formula for $a_n$ where $a_{n+1}=4-a_n-\frac1{a_n}$ and $a_1=1$? There's a sequence $a_{n+1}=4-a_n-\frac{1}{a_n}$ staring with $a_1=1$. 
Is it possible to find a general formula for $a_n$?
 A: Not a full answer, but a few steps in a possibly-fruitful direction: first of all, let's remove the inhomogeneous term.  Set $a_n=2+b_n$; then we can write $(2+b_{n+1})=4-(2+b_n)-\frac1{2+b_n}$ $=2-b_n-\frac1{2+b_n}$, or in other words, $b_{n+1}=-b_n-\frac1{2+b_n}$.
Now, we write $b_n=\frac{x_n}{y_n}$ and equate numerators and denominators; this gives $\dfrac{x_{n+1}}{y_{n+1}}=-\dfrac{x_n}{y_n}-\dfrac{y_n}{x_n+2y_n}$ $=-\dfrac{2x_ny_n+x_n^2+y_n^2}{y_n(x_n+2y_n)}$ $=-\dfrac{(x_n+y_n)^2}{y_n(x_n+2y_n)}$.  In other words, we can equate your original recurrence relation with the paired recurrences $x_{n+1}=-(x_n+y_n)^2, y_{n+1}=y_n(x_n+2y_n)$.  (Or alternately, since $x_n$ is manifestly negative, we can write $x_{n+1} = (y_n-x_n)^2, y_{n+1}=y_n(2y_n-x_n)$ and then $b_n=-\frac{x_n}{y_n}$.)
Unfortunately, from here the trail looks to peter out; we can show that each fraction is in reduced terms ($\gcd(x_n, y_n)=1$ implies that $\gcd(x_n+y_n, y_n)=1$ and then that $\gcd(x_n+y_n, x_n+2y_n)=1$, so $\gcd(x_n+y_n, y_n(x_n+2y_n))=1$) but the structure of the recurrence suggests that growth is super-exponential and such quadratic recurrences tend not to have 'nice' forms unless there's some explicit telescoping or other cancellation involved in the terms.
