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If a number is both square and triangular, then we know that $\frac{n(n+1)}{2}=m^2$. If a number is both square and pentagonal then $m^2=\frac{n(3n-1)}{2}$. So if a number is $(r-1)$-gonal and $r$-gonal, then $\frac{n((r-1)n-(r-3))}{2}=\frac{m((m-2)m-(m-4))}{2}$. Firstly, how can we derive, from the simple observation given for the square triangular case and the square pentagonal case a general formula for the $k$-th square-triangular number and the $k$-th ssquare pentagonal number?

Also, and this is the main question- is it possible to derive a general formula for the $k$-th $r$-gonal $(r+1)$-gonal number?

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    Some of the $m$'s on the right hand side of your formula should be $r$'s.2012-12-14
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    Both give rise to simple Pell equations, and in both cases we can write down the fundamental solution by inspection. I have not checked the $r-1, r$.2012-12-14

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