Recurrence relations when function call is made inside loop

int fun (int n)
{
  int x=1, k;
  if (n==1) return x;
  for (k=1; k<n; ++k)
     x = x + fun(k) * fun(n – k);
  return x;
}

What is the value of fun(5)?

I find it difficult to realize a recurrence tree when the functions are called inside loops so I decided to go with the recurrence relations.

This is what I came up with :

However, the recurrence relation given in the solution book is slightly different:

I don't understand where I went wrong. I included 1 inside the summation since x loops with every iteration. So why is the 1 in the second equation outside?

First, those two summation are not equivalet, the reason that the $1$ is outside is because this line in your code:

int x=1,k;

Try to run this "by hand" and you will understand why you are addidng $1$ only once.

$\text{fun}(1)=1\\
\text{fun}(2)=2\\
\text{fun}(3)=5\\
\text{fun}(4)=15$

$$\Longrightarrow \text{fun}(n)=\begin{cases}
1;&&&&&&&&n=1\\
\color{red}1+\displaystyle\sum\limits_{k=1}^{n-1}\text{fun}(k)\times \text{fun}(n-k)&&&&&&&&n>1
\end{cases}$$

EDIT:

Step by step for n=3:

int fun (int 3)
{
  int x=1, k;
  if (3==1) return x;
  for (k=1; k<3; ++k)
     x = x + fun(k) * fun(3 – k);
  return x;
}

First loop $k=1:$

$x=\color{green}1+\underbrace{\text{fun(1)}}_{=1}\times \underbrace{\text{fun(3-1)}}_{=2}=\color{red}3$

Second loop $k=2:$

$x=\color{red}3+\underbrace{\text{fun(2)}}_{=2}\times \underbrace{\text{fun(3-2)}}_{=1}=\color{blue}5$

StackExchange Computer Science Q#51858, answer score: 3