Input: f(2)

f(x=2,n=1):
  x>0 = True, thus do a recursive call:
    f(2 - 1/1, 2)+1
f(x=1,n=2)
  x>0 = True, thus do a recursive call:
    f(1 - 1/2, 3)+1
f(x=0.5,n=3)
  x>0 = True, thus do a recursive call:
    f(0.5 - 1/3, 4)+1
f(x=0.1666...,n=4)
  x>0 = True, thus do a recursive call:
    f(0.166... - 1/4, 5)+1
f(-0.08333...,n=5)
  x>0 = False

Now the caller goes back:
f(-0.08333...,n=5) = False, so:
  f(0.166... - 1/4, 5)+1 becomes 1 (False implicitly becomes 0, which becomes 1 when adding 1)
f(x=0.1666...,n=4) = 1, so:
  f(0.5 - 1/3, 4)+1 becomes 2
f(x=0.5,n=3) = 2, so
  f(1 - 1/2, 3)+1 becomes 3
f(x=1,n=2) = 3, so
 f(2 - 1/1, 2)+1 becomes 4
f(x=2,n=1) is therefore 4

So our result for f(2) = 4