I created a new sequence (mutated collatz sequence), where the rules goes just like the normal Collatz Sequence but after every iteration, there's a probability $\varepsilon$ that we add 1. So for a odd number N, with a probability of $\varepsilon$ (let's this the mutation rate), it becomes $3N+2$ instead of $3N+1$, and for a even number N, with a probability of $\varepsilon$, it becomes $\frac{N}{2}+1$ instead of $\frac{N}{2}$. How different is this from the normal Collatz sequence? As it turns out, in this variant, the sequence can sometimes grow uncontrollably. In contrast with normal Collatz sequence, it is conjectured that all sequences are bounded and even reach 1. Let's study the expected growth in size of two consecutive odd integer in the sequence
For a odd number N,
with a probability of $\varepsilon$, it becomes 3N+2, which is odd.
With a probability of $1-\varepsilon$, it becomes the usual $3N+1$, which is even.
For a even number N, it either becomes $N/2$ or $N/2 + 1$, but either ways, there's $50\%$ chance of being odd and $50\%$ chance of being even.
From this we can see that the consecutive odd integers in the sequence is expected to grow by a factor of $3^\varepsilon \left ( \frac{3}{4} \right)^ {1-\varepsilon}$ or $3/4 * 4^\varepsilon$. If we want the sequence to not diverge, we want this multiplicative factor to be less than $1$, and that happens when $\varepsilon < log_4(\frac{4}{3}) ≈ 0.2075187...$ Of course, all of this is just heuristic argument. My question is, is it possible for the sequence to grow uncontrollably even when $\varepsilon < log_4(\frac{4}{3})$? Does the sequence that grow uncontrollably eventually contain every positive integers?
Let's analyze the sequence even more. Let's say we have $m$ odd numbers and $n$ even numbers. We repeated apply the map to each numbers. After a iteration, the number of odd numbers would be around $\varepsilon * m + \frac{n}{2}$. Because around $\varepsilon$ of the odd integers will become odd integers again and around half of the of the even integers will become odd integers. The probability that a member is odd stabalizes when $m = \varepsilon * m + \frac{n}{2}$ or $m = \frac{n}{2(1-\varepsilon)}$. Thus, after sufficiently large enough number of iterations, the probability of encountering odd integer should be $≈ \frac{\frac{1}{2(1-\varepsilon)}}{1+\frac{1}{2(1-\varepsilon)}} = \frac{1}{3-2\varepsilon}$, which is independent of actual values of $m$ and $n$. This allows us to make predictions about the mutation rate from just the abundance of odd integers. I tested this using python and as expected, it gave a pretty good approximation of the mutation rate, again just by studying the abundance of odd integers, without knowledge of the actual size of each element. It doesn't even know where those odd integers are located in the sequence.