The first 4 terms of the sequence are:
\begin{align} P_0 &= \frac{1}{B+1} \\ P_1 &= \frac{B^2+B+\alpha^2}{(B+1)^3} \\ P_2 &= \frac{2B^4+4B^3+4B^2\alpha^2 +2B^2+4B\alpha^2+\alpha^4}{2(B+1)^5} \\ P_3 &= \frac{6B^6+18B^5+18B^4\alpha^2+18B^4+36B^3\alpha^2+6B^3+9B^2\alpha^4+18B^2\alpha^2+9B\alpha^4+\alpha^6}{6(B+1)^7} \end{align}
Note that this sequence could be reduce to: $$P_n = \frac{B^n}{(B+1)^n}$$ if one take $\alpha=0$ and, $$P_n = \frac{\alpha^{2n}}{n!}$$ if one set $B = 0$.
Is there a way to guess a closed form expression for this sequence?
For those who are familiar with quantum physics, $P_n$ is obtained when I was calculating the photon number statistics of a displaced thermal state. Where $B$ is the thermal photon number and $\alpha$ is the displacement amplitude.
The numerator is $$\sum_{k=0}^n c_k(B^2+B)^k\alpha^{2(n-k)}$$ where the sequences $c_k$ are $$1\\ 1,1\\ 1,4,2\\ 1,9,18,6\\ 1,16,72,96,24$$ which are $$\frac{(n!)^2}{((n-k)!)^2k!}$$