I think I can at least show that the only denominators are powers of two.

We have the following recurrence: $$b_n=\frac{b_{n-1}(b_{n-1}+1)}{2}\frac{b_{n-6}b_{n-10}b_{n-14}\dots}{b_{n-4}b_{n-8}b_{n-12}\dots}$$ where $b_i$ is interpreted as $1$ if $i\leqslant 0$ so that the products on top and bottom have a finite number of terms that aren't equal to $1$.

Now suppose that $p$ is an odd prime, and let $n(p)$ be the first value of $n$ such that $\nu=\nu_p(b_n)\ne 0$. Then $p$ is a prime divisor of $b_{n(p)-1}+1$, and from $n(p)$ onwards the values of $\nu_p(n)$ go as follows: $\nu,\ \nu,\ \nu,\ \nu,\ 0,\ 0,\ \nu,\ \nu,\ 0,\ 0,\ \nu,\ \nu,\dots$ until a further $\nu_p(b_{n-1}+1)$ is non-zero. But this can only happen if $\nu_p(b_{n-1})$ is zero, at which point another sequence of the same form is added in, and so on. The point is that this shows $\nu_p(b_n)$ always stays non-negative.

Finally, we need to deal with $\nu_2$. This seems more mysterious to me. The problem is to understand $\nu_2(b_n+1)$ when $b_n$ is odd.

I think I can at least show that the only denominators are powers of two.

We have the following recurrence: $$b_n=\frac{b_{n-1}(b_{n-1}+1)}{2}\frac{b_{n-6}b_{n-10}b_{n-14}\dots}{b_{n-4}b_{n-8}b_{n-12}\dots}$$ where $b_i$ is interpreted as $1$ if $i\leqslant 0$ so that the products on top and bottom have a finite number of terms that aren't equal to $1$.

Now suppose that $p$ is an odd prime, and let $n(p)$ be the first value of $n$ such that $\nu=\nu_p(b_n)\ne 0$. Then $p$ is a prime divisor of $b_{n(p)-1}+1$, and from $n(p)$ onwards the values of $\nu_p(n)$ go as follows: $\nu,\ \nu,\ \nu,\ \nu,\ 0,\ 0,\ 0,\ 0,\ 0,\ \dots$ until a further $\nu_p(b_{n-1}+1)$ is non-zero. But this can only happen if $\nu_p(b_{n-1})$ is zero, at which point another sequence of the same form is added in, and so on. The point is that this shows $\nu_p(b_n)$ always stays non-negative.

Finally, we need to deal with $\nu_2$. This seems more mysterious to me. The problem is to understand $\nu_2(b_n+1)$ when $b_n$ is odd.