Haskell: Function application with $

Question

In the following snippet, you can see my two collatz functions I wrote in Haskell. For the recursive application I used parentheses in the first example (collatz) to get the

Answer 1

: binds more strongly than $. Consider

Prelude> let f x = [x]
Prelude> 1 : f 2
[1,2]
Prelude> 1 : f $ 2

:1:5:
    Couldn't match expected type `[a0]' with actual type `t0 -> [t0]'
    In the second argument of `(:)', namely `f'
    In the expression: 1 : f
    In the expression: 1 : f $ 2

Note the "expression" 1 : f found by the parser; it sees (1 : f) $ 2 rather than 1 : (f $ 2).