What does the algorithm calculate? We have the following 

i &lt;- 0 
j &lt;- 0 
x &lt;- 1 
y &lt;- 1 
z &lt;- 1 
c &lt;- 1 
u &lt;- 1 
while i&lt;n do 
   i &lt;- i+1 
   j &lt;- i 
   x &lt;- x*i 
   z &lt;- x 
   y &lt;- i+1 
   y &lt;- x*y 
   while j&lt;2*i do 
      j &lt;- j+1 
      z &lt;- z*j 
   od 
   c &lt;- z div (x*y) 
od 


I want to find the loop invariant for both while-loops. 

I applied the algorithm some times to understand what it does: 

The inner while loop calculates the following: $$z=z\prod_{m=1}^{2i-1}(i+m)=x\prod_{m=1}^{2i-1}(i+m)$$ or not? Is this also the loop invariant of the inner while loop? 

So I got the following results: 

First loop: 

i=1 
j=i=1 
x=1 
z=x=1 
y=i+1=2 
y=x*y=2 
inner while loop: z=x(i+1)=2 
c = z div (x*y) = 2 div 2 = 1 


Second loop: 

i=2 
j=i=2 
x=1*2=2 
z=x=2 
y=i+1=3 
y=x*y=2*3=6 
inner while loop: z=x(i+1)(i+2)(i+3)=2*3*4*5 
c = z div (x*y) = 2*3*4*5/2*6=10 


Third loop: 

i=3 
j=i=3 
x=2*3=6 
z=x=6
y=i+1=4 
y=x*y=24 
inner while loop: z=x(i+1)(i+2)(i+3)(i+4)(i+5)=6*4*5*6*7*8 
c = z div (x*y) = 6*4*5*6*7*8/6*24=5*7*8=280


Is everything correct? But what exactly calculates the algorithm? What is a general formula of the results?
Note that $x=i!$. Now it's easy to see $y=(i+1)!$ and $z=(2i)!$. So $c=\frac{(2i)!}{i!(i+1)!}$, which is the $i$th Catalan number.

P.S. Your formula is wrong. The condition of the inner loop is $j=i+m\le2i$, so $m$ is from $1$ to $i$ instead of $2i-1$ or something.