Curzon numbers: Difference between revisions

=={{header|Arturo}}==

 

zero? (inc base^n) % inc base*n

 

print ["\n1000th Curzon with base" withBase "=" oneThousandth withBase]

print ""

 

{{out}}

=={{header|C++}}==

{{libheader|GMP}}

#include <iostream>

#include <vector>

}

return 0;

 

{{out}}

=={{header|Factor}}==

{{works with|Factor|0.99 2022-04-03}}

prettyprint ranges sequences ;

 

curzon 10 group simple-table. ;

 

{{out}}

<pre>

=={{header|FreeBASIC}}==

===Normal basic===

 

Function pow_mod(b As ULongInt, power As ULongInt, modulus As ULongInt) As ULongInt

 

Next

1 2 5 6 9 14 18 21 26 29

30 33 41 50 53 54 65 69 74 78

===GMP version===

{{libheader|GMP}}

 

Dim As Longint t = Len(__mpz_struct)

Loop

Next k

 

=={{header|Go}}==

{{trans|Wren}}

 

import (

fmt.Println()

}

 

{{out}}

 

'''Preliminaries'''

def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);

 

n;

 

 

'''The task'''

($k | power($n) + 1) % ($k * $n + 1) == 0;

 

"";

{{out}}

<pre>

 

=={{header|Julia}}==

 

function printcurzons(klow, khigh)

 

printcurzons(2, 10)

<pre>

Curzon numbers with k = 2:

 

It isn't clear why the task description says "generalized Curzon numbers only exist for even base integers." If k >= 3 is an odd base, then, besides the trivial solution n = 1, it can be checked that n = k^(k-1) is a Curzon number according to the given definition. It seems from the output below that Curzon numbers with an odd base are much scarcer than those with an even base.

program CurzonNumbers;

uses SysUtils;

ListCurzonNumbers(11);

end.

{{out}}

<pre>

 

=={{header|Mathematica}}/{{header|Wolfram Language}}==

CurzonNumberQ[b_Integer][n_Integer]:=PowerMod[b,n,b n+1]==b n

val=Select[Range[100000],CurzonNumberQ[2]];

val=Select[Range[100000],CurzonNumberQ[10]];

Take[val,50]

{{out}}

<pre>{1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, 54, 65, 69, 74, 78, 81, 86, 89, 90, 98, 105, 113, 114, 125, 134, 138, 141, 146, 153, 158, 165, 173, 174, 186, 189, 194, 198, 209, 210, 221, 230, 233, 245, 249, 254}

 

=={{header|Perl}}==

use warnings;

use Math::AnyNum 'ipow';

(sprintf "@{['%5d' x $upto]}", @C[0..$upto-1]) =~ s/(.{125})/$1\n/gr;

print "Thousandth: $C[-1]\n\n";

{{out}}

<pre>First 50 Curzon numbers using a base of 2:

 

=={{header|Phix}}==

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #008080;">include</span> <span style="color: #004080;">mpfr</span><span style="color: #0000FF;">.</span><span style="color: #000000;">e</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">while</span>

<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>

{{out}}

<pre>

 

=={{header|Python}}==

return (k**n + 1) % (k * n + 1) == 0

 

print(f'{c: 5,}', end='\n' if (i + 1) % 25 == 0 else '')

print(f' Thousandth Curzon with k = {k}: {curzons[999]}.\n')

<pre>

Curzon numbers with k = 2:

=={{header|Quackery}}==

 

space 4 of swap join

-5 split nip echo$ ] is rjust ( n --> )

say " ... "

-1 peek echo cr cr ]

 

{{out}}

 

=={{header|Raku}}==

 

for <2 4 6 8 10> {

$curzon[^50].batch(25)».fmt("%4s").join("\n") ~

"\nOne thousandth: " ~ $curzon[999]

{{out}}

<pre>First 50 Curzon numbers using a base of 2:

 

=={{header|Rust}}==

// rug = "1.15.0"

 

println!("1000th Curzon number with base {k}: {n}\n");

}

 

{{out}}

 

=={{header|Sidef}}==

powmod(k, n, k*n + 1).is_congruent(-1, k*n + 1) && (n > 0)

}

say 50.by {|n| is_curzon(n, k) }.join(' ')

say ("1000th term: ", 1000.th {|n| is_curzon(n,k) })

 

{{out}}

=={{header|Vlang}}==

{{trans|go}}

 

fn main() {

println('')

}

 

{{out}}

{{libheader|Wren-seq}}

{{libheader|Wren-fmt}}

 

import "./gmp" for Mpz

}

System.print()

 

{{out}}