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# Smarandache prime-digital sequence

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
```

## 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)
```

## AWK

` # syntax: GAWK -f SMARANDACHE_PRIME-DIGITAL_SEQUENCE.AWKBEGIN {    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
```

## 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
```

## F#

This task uses Extensible Prime Generator (F#)

` // Generate Smarandache prime-digital sequence. Nigel Galloway: May 31st., 2019let 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 mathmath.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.functionsmath.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:" cr25 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 trueend function dim as integer smar(1 to 100), count = 1, i = 1, digit, jsmar(1) = 2print 1, 2while 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 ifwend`
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```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

## 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 {            n = n.AddTwo()        } else {            n = n.AddOne()        }    }} 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 {    return b.AddOne().AddOne()} 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 Control.Monad (guard)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 -> Stringcommas = 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 nothingdef 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    endend 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}) endtable.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 = {}, 50000for i = 2,S do sieve[i]=true endfor i = 2,S do if sieve[i] then for j=i*i,S,i do sieve[j]=nil end end end -- TASKS:local digs, cans, spds, N = {2,3,5,7}, T{0}, T{}, 100while #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) endendprint("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] = truecomposite[1] = truefor 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:    result.add n mod 10    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 ' ', necho() var count = 0for 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#Pascalconst  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.

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
"3.6s"
```

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 22 33 54 75 236 377 538 739 22310 22711 23312 25713 27714 33715 353100 33223 `

## Raku

(formerly Perl 6)

`use Lingua::EN::Numbers;use ntheory:from<Perl5> <:all>; # Implemented as a lazy, extendable listmy \$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

` load "stdlib.ring" see "First 25 Smarandache primes:" + nl + nl num = 0limit = 26limit100 = 100for 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     oknext `
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
```

## 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  endend 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 = 1000000000var n = 0var max = 0var count = 0print("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 = 1000var spds = List.filled(limit, 0)spds[0] = 2var i = 3var count = 1while (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
```

## 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
```