For a vector $v=\left( {1 \over 2},{1\over 2},{1\over 2}\right)\;$ the following is defined
$$\begin{aligned}w_1&=(e,e+2,e-2)\\ w_n&=v\times w_{n-1}+(2,-4,2)\quad (n\geq2),\end{aligned}$$ $e$ is Euter's number.
Then what is the value of $\lim\limits_{n\to \infty} (2,-3,1)\cdot w_{2n-1}?$
I tried to calculate the first few terms of $w(n)$ to see if there is a pattern but I ended up with $$w(2)=(0,-3,3), w(3)=(5,-11/2,1/2), w(4)=(5,-7/4,-13/4)$$ and I could not see a pattern in these results.
Moreover, I did the cross product first, which is $$(v×w(n-1))+(2,-4,2).$$ Is that correct?
How can I decompose $w(2n-1)$ to get a formula that describes the whole trend? Or maybe something like representing it as a matrix. I would really appreciate it if someone can guide me on how to deal with it.
Or, a hint or a reference to a question of the same form will really be helpful to me (I could not get one).
HINT
Denote ${a} = (2,-4,2).$ Note that $v \perp a$ and $v \perp (v \times w_1).$
Thus for $n\geq 3$ we can modify the reccurent form:
$$\begin{aligned}w_2&=(-2,1,1)+a \\ w_3&=(v\times a) +a\\w_4&=v\times \big((v\times a)+ a\big ) +a\\ w_5&=\cdots=v\times(v\times w_3)+w_3\\\vdots\\ w_{2n+1}&=\color{red}{v\times(v\times w_{2n-1})}+w_3 \end{aligned}$$
Rewrite the red double cross-product with the use of formula $$x\times(y\times z)=y(x\cdot z)-z(x\cdot y).$$ We get $$\color{red}{v\times(v\times w_{2n-1})}=v\underbrace{(v\cdot w_{2n-1})}_{0\;(*)}-w_{2n-1}\underbrace{(v\cdot v)}_{3 \over 4}$$ or $$w_{2n+1}=-\frac{3}{4}w_{2n-1}+w_3$$ From this find $w_{2n+1}$ as a function of $w_3,$ then the scalar product and the limit.
(*) Verify that $v$ is orthogonal to each vector $w_k.$
EDIT - Alternative approach
Find straightforwardly $w_5.$ You will obtain $w_5=\frac{-3}{4}w_3+w_3.$ then find by induction $w_{2n+1}$ in terms of $w_3.$