Giuga numbers
- 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
FreeBASIC
<lang freebasic>Function isGiuga(m As Uinteger) As Boolean
Dim As Uinteger n = m Dim As Uinteger f = 2, l = Sqr(n) Do If n Mod f = 0 Then If ((m / f) - 1) Mod f <> 0 Then Return False n /= f If f > n Then Return True Else f += 1 If f > l Then Return False End If Loop
End Function
Dim As Uinteger n = 3, c = 0, limit = 4 Print "The first "; limit; " Giuga numbers are: "; Do
If isGiuga(n) Then c += 1: Print n; " "; n += 1
Loop Until c = limit</lang>
- Output:
The first 4 Giuga numbers are: 30 858 1722 66198
Go
I thought I'd see how long it would take to 'brute force' the fifth Giuga number and the answer (without using parallelization, Core i7) is around 2 hours 40 minutes. <lang go>package main
import (
"fmt" "rcu"
)
func main() {
const limit = 5 var giuga []int for n := 4; len(giuga) < limit; n += 2 { factors := rcu.PrimeFactors(n) if len(factors) > 2 { isSquareFree := true for i := 1; i < len(factors); i++ { if factors[i] == factors[i-1] { isSquareFree = false break } } if isSquareFree { isGiuga := true for _, f := range factors { if (n/f-1)%f != 0 { isGiuga = false break } } if isGiuga { giuga = append(giuga, n) } } } } fmt.Println("The first", limit, "Giuga numbers are:") fmt.Println(giuga)
}</lang>
- Output:
The first 5 Giuga numbers are: [30 858 1722 66198 2214408306]
J
We can brute force this task building a test for giuga numbers and checking the first hundred thousand integers:
<lang J>giguaP=: {{ (1<y)*(-.1 p:y)**/(=<.) y ((_1+%)%]) q: y }}"0</lang>
<lang J> 1+I.giguaP 1+i.1e5 30 858 1722 66198</lang>
These numbers have some interesting properties but there's an issue with guaranteeing correctness of more sophisticated approaches.
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
Pascal
Free Pascal
changing main routine at the end of Factors_of_an_integer#using_Prime_decomposition <lang> const
LMT = 24423128562;//70*1000;//
var
pPrimeDecomp :tpPrimeFac; T0:Int64; n : NativeUInt; idx,p,cnt : Int32; chk: boolean;
BEGIN
InitSmallPrimes;
T0 := GetTickCount64; n := 1; Init_Sieve(n); pPrimeDecomp:= GetNextPrimeDecomp; cnt := 0; repeat inc(n); pPrimeDecomp:= GetNextPrimeDecomp;
with pPrimeDecomp^ do begin //if prime/semi-prime if pfDivCnt <= 4 then continue; //not even if pfpotPrimIdx[0] <> 0 then continue; idx := pfMaxIdx;//always>0 for pfDivcnt>2
if pfRemain = 1 then begin //not squarefree or biggest prime factor to big if ((1 shl idx) <> pfdivcnt) or (sqr(SmallPrimes[pfpotPrimIdx[idx-1]])>=n) then continue; //check for Giuga number chk := true; For idx := idx-1 downto 0 do begin p := SmallPrimes[pfpotPrimIdx[idx]]; chk := (((n DIV p)-1)MOD p) = 0; if not(chk) then BREAK; end; end else begin if (1 shl (idx+1) <> pfdivcnt) or (sqr(pfRemain)>=n) then continue; chk := (((n DIV pfRemain)-1)MOD pfRemain) = 0; if chk then For idx := idx-1 downto 0 do begin p := SmallPrimes[pfpotPrimIdx[idx]]; chk := (((n DIV p)-1)MOD p) = 0; if not(chk) then BREAK; end; end; if chk then begin inc(cnt); writeln(cnt:3,'..',OutPots(pPrimeDecomp,n)); end; end; until n >= LMT; T0 := GetTickCount64-T0; writeln('runtime ',T0/1000:0:3,' s'); writeln('Count ',cnt); writeln;
END.</lang>
- Output:
1..30 : 8 : 2*3*5_chk_30<_SoD_72< 2..858 : 16 : 2*3*11*13_chk_858<_SoD_2016< 3..1722 : 16 : 2*3*7*41_chk_1722<_SoD_4032< 4..66198 : 32 : 2*3*11*17*59_chk_66198<_SoD_155520< 5..2214408306 : 64 : 2*3*11*23*31*47057_chk_2214408306<_SoD_5204238336< 6..24423128562 : 64 : 2*3*7*43*3041*4447_chk_24423128562<_SoD_57154166784< runtime 673.643 s Count 6 real 11m13,645s user 11m12,962s sys 0m0,247s
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
Phix
with javascript_semantics constant limit = 4 sequence giuga = {} integer n = 4 while length(giuga)<limit do sequence pf = prime_factors(n) for f in pf do if remainder(n/f-1,f) then pf={} exit end if end for if length(pf) then giuga &= n end if n += 2 end while printf(1,"The first %d Giuga numbers are: %v\n",{limit,giuga})
- Output:
The first 4 Giuga numbers are: {30,858,1722,66198}
Python
<lang python>#!/usr/bin/python
from math import sqrt
def isGiuga(m):
n = m f = 2 l = sqrt(n) while True: if n % f == 0: if ((m / f) - 1) % f != 0: return False n /= f if f > n: return True else: f += 1 if f > l: return False
if __name__ == '__main__':
n = 3 c = 0 print("The first 4 Giuga numbers are: ") while c < 4: if isGiuga(n): c += 1 print(n) n += 1</lang>
Raku
<lang perl6>my @primes = (3..60).grep: &is-prime;
print 'First four Giuga numbers: ';
put sort flat (2..4).map: -> $c {
@primes.combinations($c).map: { my $n = [×] 2,|$_; $n if all .map: { ($n / $_ - 1) %% $_ }; }
}</lang>
- Output:
First 4 Giuga numbers: 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