Giuga numbers

Revision as of 19:59, 1 July 2022 by Loren (talk | contribs) (Added XPL0 example.)

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

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

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




Julia

<lang ruby>using Primes

isGiuga(n) = all(f -> f != n && rem(n ÷ f - 1, f) == 0, factor(Vector, n))

function getGiuga(N)

   gcount = 0
   for i in 4:typemax(Int)
       if isGiuga(i)
           println(i)
           (gcount += 1) >= N && break
       end
   end

end

getGiuga(4)

</lang>

Output:
30      
858 
1722
66198

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]

XPL0

<lang XPL0>func Giuga(N0); \Return 'true' if Giuga number int N0; int N, F, Q1, Q2, L; [N:= N0; F:= 2; L:= sqrt(N); loop [Q1:= N/F;

       if rem(0) = 0 then      \found a prime factor
               [Q2:= N0/F;
               if rem((Q2-1)/F) # 0 then return false;
               N:= Q1;
               if F>N then quit;
               ]
       else    [F:= F+1;
               if F>L then return false;
               ];
       ];

return true; ];

int N, C; [N:= 3; C:= 0; loop [if Giuga(N) then

               [IntOut(0, N);  ChOut(0, ^ );
               C:= C+1;
               if C >= 4 then quit;
               ];
       N:= N+1;
       ];

]</lang>

Output:
30 858 1722 66198