Talk:Giuga numbers

Is there a simpler way for generating the squarefree numbers

Giuga numbers are squarefree and therefor are simple product of different primes n = p1*p2..*pk.|p1<p2<..<pk
How big can the last prime factor pk get:
(n div pk -1) mod pk = z=[1,2,3...] | *pk
n div pk -1 = z*pk
(p1*p2*..*p(k-1))-1 = z*pk => pk must be a multible of (product of all other prime factors -1)
Examples:
2*3-1 = 5 ->pk= 5 -> n =2*3*5=30, no need to search on three factors starting with 2,3
2*3*7-1 = 41->pk= 41 -> n =2*3*7*41=1722
2*3*11-1 = 65 == z*pk AND pk>11 AND pk is divisor of 65 => pk =13 ->2*3*11*13 =858

2*3*11*17-1 = 1121 = z*pk == 19*59 -> pk =19 and 59 are possible to check 21318 and 66198

My intention is to extend the first k-1 prime factors and check for pk and then for guiga.
--Horsth (talk) 12:33, 2 July 2022 (UTC)

found this on mersenneforum [Giuga numbers]

much faster. ~~----

Presumably you meant <lang parigp>a(n)=print("n=",n);s=p=vector(n-2);t=p[1]=p[2]=2;s[1]=1/2;\

while(t>1,p[t]=nextprime(p[t]+1);s[t]=s[t-1]+1/p[t];\ if(s[t]==1||s[t]+(n-t)/p[t]<=1,t--,\ if(t<n-2,t++;p[t]=max(p[t-1],s[t-1]/(1-s[t-1])),\ c=numerator(s[n-2]);d=denominator(s[n-2]);k=d^2+c-d;f=divisors(k);\ for(i=1,(length(f)+1)\2,h=f[i];if((h+d)%(d-c)==0&&(k/h+d)%(d-c)==0,\ r1=(h+d)/(d-c);r2=(k/h+d)/(d-c);\ if(r1>p[n-2]&&r2>p[n-2]&&r1!=r2&&isprime(r1)&&isprime(r2),\ w=d*r1*r2;print(w);write("giuga.txt",w)))))))</lang>

But I'm having problems reading that code. Specifically: for(i=1,(length(f)+1)\2,h=f[i] -- I haven't been able to find any documentation on what \2 means in this context. Do you know where I should be looking? --Rdm (talk) 13:00, 14 July 2022 (UTC)