denote the average factor by which we divide by $x$. In this model, we simply divide by $x$ each time we need to divide. The sequence stops being an integer sequence, but we artificially assign a remainder mod 4 to every element of the sequence to see when we need to divide. We want to show that our model sequence grows with probability 1.

First, we want to estimate the ratio of two consecutive numbers in the model sequence.

\[thm:ratio\] If $a$ and $b$ are two consecutive numbers in the model sequence, starting from index 3, then $b > a\frac{2+x}{(1+x)x}$ and $a > b\frac{1}{1+x}$.

Let $v$ be the element before $a$. Then $b \geq (a+v)/x > a/x$. Analogously, $b \leq a+v$, and by the previous sentence, $xa > v$. Therefore, $b < a(1+x)$, which means that $a/(1+x) < v$. Plugging this back into $b \geq (a+v)/x$, we get $b > a(2+x)/(1+x)x$.

Now we are ready to prove the theorem:

In our probabilistic model, a sequence grows with probability 1.

Suppose we have two consecutive numbers in the sequence $a$ and $b$ whose sum is divisible by 4. By Lemma \[thm:averagesteps\] we have 1, 3, or 4 terms until the following 