Permutations avoiding 132, 2134, 2341, 3412, 3421, 4123, 4312, 4321

Proof tree for permutations avoiding 132, 2134, 2341, 3412, 3421, 4123, 4312, 4321

Legend

A = Av(021, 1023, 1230, 2301, 2310, 3012, 3201, 3210)

B = Av(012, 021, 120, 210)

C = Av(021, 102, 1230, 2301, 2310, 3012, 3201, 3210)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={\frac {x \left( {x}^{5}+5{x}^{4}-2{x}^{3}-3{x}^{2}-x-1 \right) }{x-1}}$

Coefficients:
$1, 1, 2, 5, 7, 2, 1, 1, 1, 1,\ldots$

Minimal polynomial:
$ \left( x-1 \right) F \left( x \right) -{x}^{6}-5{x}^{5}+2{x}^{4}+3{x}^{3}+{x}^{2}+1$

Generating function:
${\frac {{x}^{6}+5{x}^{5}-2{x}^{4}-3{x}^{3}-{x}^{2}-1}{x-1}}$

Recurrence relation:
$-a \left( n \right) +1$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =7$
$a \left( 5 \right) =2$
$a \left( 6 \right) =1$

Closed form:
$\cases{2&$n=2$\cr 5&$n=3$\cr 7&$n=4$\cr 2&$n=5$\cr 1&otherwise\cr}$