# A 2nd order recursion with polynomial coefficients

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.

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.

• The differential function $f(t):=\sum_j q_m(j)t^j$ satisfies some second order ODE, have you tried this way? – Fedor Petrov Feb 4 '19 at 12:18
• Maple solves $q_0$ in closed form, but not $q_1$ or $q_2$. – Gerald Edgar Feb 4 '19 at 13:03

For exponential generating function, $$f_m(z) = \sum_{j=0}^\infty \frac{q_m(j)}{j!}\;z^j$$ I get $$f_m(z) = \frac{1}{(1-z)^{m+1/2}(1+z)^{1/2}}$$