I am pretty new to Haskell, so sorry for the question. But - how to get rid of the endless recursion and not been overflown. This is the code:
foo :: Integer -> Integer
foo x
| x == 1 = 1
| x <= 0 = error "negative number or zero"
| odd x = foo 3 * x + 1
| x `mod` 2 == 0 = foo x `div` 2
| otherwise = x
EDIT:
foo :: Integer -> (Integer, Integer)
foo x
| x == 1 = (1, z)
| x <= 0 = error "negative number or zero"
| odd x = foo (3 * x + 1) . inc z
| even x = foo (x `div` 2) . inc z
| otherwise = (x, z)
where z = 0
inc :: Integer -> Integer
inc i = i + 1
I believe that the code is self-explanatory, but yet : If x is even then divide it with 2 otherwise 3*x + 1. This is part of the famous Collatz problem.
Function application has higher precedence than many other operations, so foo 3 * x + 1 is actually calling foo 3, then multiplying that result by x and adding 1, which looks like where your infinite loop might be.
Changing it to the following should fix that:
| odd x = foo $ 3 * x + 1
| x `mod` 2 == 0 = foo $ x `div` 2