Nice Recurrence and Explicit Formula for A001110 (Square Triangular Numbers)
Rainer Rosenthal
r.rosenthal at web.de
Fri Mar 12 00:45:18 CET 2004
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
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