Complete sequence of sequences Who can continue (complete) the following sequence:
$$1,n-1,\frac{(n-2)(n-1)}{2},\frac{(n-4)(n-3)(n+1)}{6},\frac{(n-7)(n-4)(n-2)(n+3)}{24},\dots$$
This was emerging in the course of this question as crucial coefficients in the transformation of Fibonacci polynomials.
I am pretty sure I have seen this before in another context, but I cannot remember it exactly, I have the vague memory that it contains some more complicated structure than factorials, maybe something like rising factorials/Pochhammer-symbols. 

Edit
The next sequence of this row is numerically $$a_6 = \{-4,12,-21,-24,-9,42,154,360,702,1232\dots\}$$
This is no 5$^{th}$ degree polynomial anymore and I am unable to detect the law.. If fact it is one, this can be seen from the fifth difference sequence which gets constant (when calculated correctly).
By interpolation and factorisation I finally got the law (see below).
Edit-2
Meanwhile I found an alternative way to generate these series, but however still got stuck at the 6$^{th}$ one.
They can be expressed as the coefficients of the series expansion of polynomial fractions $f_i(x)$ at $x=0$, the functions are in detail:
\begin{array}{c|c|c|c|c|c|c}
i& 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline
a_i&1 & n-1 & \frac{(n-2)(n-1)}{2}& \frac{(n-4)(n-3)(n+1)}{6}&\frac{(n-7)(n-4)(n-2)(n+3)}{24}&v.i.&v.i.\\ \hline
f_i(x)&\frac{1}{x-1}& \frac{1}{(x-1)^2}&\frac{1}{(x-1)^3}&\frac{(x-2)x}{(x-1)^4}&\frac{(x-2)(1+x(x-3))}{(x-1)^5}&?&\\ 
\end{array}
Edit-3
So, the expression for the next sequences are:
$$ a_6 = \frac{(n-5)(n-4)(n-1)(n^2-5n-54)}{120} $$
$$ a_7 = \frac{(n-6)(n-3)(n^4-12n^3-71n^2+642n+160)}{720} $$
I am still unable to guess a general law for $f_i(x)$ or $a_i$. 
Edit-4
I will try to get some recursion relation now from the consideration of the difference series.
 A: As you reference in another question, $$F_n(x) = \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} x^k$$
If we consider $F_n(x+1)$ directly we get $$\begin{eqnarray}
F_n(x+1) & = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} (x+1)^k \\
& = & \sum_{k=0}^n \binom{\frac{n+k-1}{2}}{k} \sum_{i=0}^k \binom{k}{i} x^i \\
& = & \sum_{i=0}^n x^i \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i} \\
\end{eqnarray}$$
If we consider $F_n(x+1)$ as a sum of weighted $F_m(x)$ we get $$\begin{eqnarray}
F_n(x+1) & = & \sum_{m=0}^n \lambda_m \sum_{k=0}^m \binom{\frac{m+k-1}{2}}{k} x^k \\
& = & \sum_{k=0}^n x^k \sum_{m=k}^n \binom{\frac{m+k-1}{2}}{k} \lambda_m \\
\end{eqnarray}$$
By equating coefficients in $x^i$ we get
$$\sum_{m=i}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m = \sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}$$
On the left-hand side, discard the term $m=i$ (since $2i-1$ is odd) and extract the term $m=i+1$ from the sum, and the sum can be passed to the right:
$$\lambda_{i+1} = \left(\sum_{k=i}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{i}\right) - \sum_{m=i+2}^n \binom{\frac{m+i-1}{2}}{i} \lambda_m$$
And voilà: one recurrence.
NB your $a_j$ is $\lambda_{n-j+1}$, so 
$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=n-j+2}^n \binom{\frac{n+m-j-1}{2}}{n-j} a_{n-m+1}$$
and it makes sense to reindex at least the second sum as
$$a_j = \left(\sum_{k=n-j}^n \binom{\frac{n+k-1}{2}}{k} \binom{k}{n-j}\right) - \sum_{m=1}^{j-1} \binom{n-\frac{m+j}{2}}{n-j} a_{m}$$
