I'm working on modernizing Rosetta Code's infrastructure. Starting with communications. Please accept this time-limited open invite to RC's Slack.. --Michael Mol (talk) 20:59, 30 May 2020 (UTC)

# Substring primes

Substring primes is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Find all primes in which all substrings (in base ten) are also primes.

This can be achieved by filtering all primes below 500 (there are 95 of them), but there are better ways.

Solve by testing at most 15 numbers for primality. Show a list of all numbers tested that were not prime.

## ALGOL 68

BEGIN # find primes where all substrings of the digits are prime #
# find the primes of interest #
[]BOOL prime = PRIMESIEVE 500;
FOR p TO UPB prime DO
IF prime[ p ] THEN
INT d := 10;
BOOL is substring := TRUE;
WHILE is substring AND d <= UPB prime DO
INT n := p;
WHILE is substring AND n > 0 DO
INT sub digits = n MOD d;
is substring := IF sub digits = 0 THEN FALSE ELSE prime[ sub digits ] FI;
n OVERAB 10
OD;
d *:= 10
OD;
IF is substring THEN print( ( " ", whole( p, 0 ) ) ) FI
FI
OD
END
Output:
2 3 5 7 23 37 53 73 373

## ALGOL W

starts with a hardcoded list of 1 digit primes ( 2, 3, 5, 7 ) and constructs the remaining members of the sequence (in order) using the observations that the final digit must be prime and can't be 2 or 5 or the number wouldn't be prime. Additionally, the final digit pair cannot be 33 or 77 as these are divisible by 11.

begin % find primes where every substring of the digits is also priome %
% 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 s := i * i step ii until n do p( s ) := false
end for_i ;
end Eratosthenes ;
% it can be shown that all the required primes are under 1000, however we will %
% not assume this, so we will allow for 4 digit numbers  %
integer MAX_NUMBER, MAX_SUBSTRING;
MAX_NUMBER  := 10000;
MAX_SUBSTRING := 100; % assume there will be at most 100 such primes  %
begin
logical array prime( 1 :: MAX_NUMBER );
integer array sPrime( 1 :: MAX_SUBSTRING );
integer tCount, sCount, sPos;
% adds a substring prime to the list %
procedure addPrime ( integer value p ) ;
begin
sCount := sCount + 1;
sPrime( sCount ) := p;
writeon( i_w := 1, s_w := 0, " ", p )
% sieve the primes to MAX_NUMBER %
Eratosthenes( prime, MAX_NUMBER );
% clearly, the 1 digit primes are all substring primes %
sCount := 0;
for i := 1 until MAX_SUBSTRING do sPrime( i ) := 0;
for i := 2, 3, 5, 7 do addPrime( i );
% the subsequent primes can only have 3 or 7 as a final digit as they must end  %
% with a prime digit and 2 and 5 would mean the number was divisible by 2 or 5  %
% as all substrings on the prime must also be prime, 33 and 77 are not possible %
% final digit pairs  %
sPos := 1;
while sPrime( sPos ) not = 0 do begin
integer n3, n7;
n3 := ( sPrime( sPos ) * 10 ) + 3;
n7 := ( sPrime( sPos ) * 10 ) + 7;
if sPrime( sPos ) rem 10 not = 3 and prime( n3 ) then addPrime( n3 );
if sPrime( sPos ) rem 10 not = 7 and prime( n7 ) then addPrime( n7 );
sPos := sPos + 1
end while_sPrime_sPos_ne_0 ;
write( i_w := 1, s_w := 0, "Found ", sCount, " substring primes" )
end
end.
Output:
2 3 5 7 23 37 53 73 373
Found 9 substring primes

## AWK

# syntax: GAWK -f SUBSTRING_PRIMES.AWK
# converted from FreeBASIC
BEGIN {
start = 1
stop = 500
for (i=start; i<=stop; i++) {
if (is_substring_prime(i)) {
printf("%d ",i)
count++
}
}
printf("\nSubString 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 is_substring_prime(n) {
if (!is_prime(i)) { return(0) }
if (n < 10) { return(1) }
if (!is_prime(n%100)) { return(0) }
if (!is_prime(n%10)) { return(0) }
if (!is_prime(int(n/10))) { return(0) }
if (n < 100) { return(1) }
if (!is_prime(int(n/100))) { return(0) }
if (!is_prime(int((n%100)/10))) { return(0) }
return(1)
}

Output:
2 3 5 7 23 37 53 73 373
SubString Primes 1-500: 9

## C++

#include <iostream>
#include <vector>

std::vector<bool> prime_sieve(size_t limit) {
std::vector<bool> sieve(limit, true);
if (limit > 0)
sieve[0] = false;
if (limit > 1)
sieve[1] = false;
for (size_t i = 4; i < limit; i += 2)
sieve[i] = false;
for (size_t p = 3; ; p += 2) {
size_t q = p * p;
if (q >= limit)
break;
if (sieve[p]) {
size_t inc = 2 * p;
for (; q < limit; q += inc)
sieve[q] = false;
}
}
return sieve;
}

bool substring_prime(const std::vector<bool>& sieve, unsigned int n) {
for (; n != 0; n /= 10) {
if (!sieve[n])
return false;
for (unsigned int p = 10; p < n; p *= 10) {
if (!sieve[n % p])
return false;
}
}
return true;
}

int main() {
const unsigned int limit = 500;
std::vector<bool> sieve = prime_sieve(limit);
for (unsigned int i = 2; i < limit; ++i) {
if (substring_prime(sieve, i))
std::cout << i << '\n';
}
return 0;
}
Output:
2
3
5
7
23
37
53
73
373

## Factor

For fun, a translation of FreeBASIC.

Translation of: FreeBASIC
Works with: Factor version 0.99 2021-02-05
USING: combinators kernel math math.primes prettyprint sequences ;

:: ssp? ( n -- ? )
{
{ [ n prime? not ] [ f ] }
{ [ n 10 < ] [ t ] }
{ [ n 100 mod prime? not ] [ f ] }
{ [ n 10 mod prime? not ] [ f ] }
{ [ n 10 /i prime? not ] [ f ] }
{ [ n 100 < ] [ t ] }
{ [ n 100 /i prime? not ] [ f ] }
{ [ n 100 mod 10 /i prime? not ] [ f ] }
[ t ]
} cond ;

500 <iota> [ ssp? ] filter .
Output:
V{ 2 3 5 7 23 37 53 73 373 }

## FreeBASIC

Since this is limited to one, two, or three-digit numbers I will be a bit cheeky.

#include "isprime.bas"

function is_ssp(n as uinteger) as boolean
if not isprime(n) then return false
if n < 10 then return true
if not isprime(n mod 100) then return false
if not isprime(n mod 10) then return false
if not isprime(n\10) then return false
if n < 100 then return true
if not isprime(n\100) then return false
if not isprime( (n mod 100)\10 ) then return false
return true
end function

for i as uinteger = 1 to 500
if is_ssp(i) then print i;" ";
next i
print
Output:
2 3 5 7 23 37 53 73 373

## Go

Translation of: Wren
Library: Go-rcu

### Using a limit

package main

import (
"fmt"
"rcu"
)

func main() {
primes := rcu.Primes(499)
var sprimes []int
for _, p := range primes {
digits := rcu.Digits(p, 10)
var b1 = true
for _, d := range digits {
if !rcu.IsPrime(d) {
b1 = false
break
}
}
if b1 {
if len(digits) < 3 {
sprimes = append(sprimes, p)
} else {
b2 := rcu.IsPrime(digits[0]*10 + digits[1])
b3 := rcu.IsPrime(digits[1]*10 + digits[2])
if b2 && b3 {
sprimes = append(sprimes, p)
}
}
}
}
fmt.Println("Found", len(sprimes), "primes < 500 where all substrings are also primes, namely:")
fmt.Println(sprimes)
}
Output:
Found 9 primes < 500 where all substrings are also primes, namely:
[2 3 5 7 23 37 53 73 373]

package main

import (
"fmt"
"rcu"
)

func main() {
results := []int{2, 3, 5, 7} // number must begin with a prime digit
odigits := []int{3, 7} // other digits must be 3 or 7
tests := 4 // i.e. to obtain initial results in the first place

// check 2 digit numbers or greater
// note that 'results' is a moving feast. If the loop eventually terminates that's all there are.
for i := 0; i < len(results); i++ {
r := results[i]
for _, od := range odigits {
// the last digit of r and od must be different otherwise number would be divisible by 11
if (r % 10) != od {
n := r*10 + od
if rcu.IsPrime(n) {
results = append(results, n)
} else {
}
tests++
}
}
}
fmt.Println("There are", len(results), "primes where all substrings are also primes, namely:")
fmt.Println(results)
fmt.Println("\nThe following numbers were also tested for primality but found to be composite:")
fmt.Println("\nTotal number of primality tests =", tests)
}
Output:
There are 9 primes where all substrings are also primes, namely:
[2 3 5 7 23 37 53 73 373]

The following numbers were also tested for primality but found to be composite:
[27 57 237 537 737 3737]

Total number of primality tests = 15

## jq

Works with: jq

Works with gojq, the Go implementation of jq

In the following, we verify that there are only nine "substring primes" as defined for this task..

See e.g. Erdős-primes#jq for a suitable implementation of `is_prime`.

def emit_until(cond; stream): label \$out | stream | if cond then break
\$out else . end;

def primes:
2, (range(3;infinite;2) | select(is_prime));

def is_substring(checkPrime):
def isp: if . == "" then true else tonumber|is_prime end;
(if checkPrime then is_prime else true end)
and (tostring
| . as \$s
| all(range(0;length) as \$i | range(\$i; length+1) as \$j | [\$i,\$j];
\$s[.[0]:.[1]]|isp ));

# Output an array of the substring primes less than or equal to `.`
def substring_primes:
. as \$n
| reduce emit_until(. > \$n; primes) as \$p ( null;
if \$p | is_substring(false)
then . += [\$p]
else .
end );

# Input: an array of the substring primes less than or equal to 373.
# Output: any other substring primes.
# Comment: if there are any others, they would have to be constructed
# from the numbers in the input array, as by assumption it includes
# all substring primes less than 100.
def verify:
. as \$sp
| range(0;length) as \$i
| range(0;length) as \$j
| ([\$sp[\$i, \$j]] | map(tostring) | add | tonumber) as \$candidate
| if \$candidate | IN(\$sp[]) then empty
elif \$candidate | is_substring(true) then \$candidate
else empty
end;

500 | substring_primes
| "Verifying that the following are the only substring primes:",
.,
"...",
( [verify] as \$extra
| if \$extra == [] then "done" else \$extra end )
Output:
Verifying that the following are the only substring primes:
[2,3,5,7,23,37,53,73,373]
...
done

## Julia

using Primes

function isA085823(n, base = 10, sieve = pmask)
dig = digits(n; base=base)
for i in 1:length(dig), j in i:length(dig)
k = evalpoly(base, dig[i:j])
(k == 0 || !sieve[k]) && return false
end
return true
end

println(filter(isA085823, 1:1000))

Output:
[2, 3, 5, 7, 23, 37, 53, 73, 373]

using Primes

const nt, nons = [0], Int[]

counted_primetest(n) = (nt[1] += 1; b = isprime(n); !b && push!(nons, n); b)

const results = [2, 3, 5, 7]

# check 2 digit candidates
for n in results, i in [3, 7]
if n != i
candidate = n * 10 + i
candidate < 100 && counted_primetest(candidate) && push!(results, candidate)
end
end

# check 3 digit candidates
for n in results, i in [3, 7]
if 10 < n < 100 && n % 10 != i
candidate = n * 10 + i
counted_primetest(candidate) && push!(results, candidate)
end
end

println("Results: \$results.\nThe function isprime() was called \$(nt[1]) times.")

# Because 237, 537, and 737 are already excluded, we cannot generate any larger candidates from 373.

Output:
Results: [2, 3, 5, 7, 23, 37, 53, 73, 373].
The function isprime() was called 10 times.
Discarded candidates: [27, 57, 237, 537, 737]

## Mathematica/Wolfram Language

Select[Range[500], PrimeQ[#] && AllTrue[Subsequences[IntegerDigits[#], {1, \[Infinity]}], FromDigits /* PrimeQ] &]
Output:
{2, 3, 5, 7, 23, 37, 53, 73, 373}

## Nim

The algorithm we use solves the advanced task as it finds the substring primes with only 11 primality tests. Note that, if we limit to numbers with at most three digits, 10 tests are sufficient. As we don’t use this limitation, we need one more test to detect than 3737 is not prime.

import sequtils, strutils

type
Digit = 0..9
DigitSeq = seq[Digit]

func isOddPrime(n: Positive): bool =
## Check if "n" is an odd prime.
assert n > 10
var d = 3
while d * d <= n:
if n mod d == 0: return false
inc d, 2
return true

func toInt(s: DigitSeq): int =
## Convert a sequence of digits to an int.
for d in s:
result = 10 * result + d

var result = @[2, 3, 5, 7]
var list: seq[DigitSeq] = result.mapIt(@[Digit it])
var primeTestCount = 0

while list.len != 0:
var newList: seq[DigitSeq]
for dseq in list:
for d in [Digit 3, 7]:
if dseq[^1] != d: # New digit must be different of last digit.
inc primeTestCount
let newDseq = dseq & d
let candidate = newDseq.toInt
if candidate.isOddPrime:
list = move(newList)

echo "List of substring primes: ", result.join(" ")
echo "Number of primality tests: ", primeTestCount
Output:
List of substring primes: 2 3 5 7 23 37 53 73 373
Number of primality tests: 11

## Perl

#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Substring_primes
use warnings;

my %prime;

LOOP:
for (2 .. 500 )
{
my %substrings = ();
/.+(?{ \$prime{\$&} or \$substrings{\$&}++ })(*FAIL)/;
for my \$try ( sort { \$a <=> \$b } keys %substrings )
{
\$try < 2 and next LOOP;
\$prime{\$try} || (1 x \$try) !~ /^(11+)\1+\$/ ? \$prime{\$try}++ : next LOOP;
}
}
printf "  %d" x %prime . "\n", sort {\$a <=> \$b} keys %prime;
Output:
2  3  5  7  23  37  53  73  373

## Phix

This tests a total of just 15 numbers for primality.

with javascript_semantics
--sequence tested = {}
--function a085823(integer p=0)
--  sequence res={}
function a085823(sequence res={}, tested={}, integer p=0)
for i=(p!=0)+1 to 4 do
integer t = get_prime(i)
if t!=remainder(p,10) and (p=0 or t!=5) then
t += p*10
if is_prime(t) then
--              {res,tested} = a085823(res&t,tested,t)
{res,tested} = a085823(deep_copy(res)&t,deep_copy(tested),t)    -- [1]
--              res &= t
--              res &= a085823(t)
else
tested &= t
end if
end if
end for
return {res,tested}
--  return res
end function
sequence {res,tested} = a085823()  -- sort() if you prefer...
--sequence res = a085823()  -- sort() if you prefer...
printf(1,"There are %d such A085823 primes: %V\n",{length(res),res})
printf(1,"%d innocent bystanders falsly accused of being prime (%d tests in total): %V\n",
{length(tested),length(tested)+length(res),tested})

[1]It is possible that the compiler could be improved to avoid the need for those deep_copy().
The other seven commented out lines show a different way to avoid any use of deep_copy().

Output:
There are 9 such A085823 primes: {2,23,3,37,373,5,53,7,73}
6 innocent bystanders falsly accused of being prime (15 tests in total): {237,27,3737,537,57,737}

## Raku

my @p = (^10).grep: *.is-prime;

say gather while @p {
.take for @p;

@p = ( @p X~ <3 7> ).grep: { .is-prime and .substr(*-2,2).is-prime }
}
Output:
(2 3 5 7 23 37 53 73 373)

### Stretch Goal

my \$prime-tests = 0;
my @non-primes;
sub spy-prime (\$n) {
\$prime-tests++;
my \$is-p = \$n.is-prime;

push @non-primes, \$n unless \$is-p;
return \$is-p;
}

my @p = <2 3 5 7>;

say gather while @p {
.take for @p;

@p = ( @p X~ <3 7> ).grep: { !.ends-with(33|77) and .&spy-prime };
}
.say for :\$prime-tests, :@non-primes;
Output:
(2 3 5 7 23 37 53 73 373)
prime-tests => 11
non-primes => [27 57 237 537 737 3737]

## REXX

/*REXX program finds/shows decimal primes where all substrings are also prime,  N < 500.*/
parse arg n cols . /*obtain optional argument from the CL.*/
if n=='' | n=="," then n= 500 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
call genP /*build array of semaphores for primes.*/
w= 10 /*width of a number in any column. */
title= ' primes (base ten) where all substrings are also primes, where N < ' commas(n)
say ' index │'center(title, 1 + cols*(w+1) ) /*display the title of the output. */
say '───────┼'center("" , 1 + cols*(w+1), '─') /* " " separator " " " */
found= 0; idx= 1 /*define # substring primes found; IDX.*/
\$= /*a list of substring primes (so far). */
do j=1 for #; p= @.j; p2= substr(p, 2) /*search for primes that fit criteria. */
if verify(p, 014689, 'M')>0 then iterate /*does it contain any of these digits? */
if verify(p2, 25 , 'M')>0 then iterate /* " P2 " " " " " */
L= length(p) /*obtain the length of the P prime.*/
do k=1 for L-1 /*test for primality for all substrings*/
do m=k+1 to L; y= substr(p, k, m-1) /*extract a substring from the P prime.*/
if \!.y then iterate j /*does substring of P not prime? Skip.*/
end /*m*/
end /*k*/

found= found + 1 /*bump the number of substring primes. */
\$= \$ right( commas(p), w) /*add a substring prime ──► \$ list. */
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr(\$, 2); \$= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/

if \$\=='' then say center(idx, 7)'│' substr(\$, 2) /*display any residual numbers.*/
say '───────┴'center("" , 1 + cols*(w+1), '─') /* " a foot separator. */
say; say 'Found ' words(\$) title /* " the summary. */
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 /*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; s.#= @.# **2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do [email protected].#+2 by 2 to n-1 /*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 s.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; s.#= j*j;  !.j= 1 /*bump # of Ps; assign next P; P²; P# */
end /*j*/; return
output   when using the default inputs:
index │                    primes (base ten) where all substrings are also primes, where  N  <  500
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
1   │          2          3          5          7         23         37         53         73        373
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  9  primes (base ten) where all substrings are also primes, where  N  <  500

## Ring

see "working..." + nl
see "Numbers in which all substrings are primes:" + nl

row = 0
limit1 = 500

for n = 1 to limit1
flag = 1
strn = string(n)
for m = 1 to len(strn)
for p = 1 to len(strn)
temp = substr(strn,m,p)
if temp != ""
if isprime(number(temp))
flag = 1
else
flag = 0
exit 2
ok
ok
next
next
if flag = 1
see "" + n + " "
ok
next

see nl + "Found " + row + " numbers in which all substrings are primes" + nl
see "done..." + nl

Output:
working...
Numbers in which all substrings are primes:
2 3 5 7 23 37 53 73 373
Found 9 numbers in which all substrings are primes
done...

## Sidef

Generic solution for any base >= 2.

func split_at_indices(array, indices) {

var parts = []
var i = 0

for j in (indices) {
parts << array.slice(i, j)
i = j+1
}

parts
}

func consecutive_partitions(array, callback) {
for k in (0..array.len) {
combinations(array.len, k, {|*indices|
var t = split_at_indices(array, indices)
if (t.sum_by{.len} == array.len) {
callback(t)
}
})
}
}

func is_substring_prime(digits, base) {

for k in (^digits) {
digits.first(k+1).digits2num(base).is_prime || return false
}

consecutive_partitions(digits, {|part|
part.all { .digits2num(base).is_prime } || return false
})

return true
}

func generate_from_prefix(p, base, digits) {

var seq = [p]

for d in (digits) {
var t = [d, p...]
if (is_prime(t.digits2num(base)) && is_substring_prime(t, base)) {
seq << __FUNC__(t, base, digits)...
}
}

return seq
}

func substring_primes(base) { # finite sequence for each base >= 2

var prime_digits = (base-1 -> primes) # prime digits < base

prime_digits.map {|p| generate_from_prefix([p], base, prime_digits)... }\
.map {|t| digits2num(t, base) }\
.sort
}

for base in (2..20) {
say "base = #{base}: #{substring_primes(base)}"
}
Output:
base = 2: []
base = 3: [2]
base = 4: [2, 3, 11]
base = 5: [2, 3, 13, 17, 67]
base = 6: [2, 3, 5, 17, 23]
base = 7: [2, 3, 5, 17, 19, 23, 37]
base = 8: [2, 3, 5, 7, 19, 23, 29, 31, 43, 47, 59, 61, 157, 239, 251, 349, 379, 479, 491]
base = 9: [2, 3, 5, 7, 23, 29, 47]
base = 10: [2, 3, 5, 7, 23, 37, 53, 73, 373]
base = 11: [2, 3, 5, 7, 29, 79]
base = 12: [2, 3, 5, 7, 11, 29, 31, 41, 43, 47, 67, 71, 89, 137, 139, 359, 499, 503, 521, 569, 571, 809, 857, 859, 6043]
base = 13: [2, 3, 5, 7, 11, 29, 31, 37, 41, 67, 379]
base = 14: [2, 3, 5, 7, 11, 13, 31, 41, 47, 53, 73, 83, 101, 103, 109, 157, 167, 193, 439, 661, 1033, 2203]
base = 15: [2, 3, 5, 7, 11, 13, 37, 41, 43, 47, 107, 167, 197, 557, 617, 647]
base = 16: [2, 3, 5, 7, 11, 13, 37, 43, 53, 59, 61, 83, 179, 181, 211, 691, 947, 3389]
base = 17: [2, 3, 5, 7, 11, 13, 37, 41, 47, 53, 223, 631]
base = 18: [2, 3, 5, 7, 11, 13, 17, 41, 43, 47, 53, 59, 61, 67, 71, 97, 101, 103, 107, 131, 137, 139, 211, 239, 241, 251, 311, 313, 317, 751, 787, 859, 1069, 1103, 1109, 1213, 1223, 1283, 1289, 1759, 1831, 1861, 1871, 1931, 1933, 2371, 3803, 4349, 4523, 5639, 5647, 15467, 19867, 34807]
base = 19: [2, 3, 5, 7, 11, 13, 17, 41, 43, 59, 97, 211]
base = 20: [2, 3, 5, 7, 11, 13, 17, 19, 43, 47, 53, 59, 67, 71, 73, 79, 103, 107, 113, 151, 157, 223, 227, 233, 239, 263, 271, 277, 347, 353, 359, 383, 397, 1063, 1423, 1427, 1433, 1439, 1471, 1583, 1597, 3023, 4663, 4783, 5273, 5279, 7673, 28663]

## Wren

Library: Wren-math

### Using a limit

import "/math" for Int
var primes = Int.primeSieve(499)
var sprimes = []
for (p in primes) {
var digits = Int.digits(p)
var b1 = digits.all { |d| Int.isPrime(d) }
if (b1) {
if (digits.count < 3) {
} else {
var b2 = Int.isPrime(digits[0] * 10 + digits[1])
var b3 = Int.isPrime(digits[1] * 10 + digits[2])
}
}
}
System.print("Found %(sprimes.count) primes < 500 where all substrings are also primes, namely:")
System.print(sprimes)
Output:
Found 9 primes < 500 where all substrings are also primes, namely:
[2, 3, 5, 7, 23, 37, 53, 73, 373]

This follows the logic in the OEIS A085823 comments.

import "/math" for Int

var results = [2, 3, 5, 7] // number must begin with a prime digit
var odigits = [3, 7] // other digits must be 3 or 7 as there would be a composite substring otherwise
var tests = 4 // i.e. to obtain initial results in the first place

// check 2 digit numbers or greater
// note that 'results' is a moving feast. If the loop eventually terminates that's all there are.
for (r in results) {
for (od in odigits) {
// the last digit of r and od must be different otherwise number would be divisible by 11
if ((r % 10) != od) {
var n = r * 10 + od
tests = tests + 1
}
}
}

System.print("There are %(results.count) primes where all substrings are also primes, namely:")
System.print(results)
System.print("\nThe following numbers were also tested for primality but found to be composite:")
System.print("\nTotal number of primality tests = %(tests)")
Output:
There are 9 primes where all substrings are also primes, namely:
[2, 3, 5, 7, 23, 37, 53, 73, 373]

The following numbers were also tested for primality but found to be composite:
[27, 57, 237, 537, 737, 3737]

Total number of primality tests = 15

## 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;
];

int Digit, Huns, Tens, Ones, N;
[Digit:= [0, 2, 3, 5, 7]; \leading zeros are ok
for Huns:= 0 to 4 do
for Tens:= 0 to 4 do
if Huns+Tens=0 or IsPrime(Digit(Huns)*10+Digit(Tens)) then
for Ones:= 1 to 4 do \can't end in 0
[N:= Digit(Huns)*100 + Digit(Tens)*10 + Digit(Ones);
if N<500 & IsPrime(N) & IsPrime(Digit(Tens)*10+Digit(Ones)) then
[IntOut(0, N); ChOut(0, ^ )];
];
]
Output:
2 3 5 7 23 37 53 73 373