This question is related to my previous question: Need help finding a closed form for complicated sum.
An answer to that question led my to try and find the general term of the following recurrence:
$$t_n=(-i)^nr_n$$ Where
$$r_n=\frac{1}{n}(i(n+d-1)r_{n-1}-(n+2d-2)r_{n-2})$$ And $d$ is some positive integer.
Now this might look terrible, and it is, but somehow Mathematica manages to find closed forms for sequences for specific values of $d$.
So for example, if $d=1$, a closed form for this sequence $t_n$ is this: $$t_{d=1}(n)=F_n$$
Where $F_n$ is the $n$th fibonacci number. However, for $d=2$ things become a lot more complicated:
$$t_{d=2}(n)=-\frac{(2^{-n}(2(365 + 163\sqrt5)((1-\sqrt5)^n-(1+\sqrt[5])^n)+ 5((1-\sqrt5)^n(101+45\sqrt5)-2(1+\sqrt5)^n(132+59\sqrt[5])) n)}{25(163+73\sqrt5))}$$
In which some traces of the fibonacci sequence can still be recognised, but it is not clear exactly how this solution came to be (at least not to me).
Now the thing is, that Mathematica can generate an expression like this for every $d$. However, when I try to make it solve the 'general' case with $d$ as some variable, it cannot do it.
This general closed form solution is of course what I'm interested in. I am (again), not at all sure that such a general closed form in terms of $n$ and $d$ exists, but I would really like to try and find one.
However, I have no idea how Mathematica finds those closed forms for specific values of $d$, and I feel like knowing what it is doing might help me figure out way to tackle this problem.
So does some one have an idea of what kind of techniques can be used (maybe also other techniques than just those used by Mathematica) to find solutions to this recurrence?
Thanks.