How to find general formula for $a_n$ if $a_{n+1}=4a_n^2+a_n$ with $a_1=-0.5$?

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Does a general formula exist for $a_{n+1}=4a_n^2+a_n$ with $a_1=-0.5$?, if it does, how should one find it?

Thanks

Updates: I didn't know own effort is expected to be shown in here. Here's my own effort:

Originally, the question was this: Given: $b_{n+1} = \frac{b^2_n+3}{b_n+1}$ , where $b_1=1$ find a general formula for $b_n$.

At first glance this seems to be an easy problem solvable via the ``fixed point'' method. However, when I solved for $f(b_n)=\frac{b^2_n+3}{b_n+1}$ I realize there's only one fixed point which is 3. Had there been two (denoted $(A,B)$) I would write down $b_{n+1}-A = \frac{b^2_n+3}{b_n+1} - A$ and $b_{n+1}-B = \frac{b^2_n+3}{b_n+1} - B$ and divide the left side of eq 1 by the left side of eq 2 and right side of eq1 by right side of eq2. And convert it to a geometric progression sequence of $\frac{b_{n+1}-C}{b_{n+1}-D}$ or something similar.

That was a dead end, so through some math manipulation I converted it to the above form (which I think is more likely to be fitted into a formula), by substituting $a_n = \frac{1}{b_n-3}$. And I couldn't go any further beyond $a_{n+1}=4a_n^2+a_n$. One comment by @Alex R. refers to quadratic recurrence/map, and said it likely has no explicit solution in its current form. I am looking into that direction.

I also plotted the first 100th term of $b_n$, it converges to 3 very quickly. I did some guesses and tried to fit some logrithmatic curves and had no luck.

Plot of $b_n$, the original question