Nice Recurrence and Explicit Formula for A001110 (Square Triangular Numbers)

[Rainer Rosenthal]

In alt.math.recreational there was an interesting
discussion related to Square Triangular Numbers.
Colin Dickson alias zaphod found a remarkable
recurrence formula:

   a(n-1) * a(n+1) = (a(n)-1)^2            (1)

and asked for some explanations, Robin Chapman
provided us with a lucid posting, which culminated
in an explicit formula for the squaretriangles:

   a(n) = (t^(2n) + t^(-2n) - 2)/32        (2)

For the latter see 
Message-ID: <c2fri1$luu$1 at newsg1.svr.pol.co.uk>

Despite Robin Chapmans mighty words
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
     Using these any "mysterious" identity 
        involving these numbers can be
        sledgehammered into submission.
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I did not see an easy bridge between (1) and (2).
I asked Robin, whether he would like to add his
formula and that of Colin Dickson to the OEIS.
I don't have an answer yet. But I think this new
addendum will be of interest to nearly all SeqFans.

Best regards,
Rainer Rosenthal
r.rosenthal at web.de