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

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• ... Proposition: For every integer pair (p,k), we have V(p, (23*k+11)*3103, 4638985) = p mod 4638985 Proof [using the Sun Tzu Suan Ching]:
Message 1 of 46 , Nov 1, 2010

> > 1080905, 1739089, 1992641, 2110159, found in only a few minutes
> A few minutes later these turned up: 4013569, 4638985

Proposition: For every integer pair (p,k), we have
V(p, (23*k+11)*3103, 4638985) = p mod 4638985

Proof [using the Sun Tzu Suan Ching]:
v(p,q,n,m)=2*polcoeff(lift(Mod((p+x)/Mod(2,m),x^2+4*q-p^2)^n),0);
n=4638985;
t(p,k,m)=lift(v(p,(23*k+11)*3103,n,m)-p);
s(m)=sum(p=1,m,sum(k=1,m,t(p,k,m)));
if(issquarefree(n),fordiv(n,m,if(isprime(m),print1(s(m)" "))));
0 0 0 0 0

Comment: That gives 65 (q,n) pairs, beating Mike's 37 pairs. But
asked, on All-Hallows-Even, for more than 70,000 such pairs,
with n not Carmichael.

• ... 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.