## Re: Lucas super-pseudoprimes for Q <> 1

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• Mike had difficulty posting and asked me to post this for him: ... Yes. Here is my function of 3Nov10:- { is_eligible(n)= local(f,ind,p,res=1); f=factor(n);
Message 1 of 46 , Nov 7, 2010
Mike had difficulty posting and asked me to post this for him:

=======

> Is the above, in essence, your conjectural method?

Yes.

Here is my function of 3Nov10:-
{
is_eligible(n)=
local(f,ind,p,res=1);
f=factor(n);
for(ind=1,#f[,1],
if(f[ind,2]>1,res=0;break(1)); \\not squarefree
p=f[ind,1];
if( ((n-1)%(p-1))<>0,
if( !((((n-1)%(p+1))==0)&&(((n+1)%(p-1))==0)),res=0;break(1));
); \\end if
); \\end for ind
res;
}

Now, how to prove this is kosher?...

Mike

=====

I have already posted the "sufficient" part of the proof.
The "necessary" part is awaited ...

David (pp Mike)
• ... http://physics.open.ac.uk/~dbroadhu/cert/dbmo116.out gives my 116, in the format [n, factors, number of solutions] With n
Message 46 of 46 , Nov 9, 2010
"mikeoakes2" <mikeoakes2@...> wrote:

> > My revised count up to 2*10^10 is 116.
> My (original) count up to 2*10^10 was 105.
> So it must have missed 11, i.e. a bigger proportion.