Primes whose first and last number is 3

Find primes   n   (in base ten)   whose first and last decimal digit is   3,   and where   n   <   4,000.

Primes whose first and last number is 3 is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.
Task

Show all output here on this page.


Stretch goal

Find and show only the   number   of these types of primes   that are   <   1,000,000.

11l

Translation of: Nim
F is_prime(n)
   I n == 2
      R 1B
   I n < 2 | n % 2 == 0
      R 0B
   L(i) (3 .. Int(sqrt(n))).step(2)
      I n % i == 0
         R 0B
   R 1B

V lim = 1'000'000
V primes3x3 = [3]
V m = 100
V count = 1
L m * 3 < lim
   L(n) (3 * m + 3 .. 4 * m - 7).step(10)
      I n > lim
         L.break
      I is_prime(n)
         count++
         I n < 4000
            primes3x3.append(n)
   m *= 10

print(‘Found ’primes3x3.len‘ primes starting and ending with 3 below 4000:’)
L(n) primes3x3
   print(‘#4’.format(n), end' I (L.index + 1) % 11 == 0 {"\n"} E ‘ ’)

print("\nFound "count‘ primes starting and ending with 3 below 1000000.’)
Output:
Found 33 primes starting and ending with 3 below 4000:
   3  313  353  373  383 3023 3083 3163 3203 3253 3313
3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623
3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

Found 2251 primes starting and ending with 3 below 1000000.

Action!

INCLUDE "H6:SIEVE.ACT"

BYTE Func IsSpecialPrime(INT i BYTE ARRAY primes)
  BYTE d,first

  IF primes(i)=0 THEN
    RETURN (0)
  FI
  first=1
  WHILE i#0
  DO
    d=i MOD 10
    IF first THEN
      IF d#3 THEN
        RETURN (0)
      FI
      first=0
    FI
    i==/10
  OD
  IF d#3 THEN
    RETURN (0)
  FI
RETURN (1)

PROC Main()
  DEFINE MAX="3999"
  BYTE ARRAY primes(MAX+1)
  INT i,count=[0]

  Put(125) PutE() ;clear the screen
  Sieve(primes,MAX+1)
  FOR i=2 TO MAX
  DO
    IF IsSpecialPrime(i,primes) THEN
      PrintI(i) Put(32)
      count==+1
    FI
  OD
  PrintF("%E%EThere are %I primes",count)
RETURN
Output:

Screenshot from Atari 8-bit computer

3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463
3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

There are 33 primes

ALGOL 68

Translation of: ALGOL W

With added stretch goal. As with the Go and other samples, generates the candidate sequence.

BEGIN # find some primes whose first and last digits are 3                   #
    INT max prime = 1 000 000; # largest number to consider                  #
    # sieve the primes to max prime                                          #
    PR read "primes.incl.a68" PR
    []BOOL prime = PRIMESIEVE max prime;
    INT p3count := 0;
    # prints n, if it is prime, handles newlines                             #
    PROC p = ( INT n )VOID:
         IF prime[ n ] THEN
             print( ( " ", whole( n, - 4 ) ) );
             IF ( p3count +:= 1 ) MOD 12 = 0 THEN print( ( newline ) ) FI
         FI # p #;
    # find 3x3 primes                                                        #
    p( 3 ); p( 33 );                                # a & 2 digit 3x3 primes #
    FOR i FROM 0 BY 10 TO  90 DO p(  303 + i ) OD;  # 3 digit 3x3 primes     #
    FOR i FROM 0 BY 10 TO 990 DO p( 3003 + i ) OD;  # 4 digit 3x3 primes     #
    # 5 and 6 digit 3x3 primes                                               #
    FOR i FROM 0 BY 10 TO  9 990 DO IF prime[  30 003 + i ] THEN p3count +:= 1 FI OD;
    FOR i FROM 0 BY 10 TO 99 990 DO IF prime[ 300 003 + i ] THEN p3count +:= 1 FI OD;
    print( ( newline, "Found ", whole( p3count, 0 ), " ""3x3"" primes below 1000000", newline ) )
END
Output:
    3  313  353  373  383 3023 3083 3163 3203 3253 3313 3323
 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673
 3733 3793 3803 3823 3833 3853 3863 3923 3943
Found 2251 "3x3" primes below 1000000

ALGOL W

As with the Go and oher samples, finds the numbers by generating the candidate seuence.

begin % find some primes whose first and last digits are 3                   %
    integer MAX_PRIME;
    MAX_PRIME := 4000;
    begin
        logical array isPrime ( 1 :: MAX_PRIME );
        integer p3Count;
        % increments n by 1                                                  %
        integer procedure inc( integer value result n ) ; begin n := n + 1; n end;
        % prints n, if it is prime, handles newlines                         %
        procedure p ( integer value n ) ;
            if isPrime( n ) then begin writeon( i_w := 4, s_w := 0, " ", n ); if inc( p3Count ) rem 12 = 0 then write() end;
        % sieve the primes to MAX_PRIME                                      %
        isPrime( 1 ) := false; isPrime( 2 ) := true;
        for i := 3 step 2 until MAX_PRIME do isPrime( i ) := true;
        for i := 4 step 2 until MAX_PRIME do isPrime( i ) := false;
        for i := 3 step 2 until truncate( sqrt( MAX_PRIME ) ) do begin
            integer ii; ii := i + i;
            if isPrime( i ) then for pr := i * i step ii until MAX_PRIME do isPrime( pr ) := false
        end for_i ;
        % find the 3x3 primes                                                %
        p3Count := 0;
        % 1, 2 and 3 digit 3x3 primes                                        %
        p( 3 ); p( 33 ); for i := 0 step 10 until 90 do p( 303 + i );
        % 4 digit 3x3 primes                                                 %
        for i := 0 step 10 until 990 do p( 3003 + i );
    end
end.
Output:
    3  313  353  373  383 3023 3083 3163 3203 3253 3313 3323
 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673
 3733 3793 3803 3823 3833 3853 3863 3923 3943

Arturo

firstAndLastIs3?: function [n][
    if not? prime? n -> return false
    return and? -> 3 = first digits n
                -> 3 = last digits n
]

primesWithFirstAndLast3: select 1..4000 => firstAndLastIs3?

loop split.every: 11 primesWithFirstAndLast3 'x ->
    print map x 's -> pad to :string s 5

nofPrimesBelow1M: enumerate 1..1000000 => firstAndLastIs3?

print ""
print ["Found" nofPrimesBelow1M "primes starting and ending with 3 below 1000000."]
Output:
    3   313   353   373   383  3023  3083  3163  3203  3253  3313 
 3323  3343  3373  3413  3433  3463  3533  3583  3593  3613  3623 
 3643  3673  3733  3793  3803  3823  3833  3853  3863  3923  3943 

Found 2251 primes starting and ending with 3 below 1000000. 

AWK

# syntax: GAWK -f PRIMES_WHOSE_FIRST_AND_LAST_NUMBER_IS_3.AWK
BEGIN {
    start = 1
    stop = 3999
    for (i=start; i<=stop; i++) {
      if (is_prime(i) && i ~ /^3/ && i ~ /3$/) {
        printf("%5d%1s",i,++count1%10?"":"\n")
      }
    }
    printf("\nPrimes beginning and ending with '3' %d-%d: %d\n",start,stop,count1)
    start = 1
    stop = 999999
    for (i=start; i<=stop; i++) {
      if (is_prime(i) && i ~ /^3/ && i ~ /3$/) {
        count2++
      }
    }
    printf("Primes beginning and ending with '3' %d-%d: %d\n",start,stop,count2)
    exit(0)
}
function is_prime(x,  i) {
    if (x <= 1) {
      return(0)
    }
    for (i=2; i<=int(sqrt(x)); i++) {
      if (x % i == 0) {
        return(0)
      }
    }
    return(1)
}
Output:
    3   313   353   373   383  3023  3083  3163  3203  3253
 3313  3323  3343  3373  3413  3433  3463  3533  3583  3593
 3613  3623  3643  3673  3733  3793  3803  3823  3833  3853
 3863  3923  3943
Primes beginning and ending with '3' 1-3999: 33
Primes beginning and ending with '3' 1-999999: 2251

C

Translation of: FreeBASIC
#include<stdio.h>
#include<stdlib.h>
#include<math.h>

int isprime( int p ) {
    int i;
    if(p==2) return 1;
    if(!(p%2)) return 0;
    for(i=3; i*i<=p; i+=2) {
       if(!(p%i)) return 0;
    }
    return 1;
}

int main(void) {
    int np = 1, d, i, n;
    printf( "3  " );
    for(d=1; d<6; d++) {
        for(i=3; i<pow(10,d)-1; i+=10) {
            n = i + 3*pow(10,d);
            if(isprime(n)) {
                ++np;
                if(n<4009) {
                    printf("%d  ",n);
                    if(!(np%10)) printf("\n");
                }
            }
        }
    }
    printf( "\n\nThere were %d primes of the form 3x3 below one million.\n", np );
    return 0;
}
Output:

3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

There were 2251 primes of the form 3x3 below one million.

Delphi

Works with: Delphi version 6.0


function IsPrime(N: int64): boolean;
{Fast, optimised prime test}
var I,Stop: int64;
begin
if (N = 2) or (N=3) then Result:=true
else if (n <= 1) or ((n mod 2) = 0) or ((n mod 3) = 0) then Result:= false
else
     begin
     I:=5;
     Stop:=Trunc(sqrt(N+0.0));
     Result:=False;
     while I<=Stop do
           begin
           if ((N mod I) = 0) or ((N mod (I + 2)) = 0) then exit;
           Inc(I,6);
           end;
     Result:=True;
     end;
end;




procedure GetDigits(N: integer; var IA: TIntegerDynArray);
{Get an array of the integers in a number}
var T: integer;
begin
SetLength(IA,0);
repeat
	begin
	T:=N mod 10;
	N:=N div 10;
	SetLength(IA,Length(IA)+1);
	IA[High(IA)]:=T;
	end
until N<1;
end;



procedure ShowFirstLast3Prime(Memo: TMemo);
var I,Cnt1,Cnt2: integer;
var IA: TIntegerDynArray;
var S: string;

	function FirstLast3(N: integer): boolean;
	var I: integer;
	begin
	GetDigits(N,IA);
	Result:=(IA[0]=3) and (IA[High(IA)]=3);
	end;

begin
Cnt1:=0; Cnt2:=0;
S:='';
for I:=0 to 1000000-1 do
 if IsPrime(I) then
  if FirstLast3(I) then
	begin
	Inc(Cnt1);
	if I<4000 then
	  	begin
	  	Inc(Cnt2);
	  	S:=S+Format('%5D',[I]);
	  	If (Cnt2 mod 5)=0 then S:=S+CRLF;
	  	end;
	 end;
Memo.Lines.Add(S);
Memo.Lines.Add('Count < 1,000     = '+IntToStr(Cnt2));
Memo.Lines.Add('Count < 1,000,000 = '+IntToStr(Cnt1));
end;
Output:
    3  313  353  373  383
 3023 3083 3163 3203 3253
 3313 3323 3343 3373 3413
 3433 3463 3533 3583 3593
 3613 3623 3643 3673 3733
 3793 3803 3823 3833 3853
 3863 3923 3943
Count < 1,000     = 33
Count < 1,000,000 = 2251
Elapsed Time: 181.797 ms.


EasyLang

fastfunc isprim num .
   i = 2
   while i <= sqrt num
      if num mod i = 0
         return 0
      .
      i += 1
   .
   return 1
.
fastfunc nextprim prim .
   repeat
      prim += 1
      until isprim prim = 1
   .
   return prim
.
func digok n .
   f = n mod 10
   while n > 0
      l = n mod 10
      n = n div 10
   .
   return if f = 3 and l = 3
.
p = 2
repeat
   if digok p = 1
      write p & " "
   .
   p = nextprim p
   until p >= 4000
.
Output:
3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943 

F#

This task uses Extensible Prime Generator (F#)

//3 Sandwich Primes. Nigel Galloway: July 25th., 2021
primes32()|>Seq.takeWhile((>)4000)|>Seq.filter(fun n->n%10=3 && (n=3||(n>29 && n<40)||(n>299 && n<400)||n>2999))|>Seq.iter(printf "%d "); printfn ""
Output:
3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

Factor

Works with: Factor version 0.99 2021-06-02
USING: formatting grouping io kernel lists lists.lazy math
math.functions math.primes sequences ;

: under ( list n -- list' ) '[ _ < ] lwhile ;

: (surrounded) ( n -- list )
    [ 1list 1 lfrom ] keep dup dup
    '[ 10^ _ * _ + [ [ 10 + ] lfrom-by ] keep dup _ / + 10 - under ]
    lmap-lazy lconcat lappend-lazy ;

: surrounded ( n upto -- list )
    [ (surrounded) ] [ under ] bi* [ prime? ] lfilter ;

: surrounded. ( n -- )
    dup "Primes under 10,000 beginning and ending with %d:\n" printf
    10,000 surrounded list>array 10 group
    [ [ "%6d" printf ] each nl ] each nl ;

{ 1 3 5 7 9 } [ surrounded. ] each

3 1,000,000 surrounded llength
"Found %d primes beginning and ending with 3 under 1,000,000.\n" printf
Output:
Primes under 10,000 beginning and ending with 1:
    11   101   131   151   181   191  1021  1031  1051  1061
  1091  1151  1171  1181  1201  1231  1291  1301  1321  1361
  1381  1451  1471  1481  1511  1531  1571  1601  1621  1721
  1741  1801  1811  1831  1861  1871  1901  1931  1951

Primes under 10,000 beginning and ending with 3:
     3   313   353   373   383  3023  3083  3163  3203  3253
  3313  3323  3343  3373  3413  3433  3463  3533  3583  3593
  3613  3623  3643  3673  3733  3793  3803  3823  3833  3853
  3863  3923  3943

Primes under 10,000 beginning and ending with 5:
     5

Primes under 10,000 beginning and ending with 7:
     7   727   757   787   797  7027  7057  7127  7177  7187
  7207  7237  7247  7297  7307  7417  7457  7477  7487  7507
  7517  7537  7547  7577  7607  7687  7717  7727  7757  7817
  7867  7877  7907  7927  7937

Primes under 10,000 beginning and ending with 9:
   919   929  9029  9049  9059  9109  9199  9209  9239  9319
  9349  9419  9439  9479  9539  9619  9629  9649  9679  9689
  9719  9739  9749  9769  9829  9839  9859  9929  9949

Found 2251 primes beginning and ending with 3 under 1,000,000.

Fermat

np := 1;
!(3,'  ');
for d = 1 to 5 do
    for i = 3 to 10^d-1 by 10 do
        n:=3*10^d + i;
        if Isprime(n) = 1 then
            np:=np+1;
            if n<4000 then
                !(n,'  ');
                if Divides(10,np) then !! fi;
            fi;
        fi;
    od;
od;
!!;
!!;
!!('There were ',np,' primes of the form 3...3 below 1,000,000');
Output:

3 313 353 373 383 3023 3083 3163 3203 3253

3313 3323 3343 3373 3413 3433 3463 3533 3583 3593

3613 3623 3643 3673 3733 3793 3803 3823 3833 3853

3863 3923 3943

There were 2251 primes of the form 3...3 below 1,000,000

FreeBASIC

#include "isprime.bas"

dim as integer np = 1, d, n, i
print 3;"  ";               'three counts, but is a special case
for d = 1 to 5              'how many digits after the initial 3
    for i = 3 to 10^d-1 step 10  'the actual digits
        n = 3*10^d + i
        if isprime(n) then
            np += 1
            if n<4000 then 
                print n;"  ";
                if np mod 10 = 0 then print
            end if
        end if
    next i
next d
print : print
print "There were ";np;" 3...3 primes below 1000000"
Output:

3   313   353   373   383   3023   3083   3163   3203   3253  
3313   3323   3343   3373   3413   3433   3463   3533   3583   3593  
3613   3623   3643   3673   3733   3793   3803   3823   3833   3853  
3863   3923   3943  

There were 2251 3...3 primes below 1000000

Go

Library: Go-rcu
package main

import (
    "fmt"
    "rcu"
)

func main() {
    var primes []int
    candidates := []int{3, 33}
    for i := 303; i <= 393; i += 10 {
        candidates = append(candidates, i)
    }
    for i := 3003; i <= 3993; i += 10 {
        candidates = append(candidates, i)
    }
    for _, cand := range candidates {
        if rcu.IsPrime(cand) {
            primes = append(primes, cand)
        }
    }
    fmt.Println("Primes under 4,000 which begin and end in 3:")
    for i, p := range primes {
        fmt.Printf("%5s ", rcu.Commatize(p))
        if (i+1)%11 == 0 {
            fmt.Println()
        }
    }
    fmt.Println("\nFound", len(primes), "Such primes.")
    pc := len(primes)
    for i := 30003; i <= 39993; i += 10 {
        if rcu.IsPrime(i) {
            pc++
        }
    }
    for i := 300003; i <= 399993; i += 10 {
        if rcu.IsPrime(i) {
            pc++
        }
    }
    pcc := rcu.Commatize(pc)
    fmt.Println("\nFound", pcc, "primes under 1,000,000 which begin and end with 3.")
}
Output:
Primes under 4,000 which begin and end in 3:
    3   313   353   373   383 3,023 3,083 3,163 3,203 3,253 3,313 
3,323 3,343 3,373 3,413 3,433 3,463 3,533 3,583 3,593 3,613 3,623 
3,643 3,673 3,733 3,793 3,803 3,823 3,833 3,853 3,863 3,923 3,943 

Found 33 Such primes.

Found 2,251 primes under 1,000,000 which begin and end with 3.

Haskell

isPrime :: Int -> Bool
isPrime n 
   |n == 2 = True
   |n == 1 = False
   |otherwise = null $ filter (\i -> mod n i == 0 ) [2 .. root]
   where
      root :: Int
      root = floor $ sqrt $ fromIntegral n   

condition :: Int -> Bool
condition n = isPrime n && head numstr == '3' && last numstr == '3'
 where
  numstr :: String
  numstr = show n

solution :: [Int]
solution = filter condition [1..3999]

main :: IO ( )
main = do
   print solution 
   putStrLn ( "There are " ++ ( show $ length $ filter condition [1..999999]
            ) ++ " 3 x 3 primes below 1000000!" )
Output:
[3,313,353,373,383,3023,3083,3163,3203,3253,3313,3323,3343,3373,3413,3433,3463,3533,3583,3593,3613,3623,3643,3673,3733,3793,3803,3823,3833,3853,3863,3923,3943]
There are 2251 3 x 3 primes below 1000000!

J

   primes3x3=. 3 ;@; 10 <@(#~ 1&p:)@(30&* + 3 + 10 * i.)@^ i.

   primes3x3 3
3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

   # primes3x3 5
2251

jq

Works with: jq

Works with gojq, the Go implementation of jq

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

# For the stretch goal:
def count(s): reduce s as $x (null; .+1); 

# $n is the max number of intervening digits
def task($n):
  3,
  (range(1; $n+1) as $power
   | (3 * pow(10; $power+1) + 3) as $min
   | ("3" + ((pow(10; $power) -1)|tostring) + "4"|tonumber) as $max
   | range($min; $max; 10)
   | select(is_prime)   ) ;

task(2),
 "\nStretch goal: \(count( task(4) ))"
Output:
3
313
353
373
383
3023
3083
3163
3203
3253
3313
3323
3343
3373
3413
3433
3463
3533
3583
3593
3613
3623
3643
3673
3733
3793
3803
3823
3833
3853
3863
3923
3943

Stretch goal: 2251

Julia

using Primes

isxbyx(n, base=10, dig=3) = n ÷ prevpow(base, n) == dig && n % base == dig
p3x3(N, base=10, dig=3) = [p for p in primes(N) if isxbyx(p, base, dig)]

for d in 1:2:9
    println("\n$(d)x$d primes < 10000:")
    foreach(p -> print(rpad(last(p), 5), first(p) % 11 == 0 ? "\n" : ""),
        enumerate(p3x3(10000, 10, d)))
    println("\nTotal $(d)x$d primes less than 1,000,000: ", length(p3x3(1_000_000, 10, d)), ".")
end
Output:
        
1x1 primes < 10000:
11   101  131  151  181  191  1021 1031 1051 1061 1091 
1151 1171 1181 1201 1231 1291 1301 1321 1361 1381 1451
1471 1481 1511 1531 1571 1601 1621 1721 1741 1801 1811
1831 1861 1871 1901 1931 1951
Total 1x1 primes less than 1,000,000: 2387.

3x3 primes < 10000:
3    313  353  373  383  3023 3083 3163 3203 3253 3313
3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623
3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

Total 3x3 primes less than 1,000,000: 2251.

5x5 primes < 10000:
5
Total 5x5 primes less than 1,000,000: 1.

7x7 primes < 10000:
7    727  757  787  797  7027 7057 7127 7177 7187 7207
7237 7247 7297 7307 7417 7457 7477 7487 7507 7517 7537
7547 7577 7607 7687 7717 7727 7757 7817 7867 7877 7907
7927 7937
Total 7x7 primes less than 1,000,000: 2104.

9x9 primes < 10000:
919  929  9029 9049 9059 9109 9199 9209 9239 9319 9349
9419 9439 9479 9539 9619 9629 9649 9679 9689 9719 9739
9749 9769 9829 9839 9859 9929 9949
Total 9x9 primes less than 1,000,000: 2053.

Mathematica / Wolfram Language

Cases[NestWhileList[NextPrime, 
  2, # < 4000 &], _?(IntegerDigits[#][[{1, -1}]] === {3, 3} &)]
Output:

{3,313,353,373,383,3023,3083,3163,3203,3253,3313,3323,3343,3373,3413,3433,3463,3533,3583,3593,3613,3623,3643,3673,3733,3793,3803,3823,3833,3853,3863,3923,3943}

Nim

import strformat

func isPrime(n: Positive): bool =
  for d in countup(3, n, 2):
    if d * d > n: break
    if n mod d == 0: return false
  result = true


iterator primes3x3(lim: Natural): int =
  assert lim >= 3
  yield 3
  var m = 100
  while m * 3 < lim:
    for n in countup(3 * m + 3, 4 * m - 7, 10):
      if n > lim: break
      if n.isPrime: yield n
    m *= 10

var list: seq[int]
var count = 0
for n in primes3x3(1_000_000):
  inc count
  if n < 4000: list.add n

echo &"Found {list.len} primes starting and ending with 3 below 4_000:"
for i, n in list:
  stdout.write &"{n:4}", if (i + 1) mod 11 == 0: '\n' else: ' '

echo &"\nFound {count} primes starting and ending with 3 below 1_000_000."
Output:
Found 33 primes starting and ending with 3 below 4_000:
   3  313  353  373  383 3023 3083 3163 3203 3253 3313
3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623
3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

Found 2251 primes starting and ending with 3 below 1_000_000.

Perl

Library: ntheory
#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Primes_whose_first_and_last_number_is_3
use warnings;
use ntheory qw( primes );

my @n33 = grep /^3/ && /3$/, @{ primes( 4000 ) };
my $n33 = grep /^3/ && /3$/, @{ primes( 1_000_000 ) };
print @n33 . " under 4000\n\n@n33" =~ s/.{75}\K /\n/gr,
  "\n\n$n33 under 1000000\n";
Output:
33 under 4000

3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463
3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943

2251 under 1000000

Phix

Works for any digit, partly inspired by the Wren entry.

with javascript_semantics
function primes_with_first_and_last_digit(integer d, p10)
    -- d should be 0..9
    -- p10 should be 1 for 10, 4 for 10,000, 6 for 1,000,000, etc
    if d<0 or d>9 then crash("d is not a digit!") end if
    if d=2 or d=5 then return {d} end if
    if even(d) then return {} end if
    sequence res = filter({d},is_prime)
    for i=1 to p10-1 do
        integer p = power(10,i)
        for candidate = d*p+d to (d+1)*p-1 by 10 do
            if is_prime(candidate) then
                res &= candidate
            end if
        end for
    end for
    return res
end function

sequence res = primes_with_first_and_last_digit(3,4)
string sres = join_by(apply(true,sprintf,{{"%5d"},res}),1,10)
printf(1,"There are %d primes under 10,000 which begin and end in 3:\n%s\n",{length(res),sres})
res = primes_with_first_and_last_digit(3,6)
printf(1,"There are %d primes under 1,000,000 which begin and end in 3\n",{length(res)})
Output:
There are 33 primes under 10,000 which begin and end in 3:
    3     313     353     373     383    3023    3083    3163    3203    3253
 3313    3323    3343    3373    3413    3433    3463    3533    3583    3593
 3613    3623    3643    3673    3733    3793    3803    3823    3833    3853
 3863    3923    3943

There are 2251 primes under 1,000,000 which begin and end in 3

Python

def prime(n):
    if n == 1:
        return False
    if n == 2:
        p.append(n)
        return True
    for y in p:
        if n % y == 0:
            return False
        if y > int(n ** 0.5):
            p.append(n)
            return True


p = []
x = 1
stopper = 0
lis = []

for i in range(0, 1000000):
    x += 1
    if prime(x) == True:
        str_x = str(x)
        if str_x[0] == "3" and str_x[-1] == "3":
                lis.append(x)
    if int(x) == 4000 and stopper == 0:
        print(f'There are {len(lis)} primes that start and end with 3 less than 4000')
        for i in lis:
            print(i)

print(f'There are {len(lis)} primes that start and end with 3 less than 1000000')
Output:
There are 33 primes that start and end with 3 less than 4000
3
313
353
373
383
3023
3083
3163
3203
3253
3313
3323
3343
3373
3413
3433
3463
3533
3583
3593
3613
3623
3643
3673
3733
3793
3803
3823
3833
3853
3863
3923
3943
There are 2251 primes that start and end with 3 less than 1000000

Raku

my $upto = 1e4;

for 1,3,5,7,9 -> $bracket {
    print "\nPrimes up to $upto bracketed by $bracket - ";
    say display ^$upto .grep: { .starts-with($bracket) && .ends-with($bracket) && .is-prime }
}

sub display ($list, :$cols = 10, :$fmt = '%6d', :$title = "{+$list} matching:\n" )   {
    cache $list;
    $title ~ $list.batch($cols)».fmt($fmt).join: "\n"
}
Output:
Primes up to 10000 bracketed by 1 - 39 matching:
    11    101    131    151    181    191   1021   1031   1051   1061
  1091   1151   1171   1181   1201   1231   1291   1301   1321   1361
  1381   1451   1471   1481   1511   1531   1571   1601   1621   1721
  1741   1801   1811   1831   1861   1871   1901   1931   1951

Primes up to 10000 bracketed by 3 - 33 matching:
     3    313    353    373    383   3023   3083   3163   3203   3253
  3313   3323   3343   3373   3413   3433   3463   3533   3583   3593
  3613   3623   3643   3673   3733   3793   3803   3823   3833   3853
  3863   3923   3943

Primes up to 10000 bracketed by 5 - 1 matching:
     5

Primes up to 10000 bracketed by 7 - 35 matching:
     7    727    757    787    797   7027   7057   7127   7177   7187
  7207   7237   7247   7297   7307   7417   7457   7477   7487   7507
  7517   7537   7547   7577   7607   7687   7717   7727   7757   7817
  7867   7877   7907   7927   7937

Primes up to 10000 bracketed by 9 - 29 matching:
   919    929   9029   9049   9059   9109   9199   9209   9239   9319
  9349   9419   9439   9479   9539   9619   9629   9649   9679   9689
  9719   9739   9749   9769   9829   9839   9859   9929   9949

REXX

This REXX version allows the specification of the limit to search for these types of primes   (in base ten),   and it also
allows the specification of what (both) the leading and trailing decimal digit must be.

Also,   if a negative   cols   is specified,   only the   count   of primes found is shown.

/*REXX pgm finds and displays primes (base ten) that contain a leading and trailing  3. */
parse arg  hi cols dig .                         /*obtain optional argument from the CL.*/
if   hi=='' |   hi==","  then   hi=   4000       /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=     10       /* "      "         "   "   "     "    */
if  dig=='' |  dig==","  then  dig=      3       /* "      "         "   "   "     "    */
call genP                                        /*build array of semaphores for primes.*/
w= 10                                            /*width of a number in any column.     */
title= ' primes  N  (in base ten)  that have a leading and trailing digit of '  dig     ,
                                                             " where  N < "   commas(hi)
if cols>0  then say ' index │'center(title,   1 + cols*(w+1)     )
if cols>0  then say '───────┼'center(""   ,   1 + cols*(w+1), '─')
found= 0;                    idx= 1              /*initialize # of primes found;  IDX.  */
$=                                               /*list of primes that contain a string.*/
     do j=1  for #                               /*find primes with leading/trailing dig*/
     parse var  @.j   Ld  2  ''  -1  Td          /*obtain the leading and trailing digit*/
     if Ld\==dig           then iterate          /*does prime contain a leading  DIG ?  */      /* ◄■■■■■■■ a filter.*/
     if Td\==dig           then iterate          /*  "    "      "    " trailing  ?  ?  */      /* ◄■■■■■■■ a filter.*/
     found= found + 1                            /*bump the number of primes found.     */
     if cols<=0            then iterate          /*Build the list  (to be shown later)? */
     c= commas(@.j)                              /*maybe add commas to the number.      */
     $= $  right(c, max(w, length(c) ) )         /*add a prime  ──►  $ list, allow big #*/
     if found//cols\==0    then iterate          /*have we populated a line of output?  */
     say center(idx, 7)'│'  substr($, 2);   $=   /*display what we have so far  (cols). */
     idx= idx + cols                             /*bump the  index  count for the output*/
     end   /*j*/

if $\==''  then say center(idx, 7)"│"  substr($, 2)  /*possible display residual output.*/
if cols>0  then say '───────┴'center(""   ,   1 + cols*(w+1), '─')
say
say 'Found '       commas(found)      title
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?;  do jc=length(?)-3  to 1  by -3; ?=insert(',', ?, jc); end;  return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP:        @.1=2; @.2=3; @.3=5; @.4=7;  @.5=11 /*define some low primes.              */
                           #=5;   sq.#= @.# **2  /*number of primes so far; prime square*/
        do j=@.#+2  by 2  to hi-1                /*find odd primes from here on.        */
        parse var j '' -1 _; if     _==5  then iterate  /*J divisible by 5?  (right dig)*/
                             if j// 3==0  then iterate  /*"     "      " 3?             */
                             if j// 7==0  then iterate  /*"     "      " 7?             */
               do k=5  while sq.k<=j             /* [↓]  divide by the known odd primes.*/
               if j // @.k == 0  then iterate j  /*Is  J ÷ X?  Then not prime.     ___  */
               end   /*k*/                       /* [↑]  only process numbers  ≤  √ J   */
        #= #+1;    @.#= j;            sq.#= j*j  /*bump # of Ps; assign next P; P square*/
        end          /*j*/;           return
output   when using the default inputs:
 index │           primes  N  (in base ten)  that have a leading and trailing digit of  3  where  N <  4,000
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │          3        313        353        373        383      3,023      3,083      3,163      3,203      3,253
  11   │      3,313      3,323      3,343      3,373      3,413      3,433      3,463      3,533      3,583      3,593
  21   │      3,613      3,623      3,643      3,673      3,733      3,793      3,803      3,823      3,833      3,853
  31   │      3,863      3,923      3,943
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  33  primes  N  (in base ten)  that have a leading and trailing digit of  3  where  N <  4,000
output   when using the default inputs of:     1000000   -1
Found  2,251  primes  N  (in base ten)  that have a leading and trailing digit of  3  where  N <  1,000,000

Ring

load "stdlib.ring"
see "working..." + nl
see "Primes whose first and last number is 3" + nl
row = 0

for n = 1 to 4000
    strn = string(n)
    if left(strn,1) = "3" and right(strn,1) = "3" and isprime(n)
       see "" + n + " "
       row++
       if row%10 = 0
          see nl
       ok
    ok
next

see nl + "Found " + row + " numbers" + nl
see "done..." + nl
Output:
working...
Primes whose first and last number is 3
3 313 353 373 383 3023 3083 3163 3203 3253 
3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 
3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 
3863 3923 3943 
Found 33 numbers
done...

RPL

Uses a candidate numbers generator to speed up execution.

Works with: HP version 49
≪ { } 3
   WHILE DUP 4000 < REPEAT
     IF DUP ISPRIME? THEN SWAP OVER + SWAP END               
     10 +
     IF DUP MANT IP 3 ≠ THEN XPON 1 + ALOG 3 * 3 + END
   END DROP
≫ '33PRIMES' STO
Output:
1: { 3 313 353 373 383 3023 3083 3163 3203 3253 3313 3323 3343 3373 3413 3433 3463 3533 3583 3593 3613 3623 3643 3673 3733 3793 3803 3823 3833 3853 3863 3923 3943 }

Ruby

require 'prime'

puts "Primes below #{n=4000} which start and end with #{d=3}: "
p Prime.each(n).select{|pr| p_d = pr.digits; p_d.first == d && p_d.last == d}

print "\nCount of primes below #{n=1_000_000} which start and en with #{d=3}: "
puts Prime.each(n).count{|pr| p_d = pr.digits; p_d.first == d && p_d.last == d}
Output:
Primes below 4000 which start and end with 3: 
[3, 313, 353, 373, 383, 3023, 3083, 3163, 3203, 3253, 3313, 3323, 3343, 3373, 3413, 3433, 3463, 3533, 3583, 3593, 3613, 3623, 3643, 3673, 3733, 3793, 3803, 3823, 3833, 3853, 3863, 3923, 3943]

Count of primes below 1000000 which start and en with 3: 2251

Sidef

func numbers_with_edges(upto, base = 10, s = [3]) {
    Enumerator({|callback|
        callback(s.digits2num(base))
        for k in (0 .. base**(upto.len(base) - 2*s.len)) {

            break if (s + k.digits(base) + s -> digits2num(base) > upto)

            Inf.times { |j|

                var d = (s + k.digits(base) + j.of(0) + s)
                var n = d.digits2num(base)

                (n <= upto) ? callback(n) : break
            }
        }
    })
}

with (4e3) { |n|
    var list  = numbers_with_edges(n).grep{.is_prime}.sort
    say "There are #{list.len} primes <= #{n.commify} which begin and end in 3:"
    list.each_slice(10, {|*a| say a.map { '%5s' % _ }.join(' ') })
}

with (1e6) {|n|
    var count = numbers_with_edges(n).grep{.is_prime}.len
    say "\nThere are #{count} primes <= #{n.commify} which begin and end in 3"
}
Output:
There are 33 primes <= 4,000 which begin and end in 3:
    3   313   353   373   383  3023  3083  3163  3203  3253
 3313  3323  3343  3373  3413  3433  3463  3533  3583  3593
 3613  3623  3643  3673  3733  3793  3803  3823  3833  3853
 3863  3923  3943

There are 2251 primes <= 1,000,000 which begin and end in 3

Wren

Library: Wren-math
Library: Wren-iterate
Library: Wren-fmt

Basic task

import "./math" for Int
import "./iterate" for Stepped
import "./fmt" for Fmt

var primes = []
for (seq in [ 3..3, 33..33, Stepped.new(303..393, 10), Stepped.new(3003..3993, 10) ]) {
    for (e in seq) if (Int.isPrime(e)) primes.add(e)
}
System.print("Primes under 4,000 which begin and end in 3:")
Fmt.tprint("$,5d", primes, 11)
System.print("\nFound %(primes.count) such primes.")
Output:
Primes under 4,000 which begin and end in 3:
    3   313   353   373   383 3,023 3,083 3,163 3,203 3,253 3,313
3,323 3,343 3,373 3,413 3,433 3,463 3,533 3,583 3,593 3,613 3,623
3,643 3,673 3,733 3,793 3,803 3,823 3,833 3,853 3,863 3,923 3,943

Found 33 such primes.

More general

This version deals with primes (in base 10) beginning and ending with any specified digit and with up to a given number of digits.

import "./math" for Int
import "./iterate" for Stepped
import "./fmt" for Fmt

var getQualifyingPrimes = Fn.new { |x, d|
    if (d.type != Num || !d.isInteger || d < 1) Fiber.abort("Invalid number of digits.")
    if (x == 2 || x == 5) return [x]
    if (x % 2 == 0) return []
    var primes = []
    var candidates = [x]
    for (i in 1...d) {
        var pow = 10.pow(i)
        var start = x * (pow + 1)
        var end = start + pow - 10
        candidates.addAll(Stepped.new(start..end, 10).toList)
    }
    return candidates.where { |cand| Int.isPrime(cand) }.toList
}

var d = 4  // up to 'd' digits
for (x in [1, 2, 3, 5, 7, 9]) { // begins and ends with 'x'
    var primes = getQualifyingPrimes.call(x, d)
    var len = d + ((d-1)/3).floor
    Fmt.print("Primes under $,%(len)d which begin and end in $d:", 10.pow(d), x)
    Fmt.tprint("$,%(len)d", primes, 10)
    System.print("\nFound %(primes.count) such primes.\n")
}

d = 6
for (x in [1, 3, 7, 9]) {
    var primes = getQualifyingPrimes.call(x, d)
    Fmt.print("Found $,d primes under $,d which begin and end with $d.\n", primes.count, 10.pow(d), x)
}
Output:
Primes under 10,000 which begin and end in 1:
   11   101   131   151   181   191 1,021 1,031 1,051 1,061
1,091 1,151 1,171 1,181 1,201 1,231 1,291 1,301 1,321 1,361
1,381 1,451 1,471 1,481 1,511 1,531 1,571 1,601 1,621 1,721
1,741 1,801 1,811 1,831 1,861 1,871 1,901 1,931 1,951

Found 39 such primes.

Primes under 10,000 which begin and end in 2:
    2

Found 1 such primes.

Primes under 10,000 which begin and end in 3:
    3   313   353   373   383 3,023 3,083 3,163 3,203 3,253
3,313 3,323 3,343 3,373 3,413 3,433 3,463 3,533 3,583 3,593
3,613 3,623 3,643 3,673 3,733 3,793 3,803 3,823 3,833 3,853
3,863 3,923 3,943

Found 33 such primes.

Primes under 10,000 which begin and end in 5:
    5

Found 1 such primes.

Primes under 10,000 which begin and end in 7:
    7   727   757   787   797 7,027 7,057 7,127 7,177 7,187
7,207 7,237 7,247 7,297 7,307 7,417 7,457 7,477 7,487 7,507
7,517 7,537 7,547 7,577 7,607 7,687 7,717 7,727 7,757 7,817
7,867 7,877 7,907 7,927 7,937

Found 35 such primes.

Primes under 10,000 which begin and end in 9:
  919   929 9,029 9,049 9,059 9,109 9,199 9,209 9,239 9,319
9,349 9,419 9,439 9,479 9,539 9,619 9,629 9,649 9,679 9,689
9,719 9,739 9,749 9,769 9,829 9,839 9,859 9,929 9,949

Found 29 such primes.

Found 2,387 primes under 1,000,000 which begin and end with 1.

Found 2,251 primes under 1,000,000 which begin and end with 3.

Found 2,104 primes under 1,000,000 which begin and end with 7.

Found 2,053 primes under 1,000,000 which begin and end with 9.

XPL0

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

func Starts3(N);        \Return 'true' if most significant digit is 3
int     N;
[repeat N:= N/10;
until   N = 0;
return rem(0) = 3;
];

int Count, N;
[Count:= 0;
N:= 3;
repeat  if Starts3(N) then
            if IsPrime(N) then
                [Count:= Count+1;
                if N < 4000 then
                    [IntOut(0, N);
                    if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
                    ];
                ];
        N:= N+10;
        if N = 4003 then
            [CrLf(0);
            IntOut(0, Count);
            Text(0, " such primes found below 4000.");
            ];
until   N >= 1_000_000;
CrLf(0);
IntOut(0, Count);
Text(0, " such primes found below 1,000,000.
");
]
Output:
3       313     353     373     383     3023    3083    3163    3203    3253
3313    3323    3343    3373    3413    3433    3463    3533    3583    3593
3613    3623    3643    3673    3733    3793    3803    3823    3833    3853
3863    3923    3943    
33 such primes found below 4000.
2251 such primes found below 1,000,000.