Sequence $a_n$ is defined as follows.
$a_1=a_2=a_3=1$
For $n>3,\text{ } a_n=\frac{1}{2}(a_{n-1}+a_{n-2})-a_{n-3}$
Using excel, plotting $a_n$ against $n$ produces a sine-like curve, with the local maximums seeming to be unique and approaching (but not sequentially) some limit; same for the local minimums.
My question is:
$$\text{What is}\lim_{n\to\infty}\max{(a_1, a_2, a_3, …, a_n)}\text{ }?$$
Using excel, the limit seems to be $\frac{9}{7}$ (e.g. the largest of the first 10,000 terms is approximately 1.2857142854), but I do not know how to prove this.
(Context: I was looking at the sequence in this question and then started tweaking the sequence and seeing what happens.)
Best Answer
The characteristic polynomial $\,2t^3-t^2-t+2\,$ has roots $\,-1\,$, $\,\omega = \frac{3 + i \sqrt{7}}{4}\,$ and $\,\bar \omega\,$ where $\,|\omega|=1\,$, so $\,a_n = \alpha (-1)^n + \beta\omega^n + \bar\beta\bar\omega^n\,$. From the initial conditions it follows that $\,\alpha = 1/7\,$ and $\,\beta= (3-i\sqrt{7})/7\,$, then:
$$ 7 a_n = (-1)^n + (3-i\sqrt{7}) \omega^n + (3+i\sqrt{7}) \bar\omega^n $$
By the triangle inequality:
$$ 7 |a_n| \le 1 + 2\cdot\big|3-i \sqrt{7}\big| = 1 + 2 \cdot 4 = 9 $$
Since the argument of $\,\omega^2\,$ is an irrational multiple of $\,\pi\,$, there will exist powers $\,\omega^{2n}\,$ with argument arbitrarily close to $\,\arg\big((-1)^{2n}\big)=0\,$, which make the triangle inequality arbitrarily close to an equality, so in the end $\,\limsup a_n = 9/7\,$.