capture some of the behavior of 3-free Fibonacci sequences. We start with two numbers $a_1$ and $a_2$ and flip a coin. If the coin turns heads we add the next number $a_3 = (a_1+a_2)/5$ to the sequence. If the coin turns tails we add two more terms: $a_3 = a_1+a_2$ and $a_4 = (a_2+a_3)/5$ to the sequence. Then repeat. We expect that this sequence on average grows faster than 3-free Fibonacci sequences, because we divide by a smaller number.

Now we want to bound the maximum of the last two terms of this probabilistic sequence after two coin flips. Let $M = \max\{a_1,a_2\}$. We have the following cases:

-   After two heads, the sequence becomes: $a_1$, $a_2$, $(a_1+a_2)/5$, $(a_1+6a_2)/25$. The last two terms do not exceed $2M/5$.

-   After head, tail, the sequence becomes: $a_1$, $a_2$, $(a_1+a_2)/5$, $(a_1+6a_2)/5$, $(2a_1+7a_2)/25$. The last two terms do not exceed $7M/5$.

-   After tail, head, the sequence becomes: $a_1$, $a_2$, $a_1+a_2$, $(a_1+2a_2)/5$, $(6a_1+7a_2)/25$. The last two terms do not exceed $3M/5$.

-   After two tails the sequence becomes: $a_1$, $a_2$, $a_1+a_2$, $(a_1+2a_2)/5$, $(6a_1+7a_2)/5$, $(7a_1+9a_2)/25$. The last two terms do not exceed $13M/5$.

As each event happens with the same probability $1/4$, the average growth 