# 10001th prime

10001th prime 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.

## 11l

```V k = 10001
V n = k * 17
V primes = [1B] * n
primes[0] = primes[1] = 0B

L(i) 2 .. Int(sqrt(n))
I !primes[i]
L.continue
L(j) (i * i .< n).step(i)
primes[j] = 0B

L(i) 0 .< n
I primes[i]
I k == 1
print(i)
L.break
k--```
Output:
```104743
```

## ALGOL 68

Chebyshev showed that the limit of pi( n ) / ( n / ln(n) ) as n approaches infinity is 1 ( pi( n ) is the number of primes up to n ).
The APPROXIMATESIEVESIZEFOR operator uses this to find a rough value for size of sieve needed to contain the required number of primes.

```BEGIN # find the 10001st prime #
# construct a sieve of primes that should be large enough to contain 10001 primes #
INT required prime = 10 001;
[]BOOL prime = PRIMESIEVE APPROXIMATESIEVESIZEFOR required prime;
INT p count := 1;
FOR n FROM 3 BY 2 WHILE p count < required prime DO
IF prime[ n ] THEN
p count +:= 1;
IF p count = required prime THEN
print( ( "Prime ", whole( required prime, 0 ), " is ", whole( n, 0 ) ) )
FI
FI
OD
END```
Output:
```Prime 10001 is 104743
```

## Arturo

```primes: select 2..110000 => prime?
print primes\[10000]
```
Output:
`104743`

## AWK

#### Converted from FreeBASIC

```# syntax: GAWK -f 10001TH_PRIME.AWK

BEGIN {
printf("%s\n",main(10001))
exit(0)
}
function main(n,  p,pn) {
if (n == 1) { return(2) }
p = 3
pn = 1
while (pn < n) {
if (is_prime(p)) {
pn++
}
p += 2
}
return(p-2)
}
function is_prime(n,  d) {
d = 5
if (n < 2) { return(0) }
if (n % 2 == 0) { return(n == 2) }
if (n % 3 == 0) { return(n == 3) }
while (d*d <= n) {
if (n % d == 0) { return(0) }
d += 2
if (n % d == 0) { return(0) }
d += 4
}
return(1)
}
```

#### Idiomatic

```# awk -f 10001prime.awk

# n: odd number and n>8
function IsOddPrime(n,	i,limit) {
limit = int(sqrt(n))
for (i=3;i <= limit;i+=2)
if (n%i==0) return 0
return 1
}

# pos: positive
function PrimePosition(pos,	prime){
pos -= ( (pos==1) ? prime=2 : prime=3 ) - 1
while (pos)
if (IsOddPrime(prime+=2)) pos--
return prime
}

BEGIN { print PrimePosition(10001) }
```
Output:
```104743
```

## BASIC

### BASIC256

```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

function prime(n)
if n=1 then return 2
p = 3
pn = 1
while pn < n
if isPrime(p) then pn += 1
p += 2
end while
return p-2
end function

print prime(10001)
end```

### Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4

The GW-BASIC solution works without any changes.

### FreeBASIC

```#include "isprime.bas"
function prime( n as uinteger ) as ulongint
if n=1 then return 2
dim as integer p=3, pn=1
while pn<n
if isprime(p) then pn + = 1
p += 2
wend
return p-2
end function

print prime(10001)
```
Output:
`104743`

### Gambas

```'Use "isprime.bas"

Public Sub Main()

Print prime(10001)

End

Function prime(n As Integer) As Long

If n = 1 Then Return 2
Dim p As Integer = 3, pn As Integer = 1
While pn < n
If isPrime(p) Then pn += 1
p += 2
Wend
Return p - 2

End Function
```
Output:
`104743`

### GW-BASIC

Works with: BASICA
```10 PN = 1
20 P = 3
30 WHILE PN < 10001
40 GOSUB 100
50 IF Q = 1 THEN PN = PN + 1
60 P = P + 2
70 WEND
80 PRINT P - 2
90 END
100 REM Tests if a number is prime
110 Q = 0
120 IF P = 2 THEN Q = 1: RETURN
130 IF P = 3 THEN Q = 1: RETURN
140 I = 1
150 I = I + 2
160 IF INT(P / I) * I = P THEN RETURN
170 IF I * I <= P THEN GOTO 150
180 Q = 1
190 RETURN
```
Output:
`104743`

### PureBasic

```Procedure isPrime(v.i)
If     v <= 1    : ProcedureReturn #False
ElseIf v < 4     : ProcedureReturn #True
ElseIf v % 2 = 0 : ProcedureReturn #False
ElseIf v < 9     : ProcedureReturn #True
ElseIf v % 3 = 0 : ProcedureReturn #False
Else
Protected r = Round(Sqr(v), #PB_Round_Down)
Protected f = 5
While f <= r
If v % f = 0 Or v % (f + 2) = 0
ProcedureReturn #False
EndIf
f + 6
Wend
EndIf
ProcedureReturn #True
EndProcedure

Procedure prime(n.i)
If n = 1
ProcedureReturn 2
EndIf
p.i = 3
pn.i = 1
While pn < n
If isprime(p)
pn + 1
EndIf
p + 2
Wend
ProcedureReturn p-2
EndProcedure

OpenConsole()
n = prime(10001)
PrintN(Str(n))
CloseConsole()
```

### QB64

#### Trial Division Method

```max=10001: n=1: p=0: t=Timer ' PRIMES.bas russian DANILIN
While n <= max ' 10001 104743 0.35 seconds
f=0: j=2
While f < 1
If j >= p ^ 0.5 Then f=2
If p Mod j = 0 Then f=1
j=j+1
Wend
If f <> 1 Then n=n+1: ' Print n, p
p=p+1
Wend
Print n-1, p-1, Timer-t
```
Output:
`10001 104743 0.35 seconds`

#### More Efficient TD Method

```'JRace's results:
'Original version: 10001   104743  .21875
'This version:     10001   104743  .109375
max = 10001: n=1: p=0: t = Timer ' PRIMES.bas
While n <= max ' 10001 104743 0.35 seconds
f = 0: j = 2
pp = p^0.5 'NEW VAR: move exponentiation here, to outer loop
While f < 1
' If j >= p ^ 0.5 Then f=2 'original line
' NOTE: p does not change in this loop,
'      therefore p^0.5 doesn't change;
'      moving p^0.5 to the outer loop will eliminate a lot of FP math)
If j >= pp Then f=2 'new version with exponentiation relocated
If p Mod j = 0 Then f=1
j=j+1
Wend
If f <> 1 Then n=n+1: ' Print n, p
p=p+1
Wend
Print n-1, p-1, Timer - t
```
Output:
`10001 104743 0.11 seconds`

### True BASIC

#### Trial Division Method

Translation of: QB64
```LET max = 10001
LET n = 1
LET p = 0
DO WHILE n <= max
LET f = 0
LET j = 2
DO WHILE f < 1
IF j >= p^0.5 THEN LET f = 2
IF REMAINDER(p, j) = 0 THEN LET f = 1
LET j = j+1
LOOP
IF f <> 1 THEN LET n = n+1
LET p = p+1
LOOP
PRINT p-1
END
```
Output:
`104743`

### Yabasic

```sub isPrime(v)
if v < 2 then return False : fi
if mod(v, 2) = 0 then return v = 2 : fi
if mod(v, 3) = 0 then return v = 3 : fi
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub

sub prime(n)
if n = 1 then return 2 : fi
p = 3
pn = 1
while pn<n
if isPrime(p) then pn = pn + 1 : fi
p = p + 2
wend
return p-2
end sub

print prime(10001)
end```

## bc

```a[0] = 2
a[o = 1] = 3
for (n = 5; o < 10000; n += 2) {
for (i = 1; n % (p = a[i]) != 0; ++i) {
if (p * p > n) {
a[++o] = n
break
}
}
}
a[o]
```
Output:
`104743`

## C

```#include<stdio.h>
#include<stdlib.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 prime( int n ) {
int p, pn=1;
if(n==1) return 2;
for(p=3;pn<n;p+=2) {
if(isprime(p)) pn++;
}
return p-2;
}

int main(void) {
printf( "%d\n", prime(10001) );
return 0;
}
```
Output:
`104743`

## C#

### Sieve vs Trial Division

Comparing performance of the one-at-a-time trial division method vs the sieve of Eratosthenes method. About ten times faster for the sieve. It may appear that the sieve may be off by one, `pr[10000]` but since the array is zero based, it's the 10001st value.

```using System; class Program {

static bool isprime(uint p ) { if ((p & 1) == 0) return p == 2;
if ((p % 3) == 0) return p == 3;
for (uint i = 5, d = 4; i * i <= p; i += (d = 6 - d))
if (p % i == 0) return false; return true; }

static uint prime(uint n) { uint p, d, pn;
for (p = 5, d = 4, pn = 2; pn < n; p += (d = 6 - d))
if (isprime(p)) pn++; return p - d; }

static void Main(string[] args) {
Console.WriteLine("One-at-a-time trial division vs sieve of Eratosthenes");
var sw = System.Diagnostics.Stopwatch.StartNew();
var t = prime(10001); sw.Stop(); double e1, e2;
Console.Write("{0:n0} {1} ms", prime(10001),
e1 = sw.Elapsed.TotalMilliseconds);
sw.Restart(); uint n = 105000, i, j; var pr = new uint[10100];
pr[0] = 2; pr[1] = 3; uint pc = 2, r, d, ii;
var pl = new bool[n + 1]; r = (uint)Math.Sqrt(n);
for (i = 9; i < n; i += 6) pl[i] = true;
for (i = 5, d = 4; i <= r; i += (d = 6 - d)) if (!pl[i]) {
pr[pc++] = i; for (j = i * i, ii = i << 1; j <= n; j += ii)
pl[j] = true; }
for ( ;i <= n; i += (d = 6 - d)) if (!pl[i]) pr[pc++] = i;
t = pr[10000]; sw.Stop();
Console.Write("  {0:n0} {1} μs  {2:0.000} times faster", t,
(e2 = sw.Elapsed.TotalMilliseconds) * 1000.0, e1 / e2); } }
```
Output @ Tio.run:
```One-at-a-time trial division vs sieve of Eratosthenes
104,743 3.8943 ms  104,743 357.9 μs  10.881 times faster```

### Alternative Trial Division Method

```using System; using System.Text; // PRIME_numb.cs russian DANILIN
namespace p10001 // 1 second  10001  104743
{ class Program // rextester.com/ZBEPGE34760
{ static void Main(string[] args)
{ int max=10001; int n=1; int p=1; int f; int j; long s;
while (n <= max)
{ f=0; j=2; s=Convert.ToInt32(Math.Pow(p,0.5));
while (f < 1)
{ if (j >= s)
{ f=2; }
if (p % j == 0) { f=1; }
j++;
}
if (f != 1) { n++; } // Console.WriteLine("{0} {1}", n, p);
p++;
}
Console.Write("{0} {1}", n-1, p-1);
}}}
```
Output:
`104743`

## C++

Library: Primesieve
```#include <iostream>
#include <locale>

#include <primesieve.hpp>

int main() {
std::cout.imbue(std::locale(""));
std::cout << "The 10,001st prime is " << primesieve::nth_prime(10001) << ".\n";
}
```
Output:
```The 10,001st prime is 104,743.
```

## COBOL

Limit based on the fact that the nth prime is guaranteed to be less than n * (ln n + ln(ln n)).

```IDENTIFICATION DIVISION.
PROGRAM-ID. 10001th_prime.

DATA DIVISION.
WORKING-STORAGE SECTION.
01 n        PIC 9(3)    COMP.
01 cnt      PIC 9(5)    COMP    VALUE 1.
01 sieve.
05 isp  PIC 9               VALUE 1 OCCURS 115000 TIMES
INDEXED BY i.
01 out      PIC Z(10).

PROCEDURE DIVISION.
PERFORM VARYING n FROM 2 BY 1 UNTIL n * n > 115000
SET i TO n

IF isp(i) = 1
MULTIPLY n BY n GIVING i
PERFORM VARYING i FROM i BY n UNTIL i > 115000
SET isp(i) TO 0
END-PERFORM

END-PERFORM

SET i to 1

PERFORM UNTIL cnt = 10001
SET i UP BY 2
END-PERFORM

MOVE i TO out
DISPLAY FUNCTION TRIM (out)
STOP RUN.
```
Output:
`104743`

## Dart

```import 'dart:math';

bool isPrime(int n) {
if (n <= 1) return false;
if (n == 2) return true;
for (int i = 2; i <= sqrt(n); ++i) {
if (n % i == 0) return false;
}
return true;
}

int prime(int n) {
int p, pn = 1;
if (n == 1) return 2;
for (p = 3; pn < n; p += 2) {
if (isPrime(p)) pn++;
}
return p - 2;
}

void main() {
print(prime(10001));
}
```
Output:
`104743`

## Delphi

Works with: Delphi version 6.0
Library: none
```function IsPrime(N: integer): boolean;
{Fast, optimised prime test}
var I,Stop: integer;
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));
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;

function Find10001thPrime: integer;
{Return the 10,001th prime number}
var Count: integer;
begin
Count:=1;
Result:= 3;
while true do
begin
if IsPrime(Result) then Count:= Count+1;
if Count=10001 then exit;
Inc(Result,2);
end;
end;
```
Output:
```10,001th Prime= 104,743
```

## D

Translation of: C
```import std.stdio;

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 prime( int n ) {

if(n==1) return 2;

int p, pn=1;

for(p=3;pn<n;p+=2) {
if(isprime(p)) pn++;
}

return p-2;
}

void main()
{
writeln(prime(10_001));
}
```
Output:
`104743`

## Erlang

```-module(task_10001th_prime).

-export([main/0]).

% 2 and 3 are both primes, so we start searching at 5
main() -> search(5, 2, 2).

search(N, Inc, Count) when Count < 10_001 ->
case is_prime(N) of
true -> search(N + Inc, 6 - Inc, Count + 1);
false -> search(N + Inc, 6 - Inc, Count)
end;
search(N, Inc, _) -> N - 6 + Inc.

is_prime(P) -> is_prime(P, 5, 2).

is_prime(N, P, _) when P * P > N -> true;
is_prime(N, P, _) when N rem P =:= 0 -> false;
is_prime(N, P, Inc) -> is_prime(N, P + Inc, 6 - Inc).
```
Output:
``` timer:tc(task_10001th_prime, main, []).
{68615,104743}
```

## EasyLang

```fastfunc isprim num .
if num mod 2 = 0 and num > 2
return 0
.
i = 3
while i <= sqrt num
if num mod i = 0
return 0
.
i += 2
.
return 1
.
i = 2
repeat
if isprim i = 1
nprim += 1
.
until nprim = 10001
i += 1
.
print i```
Output:
`104743`

## F#

This task uses Extensible Prime Generator (F#)

```// 10001st prime. Nigel Galloway: November 22nd., 2021
printfn \$"%d{Seq.item 10000 (primes32())}"
```
Output:
```104743
```

## Factor

```USING: math math.primes prettyprint ;

2 10,000 [ next-prime ] times .
```
Output:
```104743
```

## Fermat

```Prime(10001);
```
Output:
`104743`

## Frink

`nth[primes[], 10001-1]`
Output:
```104743
```

## FutureBasic

```local fn IsPrime( n as NSUInteger ) as BOOL
BOOL       isPrime = YES
NSUInteger i

if n < 2        then exit fn = NO
if n = 2        then exit fn = YES
if n mod 2 == 0 then exit fn = NO
for i = 3 to int(n^.5) step 2
if n mod i == 0 then exit fn = NO
next
end fn = isPrime

local fn Prime( n as NSUInteger ) as long
long p = 3, pn = 1

if n = 1 then exit fn = 2
while ( pn < n )
if fn IsPrime( p ) then pn++
p += 2
wend
p -= 2
end fn = p

print fn Prime(10001)

HandleEvents```
Output:
```104743
```

## Go

```package main

import "fmt"

func isPrime(n int) bool {
if n == 1 {
return false
}
i := 2
for i*i <= n {
if n%i == 0 {
return false
}
i++
}
return true
}

func main() {
var final, pNum int

for i := 1; pNum < 10001; i++ {
if isPrime(i) {
pNum++
}
final = i
}
fmt.Println(final)
}
```
Output:
`104743`

## J

```p:10000      NB. the index starts at 0; p:0 = 2
```
Output:
`104743`

## Java

Uses the PrimeGenerator class from Extensible prime generator#Java.

```public class NthPrime {
public static void main(String[] args) {
System.out.printf("The 10,001st prime is %,d.\n", nthPrime(10001));
}

private static int nthPrime(int n) {
assert n > 0;
PrimeGenerator primeGen = new PrimeGenerator(10000, 100000);
int prime = primeGen.nextPrime();
while (--n > 0)
prime = primeGen.nextPrime();
return prime;
}
}
```
Output:
```The 10,001st prime is 104,743.
```

## JavaScript

Prime 10001 from Russia jdoodle.com/h/2V1

```var max = 10001, n=1, p=1; var f,j,s
while (n <= max)
{ f=0; j=2; s = parseInt(Math.pow(p, 0.5))
while (f < 1)
{ if (j >= s) f=2
if ( p % j == 0 ) f=1
j++
}
if (f != 1) n++ // { document.write(n +" "+ p +"<br>") }
p++
}
document.write("<br>"+ (n-1) +" "+ (p-1) +"<br>" )
```
Output:
```10001 104743
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

A solution without any pre-determined numbers or guesses.

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

```# Output: a stream of the primes
def primes: 2, (range(3; infinite; 2) | select(is_prime));

# jq uses an index-origin of 1 and so:
nth(10000; primes)```
Output:
```104743
```

## Julia

```julia> using Primes

julia> prime(10001)
104743
```

## Mathematica / Wolfram Language

```Prime[10001]
```
Output:
```
104743

```

## Maxima

```primes_first_n(n) := (
if n<=1 then
[]
else (
L : [2],
for i:2 thru n do (
L : append(L, [next_prime(last(L))])
),
L
)
)\$
last(primes_first_n(10001));
```
Output:
```
104743

```

## MiniScript

```isPrime = function(n)
s = floor(n ^ 0.5)
for i in range(2, s)
if n % i == 0 then return false
end for
return true
end function

tt = time
c = 2
p = 2
lastPrime = 2
while c < 10001
p += 1
if isPrime(p) then
c += 1
end if
end while
print p
```
Output:
`104743`

## Nim

```func isPrime(n: Positive): bool =
## Return true if "n" is prime.
## Assume n >= 5 and n not multiple of 2 and 3.
var d = 5
var step = 2
while d * d <= n:
if n mod d == 0:
return false
inc d, step
step = 6 - step
result = true

iterator primes(): tuple[rank, value: int] =
## Yield the primes preceded by their rank in the sequence.
yield (1, 2)
yield (2, 3)
var idx = 2
var n = 5
var step = 2
while true:
if n.isPrime:
inc idx
yield (idx, n)
inc n, step
step = 6 - step

for (i, n) in primes():
if i == 10001:
echo "10001st prime: ", n
break
```
Output:
```10001st prime: 104743
```

## OCaml

Using the function `seq_primes` from Extensible prime generator#OCaml:

```let () =
seq_primes |> Seq.drop 10000 |> Seq.take 1 |> Seq.iter (Printf.printf "%u\n")
```
Output:
`104743`

## PARI/GP

`prime(10001)`
Output:
`%1 = 104743`

## Pascal

### Free Pascal

```Program nth_prime;

Function nthprime(x : uint32): uint32;
Var primes : array Of uint32;
i,pr,count : uint32;
found : boolean;
Begin
setlength(primes, x);
primes[1] := 2;
count := 1;
i := 3;
Repeat
found := FALSE;
pr := 0;
Repeat
inc(pr);
found := i Mod primes[pr] = 0;
Until (primes[pr]*primes[pr] > i) Or found;
Begin
inc(count);
primes[count] := i;
End;
inc(i,2);
Until count = x;
nthprime := primes[x];
End;

Begin
writeln(nthprime(10001));
End.
```
Output:
```104743
```

## Perl

Library: ntheory
```use strict;
use warnings;
use feature 'say';

# the lengthy wait increases the delight in finally seeing the answer
my(\$n,\$c) = (1,0);
while () {
\$c++ if (1 x ++\$n) !~ /^(11+)\1+\$/;
\$c == 10_001 and say \$n and last;
}

# or if for some reason you want the answer quickly
use ntheory 'nth_prime';
say nth_prime(10_001);
```
Output:
```104743
104743```

## Phix

```with js ?get_prime(10001)
```
Output:
```104743
```

## Picat

```go ?=>
println(nth_prime(10001)),

% faster but probably considered cheating
nth(10001,primes(200000),P),
println(P).

nth_prime(Choosen) = P =>
nth_prime(1,0,Choosen, P).

nth_prime(Num, Choosen, Choosen, Num) :-
prime(Num).
nth_prime(Num, Ix, Choosen, P) :-
Ix < Choosen,
next_prime(Num, P2),
nth_prime(P2, Ix+1, Choosen, P).

next_prime(Num, P) :-
next_prime2(Num+1, P).
next_prime2(Num, Num) :-
prime(Num).
next_prime2(Num, P) :-
next_prime2(Num+1,P).```
Output:
```104743
104743```

## Prolog

for swi-prolog

```isPrime(2).                 % prime generator
isPrime(N):-
between(3, inf, N),
N /\ 1 > 0,             % odd
M is floor(sqrt(N)) - 1, % reverse 2*I+1
Max is M div 2,
forall(between(1, Max, I), N mod (2*I+1) > 0).

do:- Index is 10001,
findnsols(Index, N, isPrime(N), PrimeList),!,
last(PrimeList, PrimeAtIndex),
format('prime(~d) is ~d', [Index, PrimeAtIndex]), nl.
```
Output:
```?- do.
prime(10001) is 104743
true.```

## Python

### Trial Division Method

```#!/usr/bin/python

def isPrime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True

def prime(n: int) -> int:
if n == 1:
return 2
p = 3
pn = 1
while pn < n:
if isPrime(p):
pn += 1
p += 2
return p-2

if __name__ == '__main__':
print(prime(10001))
```
Output:
`104743`

### Alternative Trial Division Method

```import time; max=10001; n=1; p=1; # PRIME_numb.py russian DANILIN
while n<=max: # 78081 994271 45 seconds
f=0; j=2; s = int(p**0.5) # rextester.com/AAOHQ6342
while f < 1:
if j >= s:
f=2
if p % j == 0:
f=1
j+=1
if f != 1:
n+=1;
#print(n,p);
p+=1
print(n-1,p-1)
print(time.perf_counter())
```
Output:
`10001 104743 7 seconds`

## Quackery

The 10001st prime number is the 10000th odd prime number.

`prime` is defined at Miller–Rabin primality test#Quackery.

`  1 10000 times [ 2 + dup prime until ] echo`
Output:
`104743`

## R

```library(primes)
nth_prime(10001)
```
Output:
`104743`

## Racket

```#lang racket
(require math/number-theory)
; Index starts at 0, (nth-prime 0) is 2
(nth-prime 10000)
```
Output:
`104743`

## Raku

```say (^∞).grep( &is-prime )[10000]
```
Output:
`104743`

## REXX

The task has no stretch goal, nevertheless I will try one: showing primes higher in de sequence.

### Trial division

```/* REXX */
Parse Version v
Say v
Call Time 'R'
z=1
p.0=3
p.1=2
p.2=3
p.3=5
Do n=5 By 2 Until z=10001
If right(n,1)=5 Then Iterate
Do i=2 To p.0 Until b**2>n
b=p.i
If n//b=0 Then Leave
End
If b**2>n Then Do
z=p.0+1
p.z=n
p.0=z
End
End
Say z n time('E')
```
Output:
```H:\>rexx prime10001
REXX-ooRexx_5.0.0(MT)_64-bit 6.05 8 Sep 2020
10001 104743 0.219000

H:\>regina prime10001
REXX-Regina_3.9.4(MT) 5.00 25 Oct 2021
10001 104743 .615000

Finding the 100001th prime number:

REXX-ooRexx_5.0.0(MT)_64-bit 6.05 23 Dec 2022
100001 1299721 11.989000

REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
100001 1299721 19.117000
```

### Libraries

REXX does not have an 'include' facility nor 'arbitrary precision' mathematical functions out of the box. In previous REXX entries it was custom the have all needed routines or procedures copied into the program. Newer entries redirect to
Libraries and
Pre processor and Include clause.
At the end of each entry you'll find several 'Include' clauses, showing the libraries that the program needs (cf '#include', 'import', 'uses' or '::requires').

### Sieve

As some other REXX entries suggest, using a sieve is faster, but requires memory to store the primes. Consider following program, based on the property 'nth prime < n*(ln(n)+ln(ln(n)))'.

```/* REXX */
Parse Version v
Say v
glob. = ''
Call Time 'R'
Call Primes(105000)
n = 10001
Say glob.prime.n n time('E')
Exit

Primes:
/* Prime numbers */
procedure expose glob.
arg x
/* Validity */
if \ Whole(x) then
return 'X'
if x < 1 then
return 'X'
/* Fast values */
if x = 1 then do
glob.prime.0 = 0
return 0
end
if x < 101 then do
p = '2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 999'
do n = 1 to Words(p)
w = Word(p,n)
if w > x then do
n = n-1
leave
end
glob.prime.n = w
end
glob.prime.0 = n
return n
end
/* Wheeled sieve of Eratosthenes */
do i = 3 by 2 to x while i*i <= x
if glob.priem.i = '' then do
do j = i*3 by i+i to x
glob.priem.j = 0
end
end
end
/* Save results */
n = 1; glob.prime.1 = 2
do i = 3 by 2 to x
if glob.priem.i = '' then do
n = n+1; glob.prime.n = i
end
end
glob.prime.0 = n
return n

include Numbers
```
Output:
```As required by the task:
REXX-ooRexx_5.0.0(MT)_64-bit 6.05 23 Dec 2022
104743 10001 0.047000
REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
104743 10001 0.027000

And for some higher entries in the sequence:
REXX-ooRexx_5.0.0(MT)_64-bit 6.05 23 Dec 2022
1299721 100001 0.829000
REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
1299721 100001 0.469000

REXX-ooRexx_5.0.0(MT)_64-bit 6.05 23 Dec 2022
15485867 1000001 13.938000
REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
15485867 1000001 7.726000

REXX-ooRexx_5.0.0(MT)_64-bit 6.05 23 Dec 2022
86028157 5000001 89.021000
REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
86028157 5000001 81.339000
```

The last example, finding all primes in the range 2-87M (more than 5M primes!), took > 10GB memory on my 16GB machine, pushing it to the max. But without using the sieve solution, the last two examples would have taken hours.

## Ring

```load "stdlib.ring"
see "working..." + nl
num = 0 pr = 0 limit = 10001

while true
num++
if isprime(num) pr++ ok
if pr = limit exit ok
end

see "" + num + nl + "done..." + nl```
Output:
```working...
The 10001th prime is: 104743
done...
```

## RPL

Works with: HP version 49
```« 2
1 10000 START NEXTPRIME NEXT
```
Output:
```1: 104743
```

## Ruby

```require "prime"
puts  Prime.lazy.drop(10_000).next
```
Output:
```104743
```

## Rust

```// [dependencies]
// primal = "0.3"

fn main() {
let p = primal::StreamingSieve::nth_prime(10001);
println!("The 10001st prime is {}.", p);
}
```
Output:
```The 10001st prime is 104743.
```

## Scala

```object Prime10001 extends App {

val oddPrimes: LazyList[Int] = 3 #:: LazyList.from(5, 2)
.filter(p => oddPrimes.takeWhile(_ <= math.sqrt(p)).forall(p % _ > 0))
val primes = 2 #:: oddPrimes

val index = 10_001
println(s"prime(\$index): " + primes.drop(index - 1).take(1).head)
}
```
Output:
```prime(10001): 104743
```

## Sidef

`say 10001.prime`
Output:
```104743
```

## Wren

Library: Wren-math
Library: Wren-fmt
```import "./math" for Int
import "./fmt" for Fmt

var n = 10001
var limit = (n.log * n * 1.2).floor  // should be enough
var primes = Int.primeSieve(limit)
Fmt.print("The \$,r prime is \$,d.", n, primes[n-1])
```
Output:
```The 10,001st prime is 104,743.
```

## x86 Assembly

Using the Sieve of Eratosthenes and n * (ln n + ln(ln n)) as the upper limit for finding the nth prime.

```section .data
primes times 14375 db 255   ;N * (ln N + ln(ln N)) bits, rounded up

section .text
global main

main:
mov     ebx, 1          ;array index for outer loop
xor     ecx, ecx        ;counter

sieve_outer:
inc     ebx             ;increase index
mov     eax, ebx        ;copy to eax for squaring
mul     ebx             ;square
cmp     eax, 115000     ;check if square is > limit
jg      find10001st     ;if it is, sieve is done
bt      [primes], ebx   ;check if ebx is no prime
jnc     sieve_outer     ;if no prime, try next number
inc     ecx             ;else increase counter

sieve_inner:
btr     [primes], eax   ;set multiple to not prime
cmp     eax, 115000     ;check if multiple is < limit
jl      sieve_inner     ;if it is, continue with inner loop
jmp     sieve_outer     ;if not, continue with outer loop

find10001st:
inc     ebx             ;increase array index
bt      [primes], ebx   ;is it prime?
adc     ecx, 0          ;if yes, increase ecx
cmp     ecx, 10001      ;check if counter arrived at 10001
jl      find10001st     ;if not, continue
;done, result in ebx
```
Output:
`104743`

## XPL0

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

int C, N;
[C:= 1;         \count 2 as first prime
N:= 3;
loop    [if IsPrime(N) then
[C:= C+1;
if C = 10001 then quit;
];
N:= N+2;
];
IntOut(0, N);
]```
Output:
```104743
```

## Zig

```const std = @import("std");
const stdout = @import("std").io.getStdOut().writer();

const limit = 10001;

fn isPrime(x: usize) bool {
if (x % 2 == 0) return false;

var i: usize = 3;
while (i * i <= x) : (i += 2) {
if (x % i == 0) return false;
}
return true;
}

pub fn main() !void {
var cnt: usize = 0;
var last: usize = 0;
var n: usize = 1;

while (cnt < limit) : (n += 1) {
if (isPrime(n)) {
cnt += 1;
last = n;
}
}
try stdout.print("{d}st prime number is: {d}\n", .{ limit, last });
}
```
Output:
```10001st prime number is: 104743
```