Solving a Recursive Multivariable Equation

[BobbyOcean]

I am looking for a solution to a multivariable recursive formula as follows:

Define:
Initial Conditions:
a(1,0)=-3, a(2,0)=2, a(n,0)=0 for n≠1,2
Recursive Formula:
a(n,k)=(4n+1)a(n,k-1)-4a(n-1,k-1)

Find a formula in terms of n and k for a(n,k).


This is my first post and so I may have chosen the wrong category. This seems to be a differential equation problem since most single variable recursive formulas are solved using some series or differential equations. For this problem I am quite stuck. I have already tried to use (possibly incorrectly) mathematica to solve it for me, but the program failed to solve it.

If anyone has any insight or knowledge on solving general recursive equations, I would be most appreciative. Even a proof that no such function exists would be really helpful.

 

[omkar13]

I tried it and I found that there should be another condition w.r.t. to k too.

 

[BobbyOcean]

Thought I would post that I solved the recursive equation.

For those that wish to know,

Start by defining some infinite dimensional vectors:

$V_0=(-3,2,0,0,0,0,...)$

$V_1=(-15,30,-8,0,0,0,...)$

In general,

$V_k=(a(1,k),a(2,k),...)$

Notice, if we define the linear operator $T$ as

$T=\begin{matrix}5&0&0&0&\cdots\\-4&9&0&0&\cdots\\0&-4&13&0&\cdots\\0&0&-4&17&\cdots\\0&0&0&-4&\cdots\\ \vdots&\vdots&\vdots&\vdots&\ddots\end{matrix}$

then we have,

$V_n=T^{(n)}V_0$

Solving the entries of $T^{(n)}$ is a few simple induction arguments and thus we arrive at the final equation, which is:

$\displaystyle a(n,k)=\frac{-3}{(n-1)!}\sum_{i=0}^{n-1} \binom{n-1}{i} (4i+5)^k (-1)^i+\frac{2}{(n-2)!} \sum_{i=0}^{n-2}\binom{n-2}{i} (4i+9)^k (-1)^i$

Please excuse any typos. I verified the equation is correct in mathematica.

I was hoping the closed form solution would be nice looking. Oh well.

Well, thanks for reading my post.