Initial value for a sequence to become periodic. The following is from the previous Proofathon contest:

Let $a_{n}$ be the sequence defined by the recursion $ \sqrt{a_{n+1}}= (2(\sqrt[2014]{a_n})-1)^{2014}. $
Find all the values of $a_1$ such that $(a_n)$ becomes eventually periodic.

I couldn't solve this completely.
Setting $v_n=\sqrt[2014]{a_n}$, we have $v_{n+1}=(2v_n-1)^2$.
Playing around with this last recursion, it seems to behave very randomly for initial values in $(\frac34,1)$ ...
Besides, it is trivial to prove that $a_1>1 \implies a_n \to \infty$.
Of course, $(a_n)$ is constant when $\displaystyle a_1=\frac{1}{4^{2014}}$ and $a_1=1$.
$a_1=0$ also fits. What about other values ?
 A: I assume that we are talking about real values, so that we may suppose $a_n\ge0$.  And as you have already disposed of $a_n>1$, we may assume that $0\le a_n\le1$.  So, let your $v_n$ be $\cos^2\theta_n$, that is,
$$a_n=\cos^{4028}\theta_n\ .$$
Then
$$\cos^2\theta_{n+1}=(2\cos^2\theta_n-1)^2=\cos^2(2\theta_n)\ .$$
The sequence will be periodic if and only if
$$\cos^2(\theta_{l+k+1})=\cos^2(\theta_{l+1})$$
for some integers $k>0$, $l\ge0$, that is,
$$\cos^2(2^{l+k}\theta_1)=\cos^2(2^l\theta_1)\ ,$$
which holds if and only if
$$2^{l+k}\theta_1=m\pi\pm2^l\theta_1$$
for some $m\in\Bbb N$.  So the sequence is periodic if and only if
$$a_1=\cos^{4028}\Bigl(\frac{m\pi}{2^l(2^k\pm1)}\Bigr)$$
for some integers $k>0$, $l\ge0$, $m\ge0$.
But there's more... in fact, $k,l,m$ can be chosen so that the bracketed term is any positive rational multiple of $\pi$.  For if $p/q$ is given we can write
$$q=2^sr$$
where $r$ is odd and then
$$k=\phi(r)\ ,\quad l=s\ ,\quad m=pt$$
where $2^k-1=tr$, to give
$$\frac{m}{2^l(2^k-1)}=\frac{pt}{2^s(tr)}=\frac{p}{q}\ .$$
So to sum up: the sequence is eventually periodic if and only if
$$a_1=\cos^{4028}(\alpha\pi)$$
for some rational $\alpha$.
Comments


*

*In fact, the plus sign in the denominator is redundant as we can write
$$\frac{m\pi}{2^l(2^k+1)}=\frac{(2^k-1)m\pi}{2^l(2^{2k}-1)}
  =\frac{m'\pi}{2^l(2^{k'}-1)}\ .$$

*If we take the minus sign in the denominator and $l=0$, then $m=1$, $k=2$ gives
$$a_1=\frac{1}{4^{2014}}$$
as you mentioned in the question, while $m=3$, $k=4$ gives
$$v_1=\cos^2\Bigl(\frac{\pi}{5}\Bigr)=\frac{3+\sqrt5}{8}$$
as posted in a comment by @EwanDelanoy.

*To get $a_1$ close to $1$, say $a_1>\frac{3}{4}$ as in the OP, we'll need $\cos\theta_1$ very very close to $1$, and therefore $\theta_1$ very very small.  So to get a repeated value we will have to take a very large number of steps.  In other words, the period (even if $a_1$ does have a suitable value) will be very large, and this explains why the behaviour appeared random. 

