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.
 A: The characteristic equation  of the linear recurrence is
$$\lambda^3 - \frac{1}{2}\lambda-\frac{1}{2} = 0$$ with roots $\lambda_1=1$, $\lambda_{2,3} = \frac{1}{\sqrt{2}}\cdot(-\frac{1}{\sqrt{2}} \pm \frac{i}{\sqrt{2}})$. There exist unique $a$, $b$, $c$ such that for all $n \ge 0$ we have
$$x_n = a \cdot 1^n + b \cdot \lambda_2^n + c \cdot \lambda_3^n$$
Since $|\lambda_{2,3}|= \frac{1}{\sqrt{2}}< 1$, we have
$$\lambda_2^n \to 0, \ \ \lambda_3^n \to 0$$
for $n \to \infty$. We conclude that $x_n \to a$. Now we only need to obtain $a$ from the equalities
$$x_0 = a + b + c \\
x_1 = a + b \lambda_2 + c \lambda_3\\
x_2 = a + b \lambda_2^2 + c \lambda_3^3$$
We get  with Cramer's rule
$$ a = \frac{\left| \begin{matrix} x_0 & 1 & 1 \\ x_1 & -\frac{1}{2} + \frac{i}{2} & -\frac{1}{2} - \frac{i}{2}\\ x_2 & (-\frac{1}{2} + \frac{i}{2})^2 & (-\frac{1}{2} - \frac{i}{2})^2 \end{matrix} \right |} {\left| \begin{matrix} 1 & 1 & 1 \\ 1 & -\frac{1}{2} + \frac{i}{2} & -\frac{1}{2} - \frac{i}{2}\\ 1 & (-\frac{1}{2} + \frac{i}{2})^2 & (-\frac{1}{2} - \frac{i}{2})^2 \end{matrix} \right |} = \frac{1}{5}(x_0 + 2 x_1 + 2 x_2)$$
$\bf{Added:}$ Since we are dealing with linear recurrences it's worth getting comfortable with determinants of the form
$$D(x_0, x_1, \ldots x_{n-1}) \colon = \left| \begin{matrix} x_0 & 1 & \ldots &1 \\
                        x_1 & \mu_2& \ldots& \mu_{n} \\
                        x_2 & \mu_2^2& \ldots & \mu_{n}^2 \\
                        \ldots& & & \ldots \\
                        x_{n-1} & \mu_2^{n-1} & \ldots & \mu_{n}^{n-1} \end{matrix} \right |$$
Note that when $x_k = x^k$ for $0\le k \le n-1$ we have a Vandermonde determinant, which equals
$$V(x, \mu_2, \ldots, \mu_{n}) = V( \mu_2, \ldots, \mu_{n}) \cdot \prod_{i=2}^{n}( \mu_i - x)= \\ =(-1)^{n-1} V( \mu_2, \ldots, \mu_{n})\cdot (x-\mu_2)\cdot (x-\mu_{n}) = (-1)^{n-1} V( \mu_2, \ldots, \mu_{n})\cdot ( x^{n-1} - s_1 x^{n-2} + s_2 x^{n-3} - \cdots)$$
It follows that the determinant with first column $(x_0, x_1, \ldots x_{n-1})$ equals
$$(-1)^{n-1} V(\mu_2, \ldots, \mu_{n+1}) \cdot ( x_{n-1} - s_1 x_{n-2} + s_2 x_{n-3} - \cdots)$$
Now when solving the system we have a quotient of such determinants $D$, the extra factor $(-1)^{n-1} \cdot V$ can be omitted. We get: the coefficient in front of a power $\mu_1^n$ equals
$$\frac{P(x)/(x-\mu_1)}{P'(\mu_1)}$$
where the numerator is considered "in the umbral sense".
Example with our problem at hand:
The characteristic polynomial equals $P(x)=x^3 - \frac{1}{2} x - \frac{1}{2}$.

*

*For the numerator : divide $P(x)$ by $x-\mu_1= x-1$ and get
$\frac{1}{2}(2 x^2 + 2 x + 1)$. This umbrally is $\frac{1}{2}(2 x_2 + 2x_1 + x_0)$.


*For the denominator we have $P'(x) = 3 x^2 - \frac{1}{2}$ so $P'(1) = \frac{5}{2}$


*Set up the fraction
$$a_1 = a =\frac{\frac{1}{2}(2 x_2 + 2 x_1 + x_0)}{\frac{5}{2}} = \frac{2 x_2 + 2 x_1 + x_0}{5}$$
A: Once the convergence has been proved, one thing to try is to pass the recurrence relation to the limit, in hope that it can then be solved for the limit $\,L\,$. Here, though, this results in the tautology $\,L=L\,$ which is useless for the purpose of calculating the limit. However, in such cases there is a fair chance that there is some form of telescoping "hidden" in the recursion, which can provide an alternate way to find the limit without actually solving the recurrence.
For a linear recurrence like the one here, this amounts to adding up the consecutive relations:
$$
\require{cancel}
\begin{align*}
2x_{n} &= \color{red}{\bcancel{x_{n-2}}} + \color{blue}{\cancel{x_{n-3}}}
\\ 2x_{n-1} &= \color{blue}{\cancel{x_{n-3}}} + \color{green}{\cancel{x_{n-4}}}
\\ \color{red}{\bcancel{2}}x_{n-2} &= \color{green}{\cancel{x_{n-4}}} + \cancel{x_{n-5}}
\\ \color{blue}{\cancel{2x_{n-3}}} &= \cancel{x_{n-5}} + \cancel{x_{n-6}}
\\ \color{green}{\cancel{2x_{n-4}}} &= \cancel{x_{n-6}} + \cancel{x_{n-7}}
\\ \dots
\\ \cancel{2x_7} &= \cancel{x_5} + \color{brown}{\cancel{x_4}}
\\ \cancel{2x_6} &= \color{brown}{\cancel{x_4}} + x_3
\\ \cancel{2x_5} &= x_3 + x_2
\\ \color{brown}{\cancel{2x_4}} &= x_2 + x_1
\\ \hline
\\2 x_n + 2x_{n-1}+x_{n-2} &= x_1 + 2x_2 + 2x_3
\end{align*}
$$
Then, passing to the limit with $\,x_n, x_{n-1}, x_{n-2} \to L\,$ gives $\,5L = x_1 + 2x_2 + 2x_3\,$.
A: 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.
A: You already received excellent answers, yet let me present a matricial approach
(Fibonacci style) which might be of interest in smilar cases.
$$
\begin{array}{l}
 x_n  = \frac{{x_{n - 2}  + x_{n - 3} }}{2}\quad  \Rightarrow  \\ 
  \Rightarrow \quad \left( {\begin{array}{*{20}c}
   {x_n }  \\   {x_{n - 1} }  \\   {x_{n - 2} }  \\
\end{array}} \right) = \left( {\begin{array}{*{20}c}
   0 & {1/2} & {1/2}  \\   1 & 0 & 0  \\   0 & 1 & 0  \\
\end{array}} \right)\left( {\begin{array}{*{20}c}
   {x_{n - 1} }  \\   {x_{n - 2} }  \\   {x_{n - 3} }  \\
\end{array}} \right)\quad  \Rightarrow  \\ 
  \Rightarrow \quad {\bf x}_n  = {\bf A}\;{\bf x}_{n - 1}  \\ 
 \end{array}
$$
The matrix $\bf A$ is  (right) stochastic matrix and thus the matrix of a Markov chain.
It is easy to see that the eigenvalues are
$$
1 = e^{\,i0} ,\; - \left( {1 + i} \right)/2 = \frac{{\sqrt 2 }}{2}e^{ - i\frac{3}{4}\pi } ,\;
 - \left( {1 - i} \right)/2 = \frac{{\sqrt 2 }}{2}e^{i\frac{3}{4}\pi } 
$$
and to find the relevant eigenvectors. In particular that the eigenvector corresponding to the eigenvalue $1$
is $$\bf v =(1,1,1)^T$$ and the theory of Markov chain assures us that
$$
{\bf v} = {\bf A}\;{\bf v}
$$
is the stable vector to which the chain is going to converge, once we start from any "probabilistic" vector, i.e.
whose components are non-negative and sum to one.  Otherwise, you can just reduce to that by a suitable scaling diagonal matrix.
You can get more details on the convergence chain by diagonalizing $\bf A$., but is clear that the remaining two eigenvalues
are declining as $2^{-n/2}$ (or $2^{-(n-3)/2}$, or $2^{-(n-4)/2}$ depending on the starting point).
