Consider $P_n(x)$ polynomials defined through the recurrence relations $$P_n(x)=2(1-x)P_{n-1}(x)-(1+x)^2P_{n-2}(x),$$ with $P_0(x)=1$ and $P_1(x)=1-3x$. In fact, the explicit solution of these recurrence relations is given by the formula $$P_n(x)=\frac{1}{2}\left[ (1+i\sqrt{x})^{2n+1}+(1-i\sqrt{x})^{2n+1}\right].$$ What condition(s) should satisfy the number $n$, the polynomial $P_n(x)$ to be reducible? I have checked, that $P_n(x)$ is reducible for $n=4,7,10,12,13,16,17,19,\ldots$ What is special about these numbers?

Another question: given a number $n$, for what numbers $0<m<n$ the polynomial $P_n(x)$ is divisible by $P_m(x)$? For example, $P_4$ is divisible by $P_1$; $P_7$ is divisible by $P_1$ and $P_2$; $P_{10}$ is divisible by $P_1$ and $P_3$; $P_{12}$ is divisible by $P_2$; $P_{13}$ is divisible by $P_1$; $P_{16}$ is divisible by $P_1$ and $P_5$; $P_{17}$ is divisible by $P_2$ and $P_3$; $P_{19} $ is divisible by $P_1$ and $P_6$.

These polynomials emerged in the solution of the certain collision problem (hence the name) from the book: David Morin, Introduction to Classical Mechanics With Problems and Solutions (Cambridge University Press, 2007). See problem 5.88 on page 192. It can be also found here (the problem for Week 19. The solution is also provided there, which, however, does not use the collision polynomials):

http://www.physics.harvard.edu/academics/undergrad/problems.html