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.

  • 2
    $\begingroup$ The differential function $f(t):=\sum_j q_m(j)t^j$ satisfies some second order ODE, have you tried this way? $\endgroup$ – Fedor Petrov Feb 4 '19 at 12:18
  • $\begingroup$ Maple solves $q_0$ in closed form, but not $q_1$ or $q_2$. $\endgroup$ – 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 $$ f_m(z) = \frac{1}{(1-z)^{m+1/2}(1+z)^{1/2}} $$


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