--- In

primenumbers@yahoogroups.com,

"djbroadhurst" <d.broadhurst@...> wrote:

> Here's a plupluperfect chain of 8:

>

> p=3;for(k=1,9,print([k,factor(p)]);p=140820*p-422453)

> [1, Mat([3, 1])]

> [2, Mat([7, 1])]

> [3, Mat([563287, 1])]

> [4, Mat([79321652887, 1])]

> [5, Mat([11170075159124887, 1])]

> [6, Mat([1572969983907966164887, 1])]

> [7, Mat([221505633133919795338964887, 1])]

> [8, Mat([31192423257918585579633034964887, 1])]

> [9, [7, 1; 627502434740013603046274854822137841, 1]] \\ Bounded!

Puzzle 11: Find a plupluperfect chain of 12, i.e. a triplet

(A, B, p[1]) such that A = 1 mod 11, B is coprime to 11,

and the iteration p[n+1] = A*p[n] + B yields a chain of

12 increasing primes, p[n], for n = 1 to 12.

Comments: Clearly, p[2] = 11 and hence B = 11 - A*p[1], with

p[1] = 2, 3, 5 or 7. So this boils down to 4 sub-problems of

Carmody-scale difficulty (maybe easy for Phil; harder for mere

mortals like Kevin, Mike, or me). I did not yet find a solution.

David