPrime(Byval p As Integer) As Boolean
Dim As Integer n = floor(Log(p)/Log(10))
Dim As Integer m = 10^n, q = p
For i As Integer = 0 To n
If (q < p Or Not isPrime(q)) Then Return false
q = (q Mod m) * 10 + floor(q / m)
Next i
Return true
End Function

Dim As Integer p = 2, dp = 1, cont = 0
Print("Primeros 19 primos circulares:")
While cont < 19
If isCircularPrime(p) Then Print p;" "; : cont += 1
p += dp: dp = 2
Wend
Sleep
```
Output:
```Primeros 19 primos circulares:
2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## Go

```package main

import (
"fmt"
big "github.com/ncw/gmp"
"strings"
)

// OK for 'small' numbers.
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 repunit(n int) *big.Int {
ones := strings.Repeat("1", n)
b, _ := new(big.Int).SetString(ones, 10)
return b
}

var circs = []int{}

// binary search is overkill for a small number of elements
for _, i := range circs {
if i == n {
return true
}
}
return false
}

func isCircular(n int) bool {
nn := n
pow := 1 // will eventually contain 10 ^ d where d is number of digits in n
for nn > 0 {
pow *= 10
nn /= 10
}
nn = n
for {
nn *= 10
f := nn / pow // first digit
nn += f * (1 - pow)
return false
}
if nn == n {
break
}
if !isPrime(nn) {
return false
}
}
return true
}

func main() {
fmt.Println("The first 19 circular primes are:")
digits := [4]int{1, 3, 7, 9}
q := []int{1, 2, 3, 5, 7, 9}  // queue the numbers to be examined
fq := []int{1, 2, 3, 5, 7, 9} // also queue the corresponding first digits
count := 0
for {
f := q[0]   // peek first element
fd := fq[0] // peek first digit
if isPrime(f) && isCircular(f) {
circs = append(circs, f)
count++
if count == 19 {
break
}
}
copy(q, q[1:])   // pop first element
q = q[:len(q)-1] // reduce length by 1
copy(fq, fq[1:]) // ditto for first digit queue
fq = fq[:len(fq)-1]
if f == 2 || f == 5 { // if digits > 1 can't contain a 2 or 5
continue
}
// add numbers with one more digit to queue
// only numbers whose last digit >= first digit need be added
for _, d := range digits {
if d >= fd {
q = append(q, f*10+d)
fq = append(fq, fd)
}
}
}
fmt.Println(circs)
fmt.Println("\nThe next 4 circular primes, in repunit format, are:")
count = 0
var rus []string
for i := 7; count < 4; i++ {
if repunit(i).ProbablyPrime(10) {
count++
rus = append(rus, fmt.Sprintf("R(%d)", i))
}
}
fmt.Println(rus)
fmt.Println("\nThe following repunits are probably circular primes:")
for _, i := range []int{5003, 9887, 15073, 25031, 35317, 49081} {
fmt.Printf("R(%-5d) : %t\n", i, repunit(i).ProbablyPrime(10))
}
}
```
Output:
```The first 19 circular primes are:
[2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933]

The next 4 circular primes, in repunit format, are:
[R(19) R(23) R(317) R(1031)]

The following repunits are probably circular primes:
R(5003 ) : false
R(9887 ) : false
R(15073) : false
R(25031) : false
R(35317) : false
R(49081) : true
```

```import Math.NumberTheory.Primes (Prime, unPrime, nextPrime)
import Math.NumberTheory.Primes.Testing (isPrime, millerRabinV)
import Text.Printf (printf)

rotated :: [Integer] -> [Integer]
rotated xs
| any (< head xs) xs = []
| otherwise          = map asNum \$ take (pred \$ length xs) \$ rotate xs
where
rotate [] = []
rotate (d:ds) = ds <> [d] : rotate (ds <> [d])

asNum :: [Integer] -> Integer
asNum [] = 0
asNum n@(d:ds)
| all (==1) n = read \$ concatMap show n
| otherwise = (d * (10 ^ length ds)) + asNum ds

digits :: Integer -> [Integer]
digits 0 = []
digits n = digits d <> [r]
where (d, r) = n `quotRem` 10

isCircular :: Bool -> Integer -> Bool
isCircular repunit n
| repunit = millerRabinV 0 n
| n < 10 = True
| even n = False
| null rotations = False
| any (<n) rotations = False
| otherwise = all isPrime rotations
where
rotations = rotated \$ digits n

repunits :: [Integer]
repunits = go 2
where go n = asNum (replicate n 1) : go (succ n)

asRepunit :: Int -> Integer
asRepunit n = asNum \$ replicate n 1

main :: IO ()
main = do
printf "The first 19 circular primes are:\n%s\n\n" \$ circular primes
printf "The next 4 circular primes, in repunit format are:\n"
mapM_ (printf "R(%d) ") \$ reps repunits
printf "\n\nThe following repunits are probably circular primes:\n"
mapM_ (uncurry (printf "R(%d) : %s\n") . checkReps) [5003, 9887, 15073, 25031, 35317, 49081]
where
primes = map unPrime [nextPrime 1..]
circular = show . take 19 . filter (isCircular False)
reps = map (sum . digits). tail . take 5 . filter (isCircular True)
checkReps = (,) <\$> id <*> show . isCircular True . asRepunit
```
Output:
```The first 19 circular primes are:
[2,3,5,7,11,13,17,37,79,113,197,199,337,1193,3779,11939,19937,193939,199933]

The next 4 circular primes, in repunit format are:
R(19) R(23) R(317) R(1031)

The following repunits are probably circular primes:
R(5003) : False
R(9887) : False
R(15073) : False
R(25031) : False
R(35317) : False
R(49081) : True
./circular_primes  277.56s user 1.82s system 97% cpu 4:47.91 total
```

## J

```R=: [: ". 'x' ,~ #&'1'
assert 11111111111111111111111111111111x -: R 32

Filter=: (#~`)(`:6)

rotations=: (|."0 1~ i.@#)&.(10&#.inv)
assert 123 231 312 -: rotations 123

primes_less_than=: i.&.:(p:inv)
assert 2 3 5 7 11 -: primes_less_than 12

NB. circular y --> y is the order of magnitude.
P25=: ([: -. (0 e. 1 3 7 9 e.~ 10 #.inv ])&>)Filter primes_less_than 10^y  NB. Q25 are primes with 1 3 7 9 digits
P=: 2 5 , P25
en=: # P
group=: en # 0
next=: 1
for_i. i. # group do.
if. 0 = i { group do.       NB. if untested
j =: P i. rotations i { P   NB. j are the indexes of the rotated numbers in the list of primes
if. en e. j do.             NB. if any are unfound
j=: j -. en                 NB. prepare to mark them all as searched, and failed.
g=: _1
else.
g=: next                    NB. mark the set as found in a new group.  Because we can.
next=: >: next
end.
group=: g j} group          NB. apply the tested mark
end.
end.
group </. P
)
```

J lends itself to investigation. Demonstrate and play with the definitions.

```   circular 3  NB. the values in the long list have a composite under rotation
┌─┬─┬─┬─┬──┬─────┬─────┬──────────────────────────────────────────────────────────────────────────┬─────┬─────┬───────────┬───────────┬───────────┬───────────┐
│2│5│3│7│11│13 31│17 71│19 137 139 173 179 191 193 313 317 331 379 397 739 773 797 911 937 977 997│37 73│79 97│113 131 311│197 719 971│199 919 991│337 373 733│
└─┴─┴─┴─┴──┴─────┴─────┴──────────────────────────────────────────────────────────────────────────┴─────┴─────┴───────────┴───────────┴───────────┴───────────┘

circular 2 NB.       VV
┌─┬─┬─┬─┬──┬─────┬─────┬──┬─────┬─────┐
│2│5│3│7│11│13 31│17 71│19│37 73│79 97│
└─┴─┴─┴─┴──┴─────┴─────┴──┴─────┴─────┘

q: 91   NB. factor the lone entry
7 13

RC=: circular 8

{: RC  NB. tail
┌─────────────────────────────────────────┐
│199933 319993 331999 933199 993319 999331│
└─────────────────────────────────────────┘

(< >./)@:(#&>) Filter circular 3   NB. remove the box containing most items
┌─┬─┬─┬─┬──┬─────┬─────┬─────┬─────┬───────────┬───────────┬───────────┬───────────┐
│2│5│3│7│11│13 31│17 71│37 73│79 97│113 131 311│197 719 971│199 919 991│337 373 733│
└─┴─┴─┴─┴──┴─────┴─────┴─────┴─────┴───────────┴───────────┴───────────┴───────────┘

] CPS=: {.&> (< >./)@:(#&>) Filter RC   NB. first 19 circular primes
2 5 3 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933

# CPS   NB. yes, 19 found.
19
```

Brief investigation into repunits.

```   Note 'The current Miller-Rabin test implemented in c is insufficient for this task'
R=: ([: ". 'x' ,~ #&'1')&>
(;q:@R)&> 500
|limit error
|       (;q:@R)&>500
)

boxdraw_j_ 0
Filter=: (#~`)(`:6)

R=: ([: ". 'x' ,~ #&'1')&>
(; q:@R)&> (0 ~: 3&|)Filter >: i. 26  NB. factor some repunits
┌──┬─────────────────────────────────┐
│1 │                                 │
├──┼─────────────────────────────────┤
│2 │11                               │
├──┼─────────────────────────────────┤
│4 │11 101                           │
├──┼─────────────────────────────────┤
│5 │41 271                           │
├──┼─────────────────────────────────┤
│7 │239 4649                         │
├──┼─────────────────────────────────┤
│8 │11 73 101 137                    │
├──┼─────────────────────────────────┤
│10│11 41 271 9091                   │
├──┼─────────────────────────────────┤
│11│21649 513239                     │
├──┼─────────────────────────────────┤
│13│53 79 265371653                  │
├──┼─────────────────────────────────┤
│14│11 239 4649 909091               │
├──┼─────────────────────────────────┤
│16│11 17 73 101 137 5882353         │
├──┼─────────────────────────────────┤
│17│2071723 5363222357               │
├──┼─────────────────────────────────┤
│19│1111111111111111111              │
├──┼─────────────────────────────────┤
│20│11 41 101 271 3541 9091 27961    │
├──┼─────────────────────────────────┤
│22│11 11 23 4093 8779 21649 513239  │
├──┼─────────────────────────────────┤
│23│11111111111111111111111          │
├──┼─────────────────────────────────┤
│25│41 271 21401 25601 182521213001  │
├──┼─────────────────────────────────┤
│26│11 53 79 859 265371653 1058313049│
└──┴─────────────────────────────────┘

NB. R(2) R(19), R(23) are probably prime.
```

## Java

```import java.math.BigInteger;
import java.util.Arrays;

public class CircularPrimes {
public static void main(String[] args) {
System.out.println("First 19 circular primes:");
int p = 2;
for (int count = 0; count < 19; ++p) {
if (isCircularPrime(p)) {
if (count > 0)
System.out.print(", ");
System.out.print(p);
++count;
}
}
System.out.println();
System.out.println("Next 4 circular primes:");
int repunit = 1, digits = 1;
for (; repunit < p; ++digits)
repunit = 10 * repunit + 1;
BigInteger bignum = BigInteger.valueOf(repunit);
for (int count = 0; count < 4; ) {
if (bignum.isProbablePrime(15)) {
if (count > 0)
System.out.print(", ");
System.out.printf("R(%d)", digits);
++count;
}
++digits;
bignum = bignum.multiply(BigInteger.TEN);
}
System.out.println();
testRepunit(5003);
testRepunit(9887);
testRepunit(15073);
testRepunit(25031);
}

private static boolean isPrime(int n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (int p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}

private static int cycle(int n) {
int m = n, p = 1;
while (m >= 10) {
p *= 10;
m /= 10;
}
return m + 10 * (n % p);
}

private static boolean isCircularPrime(int p) {
if (!isPrime(p))
return false;
int p2 = cycle(p);
while (p2 != p) {
if (p2 < p || !isPrime(p2))
return false;
p2 = cycle(p2);
}
return true;
}

private static void testRepunit(int digits) {
BigInteger repunit = repunit(digits);
if (repunit.isProbablePrime(15))
System.out.printf("R(%d) is probably prime.\n", digits);
else
System.out.printf("R(%d) is not prime.\n", digits);
}

private static BigInteger repunit(int digits) {
char[] ch = new char[digits];
Arrays.fill(ch, '1');
return new BigInteger(new String(ch));
}
}
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19), R(23), R(317), R(1031)
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

jq's integer-arithmetic accuracy is only sufficient for the first task; gojq has the accuracy for the second task but is not well-suited for it. Nevertheless, the code for solving both tasks is shown.

For an implementation of `is_prime` suitable for the first task, see for example Erdős-primes#jq.

```def is_circular_prime:
def circle: range(0;length) as \$i | .[\$i:] + .[:\$i];
tostring as \$s
| [\$s|circle|tonumber] as \$c
| . == (\$c|min) and all(\$c|unique[]; is_prime);

def circular_primes:
2, (range(3; infinite; 2) | select(is_circular_prime));

# Probably only useful with unbounded-precision integer arithmetic:
def repunits:
1 | recurse(10*. + 1);```

`limit(19; circular_primes)`

```last(limit(19; circular_primes)) as \$max
| limit(4; repunits | select(. > \$max and is_prime))
| "R(\(tostring|length))"```
Output:

```2
3
5
7
11
13
17
37
79
113
197
199
337
1193
3779
11939
19937
193939
199933
```

## Julia

Note that the evalpoly function used in this program was added in Julia 1.4

```using Lazy, Primes

function iscircularprime(n)
!isprime(n) && return false
dig = digits(n)
return all(i -> (m = evalpoly(10, circshift(dig, i))) >= n && isprime(m), 1:length(dig)-1)
end

filtcircular(n, rang) = Int.(collect(take(n, filter(iscircularprime, rang))))
isprimerepunit(n) = isprime(evalpoly(BigInt(10), ones(Int, n)))
filtrep(n, rang) = collect(take(n, filter(isprimerepunit, rang)))

println("The first 19 circular primes are:\n", filtcircular(19, Lazy.range(2)))
print("\nThe next 4 circular primes, in repunit format, are: ",
mapreduce(n -> "R(\$n) ", *, filtrep(4, Lazy.range(6))))

println("\n\nChecking larger repunits:")
for i in [5003, 9887, 15073, 25031, 35317, 49081]
println("R(\$i) is ", isprimerepunit(i) ? "prime." : "not prime.")
end
```
Output:
```The first 19 circular primes are:
[2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933]

The next 4 circular primes, in repunit format, are: R(19) R(23) R(317) R(1031)

Checking larger repunits:
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
R(35317) is not prime.
R(49081) is prime.
```

## Kotlin

Translation of: C
```import java.math.BigInteger

val SMALL_PRIMES = listOf(
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,
211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293,
307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997
)

fun isPrime(n: BigInteger): Boolean {
if (n < 2.toBigInteger()) {
return false
}

for (sp in SMALL_PRIMES) {
val spb = sp.toBigInteger()
if (n == spb) {
return true
}
if (n % spb == BigInteger.ZERO) {
return false
}
if (n < spb * spb) {
//if (n > SMALL_PRIMES.last().toBigInteger()) {
//    println("Next: \$n")
//}
return true
}
}

return n.isProbablePrime(10)
}

fun cycle(n: BigInteger): BigInteger {
var m = n
var p = 1
while (m >= BigInteger.TEN) {
p *= 10
m /= BigInteger.TEN
}
return m + BigInteger.TEN * (n % p.toBigInteger())
}

fun isCircularPrime(p: BigInteger): Boolean {
if (!isPrime(p)) {
return false
}
var p2 = cycle(p)
while (p2 != p) {
if (p2 < p || !isPrime(p2)) {
return false
}
p2 = cycle(p2)
}
return true
}

fun testRepUnit(digits: Int) {
var repUnit = BigInteger.ONE
var count = digits - 1
while (count > 0) {
repUnit = BigInteger.TEN * repUnit + BigInteger.ONE
count--
}
if (isPrime(repUnit)) {
println("R(\$digits) is probably prime.")
} else {
println("R(\$digits) is not prime.")
}
}

fun main() {
println("First 19 circular primes:")
var p = 2
var count = 0
while (count < 19) {
if (isCircularPrime(p.toBigInteger())) {
if (count > 0) {
print(", ")
}
print(p)
count++
}
p++
}
println()

println("Next 4 circular primes:")
var repUnit = BigInteger.ONE
var digits = 1
count = 0
while (repUnit < p.toBigInteger()) {
repUnit = BigInteger.TEN * repUnit + BigInteger.ONE
digits++
}
while (count < 4) {
if (isPrime(repUnit)) {
print("R(\$digits) ")
count++
}
repUnit = BigInteger.TEN * repUnit + BigInteger.ONE
digits++
}
println()

testRepUnit(5003)
testRepUnit(9887)
testRepUnit(15073)
testRepUnit(25031)
testRepUnit(35317)
testRepUnit(49081)
}
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19) R(23) R(317) R(1031)
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
R(35317) is not prime.```

## Lua

```-- Circular primes, in Lua, 6/22/2020 db
local function isprime(n)
if n < 2 then return false end
if n % 2 == 0 then return n==2 end
if n % 3 == 0 then return n==3 end
for f = 5, math.sqrt(n), 6 do
if n % f == 0 or n % (f+2) == 0 then return false end
end
return true
end

local function iscircularprime(p)
local n = math.floor(math.log10(p))
local m, q = 10^n, p
for i = 0, n do
if (q < p or not isprime(q)) then return false end
q = (q % m) * 10 + math.floor(q / m)
end
return true
end

local p, dp, list, N = 2, 1, {}, 19
while #list < N do
if iscircularprime(p) then list[#list+1] = p end
p, dp = p + dp, 2
end
print(table.concat(list, ", "))
```
Output:
`2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933`

## Mathematica / Wolfram Language

```ClearAll[RepUnit, CircularPrimeQ]
RepUnit[n_] := (10^n - 1)/9
CircularPrimeQ[n_Integer] := Module[{id = IntegerDigits[n], nums, t},
AllTrue[
Range[Length[id]]
,
Function[{z},
t = FromDigits[RotateLeft[id, z]];
If[t < n,
False
,
PrimeQ[t]
]
]
]
]
Select[Range[200000], CircularPrimeQ]

res = {};
Dynamic[res]
Do[
If[CircularPrimeQ[RepUnit[n]], AppendTo[res, n]]
,
{n, 1000}
]

Scan[Print@*PrimeQ@*RepUnit, {5003, 9887, 15073, 25031, 35317, 49081}]
```
Output:
```{2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933}
{2, 19, 23, 317}
False
False
False
False
False
True```

## Nim

Translation of: Kotlin
Library: bignum
```import bignum
import strformat

const SmallPrimes = [
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97,
101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199,
211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293,
307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397,
401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499,
503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599,
601, 607, 613, 617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691,
701, 709, 719, 727, 733, 739, 743, 751, 757, 761, 769, 773, 787, 797,
809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887,
907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997]

let
One = newInt(1)
Two = newInt(2)
Ten = newInt(10)

#---------------------------------------------------------------------------------------------------

proc isPrime(n: Int): bool =

if n < Two: return false

for sp in SmallPrimes:
# let spb = newInt(sp)
if n == sp: return true
if (n mod sp).isZero: return false
if n < sp * sp: return true

result = probablyPrime(n, 25) != 0

#---------------------------------------------------------------------------------------------------

proc cycle(n: Int): Int =

var m = n
var p = 1
while m >= Ten:
p *= 10
m = m div 10
result = m + Ten * (n mod p)

#---------------------------------------------------------------------------------------------------

proc isCircularPrime(p: Int): bool =

if not p.isPrime(): return false

var p2 = cycle(p)
while p2 != p:
if p2 < p or not p2.isPrime():
return false
p2 = cycle(p2)
result = true

#---------------------------------------------------------------------------------------------------

proc testRepunit(digits: int) =

var repunit = One
var count = digits - 1
while count > 0:
repunit = Ten * repunit + One
dec count
if repunit.isPrime():
echo fmt"R({digits}) is probably prime."
else:
echo fmt"R({digits}) is not prime."

#---------------------------------------------------------------------------------------------------

echo "First 19 circular primes:"
var p = 2
var line = ""
var count = 0
while count < 19:
if newInt(p).isCircularPrime():
if count > 0: line.add(", ")
inc count
inc p
echo line

echo ""
echo "Next 4 circular primes:"
var repunit = One
var digits = 1
while repunit < p:
repunit = Ten * repunit + One
inc digits
line = ""
count = 0
while count < 4:
if repunit.isPrime():
if count > 0: line.add(' ')
inc count
repunit = Ten * repunit + One
inc digits
echo line

echo ""
testRepUnit(5003)
testRepUnit(9887)
testRepUnit(15073)
testRepUnit(25031)
testRepUnit(35317)
testRepUnit(49081)
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933

Next 4 circular primes:
R(19) R(23) R(317) R(1031)

R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
R(35317) is not prime.
R(49081) is probably prime.```

## Pascal

### Free Pascal

Only test up to 14 digit numbers.RepUnit test with gmp is to boring.
Using a base 4 downcounter to create the numbers with more than one digit.
Changed the manner of the counter, so that it counts that there is no digit smaller than the first. 199-> 333 and not 311 so a base4 counter 1_(1,3,7,9) changed to base3 3_( 3,7,9 )->base2 7_(7,9) -> base 1 = 99....99. The main speed up is reached by testing with small primes within CycleNum.

```program CircularPrimes;
//nearly the way it is done:
//http://www.worldofnumbers.com/circular.htm
//base 4 counter to create numbers with first digit is the samallest used.
//check if numbers are tested before and reduce gmp-calls by checking with prime 3,7

{\$IFDEF FPC}
{\$MODE DELPHI}{\$OPTIMIZATION ON,ALL}
uses
Sysutils,gmp;
{\$ENDIF}
{\$IFDEF Delphi}
uses
System.Sysutils,?gmp?;
{\$ENDIF}

{\$IFDEF WINDOWS}
{\$APPTYPE CONSOLE}
{\$ENDIF}
const
MAXCNTOFDIGITS = 14;
MAXDGTVAL = 3;
conv : array[0..MAXDGTVAL+1] of byte = (9,7,3,1,0);
type
tDigits = array[0..23] of byte;
tUint64 = NativeUint;
var
mpz : mpz_t;
digits,
revDigits : tDigits;
CheckNum : array[0..19] of tUint64;
Found : array[0..23] of tUint64;
Pot_ten,Count,CountNumCyc,CountNumPrmTst : tUint64;

procedure CheckOne(MaxIdx:integer);
var
Num : Uint64;
i : integer;
begin
i:= MaxIdx;
repeat
inc(CountNumPrmTst);
num := CheckNum[i];
mpz_set_ui(mpz,Num);
If mpz_probab_prime_p(mpz,3)=0then
EXIT;
dec(i);
until i < 0;
Found[Count] := CheckNum[0];
inc(count);
end;

function CycleNum(MaxIdx:integer):Boolean;
//first create circular numbers to minimize prime checks
var
cycNum,First,P10 : tUint64;
i,j,cv : integer;
Begin
i:= MaxIdx;
j := 0;
First := 0;
repeat
cv := conv[digits[i]];
dec(i);
First := First*10+cv;
revDigits[j]:= cv;
inc(j);
until i < 0;
// if num is divisible by 3 then cycle numbers also divisible by 3 same sum of digits
IF First MOD 3 = 0 then
EXIT(false);
If First mod 7 = 0 then
EXIT(false);

//if one of the cycled number must have been tested before break
P10 := Pot_ten;
i := 0;
j := 0;
CheckNum[j] := First;
cycNum := First;
repeat
inc(CountNumCyc);
cv := revDigits[i];
inc(j);
cycNum := (cycNum - cv*P10)*10+cv;
//num was checked before
if cycNum < First then
EXIT(false);
if cycNum mod 7 = 0 then
EXIT(false);
CheckNum[j] := cycNum;
inc(i);
until i >= MaxIdx;
EXIT(true);
end;

var
T0: Int64;

idx,MaxIDx,dgt,MinDgt : NativeInt;
begin
T0 := GetTickCount64;
mpz_init(mpz);

fillchar(digits,Sizeof(digits),chr(MAXDGTVAL));
Count :=0;
For maxIdx := 2 to 10 do
if maxidx in[2,3,5,7] then
begin
Found[Count]:= maxIdx;
inc(count);
end;

Pot_ten := 10;
maxIdx := 1;
idx := 0;
MinDgt := MAXDGTVAL;
repeat
if CycleNum(MaxIdx) then
CheckOne(MaxIdx);
idx := 0;
repeat
dgt := digits[idx]-1;
if dgt >=0 then
break;
digits[idx] := MinDgt;
inc(idx);
until idx >MAXCNTOFDIGITS-1;

if idx > MAXCNTOFDIGITS-1 then
BREAK;

if idx<=MaxIDX then
begin
digits[idx] := dgt;
if idx=MaxIDX then
Begin
For MinDgt := 0 to idx do
digits[MinDgt]:= dgt;
minDgt := dgt;
end;
end
else
begin
minDgt := MAXDGTVAL;
For maxidx := 0 to idx do
digits[MaxIdx] := MAXDGTVAL;
Maxidx := idx;
Pot_ten := Pot_ten*10;
writeln(idx:7,count:7,CountNumCyc:16,CountNumPrmTst:12,GetTickCount64-T0:8);
end;
until false;
writeln(idx:7,count:7,CountNumCyc:16,CountNumPrmTst:12,GetTickCount64-T0:8);
T0 := GetTickCount64-T0;

For idx := 0 to count-2 do
write(Found[idx],',');
writeln(Found[count-1]);

writeln('It took ',T0,' ms ','to check ',MAXCNTOFDIGITS,' decimals');
mpz_clear(mpz);
{\$IFDEF WINDOWS}
{\$ENDIF}
end.
```
@ Tio.Run:
``` Digits  found    cycles created  primetests  time in ms
2      9               7          10       0
3     13              43          38       0
4     15             203          89       0
5     17             879         213       0
6     19            4209         816       1
7     19           16595        1794       2
8     19           67082        4666       6
9     19          270760       13108      19
10     19         1094177       39059      63
11     19         4421415      118787     220
12     19        23728859     1078484    1505
13     19        77952009     1869814    3562
14     19       296360934     4405393   11320
#### 14     19       754020918    28736408   26129 ##### before
2,3,5,7,11,13,17,37,79,113,197,199,337,1193,3779,11939,19937,193939,199933
It took 11320 ms to check 14 decimals

only creating numbers
14      4               0           0     184
creating and cycling numbers testing not ( MOD 3=0 OR MoD 7 = 0)
14      4       247209700           0    8842
that reduces the count of gmp-prime tests to 1/6 ```

## Perl

Translation of: Raku
Library: ntheory
```use strict;
use warnings;
use feature 'say';
use List::Util 'min';
use ntheory 'is_prime';

sub rotate { my(\$i,@a) = @_; join '', @a[\$i .. @a-1, 0 .. \$i-1] }

sub isCircular {
my (\$n) = @_;
return 0 unless is_prime(\$n);
my @circular = split //, \$n;
return 0 if min(@circular) < \$circular[0];
for (1 .. scalar @circular) {
my \$r = join '', rotate(\$_,@circular);
return 0 unless is_prime(\$r) and \$r >= \$n;
}
1
}

say "The first 19 circular primes are:";
for ( my \$i = 1, my \$count = 0; \$count < 19; \$i++ ) {
++\$count and print "\$i " if isCircular(\$i);
}

say "\n\nThe next 4 circular primes, in repunit format, are:";
for ( my \$i = 7, my \$count = 0; \$count < 4; \$i++ ) {
++\$count and say "R(\$i)" if is_prime 1 x \$i
}

say "\nRepunit testing:";

for (5003, 9887, 15073, 25031, 35317, 49081) {
say "R(\$_): Prime? " . (is_prime 1 x \$_ ? 'True' : 'False');
}
```
Output:
```The first 19 circular primes are:
2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933

The next 4 circular primes, in repunit format, are:
R(19)
R(23)
R(317)
R(1031)

Repunit testing:
R(5003): Prime? False
R(9887): Prime? False
R(15073): Prime? False
R(25031): Prime? False
R(35317): Prime? False
R(49081): Prime? True```

## Phix

```with javascript_semantics
function circular(integer p)
integer len = length(sprintf("%d",p)),
pow = power(10,len-1),
p0 = p
for i=1 to len-1 do
p = pow*remainder(p,10)+floor(p/10)
if p<p0 or not is_prime(p) then return false end if
end for
return true
end function

sequence c = {}
integer n = 1
while length(c)<19 do
integer p = get_prime(n)
if circular(p) then c &= p end if
n += 1
end while
printf(1,"The first 19 circular primes are:\n%v\n\n",{c})

include mpfr.e
procedure repunit(mpz z, integer n)
mpz_set_str(z,repeat('1',n))
end procedure

c = {}
n = 7
mpz z = mpz_init()
while length(c)<4 do
repunit(z,n)
if mpz_prime(z) then
c = append(c,sprintf("R(%d)",n))
end if
n += 1
end while
printf(1,"The next 4 circular primes, in repunit format, are:\n%s\n\n",{join(c)})
```
Output:
```The first 19 circular primes are:
{2,3,5,7,11,13,17,37,79,113,197,199,337,1193,3779,11939,19937,193939,199933}

The next 4 circular primes, in repunit format, are:
R(19) R(23) R(317) R(1031)
```

### stretch

(It is probably quite unwise to throw this in the direction of pwa/p2js, I am not even going to bother trying.)

```constant t = {5003, 9887, 15073, 25031, 35317, 49081}
printf(1,"The following repunits are probably circular primes:\n")
for i=1 to length(t) do
integer ti = t[i]
atom t0 = time()
repunit(z,ti)
bool bPrime = mpz_prime(z,1)
printf(1,"R(%d) : %t (%s)\n", {ti, bPrime, elapsed(time()-t0)})
end for
```
Output:

64-bit can only cope with the first five (it terminates abruptly on the sixth)
For comparison, the above Julia code (8/4/20, 64 bit) manages 1s, 5.6s, 15s, 50s, 1 min 50s (and 1 hour 45 min 40s) on the same box.
And Perl (somehow) manages 0/0/0/55s/0/21 mins 35 secs...

```The following repunits are probably circular primes:
R(5003) : false (2.0s)
R(9887) : false (13.5s)
R(15073) : false (45.9s)
R(25031) : false (1 minute and 19s)
R(35317) : false (3 minutes and 04s)
```

32-bit is much slower and can only cope with the first four

```The following repunits are probably circular primes:
R(5003) : false (10.2s)
R(9887) : false (54.9s)
R(15073) : false (2 minutes and 22s)
R(25031) : false (7 minutes and 45s)
diag looping, error code is 1, era is #00644651
```

## PicoLisp

For small primes, I only check numbers that are made up of the digits 1, 3, 7, and 9, and I also use a small prime checker to avoid the overhead of the Miller-Rabin Primality Test. For the large repunits, one can use the code from the Miller Rabin task.

```(load "plcommon/primality.l")  # see task: "Miller-Rabin Primality Test"

(de candidates (Limit)
(let Q (0)
(nth
(sort
(make
(while Q
(let A (pop 'Q)
(when (< A Limit)
(setq Q
(cons
(+ (* 10 A) 1)
(cons
(+ (* 10 A) 3)
(cons
(+ (* 10 A) 7)
(cons (+ (* 10 A) 9) Q))))))))))
6)))

(de circular? (P0)
(and
(small-prime? P0)
(fully '((P) (and (>= P P0) (small-prime? P))) (rotations P0))))

(de rotate (L)
(let ((X . Xs) L)
(append Xs (list X))))

(de rotations (N)
(let L (chop N)
(mapcar
format
(make
(do (dec (length L))

(de small-prime? (N)  # For small prime candidates only
(if (< N 2)
NIL
(let W (1 2 2 . (4 2 4 2 4 6 2 6 .))
(for (D 2  T  (+ D (pop 'W)))
(T  (> (* D D) N)  T)
(T  (=0 (% N D))   NIL)))))

(de repunit-primes (N)
(let (Test 111111  Remaining N  K 6)
(make
(until (=0 Remaining)
(setq Test (inc (* 10 Test)))
(inc 'K)
(when (prime? Test)
(dec 'Remaining))))))

(setq Circular
(conc
(2 3 5 7)
(filter circular? (candidates 1000000))
(mapcar '((X) (list 'R X)) (repunit-primes 4))))

(prinl "The first few circular primes:")
(println Circular)
(bye)```
Output:
```The first few circular primes:
(2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933 (R 19) (R 23) (R 317) (R 1031))
```

## Prolog

Uses a miller-rabin primality tester that I wrote which includes a trial division pass for small prime factors. One could substitute with e.g. the Miller Rabin Primaliy Test task.

The circular(P) predicate generates all circular primes; for those > 1e6, it returns it as a term r(K) for repunit K.

Also the code is smart in that it only checks primes > 9 that are composed of the digits 1, 3, 7, and 9.

```?- use_module(library(primality)).

circular(N) :- member(N, [2, 3, 5, 7]).
circular(N) :-
limit(15, (
candidate(N),
N > 9,
circular_prime(N))).
circular(r(K)) :-
between(6, inf, K),
N is (10**K - 1) div 9,
prime(N).

candidate(0).
candidate(N) :-
candidate(M),
member(D, [1, 3, 7, 9]),
N is 10*M + D.

circular_prime(N) :-
K is floor(log10(N)) + 1,
circular_prime(N, N, K).
circular_prime(_, _, 0) :- !.
circular_prime(P0, P, K) :-
P >= P0,
prime(P),
rotate(P, Q), succ(DecK, K),
circular_prime(P0, Q, DecK).

rotate(N, M) :-
D is floor(log10(N)),
divmod(N, 10, Q, R),
M is R*10**D + Q.

main :-
findall(P, limit(23, circular(P)), S),
format("The first 23 circular primes:~n~w~n", [S]),
halt.

?- main.
```
Output:
```The first 23 circular primes:
[2,3,5,7,11,13,17,37,79,113,197,199,337,1193,3779,11939,19937,193939,199933,r(19),r(23),r(317),r(1031)]
```

## Python

```import random

def is_Prime(n):
"""
Miller-Rabin primality test.

A return value of False means n is certainly not prime. A return value of
True means n is very likely a prime.
"""
if n!=int(n):
return False
n=int(n)
#Miller-Rabin test for prime
if n==0 or n==1 or n==4 or n==6 or n==8 or n==9:
return False

if n==2 or n==3 or n==5 or n==7:
return True
s = 0
d = n-1
while d%2==0:
d>>=1
s+=1
assert(2**s * d == n-1)

def trial_composite(a):
if pow(a, d, n) == 1:
return False
for i in range(s):
if pow(a, 2**i * d, n) == n-1:
return False
return True

for i in range(8):#number of trials
a = random.randrange(2, n)
if trial_composite(a):
return False

return True

def isPrime(n: int) -> bool:
'''
https://www.geeksforgeeks.org/python-program-to-check-whether-a-number-is-prime-or-not/
'''
# Corner cases
if (n <= 1) :
return False
if (n <= 3) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 == 0 or n % 3 == 0) :
return False
i = 5
while(i * i <= n) :
if (n % i == 0 or n % (i + 2) == 0) :
return False
i = i + 6
return True

def rotations(n: int)-> set((int,)):
'''
>>> {123, 231, 312} == rotations(123)
True
'''
a = str(n)
return set(int(a[i:] + a[:i]) for i in range(len(a)))

def isCircular(n: int) -> bool:
'''
>>> [isCircular(n) for n in (11, 31, 47,)]
[True, True, False]
'''
return all(isPrime(int(o)) for o in rotations(n))

from itertools import product

def main():
result = [2, 3, 5, 7]
first = '137'
latter = '1379'
for i in range(1, 6):
s = set(int(''.join(a)) for a in product(first, *((latter,) * i)))
while s:
a = s.pop()
b = rotations(a)
if isCircular(a):
result.append(min(b))
s -= b
result.sort()
return result

assert [2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933] == main()

repunit = lambda n: int('1' * n)

def repmain(n: int) -> list:
'''
returns the first n repunit primes, probably.
'''
result = []
i = 2
while len(result) < n:
if is_Prime(repunit(i)):
result.append(i)
i += 1
return result

assert [2, 19, 23, 317, 1031] == repmain(5)

# because this Miller-Rabin test is already on rosettacode there's no good reason to test the longer repunits.
```

## Raku

Most of the repunit testing is relatively speedy using the ntheory library. The really slow ones are R(25031), at ~42 seconds and R(49081) at 922(!!) seconds.

Library: ntheory
```sub isCircular(\n) {
return False unless n.is-prime;
my @circular = n.comb;
return False if @circular.min < @circular[0];
for 1 ..^ @circular -> \$i {
return False unless .is-prime and \$_ >= n given @circular.rotate(\$i).join;
}
True
}

say "The first 19 circular primes are:";
say ((2..*).hyper.grep: { isCircular \$_ })[^19];

say "\nThe next 4 circular primes, in repunit format, are:";
loop ( my \$i = 7, my \$count = 0; \$count < 4; \$i++ ) {
++\$count, say "R(\$i)" if (1 x \$i).is-prime
}

use ntheory:from<Perl5> qw[is_prime];

say "\nRepunit testing:";

(5003, 9887, 15073, 25031, 35317, 49081).map: {
my \$now = now;
say "R(\$_): Prime? ", ?is_prime("{1 x \$_}"), "  {(now - \$now).fmt: '%.2f'}"
}
```
Output:
```The first 19 circular primes are:
(2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933)

The next 4 circular primes, in repunit format, are:
R(19)
R(23)
R(317)
R(1031)

Repunit testing:
R(5003): Prime? False  0.00
R(9887): Prime? False  0.01
R(15073): Prime? False  0.02
R(25031): Prime? False  41.40
R(35317): Prime? False  0.32
R(49081): Prime? True  921.73```

## REXX

```/*REXX program finds & displays circular primes (with a title & in a horizontal format).*/
parse arg N hp .                                 /*obtain optional arguments from the CL*/
if  N=='' |  N==","  then N=        19           /* "      "         "   "   "     "    */
if hp=='' | hp==","  then hip= 1000000           /* "      "         "   "   "     "    */
call genP                                        /*gen primes up to  hp      (200,000). */
q= 024568                                        /*digs that most circular P can't have.*/
found= 0;                           \$=           /*found:  circular P count; \$:  a list.*/
do j=1  until found==N;       p= @.j       /* [↓]  traipse through all the primes.*/
if p>9 & verify(p, q, 'M')>0  then iterate /*Does J contain forbidden digs?  Skip.*/
if \circP(p)                  then iterate /*Not circular?  Then skip this number.*/
found= found + 1                           /*bump the  count  of circular primes. */
\$= \$  p                                    /*add this prime number  ──►  \$  list. */
end   /*j*/                                /*at this point, \$ has a leading blank.*/

say center(' first '       found        " circular primes ",  79, '─')
say strip(\$)
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
circP: procedure expose @. !.;  parse arg x 1 ox /*obtain a prime number to be examined.*/
do length(x)-1; parse var x f 2 y /*parse  X  number, rotating the digits*/
x= y || f                         /*construct a new possible circular P. */
if x<ox  then return 0            /*is number < the original?  ¬ circular*/
if \!.x  then return 0            /* "    "   not prime?       ¬ circular*/
end   /*length(x)···*/
return 1                                  /*passed all tests,  X is a circular P.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: @.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13; @.7=17; @.8=19        /*assign Ps; #Ps*/
!.= 0; !.2=1; !.3=1; !.5=1; !.7=1; !.11=1; !.13=1; !.17=1; !.19=1 /*   " primality*/
#= 8;  sq.#= @.# **2  /*number of primes so far; prime square*/
do j=@.#+4  by 2  to hip;  parse var j  ''  -1  _ /*get last decimal digit of J. */
if     _==5  then iterate;   if j// 3==0  then iterate;   if j// 7==0  then iterate
if j//11==0  then iterate;   if j//13==0  then iterate;   if j//17==0  then iterate
do k=8  while sq.k<=j                 /*divide by some generated odd primes. */
if j // @.k==0  then iterate j        /*Is J divisible by  P?  Then not prime*/
end   /*k*/                           /* [↓]  a prime  (J)  has been found.  */
#= #+1;   !.j= 1;   sq.#= j*j;   @.#= j   /*bump P cnt;  assign P to @.  and  !. */
end       /*j*/;                 return
```
output   when using the default inputs:
```
───────────────────────── first  19  circular primes ──────────────────────────
2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## Ring

```see "working..." + nl
see "First 19 circular numbers are:" + nl
n = 0
row = 0
Primes = []

while row < 19
n++
flag = 1
nStr = string(n)
lenStr = len(nStr)
for m = 1 to lenStr
leftStr = left(nStr,m)
rightStr = right(nStr,lenStr-m)
strOk = rightStr + leftStr
nOk = number(strOk)
ind = find(Primes,nOk)
if ind < 1 and strOk != nStr
ok
if not isprimeNumber(nOk) or ind > 0
flag = 0
exit
ok
next
if flag = 1
row++
see "" + n + " "
if row%5 = 0
see nl
ok
ok
end

see nl + "done..." + nl

if (num <= 1) return 0 ok
if (num % 2 = 0) and (num != 2) return 0 ok
for i = 2 to sqrt(num)
if (num % i = 0) return 0 ok
next
return 1```
Output:
```working...
First 19 circular numbers are:
2 3 5 7 11
13 17 37 79 113
197 199 337 1193 3779
11939 19937 193939 199933
done...
```

## Ruby

It takes more then 25 minutes to verify that R49081 is probably prime - omitted here.

```require 'gmp'
require 'prime'
candidate_primes = Enumerator.new do |y|
DIGS = [1,3,7,9]
[2,3,5,7].each{|n| y << n.to_s}
(2..).each do |size|
DIGS.repeated_permutation(size) do |perm|
y << perm.join if (perm == min_rotation(perm)) && GMP::Z(perm.join).probab_prime? > 0
end
end
end

def min_rotation(ar) = Array.new(ar.size){|n| ar.rotate(n)}.min

def circular?(num_str)
chars = num_str.chars
return GMP::Z(num_str).probab_prime? > 0 if chars.all?("1")
chars.size.times.all? do
GMP::Z(chars.rotate!.join).probab_prime? > 0
# chars.rotate!.join.to_i.prime?
end
end

puts "First 19 circular primes:"
puts candidate_primes.lazy.select{|cand| circular?(cand)}.take(19).to_a.join(", "),""
puts "First 5 prime repunits:"
reps = Prime.each.lazy.select{|pr| circular?("1"*pr)}.take(5).to_a
puts  reps.map{|r| "R" + r.to_s}.join(", "), ""
[5003, 9887, 15073, 25031].each {|rep| puts "R#{rep} circular_prime ? #{circular?("1"*rep)}" }
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933

First 5 prime repunits:
R2, R19, R23, R317, R1031

R5003 circular_prime ? false
R9887 circular_prime ? false
R15073 circular_prime ? false
R25031 circular_prime ? false
```

## Rust

Translation of: C
```// [dependencies]
// rug = "1.8"

fn is_prime(n: u32) -> bool {
if n < 2 {
return false;
}
if n % 2 == 0 {
return n == 2;
}
if n % 3 == 0 {
return n == 3;
}
let mut p = 5;
while p * p <= n {
if n % p == 0 {
return false;
}
p += 2;
if n % p == 0 {
return false;
}
p += 4;
}
true
}

fn cycle(n: u32) -> u32 {
let mut m: u32 = n;
let mut p: u32 = 1;
while m >= 10 {
p *= 10;
m /= 10;
}
m + 10 * (n % p)
}

fn is_circular_prime(p: u32) -> bool {
if !is_prime(p) {
return false;
}
let mut p2: u32 = cycle(p);
while p2 != p {
if p2 < p || !is_prime(p2) {
return false;
}
p2 = cycle(p2);
}
true
}

fn test_repunit(digits: usize) {
use rug::{integer::IsPrime, Integer};
let repunit = "1".repeat(digits);
if bignum.is_probably_prime(10) != IsPrime::No {
println!("R({}) is probably prime.", digits);
} else {
println!("R({}) is not prime.", digits);
}
}

fn main() {
use rug::{integer::IsPrime, Integer};
println!("First 19 circular primes:");
let mut count = 0;
let mut p: u32 = 2;
while count < 19 {
if is_circular_prime(p) {
if count > 0 {
print!(", ");
}
print!("{}", p);
count += 1;
}
p += 1;
}
println!();
println!("Next 4 circular primes:");
let mut repunit: u32 = 1;
let mut digits: usize = 1;
while repunit < p {
repunit = 10 * repunit + 1;
digits += 1;
}
let mut bignum = Integer::from(repunit);
count = 0;
while count < 4 {
if bignum.is_probably_prime(15) != IsPrime::No {
if count > 0 {
print!(", ");
}
print!("R({})", digits);
count += 1;
}
digits += 1;
bignum = bignum * 10 + 1;
}
println!();
test_repunit(5003);
test_repunit(9887);
test_repunit(15073);
test_repunit(25031);
test_repunit(35317);
test_repunit(49081);
}
```
Output:

Execution time is about 805 seconds on my system (3.2 GHz Quad-Core Intel Core i5, macOS 10.15.4).

```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19), R(23), R(317), R(1031)
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.
R(35317) is not prime.
R(49081) is probably prime.
```

## Scala

Translation of: Java
```object CircularPrimes {
def main(args: Array[String]): Unit = {
println("First 19 circular primes:")
var p = 2
var count = 0
while (count < 19) {
if (isCircularPrime(p)) {
if (count > 0) {
print(", ")
}
print(p)
count += 1
}
p += 1
}
println()

println("Next 4 circular primes:")
var repunit = 1
var digits = 1
while (repunit < p) {
repunit = 10 * repunit + 1
digits += 1
}
var bignum = BigInt.apply(repunit)
count = 0
while (count < 4) {
if (bignum.isProbablePrime(15)) {
if (count > 0) {
print(", ")
}
print(s"R(\$digits)")
count += 1
}
digits += 1
bignum = bignum * 10
bignum = bignum + 1
}
println()

testRepunit(5003)
testRepunit(9887)
testRepunit(15073)
testRepunit(25031)
}

def isPrime(n: Int): Boolean = {
if (n < 2) {
return false
}
if (n % 2 == 0) {
return n == 2
}
if (n % 3 == 0) {
return n == 3
}
var p = 5
while (p * p <= n) {
if (n % p == 0) {
return false
}
p += 2
if (n % p == 0) {
return false
}
p += 4
}
true
}

def cycle(n: Int): Int = {
var m = n
var p = 1
while (m >= 10) {
p *= 10
m /= 10
}
m + 10 * (n % p)
}

def isCircularPrime(p: Int): Boolean = {
if (!isPrime(p)) {
return false
}
var p2 = cycle(p)
while (p2 != p) {
if (p2 < p || !isPrime(p2)) {
return false
}
p2 = cycle(p2)
}
true
}

def testRepunit(digits: Int): Unit = {
val ru = repunit(digits)
if (ru.isProbablePrime(15)) {
println(s"R(\$digits) is probably prime.")
} else {
println(s"R(\$digits) is not prime.")
}
}

def repunit(digits: Int): BigInt = {
val ch = Array.fill(digits)('1')
BigInt.apply(new String(ch))
}
}
```
Output:
```First 19 circular primes:
2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933
Next 4 circular primes:
R(19), R(23), R(317), R(1031)
R(5003) is not prime.
R(9887) is not prime.
R(15073) is not prime.
R(25031) is not prime.```

## Sidef

Translation of: Raku
```func is_circular_prime(n) {
n.is_prime || return false

var circular = n.digits
circular.min < circular.tail && return false

for k in (1 ..^ circular.len) {
with (circular.rotate(k).digits2num) {|p|
(p.is_prime && (p >= n)) || return false
}
}

return true
}

say "The first 19 circular primes are:"
say 19.by(is_circular_prime)

say "\nThe next 4 circular primes, in repunit format, are:"
{|n| (10**n - 1)/9 -> is_prob_prime }.first(4, 4..Inf).each {|n|
say "R(#{n})"
}

say "\nRepunit testing:"
[5003, 9887, 15073, 25031, 35317, 49081].each {|n|
var now = Time.micro
say ("R(#{n}) -> ", is_prob_prime((10**n - 1)/9) ? 'probably prime' : 'composite',
" (took: #{'%.3f' % Time.micro-now} sec)")
}
```
Output:
```The first 19 circular primes are:
[2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933]

The next 4 circular primes, in repunit format, are:
R(19)
R(23)
R(317)
R(1031)

Repunit testing:
R(5003) -> composite (took: 0.024 sec)
R(9887) -> composite (took: 0.006 sec)
R(15073) -> composite (took: 0.389 sec)
R(25031) -> composite (took: 54.452 sec)
R(35317) -> composite (took: 0.875 sec)
```

## Wren

Translation of: Go
Library: Wren-math
Library: Wren-big
Library: Wren-fmt
Library: Wren-str

### Wren-cli

Second part is very slow - over 37 minutes to find all four.

```import "/math" for Int
import "/big" for BigInt
import "/str" for Str

var circs = []

var isCircular = Fn.new { |n|
var nn = n
var pow = 1 // will eventually contain 10 ^ d where d is number of digits in n
while (nn > 0) {
pow = pow * 10
nn = (nn/10).floor
}
nn = n
while (true) {
nn = nn * 10
var f = (nn/pow).floor // first digit
nn = nn + f * (1 - pow)
if (circs.contains(nn)) return false
if (nn == n) break
if (!Int.isPrime(nn)) return false
}
return true
}

var repunit = Fn.new { |n| BigInt.new(Str.repeat("1", n)) }

System.print("The first 19 circular primes are:")
var digits = [1, 3, 7, 9]
var q = [1, 2, 3, 5, 7, 9]  // queue the numbers to be examined
var fq = [1, 2, 3, 5, 7, 9] // also queue the corresponding first digits
var count = 0
while (true) {
var f = q[0]   // peek first element
var fd = fq[0] // peek first digit
if (Int.isPrime(f) && isCircular.call(f)) {
count = count + 1
if (count == 19) break
}
q.removeAt(0)  // pop first element
fq.removeAt(0) // ditto for first digit queue
if (f != 2 && f != 5) { // if digits > 1 can't contain a 2 or 5
// add numbers with one more digit to queue
// only numbers whose last digit >= first digit need be added
for (d in digits) {
if (d >= fd) {
}
}
}
}
System.print(circs)

System.print("\nThe next 4 circular primes, in repunit format, are:")
count = 0
var rus = []
var primes = Int.primeSieve(10000)
for (p in primes[3..-1]) {
if (repunit.call(p).isProbablePrime(1)) {
count = count + 1
if (count == 4) break
}
}
System.print(rus)
```
Output:
```The first 19 circular primes are:
[2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933]

The next 4 circular primes, in repunit format, are:
[R(19), R(23), R(317), R(1031)]
```

### Embedded

Library: Wren-gmp

A massive speed-up, of course, when GMP is plugged in for the 'probably prime' calculations. 11 minutes 19 seconds including the stretch goal.

```/* circular_primes_embedded.wren */

import "./gmp" for Mpz
import "./math" for Int
import "./fmt" for Fmt
import "./str" for Str

var circs = []

var isCircular = Fn.new { |n|
var nn = n
var pow = 1 // will eventually contain 10 ^ d where d is number of digits in n
while (nn > 0) {
pow = pow * 10
nn = (nn/10).floor
}
nn = n
while (true) {
nn = nn * 10
var f = (nn/pow).floor // first digit
nn = nn + f * (1 - pow)
if (circs.contains(nn)) return false
if (nn == n) break
if (!Int.isPrime(nn)) return false
}
return true
}

System.print("The first 19 circular primes are:")
var digits = [1, 3, 7, 9]
var q = [1, 2, 3, 5, 7, 9]  // queue the numbers to be examined
var fq = [1, 2, 3, 5, 7, 9] // also queue the corresponding first digits
var count = 0
while (true) {
var f = q[0]   // peek first element
var fd = fq[0] // peek first digit
if (Int.isPrime(f) && isCircular.call(f)) {
count = count + 1
if (count == 19) break
}
q.removeAt(0)  // pop first element
fq.removeAt(0) // ditto for first digit queue
if (f != 2 && f != 5) { // if digits > 1 can't contain a 2 or 5
// add numbers with one more digit to queue
// only numbers whose last digit >= first digit need be added
for (d in digits) {
if (d >= fd) {
}
}
}
}
System.print(circs)

System.print("\nThe next 4 circular primes, in repunit format, are:")
count = 0
var rus = []
var primes = Int.primeSieve(10000)
var repunit = Mpz.new()
for (p in primes[3..-1]) {
repunit.setStr(Str.repeat("1", p), 10)
if (repunit.probPrime(10) > 0) {
count = count + 1
if (count == 4) break
}
}
System.print(rus)
System.print("\nThe following repunits are probably circular primes:")
for (i in [5003, 9887, 15073, 25031, 35317, 49081]) {
repunit.setStr(Str.repeat("1", i), 10)
Fmt.print("R(\$-5d) : \$s", i, repunit.probPrime(15) > 0)
}
```

Output:
```The first 19 circular primes are:
[2, 3, 5, 7, 11, 13, 17, 37, 79, 113, 197, 199, 337, 1193, 3779, 11939, 19937, 193939, 199933]

The next 4 circular primes, in repunit format, are:
[R(19), R(23), R(317), R(1031)]

The following repunits are probably circular primes:
R(5003 ) : false
R(9887 ) : false
R(15073) : false
R(25031) : false
R(35317) : false
R(49081) : true
```

## XPL0

```func IsPrime(N);        \Return 'true' if N > 2 is a prime number
int  N, I;
[if (N&1) = 0 \even number\ then return false;
for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];

func CircPrime(N0);     \Return 'true' if N0 is a circular prime
int  N0, N, Digits, Rotation, I, R;
[N:= N0;
Digits:= 0;             \count number of digits in N
repeat  Digits:= Digits+1;
N:= N/10;
until   N = 0;
N:= N0;
for Rotation:= 0 to Digits-1 do
[if not IsPrime(N) then return false;
N:= N/10;           \rotate least sig digit into high end
R:= rem(0);
for I:= 0 to Digits-2 do
R:= R*10;
N:= N+R;
if N0 > N then      \reject N0 if it has a smaller prime rotation
return false;
];
return true;
];

int Counter, N;
[IntOut(0, 2);  ChOut(0, ^ );   \show first circular prime
Counter:= 1;
N:= 3;                          \remaining primes are odd
loop    [if CircPrime(N) then
[IntOut(0, N);  ChOut(0, ^ );
Counter:= Counter+1;
if Counter >= 19 then quit;
];
N:= N+2;
];
]```
Output:
```2 3 5 7 11 13 17 37 79 113 197 199 337 1193 3779 11939 19937 193939 199933
```

## Zig

As of now (Zig release 0.8.1), Zig has large integer support, but there is not yet a prime test in the standard library. Therefore, we only find the circular primes < 1e6. As with the Prolog version, we only check numbers composed of 1, 3, 7, or 9.

```const std = @import("std");
const math = std.math;
const heap = std.heap;
const stdout = std.io.getStdOut().writer();

pub fn main() !void {
var arena = heap.ArenaAllocator.init(heap.page_allocator);
defer arena.deinit();

var candidates = std.PriorityQueue(u32).init(&arena.allocator, u32cmp);
defer candidates.deinit();

try stdout.print("The circular primes are:\n", .{});
try stdout.print("{:10}" ** 4, .{ 2, 3, 5, 7 });

var c: u32 = 4;
while (true) {
var n = candidates.remove();
if (n > 1_000_000)
break;
if (n > 10 and circular(n)) {
try stdout.print("{:10}", .{n});
c += 1;
if (c % 10 == 0)
try stdout.print("\n", .{});
}
try candidates.add(10 * n + 1);
try candidates.add(10 * n + 3);
try candidates.add(10 * n + 7);
try candidates.add(10 * n + 9);
}
try stdout.print("\n", .{});
}

fn u32cmp(a: u32, b: u32) math.Order {
return math.order(a, b);
}

fn circular(n0: u32) bool {
if (!isprime(n0))
return false
else {
var n = n0;
var d = @floatToInt(u32, @log10(@intToFloat(f32, n)));
return while (d > 0) : (d -= 1) {
n = rotate(n);
if (n < n0 or !isprime(n))
break false;
} else true;
}
}

fn rotate(n: u32) u32 {
if (n == 0)
return 0
else {
const d = @floatToInt(u32, @log10(@intToFloat(f32, n))); // digit count - 1
const m = math.pow(u32, 10, d);
return (n % m) * 10 + n / m;
}
}

fn isprime(n: u32) bool {
if (n < 2)
return false;

inline for ([3]u3{ 2, 3, 5 }) |p| {
if (n % p == 0)
return n == p;
}

const wheel235 = [_]u3{
6, 4, 2, 4, 2, 4, 6, 2,
};
var i: u32 = 1;
var f: u32 = 7;
return while (f * f <= n) {
if (n % f == 0)
break false;
f += wheel235[i];
i = (i + 1) & 0x07;
} else true;
}
```
Output:
```The circular primes are:
2         3         5         7        11        13        17        37        79       113
197       199       337      1193      3779     11939     19937    193939    199933
```