Well, Arthur, I have printed the paper, and the condition is in fact what Ultra.Power says it is, not your correction. It seems that the paper is not incorrect, but rather correct and pointless.

You have already identified the worst problem: The third condition requires incontinuity of f at all integer values. However, this problem disappears if we consider only the special case a=e. So let's do that.

We want a function f satisfying f(1)=e and f(n+1)=exp(f(n)), defined as widely as possible. We observe that n=0 and n=1 present no problems: f(0)=1 and f(-1)=0.

To extend f to all positive real numbers, it is clearly sufficient to define f on an interval of length 1, such as [0,1).

And we want f to be continuously differentiable. (The paper only requires f to be continuously differentiable on (0,1), or equivalently on (-1,0). This is obviously whacky, but we needn't worry about this because a=e.)

Next, for no obvious (or explained) reason we require that f ' is monotone on the interval [0,1]. From the functional equation f(x) = ln f(x+1) we get f '(x) = f '(x+1)/f(x+1). Hence f '(-1)=f '(0)/f(0)=f '(0), so f ' must be constant on [0,1]! 

Not too surprisingly, the function we get in this way is patched together from smaller pieces in a relatively unnatural way: f(x) = exp[x](exp(x-[x])).

Thus what I learned from the paper is that the correct solution to the problem probably has a minimum of its derivative between -1 and 0.