Limit of sequence given by $x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$? Let $x_1,x_2,x_3\in\mathbb{R}$ be three distinct real numbers. I am interested in the convergence of the sequence
$$
x_{n} = \frac{x_{n-2}+x_{n-3}}{2},\quad n = 4,5,\ldots
$$
ie,
$$
x_{4} = \frac{x_{2}+x_{1}}{2},\quad x_{5} = \frac{x_{3}+x_{2}}{2},\quad x_{6} = \frac{x_{4}+x_{3}}{2},\quad\cdots
$$
Please note that, although related to, this is not a recursive averaging sequence with two initial values, as described and studied elsewhere --- see, for instance, Smith, Scott G. "Recursive Averaging". The Mathematics Teacher, Vol. 108, No. 7 (March 2015), pp. 553-557.
In the present case, convergence is easily verified by noting that
$$
\big|\,x_{n}-x_{n-1}\,\big| = \big|\frac{x_{n-2}+x_{n-3}}{2}-\frac{x_{n-3}+x_{n-4}}{2}\big| = \big|\,\frac{1}{2} (x_{n-2}-x_{n-4})\,\big| = \big|\,\frac{1}{2^{\,n-4}}\,\big|\cdot\big|\, x_{3}-x_{1}\,\big| 
$$
Therefore the sequence is Cauchy and hence it converges in $\mathbb{R}$.
However, I was not able to find the limit to where the sequence converges. Computer experiments carried out with several different sets of initial values suggest that this limit should be a linear combination of the those initial values, as occurs in the case mentioned above. Also, the results of these experiments point to the existence of (how many?) sub-sequences, all converging to the very same limit.
I would appreciate some help in the analysis of this problem.

For the linear recurrence $$x_{n} = \frac{x_{n-2}+x_{n-3}}{2}$$ the characteristic polynomial is
$$r^3=\frac {r+1}2 \implies r_1=1 \qquad r_2=-\frac {1+i} 2\qquad r_3=-\frac {1-i} 2$$
and, as usual, $$x_n=c_1 + c_2 r_1^n+c_3 r_2^n$$
Using $x_1=a$, $x_2=b$, $x_3=c$, this would give
$$x_n=A_n \,a+B_n\, b+C_n\, c$$ with
$$A_n=\frac{1}{5} \left(1-(3-i) \left(-\frac{1}{2}-\frac{i}{2}\right)^n-(3+i)
   \left(-\frac{1}{2}+\frac{i}{2}\right)^n\right)$$
$$B_n=\frac{1}{5} \left(2-(1+3 i) \left(-\frac{1}{2}-\frac{i}{2}\right)^n-(1-3 i)
   \left(-\frac{1}{2}+\frac{i}{2}\right)^n\right)$$
$$C_n=\frac{1}{5} \left(2+(2+i) (-1-i)^n 2^{1-n}+(2-i) (-1+i)^n 2^{1-n}\right)$$
Using Euler's formulae
$$A_n=\frac{1}{5}\left(1+2^{1-\frac{n}{2}} \sin \left(\frac{3 \pi  n}{4}\right)-3\ 2^{1-\frac{n}{2}} \cos\left(\frac{3 \pi  n}{4}\right) \right)$$
$$B_n=\frac{1}{5}\left(2-3\ 2^{1-\frac{n}{2}} \sin \left(\frac{3 \pi  n}{4}\right)-2^{1-\frac{n}{2}} \cos \left(\frac{3 \pi  n}{4}\right) \right)$$
$$C_n=\frac{1}{5}\left(2+2^{2-\frac{n}{2}} \sin \left(\frac{3 \pi  n}{4}\right)+2^{3-\frac{n}{2}} \cos \left(\frac{3 \pi  n}{4}\right)  \right)$$ and now, make $n \to \infty$ for a very simple result.
