<s>[INST] How find this seqence $a_{n}$ closed form if such Strange condition 
Let $\{a_{n}\}$ such
  $$a_{0}=a_{1}=a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=4,a_{6}=10,a_{7}=18$$
  and such
  $$\begin{vmatrix}
a_{n+8}&a_{n+7}&a_{n+6}\\
a_{n+5}&a_{n+4}&a_{n+3}\\
a_{n+2}&a_{n+1}&a_{n}
\end{vmatrix}=2\cdot (n+1)!$$
  Find the $a_{n}$ colsed form

or 
$$a_{n+8}a_{n+4}a_{n}+a_{n+5}a_{n+1}a_{n+6}+a_{n+2}a_{n+3}a_{n+7}
-a_{n+6}a_{n+4}a_{n+2}-a_{n+7}a_{n+5}a_{n}-a_{n+8}a_{n+1}a_{n+3}=2\cdot(n+1)!$$
if we find $a_{n}$ colsed form,I think easy to mathematical induction solve it?
 [/INST] The first calculated terms are : $1,1,1,2,3,4,10,18,28,80,162,280,880$
So the closed formula appear to be $\bbox[5px,border:2px solid]{\begin{cases}a_0=1\\a_1=1\\a_2=1\\a_n=(n-1)a_{n−3}\quad n\ge3\end{cases}}$
Let see what the determinant look like under the recurrence hypothesis until $a_{n+7}$.
$D=\begin{array}{l}
+a_{n+8}a_{n+4}a_{n}\\
+a_{n+5}a_{n+1}a_{n+6}\\
+a_{n+2}a_{n+3}a_{n+7}\\
-a_{n+6}a_{n+4}a_{n+2}\\
-a_{n+7}a_{n+5}a_{n}\\
-a_{n+8}a_{n+1}a_{n+3}\end{array} =
\begin{array}{l}
+(n+3)a_{n+8}a_{n+1}a_{n} \\
+(n+4)(n+5)(n+2)a_{n+2}a_{n+1}a_{n} \\
+(n+2)(n+6)(n+3)a_{n+2}a_{n}a_{n+1} \\
-(n+5)(n+2)(n+3)a_{n}a_{n+1}a_{n+2} \\
-(n+6)(n+3)(n+4)a_{n+1}a_{n+2}a_{n} \\
-(n+7)(n+4)(n+2)a_{n+2}a_{n+1}a_{n}
\end{array}$

$=(n+3)a_{n+8}a_{n+1}a_{n}+[-n^3-14n^2-61n-82]\,a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+8}a_{n+1}a_{n}-[(n+3)(n+7)(n+4)-2]\,a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)(n+4)a_{n+2}]+2a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2a_{n+2}a_{n+1}a_{n}$

Let's have examine for $n=3k,\ 3k+1,\ 3k+2$ the product $a_{n+2}a_{n+1}a_{n}$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n)(n-1)a_{n-1}a_{n-2}a_{n-3}=\ldots=(n+1)(n+0)(n-1)\ldots(4)(3)(2)a_2a_1a_0=(n+1)!\times 1\times 1\times 1=(n+1)!$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(5)(4)(3)a_3a_2a_1=\frac{(n+1)!}{2}\times 2\times 1\times 1=(n+1)!$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(6)(5)(4)a_4a_3a_2=\frac{(n+1)!}{2\times 3}\times 3\times 2\times 1=(n+1)!$

So $D=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2\,(n+1)!$ 
This allows to conclude that $a_{n+8}-(n+7)a_{n+5}=0$ which terminates the recurrence.
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