Hello, here is a different version of the parametric equations more easy to use to find predecessors in other than base 2. It is also more simple since it merge both parametric eq. into one.
Still using the Crandall function, let e, X_n, X_{n+1} be known (remember that X_{n+1} is + or -1(mod6) (-1 (mod6) or rather 5(mod6)). Then it is easy to prove that
from C_e(X_n)=\frac{3X_n+1}{2^e}=X_{n+1}
then we go to
C_{e+2k}[X_n+ (2^e X_{n+1} \frac{4^k-1}{3})]=X_{n+1}
using the identity 2^eX_{n+1}=3X_n+1.
Also since
C_{e+2k}(\frac{(3X_n+1)4^k}{2^e4^k}=X_{n+1}
and using the following identity
(3X_n+1)4^k= 3(4^kX_n+\sum_0^{k-1}4^j)+1
we finally get
C_{e+2k}(\frac{3[4^kX_n+\sum_0^{k-1}4^j]+1}{2^e4^k})=X_{n+1}.
Note that these equations do not depend on the the parity of the exponent e.
Next topic, we will go back to some specific use of the parametric function with e=1.