10001th prime

From Rosetta Code
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.
Task


Find and show on this page the 10001st prime number.

ALGOL 68[edit]

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 #
    PR read "primes.incl.a68" PR
    # 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[edit]

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

AWK[edit]

# syntax: GAWK -f 10001TH_PRIME.AWK
# converted from FreeBASIC
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)
}
Output:
104743

BASIC[edit]

BASIC256[edit]

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

PureBasic[edit]

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()

Yabasic[edit]

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


C[edit]

#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#[edit]

Sieve vs Trial Division[edit]

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[edit]

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);
Console.ReadKey(); 
}}}
Output:
104743

C++[edit]

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.

F#[edit]

This task uses Extensible Prime Generator (F#)

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

Factor[edit]

USING: math math.primes prettyprint ;

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

Fermat[edit]

Prime(10001);
Output:
104743

FreeBASIC[edit]

#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

Frink[edit]

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


GW-BASIC[edit]

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

Go[edit]

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[edit]

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

Java[edit]

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.

jq[edit]

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));

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

Julia[edit]

julia> using Primes

julia> prime(10001)
104743

Mathematica / Wolfram Language[edit]

Prime[10001]
Output:

104743

PARI/GP[edit]

prime(10001)
Output:
%1 = 104743

Perl[edit]

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[edit]

with js ?get_prime(10001)
Output:
104743


Picat[edit]

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[edit]

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[edit]

Trial Division Method[edit]

#!/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[edit]

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

QB64[edit]

Trial Division Method[edit]

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[edit]

'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

R[edit]

library(primes)
nth_prime(10001)
Output:
104743

Racket[edit]

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

Raku[edit]

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

REXX[edit]

/* 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

Ring[edit]

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...

Ruby[edit]

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

Rust[edit]

// [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[edit]

Scala3 ready

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[edit]

say 10001.prime
Output:
104743

Wren[edit]

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.

XPL0[edit]

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