- --- In primenumbers@yahoogroups.com,

"mikeoakes2" <mikeoakes2@...> wrote:

> As you know, my "4" came from the fact that the 3rd and 4th

Those terms are also coprime to 3, so you might, with

> terms of the sequence are known to be odd

equal arbitrariness, have set C = (2/1)^2*(3/2)^2 = 9,

as I did on first seeing your "exercise".

The asymptotic constant seems to lie between 4 and 9,

but is hard (for me) to estimate accurately.

David (over-exercised :-) - --- In primenumbers@yahoogroups.com,

Kevin Acres <research@...> wrote:

> x=a*x+b either is a square or has a square as a major factor.

Suppose that we want to start with a square and get a square.

Then we must solve the Diophantine equation

y^2 = a*x^2 + b

For any pair (a,b), Dario will tell us all the solutions:

http://www.alpertron.com.ar/QUAD.HTM

David