Another way to look at it is that it happens because the conjugate 5-2√5 has absolute value <1.

If we let A, B be the solutions to x² - 10x + 5 (equivalent to x² = 10x - 5) (equivalent to we can write any series that follows the recursion

s(n+2) = 10 s(n+1) - 5 s(n)

as a sum of powers of A and B.

If we start such a recursion with A⁰+B⁰, A¹+B¹ = 2, 10, we can continue it as 2, 10, 90, 850, ... The key point is that in each case A^n + B^n has to be an integer (since the recursion multiplies and adds integers). Since B^n keeps getting smaller, A^n has to keep getting closer and closer to the integer.

There is also a parallel series: if you subtract instead of add then the first two terms of the series are 0, 4√5. Thus, if you take the series A^n/√5. You will get a similar convergence.

This is a a particularly fun game to play with the roots of x²-x-1 or x²-2x-1.

Jonathan

On Sunday, June 19, 2022 at 9:33:32 PM UTC-4, henh...@gmail.com wrote:
> where
> A = 9.47213595499958...
> or
> A = (5 + 2 (sqrt 5))

9.47213595499958 * (10 - 9.47213595499958) ## 4.99999999999999456920

> and i do : A^2, A^3, A^4, . . .

> by A^8 (or so), the accumulation of 9's is obvious....
> (as ........49 . 999.......... )

> -- is there a simple explanation for why this happens?

> -- Does this happen with (sqrt 2) ? (sqrt 3) ?

> -- Do we ever get similar accumulations of 2's, 3's, ... ?

> ( i wish i knew this in high school ! )