Permutations avoiding 123, 1432, 2143, 2413, 3214, 3412, 4231, 4321

Proof tree for permutations avoiding 123, 1432, 2143, 2413, 3214, 3412, 4231, 4321

Legend

A = Av(012, 0321, 1032, 1302, 2103, 2301, 3120, 3210)

B = Av(01, 210)

C = Av(012, 201, 210, 1032)

D = Av(10, 012)

E = Av(01, 10)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{12}}+{ F_{14}}$
${ F_{12}}={x}^{2}+x+1$
${ F_{14}}=7{x}^{3}+4{x}^{2}+x$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 5, 7, 0, 0, 0, 0, 0,\ldots$

Minimal polynomial:
$-7{x}^{4}-5{x}^{3}-2{x}^{2}+F \left( x \right) -x-1$

Generating function:
$7{x}^{4}+5{x}^{3}+2{x}^{2}+x+1$

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

Closed form:
$\cases{1&$n=0$\cr 1&$n=1$\cr 2&$n=2$\cr 5&$n=3$\cr 7&$n=4$\cr 0&otherwise\cr}$