I have been playing on the Syracuse problem as a incentive to study of number theory. So here is some basic stuff that I have found. Here are some basic equations (which I call parametric equations) that you can deduce from congruence using Crandall function or a little modified version of it which operates only on odd numbers:
C_e(X_n)=\frac{3X_n+1}{2^{e(X_n)}}=X_{n+1}
where e is a function of X_n and X_n being an odd number. Of course e is such a way that X_{n+1} is also an odd number. Then
2^eX_{n+1}−1=3X_n
or
2^eX_{n+1}\equiv 1(mod3)
using also the fact that odd numbers are + or - 1(mod2) will lead to the two following parametric equations
if e=2m, m a natural number we get (r belong to N_0)
C_{2m}(2^{2m+1}r+∑^{m−1}_{j=0}2^{2j})=6r+1
if e=2m-1, m a natural number we get
C_{2m-1}(2^{2m}r+2^{2m-1}+∑^{m−1}_{j=0}2^{2j})=6r+5
The last equation with m=1 (or e=1) describe the only operation giving a rising trend to a trajectory. Considering both equations, it also show that after a first step any odd number in a trajectory is not a multiple of 3. Consequences are:
- Any multiple of 3 do not have any predecessors
- Any other odd number does have an infinity of at least one upper predecessor.
- As for parity or encoding vectors the numbers of zeros are explicitly shown by the parametric equations.
- Any proper analysis of the problem by congruence must be in mod 6n.
One more thing, you can use the below identity as a short hand
∑^{m−1}_{j=0}2^{2j}=∑^{m−1}_{j=0}4^j=\frac{4^m−1}3
Last remark, the above identity is directly leading to 1 so it is a set of attractors meaning that any number going to 1 has to pass one of these number (just set r to zero in the first parametric equation).
To find merging or coalescent numbers, there are other ways to derive similar equations away from binary numbers formulation but that will be for a next topic.