Consider the convolutional recurrence relation below: $$T(n)=\begin{cases}\displaystyle\sum_{k=1}^{n−1} k\cdot T(k)\cdot T(n−k) & n\ge2\\ 1 &n=1 \end{cases}$$
Does it have a closed form? Or is it possible to get a tight upper bound of $T(n)$?
2026-04-01 02:47:57.1775011677
A recurrence relation similar to Catalan number
77 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in RECURRENCE-RELATIONS
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