Solving a 2 independent variables (2nd degree) recurrence relation Changes to the recurrences and definition are changed! See here: $f(n, 1) = 2n^2 $ and $f (n, k) = 0$ for $k \geq 2n$ and for $k < 0$ and $f(n, 2n-1) = 1$ for all $n$.

Question: Is it possible to solve the following recurrence? If so in what manner should I approach it? 
Conditions Let $f(n,k)$ be the 2-independent-variable function, where $f(n, 0) = 1$, $f(n, 1) = 2 n^2 $ and $f (n, k) = 0$ for $k \geq 2n$ and for $k < 0$ and $f(n, 2n-1) = 1$ for all $n$. All $n, k$ are positive integers. 
Solve the recurrence (giving either a generating function, hopefully also an explicit formula): 
$f(n,k) = f(n-1, k) + 2(2n-k) \cdot f(n-1, k-1) + (2n-k+1)(2n-k) \cdot f(n-1, k-2).$

I read the following link on gen. functions and tried to come up with something as follows: 

Multiplying throughout by $x^k$ and summing the RHS over $k\geq 0$, define the generating function $B_n(x) = \sum_{k\geq0} f(n,k)x^k$, with $n \geq 1, B_0(x) = 0$. 
Let the notation $[x^k]g(x)$ denote the $k^{\text{th}}$ coefficients of the function $g(x)$. 

Then if we do the above for the LHS too and equate both sides, we yield: 

$ B_n(x) = B_{n-1}(x) + \sum_{k\geq0} 2(2n-k) \cdot [x^k](xB_{n-1}(x)) + \sum_{k\geq0} (2n-k+1)(2n-k) \cdot [x^k](x^2 B_{n-1}(x)) $
where the notation $\sum_{k\geq0} 2(2n-k) \cdot [x^k](xB_{n-1}(x))$ means $2(2n-k)$ multiplied by the $k^{th}$ coefficient of the function $(xB_{n-1}(x))$. 

Notice this simplifies to: 

$  B_n(x) = B_{n-1}(x) + \sum_{k\geq0} 2(n-k) \cdot [x^{k-1}](B_{n-1}(x)) + \sum_{k\geq0} (n-k+1)(n-k) \cdot [x^{k-2}](B_{n-1}(x)) $

Is the above workings correct/logical? If it is, is there a way to simplify the $\sum_{k\geq0} (2n-k+1)(2n-k)$ expression (which is something multiply by the $k^{th}$ coefficient)? Any other ideas to go about solving this recurrence function? Thanks! 
 A: For any fixed $k>1$ $f(n,k)$ as a function of $n$ (for $n\ge k$) is some polynomial $P_{2k-1}(n)$ оf order $2k-1$. Indeed, for $k=2$ we have 
$$
f(n,2) = f(n-1, 2) + 2(2n-2) + (2n-k+1)(2n-k).
$$
Every next value is obtained by adding a polynomial of the second order. Summing up will give some polynomial $P_3(n)$. By induction
$$
f(n,k) = f(n-1, k) + 2(2n-k) P_{2k-1}(n-1) + (2n-k+1)(2n-k) P_{2k-3}(n-1).
$$
The poly in the rhs is of order $2k$. Summing it up gives a $2k+1$ order polynomial.
The first several $P_{2k-1}$ are
$$
\small
\begin{eqnarray}
f[n,2]&=&\frac{1}{3} \left(4 n^3+3 n^2-7 n-27\right), \\
f[n,3]&=& \frac{1}{15} \left(16 n^5-35 n^4-20 n^3-190 n^2+499 n-285\right), \\
 f[n,4]&=&\frac{1}{630} \left(384 n^7-2212 n^6+3948 n^5-11305 n^4+52206 n^3-110593 n^2+104112 n-367290\right), \\
f[n,5]&=&\frac{1}{1890} \left(512 n^9-5478 n^8+22824 n^7-66276 n^6+249438 n^5-873747 n^4+\right.\\
&&\left. 1907296 n^3-4371069 n^2+9532260 n-17915310\right).
\end{eqnarray}
$$
It seems quite a possibility that there is no closed form for $f(n,k)$.
As for generating function of the form $\sum_{n,k}f(n,k)x^ny^k$ it probably has a convergence radius zero, because the polynomials coefficients probably grow fast enough. 
One can consider, of course, an exponential generating function 
$$
B[x,y]=\sum_{n,k}f(n,k)\frac{x^ny^k}{n!k!}=
\ldots+\sum_{k=2}^\infty \frac{y^k}{k!}\sum_{n=k}^\infty P_{2k-1}(n)\frac{x^n}{n!},
$$
but, given the expressions for $P_{2k-1}$, it seems to be  at least not evident that it has a closed form.
