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Bill Allombert on Wed, 09 Jul 2025 22:14:13 +0200
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Re: Implementation of functions for "Integer partitions detect the primes" paper
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- To: pari-users@pari.math.u-bordeaux.fr
- Subject: Re: Implementation of functions for "Integer partitions detect the primes" paper
- From: Bill Allombert <Bill.Allombert@math.u-bordeaux.fr>
- Date: Wed, 9 Jul 2025 22:14:09 +0200
- Delivery-date: Wed, 09 Jul 2025 22:14:14 +0200
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- In-reply-to: <3e138c65ca1b1611cd64330ceeef8906@stamm-wilbrandt.de>
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- References: <0e8da5d2c65f10fb56ad0e7e5d20f15f@stamm-wilbrandt.de> <aG16_dPMbQ3PewcO@seventeen> <3e138c65ca1b1611cd64330ceeef8906@stamm-wilbrandt.de>
On Wed, Jul 09, 2025 at 10:00:15PM +0200, hermann@stamm-wilbrandt.de wrote:
> On 2025-07-08 22:09, Bill Allombert wrote:
> > You should do
> > M1(n)=my(s=0);fordiv(n,d,s+=d);s;
> >
> > But this function is just sigma(n)
> >
> Ok.
>
> > > ? M2(n)=s=0;for(m=1,n,forpart(v=m,if(v[1]<v[2],for(d=1,n\v[1],r=n-d*v[1];if(r%v[2]==0,s+=d*(r\v[2])))),[1,m],[2,2]));s;
> >
> > As I understand this function is
> > (sigma(n,3)-(2*n-1)*sigma(n))/8
> > which is much faster to compute.
> >
> That does not work:
>
> ? M1(n)=sigma(n);
> ? M2(n)=(sigma(n,3)-(2*n-1)*sigma(n))/8;
> ? T1(n)=(n^2 - 3^n + 2)*M1(n) - 8*M2(n);
^^^
This should be 3*n.
Cheers,
Bill