---
description: A simple ECF test using the table
---

# Testing an ECF

We have seen a relationship of numbers in [different rows within a column](../navigating#row-skipping) in the previous section. This actually provides a rather handy method to test whether some ECF may be valid for some trajectory that starts with a given number.

Let us re-write the formula from before:

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

Now, let us reinterpret this by rewriting $$n=1$$ and $$k=n-1$$ as follows:

$$
\beta(n, \rho)=\beta(1, \rho)+(n-1)\frac{3^m}{2^{p_m}}
$$

Recall that if ECF belongs to the Collatz sequence of a number, then the terminating number would be 1. So, we could assume $$\beta(n, \rho)=1$$ and then rewrite the equation above:

$$
1=\beta(1, \rho)+(n-1)\frac{3^m}{2^{p_m}}
$$

$$
n=\frac{2^{p_m}}{3^m}\left(1-\beta(1,\rho)\right)+1
$$

If this equation holds, then $$\rho$$ is indeed the ECF of the Collatz sequence for $$n$$.&#x20;

{% hint style="danger" %}
This method is does not make sense until we find an efficient way to calculate $$\beta(1,\rho)$$. **This part of the research remains open.**
{% endhint %}
