$\text{fun}(1) = 1$;
$\text{fun}(2) = 1 + \text{fun}(1) * \text{fun}(1) = 1 + 1 = 2$;
$\text{fun}(3) = 1 + \text{fun}(1) * \text{fun}(2) + \text{fun}(2) * \text{fun}(1) = 5$;
$\text{fun}(4) = 1 + \text{fun}(1) * \text{fun}(3) + \text{fun}(2) * \text{fun}(2) + \text{fun}(3) * \text{fun}(1)$
$\qquad = 1 + 5 + 4 + 5 = 15$;
$\text{fun}(5) = 1 + \text{fun}(1) * \text{fun}(4) + \text{fun}(2) * \text{fun}(3) + \text{fun}(3) * \text{fun}(2) + \text{fun}(4) * \text{fun}(1)$
$\qquad = 1 + 15 + 10 + 10 + 15 = 51$;
More formal way:
The recurrence relation is
$f(n) = \begin{cases} 1, n = 1 \\ 1 + \sum_{i=1}^{n-1} f(i) \times f(n-i), n > 1 \end{cases}$
$f(1) = 1$
$f(2) = 1+ f(1).f(1)$
$\qquad = 1 + 1.1 = 2$
$f(3) = 1 + f(1). f(2) + f(2).f(1)$
$\qquad= 1+ 1.2 + 2.1 = 5$
$f(4) = 1 + f(1).f(3) + f(2).f(2) + f(3).f(1)$
$ \qquad= 1+ 1.5 + 2.2 + 5.1 = 15$
$f(5) = 1 + f(1).f(4) + f(2).f(3) + f(3). f(2)+ f(4).f(1)$
$\qquad = 1 + 1.15 + 2.5 + 5.2 + 15.1 = 51$