# Smarandache prime-digital sequence

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Smarandache prime-digital sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Smarandache prime-digital sequence (SPDS for brevity) is the sequence of primes whose digits are themselves prime.

For example 257 is an element of this sequence because it is prime itself and its digits: 2, 5 and 7 are also prime.

• Show the first 25 SPDS primes.
• Show the hundredth SPDS prime.

## 11l

Translation of: Python
F divisors(n)
V divs = [1]
L(ii) 2 .< Int(n ^ 0.5) + 3
I n % ii == 0
divs.append(ii)
divs.append(Int(n / ii))
divs.append(n)
R Array(Set(divs))

F is_prime(n)
R divisors(n).len == 2

F digit_check(n)
I String(n).len < 2
R 1B
E
L(digit) String(n)
I !is_prime(Int(digit))
R 0B
R 1B

F sequence(max_n)
V ii = 0
V n = 0
[Int] r
L
ii++
I is_prime(ii)
I n > max_n
L.break
I digit_check(ii)
n++
r.append(ii)
R r

V seq = sequence(100)
print(‘First 25 SPDS primes:’)
L(item) seq[0.<25]
print(item, end' ‘ ’)
print()
print(‘Hundredth SPDS prime: ’seq[99])
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
Hundredth SPDS prime: 33223

## Action!

INCLUDE "D2:REAL.ACT" ;from the Action! Tool Kit

BYTE FUNC IsZero(REAL POINTER a)
CHAR ARRAY s(10)

StrR(a,s)
IF s(0)=1 AND s(1)='0 THEN
RETURN (1)
FI
RETURN (0)

CARD FUNC MyMod(CARD a,b)
REAL ar,br,dr
CARD d,m

IF a>32767 THEN
;Built-in DIV and MOD
;do not work properly
;for numbers greater than 32767
IntToReal(a,ar)
IntToReal(b,br)
RealDiv(ar,br,dr)
d=RealToInt(dr)
m=a-d*b
ELSE
m=a MOD b
FI
RETURN (m)

BYTE FUNC IsPrime(CARD a)
CARD i

IF a<=1 THEN
RETURN (0)
FI

i=2
WHILE i*i<=a
DO
IF MyMod(a,i)=0 THEN
RETURN (0)
FI
i==+1
OD
RETURN (1)

BYTE FUNC AllDigitsArePrime(CARD a)
BYTE i
CHAR ARRAY s
CHAR c

StrC(a,s)
FOR i=1 TO s(0)
DO
c=s(i)
IF c#'2 AND c#'3 AND c#'5 AND c#'7 THEN
RETURN (0)
FI
OD
RETURN (1)

PROC Main()
BYTE count
CARD a

Put(125) PutE() ;clear screen
PrintE("Sequence from 1st to 25th:")
count=0 a=1
DO
IF AllDigitsArePrime(a)=1 AND IsPrime(a)=1 THEN
count==+1
IF count<=25 THEN
PrintC(a) Put(32)
ELSEIF count=100 THEN
PrintF("%E%E100th: %U%E",a)
EXIT
FI
FI
a==+1
OD
RETURN
Output:
Sequence from 1st to 25th:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273

100th: 33223

## ALGOL 68

Uses a sieve to find primes. Requires --heap 256m for Algol 68G.
Uses the optimisations of the Factor, Phix, etc. samples.

# find elements of the Smarandache prime-digital sequence - primes whose    #
# digits are all primes                                                     #
# Uses the observations that the final digit of 2 or more digit Smarandache #
# primes must be 3 or 7 and the only prime digits are 2, 3, 5 and 7         #
# Needs --heap 256m for Algol 68G                                           #
BEGIN
# construct a sieve of primes up to 10 000 000                          #
INT prime max = 10 000 000;
[ prime max ]BOOL prime; FOR i TO UPB prime DO prime[ i ] := TRUE OD;
FOR s FROM 2 TO ENTIER sqrt( prime max ) DO
IF prime[ s ] THEN
FOR p FROM s * s BY s TO prime max DO prime[ p ] := FALSE OD
FI
OD;
# consruct the Smarandache primes up to 10 000 000                      #
[ prime max ]BOOL smarandache; FOR i TO UPB prime DO smarandache[ i ] := FALSE OD;
[   ]INT prime digits = ( 2, 3, 5, 7 );
[ 7 ]INT digits      := ( 0, 0, 0, 0, 0, 0, 0 );
# tests whether the current digits form a Smarandache prime             #
PROC try smarandache = VOID:
BEGIN
INT possible prime := 0;
FOR i TO UPB digits DO
possible prime *:= 10 +:= digits[ i ]
OD;
smarandache[ possible prime ] := prime[ possible prime ]
END # try smarandache # ;
# tests whether the current digits plus 3 or 7 form a Smarandache prime #
PROC try smarandache 3 or 7 = VOID:
BEGIN
digits[ UPB digits ] := 3;
try smarandache;
digits[ UPB digits ] := 7;
try smarandache
END # try smarandache 3 or 7 # ;
# the 1 digit primes are all Smarandache primes                         #
FOR d7 TO UPB prime digits DO smarandache[ prime digits[ d7 ] ] := TRUE OD;
# try the possible 2, 3, etc. digit numbers composed of prime digits    #
FOR d6 TO UPB prime digits DO
digits[ 6 ] := prime digits[ d6 ];
try smarandache 3 or 7;
FOR d5 TO UPB prime digits DO
digits[ 5 ] := prime digits[ d5 ];
try smarandache 3 or 7;
FOR d4 TO UPB prime digits DO
digits[ 4 ] := prime digits[ d4 ];
try smarandache 3 or 7;
FOR d3 TO UPB prime digits DO
digits[ 3 ] := prime digits[ d3 ];
try smarandache 3 or 7;
FOR d2 TO UPB prime digits DO
digits[ 2 ] := prime digits[ d2 ];
try smarandache 3 or 7;
FOR d1 TO UPB prime digits DO
digits[ 1 ] := prime digits[ d1 ];
try smarandache 3 or 7
OD;
digits[ 1 ] := 0
OD;
digits[ 2 ] := 0
OD;
digits[ 3 ] := 0
OD;
digits[ 4 ] := 0
OD;
digits[ 5 ] := 0
OD;
# print some Smarandache primes                                           #
INT count  := 0;
INT s100   := 0;
INT s1000  := 0;
INT s last := 0;
INT p last := 0;
print( ( "First 25 Smarandache primes:", newline ) );
FOR i TO UPB smarandache DO
IF smarandache[ i ] THEN
count +:= 1;
s last := i;
p last := count;
IF count <= 25 THEN
print( ( " ", whole( i, 0 ) ) )
ELIF count = 100 THEN
s100  := i
ELIF count = 1000 THEN
s1000 := i
FI
FI
OD;
print( ( newline ) );
print( ( "100th Smarandache prime: ", whole( s100, 0 ), newline ) );
print( ( "1000th Smarandache prime: ", whole( s1000, 0 ), newline ) );
print( ( "Largest Smarandache prime under "
, whole( prime max, 0 )
, ": "
, whole( s last, 0 )
, " (Smarandache prime "
, whole( p last, 0 )
, ")"
, newline
)
)
END
Output:
First 25 Smarandache primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100th Smarandache prime: 33223
1000th Smarandache prime: 3273527
Largest Smarandache prime under 10000000: 7777753 (Smarandache prime 1903)

## Arturo

spds: 2..∞ | select.first:100 'x ->
and? -> prime? x
-> every? digits x => prime?

print "First 25 SPDS primes:"
print first.n: 25 spds

print ""
print ["100th SPDS prime:" last spds]
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273

100th SPDS prime: 33223

## AWK

# syntax: GAWK -f SMARANDACHE_PRIME-DIGITAL_SEQUENCE.AWK
BEGIN {
limit = 25
printf("1-%d:",limit)
while (1) {
if (is_prime(++n)) {
if (all_digits_prime(n) == 1) {
if (++count <= limit) {
printf(" %d",n)
}
if (count == 100) {
printf("\n%d: %d\n",count,n)
break
}
}
}
}
exit(0)
}
function all_digits_prime(n, i) {
for (i=1; i<=length(n); i++) {
if (!is_prime(substr(n,i,1))) {
return(0)
}
}
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:
1-25: 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100: 33223

## BASIC256

Translation of: FreeBASIC
arraybase 1
dim smar(100)
smar[1] = 2

cont = 1
i = 1

print 1, 2
while cont < 100
i += 2
if not isPrime(i) then continue while
for j = 1 to length(string(i))
digit = int(mid(string(i),j,1))
if not isPrime(digit) then continue while
next j
cont += 1
smar[cont] = i
if cont = 100 or cont <= 25 then print cont, smar[cont]
end while
end

function isPrime(v)
if v < 2 then return False
if v mod 2 = 0 then return v = 2
if v mod 3 = 0 then return v = 3
d = 5
while d * d <= v
if v mod d = 0 then return False else d += 2
end while
return True
end function
Output:
Igual que la entrada de FreeBASIC.

## C

Translation of: C++
#include <locale.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>

typedef uint32_t integer;

integer next_prime_digit_number(integer n) {
if (n == 0)
return 2;
switch (n % 10) {
case 2:
return n + 1;
case 3:
case 5:
return n + 2;
default:
return 2 + next_prime_digit_number(n/10) * 10;
}
}

bool is_prime(integer n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
if (n % 5 == 0)
return n == 5;
static const integer wheel[] = { 4,2,4,2,4,6,2,6 };
integer p = 7;
for (;;) {
for (int i = 0; i < 8; ++i) {
if (p * p > n)
return true;
if (n % p == 0)
return false;
p += wheel[i];
}
}
}

int main() {
setlocale(LC_ALL, "");
const integer limit = 1000000000;
integer n = 0, max = 0;
printf("First 25 SPDS primes:\n");
for (int i = 0; n < limit; ) {
n = next_prime_digit_number(n);
if (!is_prime(n))
continue;
if (i < 25) {
if (i > 0)
printf(" ");
printf("%'u", n);
}
else if (i == 25)
printf("\n");
++i;
if (i == 100)
printf("Hundredth SPDS prime: %'u\n", n);
else if (i == 1000)
printf("Thousandth SPDS prime: %'u\n", n);
else if (i == 10000)
printf("Ten thousandth SPDS prime: %'u\n", n);
max = n;
}
printf("Largest SPDS prime less than %'u: %'u\n", limit, max);
return 0;
}
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2,237 2,273
Hundredth SPDS prime: 33,223
Thousandth SPDS prime: 3,273,527
Ten thousandth SPDS prime: 273,322,727
Largest SPDS prime less than 1,000,000,000: 777,777,773

## C++

#include <iostream>
#include <cstdint>

using integer = uint32_t;

integer next_prime_digit_number(integer n) {
if (n == 0)
return 2;
switch (n % 10) {
case 2:
return n + 1;
case 3:
case 5:
return n + 2;
default:
return 2 + next_prime_digit_number(n/10) * 10;
}
}

bool is_prime(integer n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
if (n % 5 == 0)
return n == 5;
constexpr integer wheel[] = { 4,2,4,2,4,6,2,6 };
integer p = 7;
for (;;) {
for (integer w : wheel) {
if (p * p > n)
return true;
if (n % p == 0)
return false;
p += w;
}
}
}

int main() {
std::cout.imbue(std::locale(""));
const integer limit = 1000000000;
integer n = 0, max = 0;
std::cout << "First 25 SPDS primes:\n";
for (int i = 0; n < limit; ) {
n = next_prime_digit_number(n);
if (!is_prime(n))
continue;
if (i < 25) {
if (i > 0)
std::cout << ' ';
std::cout << n;
}
else if (i == 25)
std::cout << '\n';
++i;
if (i == 100)
std::cout << "Hundredth SPDS prime: " << n << '\n';
else if (i == 1000)
std::cout << "Thousandth SPDS prime: " << n << '\n';
else if (i == 10000)
std::cout << "Ten thousandth SPDS prime: " << n << '\n';
max = n;
}
std::cout << "Largest SPDS prime less than " << limit << ": " << max << '\n';
return 0;
}
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2,237 2,273
Hundredth SPDS prime: 33,223
Thousandth SPDS prime: 3,273,527
Ten thousandth SPDS prime: 273,322,727
Largest SPDS prime less than 1,000,000,000: 777,777,773

## Delphi

Works with: Delphi version 6.0

Uses the Delphi Prime-Generator Object

procedure ShowSmarandachePrimes(Memo: TMemo);
{Show primes where all digits are also prime}
var Sieve: TPrimeSieve;
var I,J,P,Count: integer;
var S: string;

function AllDigitsPrime(N: integer): boolean;
{Test all digits on N to see if they are prime}
var I,Count: integer;
var IA: TIntegerDynArray;
begin
Result:=False;
GetDigits(N,IA);
for I:=0 to High(IA) do
if not Sieve.Flags[IA[I]] then exit;
Result:=True;
end;

begin
Sieve:=TPrimeSieve.Create;
try
{Build 1 million primes}
Sieve.Intialize(1000000);
Count:=0;
{Test if all digits of the number are prime}
for I:=0 to Sieve.PrimeCount-1 do
begin
P:=Sieve.Primes[I];
if AllDigitsPrime(P) then
begin
Inc(Count);
if Count<=25 then Memo.Lines.Add(IntToStr(Count)+' - '+IntToStr(P));
if Count=100 then
begin
break;
end;
end;
end;
finally Sieve.Free; end;
end;
Output:
1 - 2
2 - 3
3 - 5
4 - 7
5 - 23
6 - 37
7 - 53
8 - 73
9 - 223
10 - 227
11 - 233
12 - 257
13 - 277
14 - 337
15 - 353
16 - 373
17 - 523
18 - 557
19 - 577
20 - 727
21 - 733
22 - 757
23 - 773
24 - 2237
25 - 2273
100th = 33223
Elapsed Time: 150.037 ms.

## EasyLang

fastfunc isprim num .
i = 2
while i <= sqrt num
if num mod i = 0
return 0
.
i += 1
.
return 1
.
n = 2
repeat
if isprim n = 1
h = n
while h > 0
d = h mod 10
if d < 2 or d = 4 or d = 6 or d > 7
break 1
.
h = h div 10
.
if h = 0
cnt += 1
if cnt <= 25
write n & " "
.
.
.
until cnt = 100
n += 1
.
print ""
print n
Output:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
33223

## F#

This task uses Extensible Prime Generator (F#)

// Generate Smarandache prime-digital sequence. Nigel Galloway: May 31st., 2019
let rec spds g=seq{yield! g; yield! (spds (Seq.collect(fun g->[g*10+2;g*10+3;g*10+5;g*10+7]) g))}|>Seq.filter(isPrime)
spds [2;3;5;7] |> Seq.take 25 |> Seq.iter(printfn "%d")
printfn "\n\n100th item of this sequence is %d" (spds [2;3;5;7] |> Seq.item 99)
printfn "1000th item of this sequence is %d" (spds [2;3;5;7] |> Seq.item 999)
Output:
2
3
5
7
23
37
53
73
223
227
233
257
277
337
353
373
523
557
577
727
733
757
773
2237
2273

100th item of this sequence is 33223
1000th item of this sequence is 3273527

## Factor

### Naive

USING: combinators.short-circuit io lists lists.lazy math
math.parser math.primes prettyprint sequences ;
IN: rosetta-code.smarandache-naive

: smarandache? ( n -- ? )
{
[ number>string string>digits [ prime? ] all? ]
[ prime? ]
} 1&& ;

: smarandache ( -- list ) 1 lfrom [ smarandache? ] lfilter ;

: smarandache-demo ( -- )
"First 25 members of the Smarandache prime-digital sequence:"
print 25 smarandache ltake list>array .
"100th member: " write smarandache 99 [ cdr ] times car . ;

MAIN: smarandache-demo
Output:
First 25 members of the Smarandache prime-digital sequence:
{
2
3
5
7
23
37
53
73
223
227
233
257
277
337
353
373
523
557
577
727
733
757
773
2237
2273
}
100th member: 33223

### Optimized

USING: combinators generalizations io kernel math math.functions
math.primes prettyprint sequences ;
IN: rosetta-code.smarandache

! Observations:
! * For 2-digit numbers and higher, only 3 and 7 are viable in
!   the ones place.
! * Only 2, 3, 5, and 7 are viable anywhere else.
! * It is possible to use this information to drastically
!   reduce the amount of numbers to check for primality.
! * For instance, by these rules we can tell that the next
!   potential Smarandache prime digital after 777 is 2223.

: next-one ( n -- n' ) 3 = 7 3 ? ; inline

: next-ten ( n -- n' )
{ { 2 [ 3 ] } { 3 [ 5 ] } { 5 [ 7 ] } [ drop 2 ] } case ;

: inc ( seq quot: ( n -- n' ) -- seq' )
[ 0 ] 2dip [ change-nth ] curry keep ; inline

: inc1  ( seq -- seq' ) [ next-one ] inc ;
: inc10 ( seq -- seq' ) [ next-ten ] inc ;

: inc-all ( seq -- seq' )
inc1 [ zero? not [ next-ten ] when ] V{ } map-index-as ;

: carry ( seq -- seq' )
dup [ 7 = not ] find drop {
{ 0 [ inc1 ] }
{ f [ inc-all 2 suffix! ] }
[ cut [ inc-all ] [ inc10 ] bi* append! ]
} case ;

: digits>integer ( seq -- n ) [ 10 swap ^ * ] map-index sum ;

: next-smarandache ( seq -- seq' )
[ digits>integer prime? ] [ carry dup ] do until ;

: .sm ( seq -- ) <reversed> [ pprint ] each nl ;

: first25 ( -- )
2 3 5 7 [ . ] 4 napply V{ 7 } clone
21 [ next-smarandache dup .sm ] times drop ;

: nth-smarandache ( n -- )
4 - V{ 7 } clone swap [ next-smarandache ] times .sm ;

: smarandache-demo ( -- )
"First 25 members of the Smarandache prime-digital sequence:"
print first25 nl { 100 1000 10000 100000 } [
dup pprint "th member: " write nth-smarandache
] each ;

MAIN: smarandache-demo
Output:
First 25 members of the Smarandache prime-digital sequence:
2
3
5
7
23
37
53
73
223
227
233
257
277
337
353
373
523
557
577
727
733
757
773
2237
2273

100th member: 33223
1000th member: 3273527
10000th member: 273322727
100000th member: 23325232253

## Forth

: is_prime? ( n -- flag )
dup 2 < if drop false exit then
dup 2 mod 0= if 2 = exit then
dup 3 mod 0= if 3 = exit then
5
begin
2dup dup * >=
while
2dup mod 0= if 2drop false exit then
2 +
2dup mod 0= if 2drop false exit then
4 +
repeat
2drop true ;

: next_prime_digit_number ( n -- n )
dup 0= if drop 2 exit then
dup 10 mod
dup 2 = if drop 1+ exit then
dup 3 = if drop 2 + exit then
5 = if 2 + exit then
10 / recurse 10 * 2 + ;

: spds_next ( n -- n )
begin
next_prime_digit_number
dup is_prime?
until ;

: spds_print ( n -- )
0 swap 0 do
spds_next dup .
loop
drop cr ;

: spds_nth ( n -- n )
0 swap 0 do spds_next loop ;

." First 25 SPDS primes:" cr
25 spds_print

." 100th SPDS prime: "
100 spds_nth . cr

." 1000th SPDS prime: "
1000 spds_nth . cr

bye
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100th SPDS prime: 33223
1000th SPDS prime: 3273527

## FreeBASIC

function isprime( n as ulongint ) as boolean
if n < 2 then return false
if n = 2 then return true
if n mod 2 = 0 then return false
for i as uinteger = 3 to int(sqr(n))+1 step 2
if n mod i = 0 then return false
next i
return true
end function

dim as integer smar(1 to 100), count = 1, i = 1, digit, j
smar(1) = 2
print 1, 2
while count < 100
i += 2
if not isprime(i) then continue while
for j = 1 to len(str(i))
digit = val(mid(str(i),j,1))
if not isprime(digit) then continue while
next j
count += 1
smar(count) = i
if count = 100 orelse count <=25 then
print count, smar(count)
end if
wend
Output:
1             2
2             3
3             5
4             7
5             23
6             37
7             53
8             73
9             223
10            227
11            233
12            257
13            277
14            337
15            353
16            373
17            523
18            557
19            577
20            727
21            733
22            757
23            773
24            2237
25            2273
100           33223

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Solution

Case 1. Show the first 25 SPDS primes

Case 2. Show the hundredth SPDS prime

Additional cases. Show the 1000-th, 10,000-th and 100,000th SPDS primes

## Go

### Basic

package main

import (
"fmt"
"math/big"
)

var b = new(big.Int)

func isSPDSPrime(n uint64) bool {
nn := n
for nn > 0 {
r := nn % 10
if r != 2 && r != 3 && r != 5 && r != 7 {
return false
}
nn /= 10
}
b.SetUint64(n)
if b.ProbablyPrime(0) { // 100% accurate up to 2 ^ 64
return true
}
return false
}

func listSPDSPrimes(startFrom, countFrom, countTo uint64, printOne bool) uint64 {
count := countFrom
for n := startFrom; ; n += 2 {
if isSPDSPrime(n) {
count++
if !printOne {
fmt.Printf("%2d. %d\n", count, n)
}
if count == countTo {
if printOne {
fmt.Println(n)
}
return n
}
}
}
}

func main() {
fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:")
fmt.Println(" 1. 2")
n := listSPDSPrimes(3, 1, 25, false)
fmt.Println("\nHigher terms:")
indices := []uint64{25, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000}
for i := 1; i < len(indices); i++ {
fmt.Printf("%6d. ", indices[i])
n = listSPDSPrimes(n+2, indices[i-1], indices[i], true)
}
}
Output:
The first 25 terms of the Smarandache prime-digital sequence are:
1. 2
2. 3
3. 5
4. 7
5. 23
6. 37
7. 53
8. 73
9. 223
10. 227
11. 233
12. 257
13. 277
14. 337
15. 353
16. 373
17. 523
18. 557
19. 577
20. 727
21. 733
22. 757
23. 773
24. 2237
25. 2273

Higher terms:
100. 33223
200. 223337
500. 723337
1000. 3273527
2000. 22332337
5000. 55373333
10000. 273322727
20000. 727535273
50000. 3725522753
100000. 23325232253

### Optimized

This version is inspired by the optimizations used in the Factor and Phix entries which are expressed here as a kind of base-4 arithmetic using a digits set of {2, 3, 5, 7} where leading '2's are significant.

This is more than 30 times faster than the above version (runs in about 12.5 seconds on my Celeron @1.6GHx) and could be quickened up further (to around 4 seconds) by using a wrapper for GMP rather than Go's native big.Int type.

package main

import (
"fmt"
"math/big"
)

type B2357 []byte

var bi = new(big.Int)

func isSPDSPrime(b B2357) bool {
bi.SetString(string(b), 10)
return bi.ProbablyPrime(0) // 100% accurate up to 2 ^ 64
}

func listSPDSPrimes(startFrom B2357, countFrom, countTo uint64, printOne bool) B2357 {
count := countFrom
n := startFrom
for {
if isSPDSPrime(n) {
count++
if !printOne {
fmt.Printf("%2d. %s\n", count, string(n))
}
if count == countTo {
if printOne {
fmt.Println(string(n))
}
return n
}
}
if printOne {
} else {
}
}
}

func incDigit(digit byte) byte {
switch digit {
case '2':
return '3'
case '3':
return '5'
case '5':
return '7'
default:
return '9' // say
}
}

func (b B2357) AddOne() B2357 {
le := len(b)
b[le-1] = incDigit(b[le-1])
for i := le - 1; i >= 0; i-- {
if b[i] < '9' {
break
} else if i > 0 {
b[i] = '2'
b[i-1] = incDigit(b[i-1])
} else {
b[0] = '2'
nb := make(B2357, le+1)
copy(nb[1:], b)
nb[0] = '2'
return nb
}
}
return b
}

func (b B2357) AddTwo() B2357 {
}

func main() {
fmt.Println("The first 25 terms of the Smarandache prime-digital sequence are:")
n := listSPDSPrimes(B2357{'2'}, 0, 4, false)
n = listSPDSPrimes(n.AddOne(), 4, 25, false)
fmt.Println("\nHigher terms:")
indices := []uint64{25, 100, 200, 500, 1000, 2000, 5000, 10000, 20000, 50000, 100000}
for i := 1; i < len(indices); i++ {
fmt.Printf("%6d. ", indices[i])
n = listSPDSPrimes(n.AddTwo(), indices[i-1], indices[i], true)
}
}
Output:
Same as before.

Using the optimized approach of generated numbers from prime digits and testing for primality.

{-# LANGUAGE NumericUnderscores #-}
import Math.NumberTheory.Primes.Testing (isPrime)
import Data.List.Split (chunksOf)
import Data.List (intercalate)
import Text.Printf (printf)

smarandache :: [Integer]
smarandache = [2,3,5,7] <> s [2,3,5,7] >>= \x -> guard (isPrime x) >> [x]
where s xs = r <> s r where r = xs >>= \x -> [x*10+2, x*10+3, x*10+5, x*10+7]

nextSPDSTerms :: [Int] -> [(String, String)]
nextSPDSTerms = go 1 smarandache
where
go _ _ [] = []
go c (x:xs) terms
| c `elem` terms = (commas c, commas x) : go nextCount xs (tail terms)
| otherwise      = go nextCount xs terms
where nextCount = succ c

commas :: Show a => a -> String
commas = reverse . intercalate "," . chunksOf 3 . reverse . show

main :: IO ()
main = do
printf "The first 25 SPDS:\n%s\n\n" \$ f smarandache
mapM_ (uncurry (printf "The %9sth SPDS: %15s\n")) \$
nextSPDSTerms [100, 1_000, 10_000, 100_000, 1_000_000]
where f = show . take 25
Output:
The first 25 SPDS:
[2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273]

The       100th SPDS:          33,223
The     1,000th SPDS:       3,273,527
The    10,000th SPDS:     273,322,727
The   100,000th SPDS:  23,325,232,253
The 1,000,000th SPDS: 753,373,253,723
./smarandache_optimized  15.25s user 0.45s system 98% cpu 15.938 total

## J

Prime numbers have a built-in verb p: . It's easy and quick to get a list of prime numbers and determine which are composed entirely of the appropriate digits.

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

NB. given a prime y, smarandache y is 1 iff it's a smarandache prime
smarandache=: [: -. (0 e. (p:i.4) e.~ 10 #.inv ])&>

SP=: smarandache Filter p: i. 1000000

SP {~ i. 25   NB. first 25 Smarandache primes
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273

99 _1 { SP    NB. 100th and largest Smarandache prime of the first million primes
33223 7777753

# SP          NB. Tally of Smarandache primes in the first million primes
1903

Graph by index of Smarandache primes in index of primes through two millionth prime. The graph shows jumps where, I suppose, the most significant digit is 8, 9, then 1. https://imgur.com/a/hvbhf2S

## Java

Generate next in sequence directly from previous, inspired by previous solutions.

public class SmarandachePrimeDigitalSequence {

public static void main(String[] args) {
long s = getNextSmarandache(7);
System.out.printf("First 25 Smarandache prime-digital sequence numbers:%n2 3 5 7 ");
for ( int count = 1 ; count <= 21 ; s = getNextSmarandache(s) ) {
if ( isPrime(s) ) {
System.out.printf("%d ", s);
count++;
}
}
System.out.printf("%n%n");
for (int i = 2 ; i <=5 ; i++ ) {
long n = (long) Math.pow(10, i);
System.out.printf("%,dth Smarandache prime-digital sequence number = %d%n", n, getSmarandachePrime(n));
}
}

private static final long getSmarandachePrime(long n) {
if ( n < 10 ) {
switch ((int) n) {
case 1:  return 2;
case 2:  return 3;
case 3:  return 5;
case 4:  return 7;
}
}
long s = getNextSmarandache(7);
long result = 0;
for ( int count = 1 ; count <= n-4 ; s = getNextSmarandache(s) ) {
if ( isPrime(s) ) {
count++;
result = s;
}
}
return result;
}

private static final boolean isPrime(long test) {
if ( test % 2 == 0 ) return false;
for ( long i = 3 ; i <= Math.sqrt(test) ; i += 2 ) {
if ( test % i == 0 ) {
return false;
}
}
return true;
}

private static long getNextSmarandache(long n) {
//  If 3, next is 7
if ( n % 10 == 3 ) {
return n+4;
}
long retVal = n-4;

//  Last digit 7.  k = largest position from right where we have a 7.
int k = 0;
while ( n % 10 == 7 ) {
k++;
n /= 10;
}

//  Determine first digit from right where digit != 7.
long digit = n % 10;

//  Digit is 2, 3, or 5.  3-2 = 1, 5-3 = 2, 7-5 = 2, so digit = 2, coefficient = 1, otherwise 2.
long coeff = (digit == 2 ? 1 : 2);

//  Compute next value
retVal += coeff * Math.pow(10, k);

//  Subtract values for digit = 7.
while ( k > 1 ) {
retVal -= 5 * Math.pow(10, k-1);
k--;
}

//  Even works for 777..777 --> 2222...223
return retVal;
}

}
Output:
First 25 Smarandache prime-digital sequence numbers:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273

100th Smarandache prime-digital sequence number = 33223
1,000th Smarandache prime-digital sequence number = 3273527
10,000th Smarandache prime-digital sequence number = 273322727
100,000th Smarandache prime-digital sequence number = 23325232253

## jq

Works with: jq

Works with gojq, the Go implementation of jq

See the preamble to the Julia entry for the rationale behind the following implementation.

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

def Smarandache_primes:
# Output: a naively constructed stream of candidate strings of length >= 1
def Smarandache_candidates:
def unconstrained(\$length):
if \$length==1 then "2", "3", "5", "7"
else ("2", "3", "5", "7") as \$n
| \$n + unconstrained(\$length -1 )
end;
unconstrained(. - 1) as \$u
|  ("3", "7") as \$tail
| \$u + \$tail ;

2,3,5,7,
(range(2; infinite) | Smarandache_candidates | tonumber | select(is_prime));

# Override jq's incorrect definition of nth/2
# Emit the \$n-th value of the stream, counting from 0; or emit nothing
def nth(\$n; s):
if \$n < 0 then error("nth/2 doesn't support negative indices")
else label \$out
| foreach s as \$x (-1; .+1; select(. >= \$n) | \$x, break \$out)
end;

"First 25:",
[limit(25; Smarandache_primes)],

# jq counts from 0 so:
"\nThe hundredth: \(nth(99; Smarandache_primes))"
Output:
jq -nrc -f rc-smarandache-primes.jq
First 25:
[2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273]

The hundredth: 33223

## Julia

The prime single digits are 2, 3, 5, and 7. Except for 2 and 5, any number ending in 2 or 5 is not prime. So we start with [2, 3, 5, 7] and then add numbers that end in 3 or 7 and that only contain 2, 3, 5, and 7. This can be done via permutations of combinations with repetition.

using Combinatorics, Primes

combodigits(len) = sort!(unique(map(y -> join(y, ""), with_replacement_combinations("2357", len))))

function getprimes(N, maxdigits=9)
ret = [2, 3, 5, 7]
perms = Int[]
for i in 1:maxdigits-1, combo in combodigits(i), perm in permutations(combo)
n = parse(Int64, String(perm)) * 10
push!(perms, n + 3, n + 7)
end
for perm in sort!(perms)
if isprime(perm) && !(perm in ret)
push!(ret, perm)
if length(ret) >= N
return ret
end
end
end
end

const v = getprimes(10000)
println("The first 25 Smarandache primes are: ", v[1:25])
println("The 100th Smarandache prime is: ", v[100])
println("The 10000th Smarandache prime is: ", v[10000])
Output:
The first 25 Smarandache primes are: [2, 3, 5, 7, 23, 37, 53, 73, 223, 227, 233, 257, 277, 337, 353, 373, 523, 557, 577, 727, 733, 757, 773, 2237, 2273]
The 100th Smarandache prime is: 33223
The 10000th Smarandache prime is: 273322727

## Lua

-- FUNCS:
local function T(t) return setmetatable(t, {__index=table}) end
table.firstn = function(t,n) local s=T{} n=n>#t and #t or n for i = 1,n do s[i]=t[i] end return s end

-- SIEVE:
local sieve, S = {}, 50000
for i = 2,S do sieve[i]=true end
for i = 2,S do if sieve[i] then for j=i*i,S,i do sieve[j]=nil end end end

local digs, cans, spds, N = {2,3,5,7}, T{0}, T{}, 100
while #spds < N do
local c = cans:remove(1)
for _,d in ipairs(digs) do cans:insert(c*10+d) end
if sieve[c] then spds:insert(c) end
end
print("1-25 :  " .. spds:firstn(25):concat(" "))
print("100th:  " .. spds[100])
Output:
1-25 :  2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100th:  33223

## Mathematica / Wolfram Language

ClearAll[SmarandachePrimeQ]
SmarandachePrimeQ[n_Integer] := MatchQ[IntegerDigits[n], {(2 | 3 | 5 | 7) ..}] \[And] PrimeQ[n]
s = Select[Range[10^5], SmarandachePrimeQ];
Take[s, UpTo[25]]
s[[100]]
Output:
{2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273}
33223

## Nim

import math, strformat, strutils

const N = 35_000

# Sieve.
var composite: array[0..N, bool]  # Default is false and means prime.
composite[0] = true
composite[1] = true
for n in 2..sqrt(N.toFloat).int:
if not composite[n]:
for k in countup(n * n, N, n):
composite[k] = true

func digits(n: Positive): seq[0..9] =
var n = n.int
while n != 0:
n = n div 10

proc isSPDS(n: int): bool =
if composite[n]: return false
result = true
for d in n.digits:
if composite[d]: return false

iterator spds(maxCount: Positive): int {.closure.} =
yield 2
var count = 1
var n = 3
while count != maxCount and n <= N:
if n.isSPDS:
inc count
yield n
inc n, 2
if count != maxCount:
quit &"Too few values ({count}). Please, increase value of N.", QuitFailure

stdout.write "The first 25 SPDS are:"
for n in spds(25):
stdout.write ' ', n
echo()

var count = 0
for n in spds(100):
inc count
if count == 100:
echo "The 100th SPDS is: ", n
Output:
The first 25 SPDS are: 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
The 100th SPDS is: 33223

## Pascal

Works with: Free Pascal

uses [[1]]
Simple Brute force.Testing for prime takes most of the time.

program Smarandache;

uses
sysutils,primsieve;// http://rosettacode.org/wiki/Extensible_prime_generator#Pascal
const
Digits : array[0..3] of Uint32 = (2,3,5,7);
var
i,j,pot10,DgtLimit,n,DgtCnt,v,cnt,LastPrime,Limit : NativeUint;

procedure Check(n:NativeUint);
var
p : NativeUint;
Begin
p := LastPrime;
while p< n do
p := nextprime;
if p = n then
begin
inc(cnt);
IF (cnt <= 25) then
Begin
IF cnt = 25 then
Begin
writeln(n);
Limit := 100;
end
else
Write(n,',');
end
else
IF cnt = Limit then
Begin
Writeln(cnt:9,n:16);
Limit *=10;
if Limit > 10000 then
HALT;
end;
end;
LastPrime := p;
end;

Begin
Limit := 25;
LastPrime:=1;

//Creating the numbers not the best way but all upto 11 digits take 0.05s
//here only 9 digits
i := 0;
pot10 := 1;
DgtLimit := 1;
v := 4;
repeat
repeat
j := i;
DgtCnt := 0;
pot10 := 1;
n := 0;
repeat
n += pot10*Digits[j MOD 4];
j := j DIV 4;
pot10 *=10;
inc(DgtCnt);
until DgtCnt = DgtLimit;
Check(n);
inc(i);
until i=v;
//one more digit
v *=4;
i :=0;
inc(DgtLimit);
until DgtLimit= 12;
end.
Output:
2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273
100           33223
1000         3273527
10000       273322727
real	0m0,171s

## Perl

Library: ntheory
use strict;
use warnings;
use feature 'say';
use feature 'state';
use ntheory qw<is_prime>;
use Lingua::EN::Numbers qw(num2en_ordinal);

my @prime_digits = <2 3 5 7>;
my @spds = grep { is_prime(\$_) && /^[@{[join '',@prime_digits]}]+\$/ } 1..100;
my @p    = map { \$_+3, \$_+7 } map { 10*\$_ } @prime_digits;

while (\$#spds < 100_000) {
state \$o++;
my \$oom = 10**(1+\$o);
my @q;
for my \$l (@prime_digits) {
push @q, map { \$l*\$oom + \$_ } @p;
}
push @spds, grep { is_prime(\$_) } @p = @q;
}

say 'Smarandache prime-digitals:';
printf "%22s: %s\n", ucfirst(num2en_ordinal(\$_)), \$spds[\$_-1] for 1..25, 100, 1000, 10_000, 100_000;
Output:
First: 2
Second: 3
Third: 5
Fourth: 7
Fifth: 23
Sixth: 37
Seventh: 53
Eighth: 73
Ninth: 223
Tenth: 227
Eleventh: 233
Twelfth: 257
Thirteenth: 277
Fourteenth: 337
Fifteenth: 353
Sixteenth: 373
Seventeenth: 523
Eighteenth: 557
Nineteenth: 577
Twentieth: 727
Twenty-first: 733
Twenty-second: 757
Twenty-third: 773
Twenty-fourth: 2237
Twenty-fifth: 2273
One hundredth: 33223
One thousandth: 3273527
Ten thousandth: 273322727
One hundred thousandth: 23325232253

## Phix

Library: Phix/mpfr

Optimised. As noted on the Factor entry, candidates>10 must end in 3 or 7 (since they would not be prime if they ended in 2 or 5), which we efficiently achieve by alternately adding {4,-4}. Digits to the left of that must all be 2/3/5/7, so we add {1,2,2,-5}*10^k to cycle round those digits. Otherwise it is exactly like counting by adding 1 to each digit and carrying 1 left when we do a 9->0, or in this case 7->2|3.

I had planned to effectively merge a list of potential candidates with a list of all prime numbers, but because of the massive gaps (eg between 777,777,777 and 2,222,222,223) it proved much faster to test each candidate for primality individually. Timings below show just how much this improves things.

with javascript_semantics
atom t0 = time()
sequence spds = {2,3,5,7}
atom nxt_candidate = 23

include mpfr.e
mpz zprime = mpz_init()

procedure populate_spds(integer n)
while length(spds)<n do
mpz_set_d(zprime,nxt_candidate)
if mpz_prime(zprime) then
spds &= nxt_candidate
end if
exit
end if
-- (this is eg 777, by now 223 carry 1, -> 2223)
exit
end if
end for
end while
end procedure

populate_spds(25)
printf(1,"spds[1..25]:%v\n",{spds[1..25]})
for n=2 to 5 do
integer p = power(10,n)
populate_spds(p)
printf(1,"spds[%,d]:%,d\n",{p,spds[p]})
end for
for n=7 to 10 do
atom p = power(10,n),
dx = abs(binary_search(p,spds))-1
printf(1,"largest spds prime less than %,15d:%,14d\n",{p,spds[dx]})
end for
?elapsed(time()-t0)
Output:
spds[1..25]:{2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273}
spds[100]:33,223
spds[1,000]:3,273,527
spds[10,000]:273,322,727
spds[100,000]:23,325,232,253
largest spds prime less than      10,000,000:     7,777,753
largest spds prime less than     100,000,000:    77,777,377
largest spds prime less than   1,000,000,000:   777,777,773
largest spds prime less than  10,000,000,000: 7,777,777,577
"4.6s"

For comparison, on the same machine:
Factor (as optimised) took 45s to calculate the 100,000th number.
Go took 1 min 50 secs to calculate the 100,000th number - the optimised version got that down to 5.6s
Julia crashed when the limit was changed to 100,000, however it took 11s just to calculate the 10,000th number anyway.
The original Raku version was by far the slowest of all I tried, taking 1 min 15s just to calculate the 10,000th number, however it has since been replaced (I cannot actually run the latest Raku version, but I assume it is similar to the Perl one) and that completes near-instantly. Adding two 0 to limit in the C entry above gets a matching 777777773 on tio/clang in 27s, not directly comparable, and obviously you cannot add a 3rd 0 without changing those uint32.

## Python

def divisors(n):
divs = [1]
for ii in range(2, int(n ** 0.5) + 3):
if n % ii == 0:
divs.append(ii)
divs.append(int(n / ii))
divs.append(n)
return list(set(divs))

def is_prime(n):
return len(divisors(n)) == 2

def digit_check(n):
if len(str(n))<2:
return True
else:
for digit in str(n):
if not is_prime(int(digit)):
return False
return True

def sequence(max_n=None):
ii = 0
n = 0
while True:
ii += 1
if is_prime(ii):
if max_n is not None:
if n>max_n:
break
if digit_check(ii):
n += 1
yield ii

if __name__ == '__main__':
generator = sequence(100)
for index, item in zip(range(1, 16), generator):
print(index, item)
for index, item in zip(range(16, 100), generator):
pass
print(100, generator.__next__())

Output

1 2
2 3
3 5
4 7
5 23
6 37
7 53
8 73
9 223
10 227
11 233
12 257
13 277
14 337
15 353
100 33223

## Quackery

isprime is defined at Primality by trial division#Quackery.

### Naive

[ true swap
[ 10 /mod
[ table 1 1 0 0 1 0 1 0 1 1 ]
iff [ dip not ] done
dup 0 = until ]
drop ]                           is digitsprime ( n --> b )

[ temp put [] 0
[ dup digitsprime if
[ dup isprime if
[ dup dip join ] ]
1+
over size temp share = until ]
drop ]                           is spds        ( n --> [ )

100 spds
25 split swap echo
cr cr
-1 peek echo
Output:
[ 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 ]

33223

### Optimised

Not the same as the Factor and Factor inspired solutions, which count in base 4 with leading zeros like a telescoping pedometer; this skips over base 5 numbers with zeros in them.

[ 0 over
[ 5 /mod 0 = while
dip [ 5 * 1+ ]
again ]
drop + ]             is skipzeros   ( n --> n )

[ [] swap
[ 5 /mod
[ table 0 2 3 5 7 ]
rot join swap
dup 0 = until ]
swap witheach
[ swap 10 * + ] ]     is primedigits ( n --> n )

[ temp put [] 0
[ 1+ skipzeros
dup primedigits
dup isprime iff
[ swap dip join ]
else drop
over size
temp share = until ]
temp release drop  ]   is spds        ( n --> [ )

100 spds
25 split swap echo
cr cr
-1 peek echo
Output:
[ 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 ]

33223

## Raku

(formerly Perl 6)

Library: ntheory
use Lingua::EN::Numbers;
use ntheory:from<Perl5> <:all>;

# Implemented as a lazy, extendable list
my \$spds = grep { .&is_prime }, flat [2,3,5,7], [23,27,33,37,53,57,73,77], -> \$p
{ state \$o++; my \$oom = 10**(1+\$o); [ flat (2,3,5,7).map: -> \$l { (|\$p).map: \$l×\$oom + * } ] } … *;

say 'Smarandache prime-digitals:';
printf "%22s: %s\n", ordinal(1+\$_).tclc, comma \$spds[\$_] for flat ^25, 99, 999, 9999, 99999;
Output:
Smarandache prime-digitals:
First: 2
Second: 3
Third: 5
Fourth: 7
Fifth: 23
Sixth: 37
Seventh: 53
Eighth: 73
Ninth: 223
Tenth: 227
Eleventh: 233
Twelfth: 257
Thirteenth: 277
Fourteenth: 337
Fifteenth: 353
Sixteenth: 373
Seventeenth: 523
Eighteenth: 557
Nineteenth: 577
Twentieth: 727
Twenty-first: 733
Twenty-second: 757
Twenty-third: 773
Twenty-fourth: 2,237
Twenty-fifth: 2,273
One hundredth: 33,223
One thousandth: 3,273,527
Ten thousandth: 273,322,727
One hundred thousandth: 23,325,232,253

## REXX

The prime number generator has been simplified and very little optimization was included.

/*REXX program lists a  sequence of  SPDS  (Smarandache prime-digital sequence)  primes.*/
parse arg n q                                    /*get optional number of primes to find*/
if n=='' | n==","  then n=  25                   /*Not specified?  Then use the default.*/
if q=''            then q= 100  1000             /* "      "         "   "   "     "    */
say '═══listing the first'     n     "SPDS primes═══"
call spds n
do i=1  for words(q)+1;     y=word(q, i);    if y=='' | y==","   then iterate
say
say '═══listing the last of '    y     "SPDS primes═══"
call spds -y
end   /*i*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
spds: parse arg x 1 ox;  x= abs(x)               /*obtain the limit to be used for list.*/
c= 0                                       /*C  number of SPDS primes found so far*/
#= 0                                       /*#  number of      primes found so far*/
do j=1  by 2  while  c<x;    z= j    /*start: 1st even prime, then use odd. */
if z==1  then z= 2                   /*handle the even prime (special case) */
/* [↓]  divide by the primes.   ___    */
do k=2  to #  while  k*k<=z  /*divide  Z  with all primes ≤ √ Z     */
if z//@.k==0  then iterate j /*÷ by prev. prime?  ¬prime     ___    */
end   /*j*/                  /* [↑]   only divide up to     √ Z     */
#= # + 1;             @.#= z         /*bump the prime count;  assign prime #*/
if verify(z, 2357)>0  then iterate j /*Digits ¬prime?  Then skip this prime.*/
c= c + 1                             /*bump the number of SPDS primes found.*/
if ox<0  then iterate                /*don't display it, display the last #.*/
say right(z, 21)                     /*maybe display this prime ──► terminal*/
end   /*j*/                          /* [↑]  only display N number of primes*/
if ox<0  then say right(z, 21)             /*display one  (the last)  SPDS prime. */
return
output   when using the default inputs:
═══listing the first 25 SPDS primes═══
2
3
5
7
23
37
53
73
223
227
233
257
277
337
353
373
523
557
577
727
733
757
773
2237
2273

═══listing the last of  100 SPDS primes═══
33223

═══listing the last of  1000 SPDS primes═══
3273527

## Ring

see "First 25 Smarandache primes:" + nl + nl

num = 0
limit = 26
limit100 = 100
for n = 1 to 34000
flag = 0
nStr = string(n)
for x = 1 to len(nStr)
nx = number(nStr[x])
if isprime(n) and isprime(nx)
flag = flag + 1
else
exit
ok
next
if flag = len(nStr)
num = num + 1
if num < limit
see "" + n + " "
ok
if num = limit100
see nl + nl + "100th Smarandache prime: " + n + nl
ok
ok
next
Output:
First 25 Smarandache primes:

2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273

100th Smarandache prime: 33223

## RPL

Brute force being not an option for the slow machines that can run RPL, optimisation is based on a prime-digital number generator returning possibly prime numbers, e.g. not ending by 2 or 5. PRIM? is defined at Primality by trial division.

RPL code Comment
≪ { "2" "3" "5" "7" } DUP SIZE → digits base
≪ DUP SIZE 1 CF 2 SF
DO
IF DUP NOT THEN digits 1 GET ROT + SWAP 1 CF
ELSE
DUP2 DUP SUB digits SWAP POS
IF 2 FS?C THEN 2 == 4 ≪ 1 SF 2 ≫ IFTE
ELSE IF DUP base == THEN
SIGN 1 SF ELSE 1 + 1 CF END
END
digits SWAP GET REPL
LASTARG ROT DROP2 1 - END
UNTIL 1 FC? END DROP
≫ ≫ ‘NSPDP’ STO
NSPDP ( "PDnumber" → "nextPDnumber" )
rank = units position, carry = 0, 1st pass = true
Loop
If rank = 0 then add a new digit rank
Else
digit = ord[rank]
If first pass then next digit = 3 or 7
Else if digit is the last of the series
Then next digit = 1 otherwise = ++digit

Replace digit with next_digit
Rank--
Until no carry
return number as a string
≪ { 2 3 5 } 7
WHILE OVER SIZE 100 < REPEAT
IF DUP PRIM? THEN SWAP OVER + SWAP END
STR→ NSPDP STR→ END DROP ≫ EVAL
DUP 1 25 SUB
SWAP 100 GET
Output:
2: { 2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273 }
1: 33223

task needs 3 min 45 s to run on a HP-48G.

## Rust

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;
}
if n % 5 == 0 {
return n == 5;
}
let mut p = 7;
const WHEEL: [u32; 8] = [4, 2, 4, 2, 4, 6, 2, 6];
loop {
for w in &WHEEL {
if p * p > n {
return true;
}
if n % p == 0 {
return false;
}
p += w;
}
}
}

fn next_prime_digit_number(n: u32) -> u32 {
if n == 0 {
return 2;
}
match n % 10 {
2 => n + 1,
3 | 5 => n + 2,
_ => 2 + next_prime_digit_number(n / 10) * 10,
}
}

fn smarandache_prime_digital_sequence() -> impl std::iter::Iterator<Item = u32> {
let mut n = 0;
std::iter::from_fn(move || {
loop {
n = next_prime_digit_number(n);
if is_prime(n) {
break;
}
}
Some(n)
})
}

fn main() {
let limit = 1000000000;
let mut seq = smarandache_prime_digital_sequence().take_while(|x| *x < limit);
println!("First 25 SPDS primes:");
for i in seq.by_ref().take(25) {
print!("{} ", i);
}
println!();
if let Some(p) = seq.by_ref().nth(99 - 25) {
println!("100th SPDS prime: {}", p);
}
if let Some(p) = seq.by_ref().nth(999 - 100) {
println!("1000th SPDS prime: {}", p);
}
if let Some(p) = seq.by_ref().nth(9999 - 1000) {
println!("10,000th SPDS prime: {}", p);
}
if let Some(p) = seq.last() {
println!("Largest SPDS prime less than {}: {}", limit, p);
}
}
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100th SPDS prime: 33223
1000th SPDS prime: 3273527
10,000th SPDS prime: 273322727
Largest SPDS prime less than 1000000000: 777777773

## Ruby

Attaching 3 and 7 to permutations of 2,3,5 and 7

require "prime"

smarandache = Enumerator.new do|y|
prime_digits = [2,3,5,7]
prime_digits.each{|pr| y << pr} # yield the below-tens
(1..).each do |n|
prime_digits.repeated_permutation(n).each do |perm|
c = perm.join.to_i * 10
y << c + 3 if (c+3).prime?
y << c + 7 if (c+7).prime?
end
end
end

seq = smarandache.take(100)
p seq.first(25)
p seq.last
Output:
[2, 3, 5, 7, 23, 37, 53, 73, 223, 227, 233, 257, 277, 337, 353, 373, 523, 557, 577, 727, 733, 757, 773, 2237, 2273]
33223

Calculating the 10,000th Smarandache number takes about 1.2 seconds.

## Sidef

func is_prime_digital(n) {
n.is_prime && n.digits.all { .is_prime }
}

say is_prime_digital.first(25).join(',')
say is_prime_digital.nth(100)
Output:
2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273
33223

## Swift

Translation of: C++
func isPrime(number: Int) -> Bool {
if number < 2 {
return false
}
if number % 2 == 0 {
return number == 2
}
if number % 3 == 0 {
return number == 3
}
if number % 5 == 0 {
return number == 5
}
var p = 7
let wheel = [4,2,4,2,4,6,2,6]
while true {
for w in wheel {
if p * p > number {
return true
}
if number % p == 0 {
return false
}
p += w
}
}
}

func nextPrimeDigitNumber(number: Int) -> Int {
if number == 0 {
return 2
}
switch number % 10 {
case 2:
return number + 1
case 3, 5:
return number + 2
default:
return 2 + nextPrimeDigitNumber(number: number/10) * 10
}
}

let limit = 1000000000
var n = 0
var max = 0
var count = 0
print("First 25 SPDS primes:")
while n < limit {
n = nextPrimeDigitNumber(number: n)
if !isPrime(number: n) {
continue
}
if count < 25 {
print(n, terminator: " ")
} else if count == 25 {
print()
}
count += 1
if (count == 100) {
print("Hundredth SPDS prime: \(n)")
} else if (count == 1000) {
print("Thousandth SPDS prime: \(n)")
} else if (count == 10000) {
print("Ten thousandth SPDS prime: \(n)")
}
max = n
}
print("Largest SPDS prime less than \(limit): \(max)")
Output:
First 25 SPDS primes:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
Hundredth SPDS prime: 33223
Thousandth SPDS prime: 3273527
Ten thousandth SPDS prime: 273322727
Largest SPDS prime less than 1000000000: 777777773

## Wren

Library: Wren-math

Simple brute-force approach.

import "./math" for Int

var limit = 1000
var spds = List.filled(limit, 0)
spds[0] = 2
var i = 3
var count = 1
while (count < limit) {
if (Int.isPrime(i)) {
var digits = i.toString
if (digits.all { |d| "2357".contains(d) }) {
spds[count] = i
count = count + 1
}
}
i = i + 2
if (i > 10) {
var j = i % 10
if (j == 1 || j == 5) {
i = i + 2
} else if (j == 9) {
i = i + 4
}
}
}
System.print("The first 25 SPDS primes are:")
System.print(spds.take(25).toList)
System.print("\nThe 100th SPDS prime is %(spds[99])")
System.print("\nThe 1,000th SPDS prime is %(spds[999])")
Output:
The first 25 SPDS primes are:
[2, 3, 5, 7, 23, 37, 53, 73, 223, 227, 233, 257, 277, 337, 353, 373, 523, 557, 577, 727, 733, 757, 773, 2237, 2273]

The 100th SPDS prime is 33223

The 1,000th SPDS prime is 3273527

## XPL0

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

func PrimeDigits(N);    \Return 'true' if all digits are prime
int  N;
[repeat N:= N/10;
case rem(0) of
0, 1, 4, 6, 8, 9: return false
other [];
until   N = 0;
return true;
];

int C, N;
[C:= 0;  N:= 2;
loop    [if IsPrime(N) then
if PrimeDigits(N) then
[C:= C+1;
if C <= 25 then
[IntOut(0, N);  ChOut(0, ^ )];
if C = 100 then
[Text(0, "^m^j100th: ");  IntOut(0, N)];
if C = 1000 then quit;
];
N:= N+1;
];
Text(0, "^m^j1000th: ");  IntOut(0, N);  CrLf(0);
]
Output:
2 3 5 7 23 37 53 73 223 227 233 257 277 337 353 373 523 557 577 727 733 757 773 2237 2273
100th: 33223
1000th: 3273527

## Yabasic

Translation of: Ring
num = 0
limit = 26
limit100 = 100

print "First 25 Smarandache primes:\n"
for n = 1 to 34000
flag = 0
nStr\$ = str\$(n)
for x = 1 to len(nStr\$)
nx = val(mid\$(nStr\$,x,1))
if isPrime(n) and isPrime(nx) then
flag = flag + 1
else
break
end if
next
if flag = len(nStr\$) then
num = num + 1
if num < limit  print "", n, " ";
if num = limit100  print "\n\n100th Smarandache prime: ", n
end if
next n
end

sub isPrime(v)
if v < 2  return False
if mod(v, 2) = 0  return v = 2
if mod(v, 3) = 0  return v = 3
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub
Output:
Igual que la entrada de Ring.

## zkl

Library: GMP

GNU Multiple Precision Arithmetic Library

Using GMP ( probabilistic primes), because it is easy and fast to generate primes.

Extensible prime generator#zkl could be used instead.

var [const] BI=Import("zklBigNum");  // libGMP

spds:=Walker.zero().tweak(fcn(ps){
var [const] nps=T(0,0,1,1,0,1,0,1,0,0);  // 2,3,5,7
p:=ps.nextPrime().toInt();
if(p.split().filter( fcn(n){ 0==nps[n] }) ) return(Void.Skip);
p   //  733 --> (7,3,3) --> () --> good,       29 --> (2,9) --> (9) --> bad
}.fp(BI(1)));

Or

spds:=Walker.zero().tweak(fcn(ps){
var [const] nps="014689".inCommon;
p:=ps.nextPrime().toInt();
if(nps(p.toString())) return(Void.Skip);
p   //  733 --> "" --> good,       29 --> "9" --> bad
}.fp(BI(1)));
println("The first 25 terms of the Smarandache prime-digital sequence are:");
spds.walk(25).concat(",").println();

println("The hundredth term of the sequence is: ",spds.drop(100-25).value);
println("1000th item of this sequence is : ",spds.drop(1_000-spds.n).value);
Output:
The first 25 terms of the Smarandache prime-digital sequence are:
2,3,5,7,23,37,53,73,223,227,233,257,277,337,353,373,523,557,577,727,733,757,773,2237,2273
The hundredth term of the sequence is: 33223
1000th item of this sequence is : 3273527