We have
3 L (L+1) = (P1)(P+1)
which seems to offer some promise. For the moment though, I prefer pursuing another lead, namely
3L^2 + 3L  P^2 +1 = 0
using the quadratic equation, we see L is sqr(12 P^2  3)/6  1/2.
Clearly the term inside the square root sign must be a perfect square,
3(4P^2  1) = M^2
but the left hand side is divisble by 3, so we can right M = 3Q and substitute R = 2P giving
R^2  3Q^2 = 1
This is Pell's equation with fundamental solution R[0]=2, Q[0]=1. All other solutions are known to be given by the sequence
R[N] = 2 R[N1] + 3 Q[N1]
Q[N] = R[N1] + 2 Q[N1].
Recapituatling results to this point, we see that 2P = R and Q = 2L +1. Notice that elements in the Q sequence are alternately even and odd, parity considerations lead us to conclude that only even N correspond to a solution in P,L.
Having examined P and L with their clothes off, so to speak, thereby inheriting the full knowledge set of the Pell equation, let's dress the back up again as P and L, remembering that only the even terms in the N series are valid:
P[J+1] = 7 P[J] + 12 L[J] + 6
L[J+1] = 4 P[J] + 7 L[J] + 3
It is convenient to combine these equations to define T[J] = P[J]  sqr(3) L[J] then
T[J] = (7  4 sqr(3)) T[J1] + 6  3 sqr(3) leading to
2 T[J] = {2  sqr(3)} {7  4 sqr(3)}^J + sqr(3)
we identify P[J] as the sum of all terms not including a multiple of sqr(3). This is easily computed using the binomial expansion:
P[J] = Sum 7^(J2L)*48^L *J!/[(J2L)! (2L)!]
+ 6* Sum 7^(J2L1)*48^L *J!/[(J2L1)!(2L+1)!]
where both sums from from L=0 up to [J/2]
... more to come
Edited on April 10, 2008, 11:34 am

Posted by FrankM
on 20080408 22:58:08 