There are several equations we can derive from the table, that allow various methods of navigation within the table from one cell to another. We have the following navigation methods, considering a row n and column ρ which maps to ECF {p0,p1,…,pm}:
Changing rows: β(n,ρ)→β(n+k,ρ).
Doubling: β(n,ρ)→β(n,2ρ).
Appending: β(n,ρ)→β(n,ρ+2q) where 2q>ρ.
Prepending: β(n,ρ)→β(n,ρ+1) where ρ≡0mod2.
Changing Rows
Consider β(n,ρ) and β(n+k,ρ) with ρ being the ECF {p0,p1,…,pm}.
n=β(n,ρ)3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 n+k=β(n+k,ρ)3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 Almost all the terms on the right hand-side are equal, expect the leftmost term. When we subtract the first equation from the second we get:
k=(β(n+k,ρ)−β(n,ρ))3m2pm So, the difference between two cells in the same column have a really neat formula:
β(n+k,ρ)=β(n,ρ)+2pm3m(k) Doubling
Going from β(n,ρ) to β(n,2ρ) is also easy to calculate. Let us write down their ICFs:
n=β(n,ρ)3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 n=β(n,2ρ)3m2pm+1−3m2pm−1+1−3m−12pm−2+1−…−322p1+1−312p0+1 When we subtract the first equation from the second, we get:
0=(2β(n,2ρ)−β(n,ρ))3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 0=(2β(n,2ρ)−β(n,ρ))3m2pm+n−β(n,ρ)3m2pm (β(n,ρ)−β(n,2ρ))3m2pm+1=n Finally, we can find the difference as:
β(n,2ρ)=β(n,ρ)−2pm3m(2n) Appending
Consider ρ as the ECF {p0,p1,…,pm} for some number n. Then, consider another ECF {p0,p1,…,pm,q} where q>pm again for number n. Note that this ECF maps to ρ+2q.
Well, this is equivalent to continuing a trajectory, as in doing an imprecise reduced Collatz functionR(β(n,ρ),q−pm). In short:
β(n,ρ+2q)=2q−pm3β(n,ρ)+1 Prepending
Consider ρ as the ECF {p0,p1,…,pm}. This time, suppose that p0>q; and consider another ECF {q,p0,p1,…,pm} which maps to ρ+2q. Let us look at the ICFs of these for number n.
n=β(n,ρ)3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 n=β(n,ρ+2q)3m+12pm+q−3m+12pm−1+q−3m2pm−2+q−…−332p1+q−322p0+q−312q Lets do a 32qn+1 operation on the second equation:
32qn+1=β(n,ρ+2q)3m2pm−3m2pm−1−3m−12pm−2−…−322p1−312p0 Now, we subtract the first equation from this new equation:
32qn+1−n=(β(n,ρ+2q)−β(n,ρ))3m2pm This finally gives us:
β(n,ρ+2q)=β(n,ρ)+2pm3m(32qn+1−n) For q=0 we get a nice result:
β(n,ρ+1)=β(n,ρ)+2pm3m(2n+1) Identities
To summarize all of these navigation methods for some number n with ECF ρ that maps to {p0,p1,…,pm}:
Operation | Equation | Condition |
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| β(n+k,ρ)=β(n,ρ)+2pm3m(k) | |
| β(n,2ρ)=β(n,ρ)−2pm3m(2n) | |
| β(n,ρ+2q)=2q−pm3β(n,ρ)+1 | |
| β(n,ρ+1)=β(n,ρ)+2pm3m(2n+1) | ρ≡0mod2 |
Looking at the table, we may connect some navigation methods together and obtain cool identity functions!
For n≡0mod2 we have:
β(n/2,ρ)=β(n,2ρ) For ρ≡0mod2 we have:
β(3n+1,ρ)=β(n,ρ+1) With the first identity, we can also derive for k∈N:
β(1,ρ)=β(2k,2kρ)