Navigating

Navigating the table

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 nn and column ρ\rho which maps to ECF {p0,p1,…,pm}\{p_0, p_1, \ldots, p_m\}:

  • Changing rows: β(n,ρ)→β(n+k,ρ)\beta(n,\rho) \to \beta(n+k, \rho).

  • Doubling: β(n,ρ)→β(n,2ρ)\beta(n, \rho) \to \beta(n, 2\rho).

  • Appending: β(n,ρ)→β(n,ρ+2q)\beta(n, \rho) \to \beta(n, \rho + 2^q) where 2q>ρ2^q > \rho.

  • Prepending: β(n,ρ)→β(n,ρ+1)\beta(n, \rho) \to \beta(n, \rho+1) where ρ≡0 mod 2\rho \equiv 0 \bmod 2.

Changing Rows

Consider β(n,ρ)\beta(n,\rho) and β(n+k,ρ)\beta(n+k,\rho) with ρ\rho being the ECF {p0,p1,…,pm}\{p_0, p_1, \ldots, p_m\}.

n=β(n,ρ)2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p031n = \beta(n, \rho)\frac{2^{p_{m}}}{3^{m}} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}
n+k=β(n+k,ρ)2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p031n+k = \beta(n+k, \rho)\frac{2^{p_{m}}}{3^{m}} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}

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))2pm3mk = (\beta(n + k, \rho) - \beta(n, \rho))\frac{2^{p_m}}{3^m}

So, the difference between two cells in the same column have a really neat formula:

β(n+k,ρ)=β(n,ρ)+3m2pm(k)\beta(n + k, \rho) = \beta(n, \rho) +\frac{3^{m}}{2^{p_m}}(k)

Doubling

Going from β(n,ρ)\beta(n, \rho) to β(n,2ρ)\beta(n,2\rho) is also easy to calculate. Let us write down their ICFs:

n=β(n,ρ)2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p031n = \beta(n, \rho)\frac{2^{p_{m}}}{3^{m}} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}
n=β(n,2ρ)2pm+13m−2pm−1+13m−2pm−2+13m−1−…−2p1+132−2p0+131n = \beta(n, 2\rho)\frac{2^{p_{m}+1}}{3^{m}} - \frac{2^{p_{m-1}+1}}{3^{m}} - \frac{2^{p_{m-2}+1}}{3^{m-1}} - \ldots - \frac{2^{p_1+1}}{3^2} - \frac{2^{p_0+1}}{3^1}

When we subtract the first equation from the second, we get:

0=(2β(n,2ρ)−β(n,ρ))2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p0310 = \left(2\beta(n, 2\rho)-\beta(n, \rho)\right)\frac{2^{p_m}}{3^m} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}
0=(2β(n,2ρ)−β(n,ρ))2pm3m+n−β(n,ρ)2pm3m0 = \left(2\beta(n, 2\rho)-\beta(n, \rho)\right)\frac{2^{p_m}}{3^m} + n - \beta(n, \rho)\frac{2^{p_m}}{3^m}
(β(n,ρ)−β(n,2ρ))2pm+13m=n\left(\beta(n, \rho)-\beta(n, 2\rho)\right)\frac{2^{p_m+1}}{3^m} = n

Finally, we can find the difference as:

β(n,2ρ)=β(n,ρ)−3m2pm(n2)\beta(n, 2\rho) = \beta(n, \rho) - \frac{3^m}{2^{p_m}}\left(\frac{n}{2}\right)

Appending

Consider ρ\rho as the ECF {p0,p1,…,pm}\{p_0, p_1, \ldots, p_m\} for some number nn. Then, consider another ECF {p0,p1,…,pm,q}\{p_0, p_1, \ldots, p_m, q\} where q>pmq > p_m again for number nn. Note that this ECF maps to ρ+2q\rho+2^q.

Well, this is equivalent to continuing a trajectory, as in doing an imprecise reduced Collatz functionR(β(n,ρ),q−pm)\mathcal{R}(\beta(n, \rho), q-p_m). In short:

β(n,ρ+2q)=3β(n,ρ)+12q−pm\beta(n, \rho+2^q) = \frac{3\beta(n, \rho)+1}{2^{q-p_m}}

Prepending

Consider ρ\rho as the ECF {p0,p1,…,pm}\{p_0, p_1, \ldots, p_m\}. This time, suppose that p0>qp_0 > q; and consider another ECF {q,p0,p1,…,pm}\{q, p_0, p_1, \ldots, p_m\} which maps to ρ+2q\rho+2^q. Let us look at the ICFs of these for number nn.

n=β(n,ρ)2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p031n = \beta(n, \rho)\frac{2^{p_{m}}}{3^{m}} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}
n=β(n,ρ+2q)2pm+q3m+1−2pm−1+q3m+1−2pm−2+q3m−…−2p1+q33−2p0+q32−2q31n = \beta(n, \rho + 2^q)\frac{2^{p_{m}+q}}{3^{m+1}} - \frac{2^{p_{m-1}+q}}{3^{m+1}} - \frac{2^{p_{m-2}+q}}{3^{m}} - \ldots - \frac{2^{p_1+q}}{3^3} - \frac{2^{p_0+q}}{3^2} - \frac{2^q}{3^1}

Lets do a 3n2q+13\frac{n}{2^q}+1 operation on the second equation:

3n2q+1=β(n,ρ+2q)2pm3m−2pm−13m−2pm−23m−1−…−2p132−2p0313\frac{n}{2^q}+1 = \beta(n, \rho+2^q)\frac{2^{p_{m}}}{3^{m}} - \frac{2^{p_{m-1}}}{3^{m}} - \frac{2^{p_{m-2}}}{3^{m-1}} - \ldots - \frac{2^{p_1}}{3^2} - \frac{2^{p_0}}{3^1}

Now, we subtract the first equation from this new equation:

3n2q+1−n=(β(n,ρ+2q)−β(n,ρ))2pm3m3\frac{n}{2^q}+1-n = (\beta(n, \rho+2^q) - \beta(n, \rho))\frac{2^{p_m}}{3^m}

This finally gives us:

β(n,ρ+2q)=β(n,ρ)+3m2pm(3n2q+1−n)\beta(n, \rho+2^q) = \beta(n, \rho) + \frac{3^m}{2^{p_m}}(3\frac{n}{2^q}+1-n)

For q=0q = 0 we get a nice result:

β(n,ρ+1)=β(n,ρ)+3m2pm(2n+1)\beta(n, \rho+1) = \beta(n, \rho) + \frac{3^m}{2^{p_m}}(2n+1)

Identities

To summarize all of these navigation methods for some number nn with ECF ρ\rho that maps to {p0,p1,…,pm}\{p_0, p_1, \ldots, p_m\}:

OperationEquationCondition

Changing Rows

β(n+k,ρ)=β(n,ρ)+3m2pm(k)\beta(n + k, \rho) = \beta(n, \rho) +\frac{3^{m}}{2^{p_m}}(k)

-

Doubling

β(n,2ρ)=β(n,ρ)−3m2pm(n2)\beta(n, 2\rho) = \beta(n, \rho) - \frac{3^m}{2^{p_m}}\left(\frac{n}{2}\right)

-

Appending

β(n,ρ+2q)=3β(n,ρ)+12q−pm\beta(n, \rho+2^q) = \frac{3\beta(n, \rho)+1}{2^{q-p_m}}

2q>ρ2^q > \rho

Prepending

β(n,ρ+1)=β(n,ρ)+3m2pm(2n+1)\beta(n, \rho+1) = \beta(n, \rho) + \frac{3^m}{2^{p_m}}(2n+1)

ρ≡0 mod 2\rho \equiv 0 \bmod 2

Looking at the table, we may connect some navigation methods together and obtain cool identity functions!

For n≡0 mod 2n \equiv 0 \bmod 2 we have:

β(n/2,ρ)=β(n,2ρ)\beta(n/2,\rho)=\beta(n,2\rho)

For ρ≡0 mod 2\rho \equiv 0 \bmod 2 we have:

β(3n+1,ρ)=β(n,ρ+1)\beta(3n+1, \rho)=\beta(n, \rho+1)

With the first identity, we can also derive for k∈Nk \in \mathbb{N}:

β(1,ρ)=β(2k,2kρ)\beta(1, \rho) = \beta(2^k, 2^k\rho)
PreviousOperation TableNextTesting an ECF

Last updated