We prove that $\alpha$ is irrational. Suppose $\alpha$ is a rational number, i.e. the sequence $a_1, a_2, a_3, \ldots, a_k, \ldots$ is periodic from some point onwards. Hence there exist $k_0$ and $T$ such that for any $k > k_0$ we have $a_k = a_{k+T}$.

Choose a positive integer $m$ for which $mT > k_0$ and $mT$ is a perfect square. If $T = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_s^{\alpha_s}$ then $m = p_1^{\beta_1} p_2^{\beta_2} \cdots p_s^{\beta_s}$, where $\alpha_i + \beta_i$ is even for all $i = 1, 2, \ldots, s$ and $\beta_i$ are large enough positive integers.

Choose a prime number $p > 2007$, $p \neq p_i$, $i = 1, 2, \ldots, s$. Since $pmT - mT$ is divisible by $T$ we have that $a_{mT} = a_{pmT}$.

Denote by $\tau(k)$ the number of divisors of $k$ and by $f(k)$ the number of divisors of $k$ that are greater than $2007$. We have $f(pmT) = f(mT) + \tau(mT)$ and since $\tau(mT)$ is odd number we conclude that $f(pmT)$ and $f(mT)$ are of the same parity, a contradiction.