A unique question in combinatorics and non-commutative variables. Let $x,y$ be two variables that satisfy:
$xy=yx+1$ (they are not commutative).
Find $(xy)^2 ,(xy)^3, (yx)^2, (yx)^3$ as a linear  combination in terms of $y^jx^j$.
Then find a formula for $(xy)^n$ and $(yx)^n$.
After some calculations we get:
$(xy)^2=y^2x^2+3yx+1$.
$(xy)^3=y^3x^3+6y^2x^2+7yx+1$.
$(xy)^4=y^4x^4+10y^3x^3+25y^2x^2+15yx+1$.
$(yx)^2=y^2x^2+yx$.
$(yx)^3=y^3x^3+3y^2x^2+y$
Now, given $xy=yx+1$ we can use the binomial formula:
$(xy)^n=\sum_{k=0}^{n} {n \choose k} (yx)^k$.
Denote:
$(yx)^{n}=\sum a_{n,k} y^kx^k$.
Then,
$(yx+1)^n=(yx)(yx)^n=\sum {n\choose k} yxy^kx^k$.
After computing we get,
$(yx)^{n+1}= \sum a_{n,k} (ky^kx^k+y^{k+1}x^{k+1})$.
By comparing the coefficient of $[y^kx^k]$ in $(yx)^{n+1}$:
$a_{n,k}=ka_{n-1,k}+a_{n-1,k-1}$

I found different formulas, namely
\begin{align*}
(xy)^1 & = yx+1 \\
(xy)^2 & = y^2x^2+3yx+1\\
(yx)^3 & = y^3x^3+6y^2x^2+7yx+1\\
(yx)^4 & = y^4x^4+10y^3x^3+25y^2x^2+15yx+1\\
(yx)^5 & = y^5x^5+15y^4x^4+65y^3x^3+90y^2x^2+31yx+1
\end{align*}
if correct, can you guess a general formula?
