Palindromic primes
- Task
Find and show all palindromic primes n, where n < 1000
ALGOL 68
Generates the palindrmic 3 digit numbers and uses the observations that all 1 digit primes are palindromic and that for 2 digit numbers, only multiples of 11 are palindromic and hence 11 is the only two digit palindromic prime.
<lang algol68>BEGIN # find primes that are palendromic in base 10 #
INT max prime = 999; # sieve the primes to max prime # PR read "primes.incl.a68" PR []BOOL prime = PRIMESIEVE max prime; # print the palendromic primes in the base 10 # # all 1 digit primes are palindromic # FOR n TO 9 DO IF prime[ n ] THEN print( ( " ", whole( n, 0 ) ) ) FI OD; # the only palindromic 2 digit numbers are multiples of 11 # # so 11 is the only possible 2 digit palindromic prime # IF prime[ 11 ] THEN print( ( " 11" ) ) FI; # three digit numbers, the first and last digits must be odd # # and cannot be 5 (as the number would be divisible by 5) # FOR fl BY 2 TO 9 DO IF fl /= 5 THEN FOR m FROM 0 TO 9 DO INT n = ( ( ( fl * 10 ) + m ) * 10 ) + fl; IF prime[ n ] THEN # have a palindromic prime # print( ( " ", whole( n, 0 ) ) ) FI OD FI OD; print( ( newline ) )
END</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Arturo
<lang rebol>loop split.every: 10 select 2..1000 'x [
and? prime? x x = to :integer reverse to :string x
] 'a -> print map a => [pad to :string & 4]</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
AWK
<lang AWK>
- syntax: GAWK -f PALINDROMIC_PRIMES.AWK
BEGIN {
start = 1 stop = 999 for (i=start; i<=stop; i++) { if (is_prime(i) && reverse(i) == i) { printf("%d ",i) count++ } } printf("\nPalindromic 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 reverse(str, i,rts) {
for (i=length(str); i>=1; i--) { rts = rts substr(str,i,1) } return(rts)
} </lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 Palindromic primes 1-999: 20
Factor
Simple
A simple solution that suffices for the task:
<lang factor>USING: kernel math.primes present prettyprint sequences ;
1000 primes-upto [ present dup reverse = ] filter stack.</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Fast
A much more efficient solution that generates palindromic numbers directly and filters primes from them:
<lang factor>USING: io kernel lists lists.lazy math math.functions math.primes math.ranges prettyprint sequences tools.memory.private ;
! Create a palindrome from its base natural number.
- create-palindrome ( n odd? -- m )
dupd [ 10 /i ] when swap [ over 0 > ] [ 10 * [ 10 /mod ] [ + ] bi* ] while nip ;
! Create an ordered infinite lazy list of palindromic numbers.
- lpalindromes ( -- l )
0 lfrom [ 10 swap ^ dup 10 * [a,b) [ [ t create-palindrome ] map ] [ [ f create-palindrome ] map ] bi [ sequence>list ] bi@ lappend ] lmap-lazy lconcat ;
- lpalindrome-primes ( -- list )
lpalindromes [ prime? ] lfilter ;
"10,000th palindromic prime:" print 9999 lpalindrome-primes lnth commas print nl
"Palindromic primes less than 1,000:" print lpalindrome-primes [ 1000 < ] lwhile [ . ] leach</lang>
- Output:
10,000th palindromic prime: 13,649,694,631 Palindromic primes less than 1,000: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
FreeBASIC
<lang freebasic>#include "isprime.bas"
function is_pal( s as string ) as boolean
dim as integer i, n = len(s) for i = 1 to n\2 if mid(s,i,1)<>mid(s,n-i+1,1) then return false next i return true
end function
for i as uinteger = 2 to 999
if is_pal( str(i) ) andalso isprime(i) then print i;" ";
next i : print</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Go
<lang go>package main
import (
"fmt" "rcu"
)
func reversed(n int) int {
rev := 0 for n > 0 { rev = rev*10 + n%10 n /= 10 } return rev
}
func main() {
primes := rcu.Primes(99999) var pals []int for _, p := range primes { if p == reversed(p) { pals = append(pals, p) } } fmt.Println("Palindromic primes under 1,000:") var smallPals, bigPals []int for _, p := range pals { if p < 1000 { smallPals = append(smallPals, p) } else { bigPals = append(bigPals, p) } } rcu.PrintTable(smallPals, 10, 3, false) fmt.Println() fmt.Println(len(smallPals), "such primes found.")
fmt.Println("\nAdditional palindromic primes under 100,000:") rcu.PrintTable(bigPals, 10, 6, true) fmt.Println() fmt.Println(len(bigPals), "such primes found,", len(pals), "in all.")
}</lang>
- Output:
Same as Wren entry.
Haskell
<lang haskell>import Data.Numbers.Primes
palindromicPrimes :: [Integer] palindromicPrimes =
filter (((==) <*> reverse) . show) primes
main :: IO () main =
mapM_ print $ takeWhile (1000 >) palindromicPrimes</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
jq
Works with gojq, the Go implementation of jq
In this entry, we define both a naive generate-and-test generator of the palindromic primes, and a more sophisticated one that is well-suited for generating very large numbers of such primes, as illustrated by counting the number less than 10^10.
For a suitable implementation of `is_prime` as used here, see Erdős-primes#jq.
Preliminaries <lang jq>def count(s): reduce s as $x (null; .+1);
def emit_until(cond; stream): label $out | stream | if cond then break $out else . end;</lang> Naive version <lang jq> def primes:
2, (range(3;infinite;2) | select(is_prime));
def palindromic_primes_slowly:
primes | select( tostring|explode | (. == reverse));
</lang> Less naive version <lang jq># Output: an unbounded stream of palindromic primes def palindromic_primes:
# Output: a naively constructed stream of palindromic strings of length >= 2 def palindromic_candidates: def rev: # reverse a string explode|reverse|implode; def unconstrained($length): if $length==1 then range(0;10) | tostring else (range(0;10)|tostring) | . + unconstrained($length -1 ) end; def middle($length): # $length > 0 if $length==1 then range(0;10) | tostring elif $length % 2 == 1 then (($length -1) / 2) as $len | unconstrained($len) as $left | (range(0;10) | tostring) as $mid | $left + $mid + ($left|rev) else ($length / 2) as $len | unconstrained($len) as $left | $left + ($left|rev) end; # palindromes with an even number of digits are divisible by 11
range(1;infinite;2) as $mid | ("1", "3", "7", "9") as $start | $start + middle($mid) + $start ; 2, 3, 5, 7, 11, (palindromic_candidates | tonumber | select(is_prime));</lang>
Demonstrations <lang jq>"Palindromic primes <= 1000:", emit_until(. >= 1000; palindromic_primes),
((range(5;10) | pow(10;.)) as $n
| "\nNumber of palindromic primes <= \($n): \(count(emit_until(. >= $n; palindromic_primes)))" )</lang>
- Output:
Palindromic primes <= 1000: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 Number of palindromic primes <= 100000: 113 Number of palindromic primes <= 1000000: 113 Number of palindromic primes <= 10000000: 781 Number of palindromic primes <= 100000000: 781 Number of palindromic primes <= 1000000000: 5953 Number of palindromic primes <= 10000000000: 5953
Julia
Generator method. <lang julia>using Primes
parray = [2, 3, 5, 7, 9, 11]
results = vcat(parray, filter(isprime, [100j + 10i + j for i in 0:9, j in 1:9]))
println(results)
</lang>
- Output:
[2, 3, 5, 7, 9, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929]
Nim
<lang Nim>import strutils
const N = 999
func isPrime(n: Positive): bool =
if (n and 1) == 0: return n == 2 var m = 3 while m * m <= n: if n mod m == 0: return false inc m, 2 result = true
func reversed(n: Positive): int =
var n = n.int while n != 0: result = 10 * result + n mod 10 n = n div 10
func isPalindromic(n: Positive): bool =
n == reversed(n)
var result: seq[int] for n in 2..N:
if n.isPrime and n.isPalindromic: result.add n
for i, n in result:
stdout.write ($n).align(3) stdout.write if (i + 1) mod 10 == 0: '\n' else: ' '</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Perl
<lang Perl>#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Palindromic_primes use warnings;
$_ == reverse and (1 x $_ ) !~ /^(11+)\1+$/ and print "$_\n" for 2 .. 1e3;</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Phix
filter primes for palindromicness
function palindrome(string s) return s=reverse(s) end function for l=3 to 5 by 2 do integer limit = power(10,l) -- 1000 then 100000 sequence res = get_primes_le(limit) res = apply(true,sprintf,{{"%d"},res}) res = filter(res,palindrome) string s = join(shorten(res,"",5)) printf(1,"found %d < %,d: %s\n",{length(res),limit,s}) end for
- Output:
found 20 < 1,000: 2 3 5 7 11 ... 757 787 797 919 929 found 113 < 100,000: 2 3 5 7 11 ... 97379 97579 97879 98389 98689
filter palindromes for primality
sequence r = {} for l=2 to 3 do for i=1 to power(10,l) do string s = sprintf("%d",i) integer t = to_number(s&reverse(s[1..$-1])), u = to_number(s&reverse(s)) if is_prime(t) then r &= t end if if is_prime(u) then r &= u end if end for r = unique(r) string s = join(shorten(apply(true,sprintf,{{"%d"},r}),"",5)) printf(1,"found %d < %,d: %s\n",{length(r),power(10,l*2-1),s}) end for
Same output. Didn't actually test if this way was any faster, but expect it would be.
Python
A non-finite generator of palindromic primes – one of many approaches to solving this problem in Python. <lang python>Palindromic primes
from itertools import takewhile
- palindromicPrimes :: Generator [Int]
def palindromicPrimes():
An infinite stream of palindromic primes def p(n): s = str(n) return s == s[::-1] return (n for n in primes() if p(n))
- ------------------------- TEST -------------------------
def main():
Palindromic primes below 1000 print('\n'.join( str(x) for x in takewhile( lambda n: 1000 > n, palindromicPrimes() ) ))
- ----------------------- GENERIC ------------------------
- primes :: [Int]
def primes():
Non finite sequence of prime numbers. n = 2 dct = {} while True: if n in dct: for p in dct[n]: dct.setdefault(n + p, []).append(p) del dct[n] else: yield n dct[n * n] = [n] n = 1 + n
- MAIN ---
if __name__ == '__main__':
main()
</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
Raku
<lang perl6>say "{+$_} matching numbers:\n{.batch(10)».fmt('%3d').join: "\n"}"
given (^1000).grep: { .is-prime and $_ eq .flip };</lang>
- Output:
20 matching numbers: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929
REXX
<lang rexx>/*REXX program finds and displays palindromic primes for all N < 1,000. */ parse arg hi cols . /*obtain optional argument from the CL.*/ if hi== | hi=="," then hi= 1000 /*Not specified? Then use the default.*/ if cols== | cols=="," then cols= 10 /* " " " " " " */ call genP /*build array of semaphores for primes.*/ w= max(8, length( commas(hi) ) ) /*max width of a number in any column. */
title= ' palindromic primes that are < ' commas(hi)
if cols>0 then say ' index │'center(title, 1 + cols*(w+1) ) if cols>0 then say '───────┼'center("", 1 + cols*(w+1), '─') finds= 0; idx= 1 /*define # of palindromic primes & idx.*/ $= /*a list of palindromic primes (so far)*/
do j=1 for hi; if \!.j then iterate /*Is this number not prime? Then skip.*/ /* ◄■■■■■■■■ a filter. */ if j\==reverse(j) then iterate /*Not a palindromic prime? " " */ /* ◄■■■■■■■■ a filter. */ finds= finds + 1 /*bump the number of palindromic primes*/ if cols<0 then iterate /*Build the list (to be shown later)? */ $= $ right( commas(j), w) /*add a palindromic prime $ list.*/ if finds//cols\==0 then iterate /*have we populated a line of output? */ say center(idx, 7)'|' substr($, 2); $= /*display what we have so for (cols). */ idx= idx + cols /*bump the index count for the output*/ end /*j*/
if $\== then say center(idx, 7)'|' substr($, 2) /*possibly display residual output.*/ if cols>0 then say '───────┴'center("", 1 + cols*(w+1), '─') say say 'Found ' commas(finds) title 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: !.= 0; hip= max(hi, copies(9,length(hi))) /*placeholders for primes (semaphores).*/
@.1=2; @.2=3; @.3=5; @.4=7; @.5=11 /*define some low primes. */ !.2=1; !.3=1; !.5=1; !.7=1; !.11=1 /* " " " " flags. */ #=5; sq.#= @.# **2 /*number of primes so far; prime². */ /* [↓] generate more primes ≤ high.*/ do j=@.#+2 by 2 to hip /*find odd primes from here on. */ parse var j -1 _; if _==5 then iterate /*J ÷ by 5? (right digit).*/ if j//3==0 then iterate; if j//7==0 then iterate /*" " " 3? J ÷ by 7? */ do k=5 while sq.k<=j /* [↓] divide by the known odd primes.*/ if j // @.k == 0 then iterate j /*Is J ÷ X? Then not prime. ___ */ end /*k*/ /* [↑] only process numbers ≤ √ J */ #= #+1; @.#= j; sq.#= j*j; !.j= 1 /*bump # of Ps; assign next P; P²; P# */ end /*j*/; return</lang>
- output when using the default inputs:
index │ palindromic primes that are < 1,000 ───────┼─────────────────────────────────────────────────────────────────────────────────────────── 1 │ 2 3 5 7 11 101 131 151 181 191 11 │ 313 353 373 383 727 757 787 797 919 929 ───────┴─────────────────────────────────────────────────────────────────────────────────────────── Found 20 palindromic primes that are < 1,000
- output when using the input of: 100000
index │ palindromic primes that are < 100,000 ───────┼─────────────────────────────────────────────────────────────────────────────────────────── 1 │ 2 3 5 7 11 101 131 151 181 191 11 │ 313 353 373 383 727 757 787 797 919 929 21 │ 10,301 10,501 10,601 11,311 11,411 12,421 12,721 12,821 13,331 13,831 31 │ 13,931 14,341 14,741 15,451 15,551 16,061 16,361 16,561 16,661 17,471 41 │ 17,971 18,181 18,481 19,391 19,891 19,991 30,103 30,203 30,403 30,703 51 │ 30,803 31,013 31,513 32,323 32,423 33,533 34,543 34,843 35,053 35,153 61 │ 35,353 35,753 36,263 36,563 37,273 37,573 38,083 38,183 38,783 39,293 71 │ 70,207 70,507 70,607 71,317 71,917 72,227 72,727 73,037 73,237 73,637 81 │ 74,047 74,747 75,557 76,367 76,667 77,377 77,477 77,977 78,487 78,787 91 │ 78,887 79,397 79,697 79,997 90,709 91,019 93,139 93,239 93,739 94,049 101 │ 94,349 94,649 94,849 94,949 95,959 96,269 96,469 96,769 97,379 97,579 111 │ 97,879 98,389 98,689 ───────┴─────────────────────────────────────────────────────────────────────────────────────────── Found 113 palindromic primes that are < 100,000
Ring
<lang ring> load "stdlib.ring"
see "working..." + nl see "Palindromic primes are:" + nl row = 0 limit1 = 1000 limit2 = 100000
palindromicPrimes(limit1)
see "Found " + row + " palindromic primes" + nl + nl see "palindromic primes that are < 100,000" + nl
palindromicPrimes(limit2)
see nl + "Found " + row + " palindromic primes that are < 100,000" + nl see "done..." + nl
func palindromicPrimes(limit)
row = 0 for n = 1 to limit strn = string(n) if ispalindrome(strn) and isprime(n) row = row + 1 see "" + n + " " if row%5 = 0 see nl ok ok next
</lang>
- Output:
working... Palindromic primes are: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 Found 20 palindromic primes palindromic primes that are < 100,000 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 10301 10501 10601 11311 11411 12421 12721 12821 13331 13831 13931 14341 14741 15451 15551 16061 16361 16561 16661 17471 17971 18181 18481 19391 19891 19991 30103 30203 30403 30703 30803 31013 31513 32323 32423 33533 34543 34843 35053 35153 35353 35753 36263 36563 37273 37573 38083 38183 38783 39293 70207 70507 70607 71317 71917 72227 72727 73037 73237 73637 74047 74747 75557 76367 76667 77377 77477 77977 78487 78787 78887 79397 79697 79997 90709 91019 93139 93239 93739 94049 94349 94649 94849 94949 95959 96269 96469 96769 97379 97579 97879 98389 98689 Found 113 palindromic primes that are < 100,000 done...
Rust
This includes a solution for the similar task Palindromic primes in base 16. <lang rust>// [dependencies] // primal = "0.3" // radix_fmt = "1.0"
fn reverse(base: u64, mut n: u64) -> u64 {
let mut rev = 0; while n > 0 { rev = rev * base + n % base; n /= base; } rev
}
fn palindromes(base: u64) -> impl std::iter::Iterator<Item = u64> {
let mut lower = 1; let mut upper = base; let mut next = 0; let mut even = false; std::iter::from_fn(move || { next += 1; if next == upper { if even { lower = upper; upper *= base; } next = lower; even = !even; } Some(match even { true => next * upper + reverse(base, next), _ => next * lower + reverse(base, next / base), }) })
}
fn print_palindromic_primes(base: u64, limit: u64) {
let width = (limit as f64).log(base as f64).ceil() as usize; let mut count = 0; let columns = 80 / (width + 1); println!("Base {} palindromic primes less than {}:", base, limit); for p in palindromes(base) .take_while(|x| *x < limit) .filter(|x| primal::is_prime(*x)) { count += 1; print!( "{:>w$}", radix_fmt::radix(p, base as u8).to_string(), w = width ); if count % columns == 0 { println!(); } else { print!(" "); } } if count % columns != 0 { println!(); } println!("Count: {}", count);
}
fn count_palindromic_primes(base: u64, limit: u64) {
let c = palindromes(base) .take_while(|x| *x < limit) .filter(|x| primal::is_prime(*x)) .count(); println!( "Number of base {} palindromic primes less than {}: {}", base, limit, c );
}
fn main() {
print_palindromic_primes(10, 1000); println!(); print_palindromic_primes(10, 100000); println!(); count_palindromic_primes(10, 1000000000); println!(); print_palindromic_primes(16, 500);
}</lang>
- Output:
Base 10 palindromic primes less than 1000: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 Count: 20 Base 10 palindromic primes less than 100000: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 10301 10501 10601 11311 11411 12421 12721 12821 13331 13831 13931 14341 14741 15451 15551 16061 16361 16561 16661 17471 17971 18181 18481 19391 19891 19991 30103 30203 30403 30703 30803 31013 31513 32323 32423 33533 34543 34843 35053 35153 35353 35753 36263 36563 37273 37573 38083 38183 38783 39293 70207 70507 70607 71317 71917 72227 72727 73037 73237 73637 74047 74747 75557 76367 76667 77377 77477 77977 78487 78787 78887 79397 79697 79997 90709 91019 93139 93239 93739 94049 94349 94649 94849 94949 95959 96269 96469 96769 97379 97579 97879 98389 98689 Count: 113 Number of base 10 palindromic primes less than 1000000000: 5953 Base 16 palindromic primes less than 500: 2 3 5 7 b d 11 101 151 161 191 1b1 1c1 Count: 13
Sidef
<lang ruby>func palindromic_primes(upto, base = 10) {
var list = [] for (var p = 2; p <= upto; p = p.next_palindrome(base)) { list << p if p.is_prime } return list
}
say palindromic_primes(1000)
for n in (1..10) {
var count = palindromic_primes(10**n).len say "There are #{count} palindromic primes <= 10^#{n}"
}</lang>
- Output:
[2, 3, 5, 7, 11, 101, 131, 151, 181, 191, 313, 353, 373, 383, 727, 757, 787, 797, 919, 929] There are 4 palindromic primes <= 10^1 There are 5 palindromic primes <= 10^2 There are 20 palindromic primes <= 10^3 There are 20 palindromic primes <= 10^4 There are 113 palindromic primes <= 10^5 There are 113 palindromic primes <= 10^6 There are 781 palindromic primes <= 10^7 There are 781 palindromic primes <= 10^8 There are 5953 palindromic primes <= 10^9 There are 5953 palindromic primes <= 10^10
Wren
<lang ecmascript>import "/math" for Int import "/fmt" for Fmt import "/seq" for Lst
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(99999) var pals = [] for (p in primes) {
if (p == reversed.call(p)) pals.add(p)
} System.print("Palindromic primes under 1,000:") var smallPals = pals.where { |p| p < 1000 }.toList for (chunk in Lst.chunks(smallPals, 10)) Fmt.print("$3d", chunk) System.print("\n%(smallPals.count) such primes found.")
System.print("\nAdditional palindromic primes under 100,000:") var bigPals = pals.where { |p| p >= 1000 }.toList for (chunk in Lst.chunks(bigPals, 10)) Fmt.print("$,6d", chunk) System.print("\n%(bigPals.count) such primes found, %(pals.count) in all.")</lang>
- Output:
Palindromic primes under 1,000: 2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929 20 such primes found. Additional palindromic primes under 100,000: 10,301 10,501 10,601 11,311 11,411 12,421 12,721 12,821 13,331 13,831 13,931 14,341 14,741 15,451 15,551 16,061 16,361 16,561 16,661 17,471 17,971 18,181 18,481 19,391 19,891 19,991 30,103 30,203 30,403 30,703 30,803 31,013 31,513 32,323 32,423 33,533 34,543 34,843 35,053 35,153 35,353 35,753 36,263 36,563 37,273 37,573 38,083 38,183 38,783 39,293 70,207 70,507 70,607 71,317 71,917 72,227 72,727 73,037 73,237 73,637 74,047 74,747 75,557 76,367 76,667 77,377 77,477 77,977 78,487 78,787 78,887 79,397 79,697 79,997 90,709 91,019 93,139 93,239 93,739 94,049 94,349 94,649 94,849 94,949 95,959 96,269 96,469 96,769 97,379 97,579 97,879 98,389 98,689 93 such primes found, 113 in all.
XPL0
<lang 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); int N, M; [M:= 0; repeat N:= N/10;
M:= M*10 + rem(0);
until N=0; return M; ];
int Count, N; [Count:= 0; for N:= 1 to 1000-1 do
if N=Reverse(N) & IsPrime(N) then [IntOut(0, N); Count:= Count+1; if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\); ];
]</lang>
- Output:
2 3 5 7 11 101 131 151 181 191 313 353 373 383 727 757 787 797 919 929