Giuga numbers: Difference between revisions

From Rosetta Code
Content added Content deleted
No edit summary
Line 24: Line 24:
* [[oeis:A007850|OEIS:A007850 - Giuga numbers]]
* [[oeis:A007850|OEIS:A007850 - Giuga numbers]]
<br><br>
<br><br>

=={{header|Perl}}==
<lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Giuga_numbers
use warnings;
use ntheory qw( factor forcomposites );
use List::Util qw( all );

forcomposites
{
my $n = $_;
all { ($n / $_ - 1) % $_ == 0 } factor $n and print "$n\n";
} 4, 67000;</lang>
{{out}}
<pre>
30
858
1722
66198
</pre>


=={{header|Wren}}==
=={{header|Wren}}==

Revision as of 17:40, 1 July 2022

Giuga numbers 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.
Definition

A Giuga number is a composite number n which is such that each of its distinct prime factors f divide (n/f - 1) exactly.

All known Giuga numbers are even though it is not known for certain that there are no odd examples.

Example

30 is a Giuga number because its distinct prime factors are 2, 3 and 5 and:

  • 30/2 - 1 = 14 is divisible by 2
  • 30/3 - 1 = 9 is divisible by 3
  • 30/5 - 1 = 5 is divisible by 5


Task

Determine and show here the first four Giuga numbers.

Stretch

Determine the fifth Giuga number and any more you have the patience for.

References



Perl

<lang perl>#!/usr/bin/perl

use strict; # https://rosettacode.org/wiki/Giuga_numbers use warnings; use ntheory qw( factor forcomposites ); use List::Util qw( all );

forcomposites

 {
 my $n = $_;
 all { ($n / $_ - 1) % $_ == 0 } factor $n and print "$n\n";
 } 4, 67000;</lang>
Output:
30
858
1722
66198

Wren

Library: Wren-math
Library: Wren-seq

Simple brute force but assumes all Giuga numbers will be even, must be square-free and can't be semi-prime.

Takes only about 0.1 seconds to find the first four Giuga numbers but finding the fifth would take many hours using this approach, so I haven't bothered. Hopefully, someone can come up with a better method. <lang ecmascript>import "./math" for Int import "./seq" for Lst

var limit = 4 var giuga = [] var n = 4 while (giuga.count < limit) {

   var factors = Int.primeFactors(n)
   var c = factors.count
   if (c > 2) {
       var factors2 = Lst.prune(factors)
       if (c == factors2.count && factors2.all { |f| (n/f - 1) % f == 0 }) {
           giuga.add(n)
       }
   }
   n = n + 2

} System.print("The first %(limit) Giuga numbers are:") System.print(giuga)</lang>

Output:
The first 4 Giuga numbers are:
[30, 858, 1722, 66198]