Primalty test with "Numerators of continued fraction convergents to sqrt(2)"?

[Creighton Dement]

Sorry, now I'm confused. Should I have written "for the prime case" in
lieu of "for the odd case", below? 

It seems that for prime p, p | A000204(p) - 1 and p | A001333(p) - 1 
(but not for all odd numbers- for example 9). I've still got Jack
Brennen's primalty test for A000129 in the back of my mind, but I don't
know how it is related at the moment: 

The Pell primality test is "If N is an odd prime, then
              P(N)-kronecker(2,N) is divisible by N". "Most" composite
numbers
              fail this test, so it makes a useful pseudoprimality test.
The odd
              composite numbers which are Pell pseudoprimes (i.e. that
pass the
              above test) are in A099011. - Jack Brennen
(jb(AT)brennen.net), Nov
              13, 2004

Ex.  
Lucas(197) - 1 = (2) (5) (13) (17) (29) (89) (97) (197) (599786069)
(18546805133) (6168709) (19801)

and 

A001333(197) - 1=  (2)^3 (5) ((13))^2 (197) (239) (293) (1471)
(8109013290449) (40710764977973) (93052705802012471592001) (52734529)
(5741)

> For the odd case, I conjecture that
> 
> tes[(a + b)^(2n+1)] = tes[b^(2n+1) + (2n+1)tes[b^(2n+1)a] +
> (2n+1)(tes...) + (2n+1)(tes...) +
> 
> Since the total number of terms in tes[(a + b)^n] was seen to be
> Lucas(n), if the conjecture is correct a direct corollary would be
> that 
> 2n+1 divides Lucas(2n+1) - 1 = A000204(2n+1) - 1. This appears to
> check (I'm sure this is already known, though I didn't seem to find it
> directly under A000204 at first glance).
> 

[snip]

> The rest of the problem, at least for odd n, seems to boil down to
> figuring out how many b's are in the various terms
> of tes[(a + b)^(2n+1)], any ideas?
> 
> Many thanks,
> Creighton
> 
>