I'm hoping to find "exact" an solution to the following simple recursion:


$q\_m(j+1) = m \cdot q\_m(j) + j(j+m)\cdot q\_m(j-1)$ 


with initial data $q\_m(0) = 1$, $q\_m(1)=m$, where $m \geq 0$ is an integer parameter. 




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Experimentally, the behavior seems to be slightly different based on the parity of $m$. For example, for even $m$ it seems as if $q\_m(2j+1) = (2j+1)!! (2j+1+m)!! F\_m(j)$ where $F\_m$ is a polynomial in $j$ whose degree may depend on $m$. I can even write a recursive equation for $F\_m$ that looks like


$(2j+1)(2j+1+m) F\_m(j+1) =(m^2 + 2j(2j+m) + (2j-1)(2j-1+m))F\_m(j) - ((2 j - 2) (2 j - 2 + m) F\_m(j-1)$ 


with initial conditions $F(1) = m$ and $F(2) = (m^3+3m^2+5m)/(3(m+3))$, but it seems like a miracle that the solution ends up being a polynomial in the first place. 


I would be also interested in relating solutions for different $m$'s, if an explicit description is out of reach.



A 2nd order recursion with polynomial coefficients
For exponential generating function,
$$
f\_m(z) = \sum\_{j=0}^\infty \frac{q\_m(j)}{j!}\;z^j
$$
I get
$$
f\_m(z) = \frac{1}{(1-z)^{m+1/2}(1+z)^{1/2}}
$$


