Giuga numbers: Difference between revisions
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The first 4 Giuga numbers are: |
The first 4 Giuga numbers are: |
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[30, 858, 1722, 66198] |
[30, 858, 1722, 66198] |
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</pre> |
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=={{header|XPL0}}== |
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<lang XPL0>func Giuga(N0); \Return 'true' if Giuga number |
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int N0; |
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int N, F, Q1, Q2, L; |
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[N:= N0; F:= 2; L:= sqrt(N); |
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loop [Q1:= N/F; |
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if rem(0) = 0 then \found a prime factor |
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[Q2:= N0/F; |
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if rem((Q2-1)/F) # 0 then return false; |
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N:= Q1; |
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if F>N then quit; |
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] |
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else [F:= F+1; |
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if F>L then return false; |
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]; |
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]; |
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return true; |
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]; |
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int N, C; |
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[N:= 3; C:= 0; |
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loop [if Giuga(N) then |
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[IntOut(0, N); ChOut(0, ^ ); |
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C:= C+1; |
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if C >= 4 then quit; |
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]; |
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N:= N+1; |
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]; |
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]</lang> |
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{{out}} |
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<pre> |
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30 858 1722 66198 |
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</pre> |
</pre> |
Revision as of 19:59, 1 July 2022
- 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
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
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