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Ive had some experience solving recurrences but i think they have been more simple than this one. This is what i have so far: \begin{array}{rcl}T[1] & = & 1 \\ T[n] & = & \frac{n}{T[n-1]}\\ \frac{n}{T[n-1]} & = & \frac {nT[n-2]}{n-1}\\ \frac{nT[n-2]}{n-1} & = & \frac{n^2 -2n}{nT[n-3]-T[n-3]}\end{array} I could keep going but I don't think it will help. The Problem I'm finding is finding a general formula for the recurrence; the fact that $T[\_]$ alternates each time is making it hard for me to get any further.

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  • $\begingroup$ Just imagine it : $T(n) = \frac{n}{T(n-1)} = \frac{(n!!)^{2}}{n!}$ $\endgroup$ – openspace May 13 '16 at 20:15
  • $\begingroup$ I guess you have a mistake in last step $\endgroup$ – openspace May 13 '16 at 20:26
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Note $T_{n+2}=\frac{n+2}{T_{n+1}}=\frac{n+2}{n+1}T_{n}$, $T_{1}=1,T_{2}=2$. So $T_{2n}=\frac{2.4..2n}{1.3..(2n-1)},T_{2n+1}=\frac{1.3..(2n+1)}{2.4..2n}$ You can find a neat way to write this as Wallis Integrals

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