The system has the form


$$
(n-2)f\_n^{(1)}=n(f\_{n-1}^{(1)}+1),
$$
$$
(n-2\cdot 2) f\_n^{(2)}=n(f\_{n-1}^{(2)}+f\_{n-1}^{(1)}),
$$
$$
\ldots 
$$
$$
(n-2k)f\_n^{(k)}=n(f\_{n-1}^{(k)}+f\_{n-1}^{(k-1)}),
$$
for the unknown sequences $f\_n^{(1)},f\_n^{(2)},\ldots,f\_n^{(k)}$ with the initial conditions $f^{(i)}\_k=0,$ for all $k=0,1,\ldots,2i.$


By direct calculation I have got 


$$
f\_n^{(1)}=n(n-2),
$$
$$
f\_n^{(2)}=\frac{1}{2!} (n-4) { n \choose 2} (3n-7),
$$
$$
f^{(3)}\_n=\frac{1}{3!} (n-6) { n \choose 3} \left( 19{n}^{2}-141n+254 \right),
$$
$$
f^{(4)}\_n=\frac{1}{4!} (n-8) { n \choose 4} \left( 
211{n}^{3}-3258{n}^{2}+16481n-27306 \right),
$$
$$
f^{(5)}\_n=\frac{1}{5!} (n-10) { n \choose 5} \left( 3651{n}^{4}-96550{n}^{3}+946185{n}^{2}-4071950n+
6492024 \right) 
$$


**Question.** What is a general expression for $f\_n^{(i)}$?


The ordinary generating function for the above sequences has the form
$$
G(f\_n^{(1)},z)={\frac {{z}^{3} \left( 3-z \right) }{ \left( 1-z \right) ^{3}}}=3{z}^{3}+8{z}^{4}+15{z}^{5}+\cdots
$$
$$
G(f\_n^{(2)},z)={\frac {z^5(-3{z}^{3}+16{z}^{2}-35z+40)}{ \left( 1-z \right) ^{5}}},
$$
$$
G(f\_n^{(3)},z)={\frac {{z}^{7} \left( -40{z}^{5}+288{z}^{4}-897{z}^{3}+1575{z}
^{2}-1701z+1155 \right) }{ \left( 1-z \right) ^{7}}}
$$



How to solve the system of recurrence equations
This is not a full solution, but it reduces the system to another recurrence relation which involves only coefficients. Let
$$f\_n^{(k)}=a\_0^k n(n-1)...(n-k+1)+a\_1^k n(n-1)...(n-k)+...+a\_k^k n(n-1)...(n-2k+1).$$
Then the above system reduces to
$$a\_i^k={a\_i^{k-1}\over i-k}$$
for $i < k$ and $a\_k^k$ is determined by the condition that $f\_{2k}^{(k)}=0$. Clearly the determination of $a\_k^k$ is the difficult part. But if the sequence 3,19 etc. shows up
elsewhere, that should be suggestive.


Addendum: If all the $a\_i^k$ are expressed in terms of $a\_i^i$, then the recurrence relation for the
$a\_k^k$ becomes the same as for the Taylor coefficients of the reciprocal Bessel function mentioned in Barry Cipra's link.


