If 64 bits represent an ordering, and you split it in two for each insert - necessarily, to permit future insertion on either side of the new insert - then at least one side has no more than 63 bits left to store its ordering information for all future inserts on its side. You can't avoid this even with tricks that represent fractional bits.

The problem is that in order to avoid renumbering (which redistributes ordering state), the bisection needs to pre-allocate state space for future ordering information. The best it can do (without having more information about future insert order) is to split the state space perfectly evenly.

In the article, there's a big blue Update box which mentions a sequence of L/R path flips which make the fraction into a Fibonacci sequence, where you can go to 46 in sequence before exceeding a 32-bit integer. You'll observe than 46 is less than 64.