- If I understand the proposition correctly: if A(n),B(n) are prime then either n is prime (odd) or n is a power of 3.

Maybe not very helpfull, because it can't provide a counterexample, but I found that for J(n): J(3n)=(J(n)-sqrt(3)) J(n) (J(n)+sqrt(3)), and J(1)=2*sqrt(3) gives for C(n)=J(n)/sqrt(3):

C(3n)=3 (C(n)-1) C(n) (C(n)+1), C(1)=2. I tested the first powers of 3 untill 3^10 and didn't find any other prime values for C(n)+/-1=A(n) resp. B(n).

[I used J(n)J(k)=J(k+n)+J(k-n), with k=2n and k=n and J(0)=2].

Jan van Delden

[Non-text portions of this message have been removed] - --- In primenumbers@yahoogroups.com, "David Broadhurst" <d.broadhurst@...> wrote:
>

Congrats! That size of 15537 digits is nearly 25% bigger than the next 12 on that "Top-20" Lehmer Primitive Part list, which all date from more than 3 years ago. So, a massive improvement!

> --- In primenumbers@yahoogroups.com,

> "David Broadhurst" <d.broadhurst@> wrote:

>

> > The Society for Suppression of Square Roots hopes to

> > be able to announce, within a few days, the proof of

> > a unique Lehmer prime with more than 15000 digits

> > (wenn die Frau GĂ¶ttin probiert hat).

> > If proven, it will also become the largest known prime at

> > http://primes.utm.edu/top20/page.php?id=68

>

> Consummatus est in brevi explevit tempora multa [Wisdom:4:13]

>

> http://primes.utm.edu/primes/page.php?id=88162#comments

Mike