${\displaystyle -\frac{2x^7-3x^6-2x^4-2x^3+8x^2-5x+1}{(x^2-3x+1)(x^2+x-1){(x-1)}^2}}$

1, 1, 2, 6, 20, 62, 178, 491, 1324, 3526, 9324, 24556, 64518, 169275, 443750, ...

${\displaystyle (x^2-3x+1)(x^2+x-1){(x-1)}^{2}F\left(x\right)+2x^7-3x^6-2x^4-2x^3+8x^2-5x+1=0}$

${\displaystyle a\left(0\right)=1}$
${\displaystyle a\left(1\right)=1}$
${\displaystyle a\left(2\right)=2}$
${\displaystyle a\left(3\right)=6}$
${\displaystyle a\left(4\right)=20}$
${\displaystyle a\left(5\right)=62}$
${\displaystyle a\left(6\right)=178}$
${\displaystyle a\left(7\right)=491}$
${\displaystyle a(n+4)=a\left(n\right)-2a(n+1)-3a(n+2)+4a(n+3)-n+2,n\ge 8}$

${\displaystyle \{\begin{array}{cc}1& n=0\text{ or }n=1\\ \frac{(-19\sqrt{5}+55){(\frac{3}{2}-\frac{\sqrt{5}}{2})}^{-n}}{20}+\frac{(19\sqrt{5}+55){(\frac{3}{2}+\frac{\sqrt{5}}{2})}^{-n}}{20}+\\ \frac{(3\sqrt{5}-15){(-\frac{\sqrt{5}}{2}-\frac{1}{2})}^{-n}}{20}+\frac{(-3\sqrt{5}-15){(\frac{\sqrt{5}}{2}-\frac{1}{2})}^{-n}}{20}-n+2& \text{otherwise}\end{array}}$