Permutations avoiding 123, 1432, 2413, 2431, 3142, 3214, 3412, 3421, 4213, 4231

Proof tree for permutations avoiding 123, 1432, 2413, 2431, 3142, 3214, 3412, 3421, 4213, 4231

Legend

A = Av(012, 0321, 1302, 1320, 2031, 2103, 2301, 2310, 3102, 3120)

B = Av(01, 210)

C = Av(012, 021, 120, 201, 2103)

D = Av(01, 10)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{12}}+{ F_{14}}$
${ F_{12}}={x}^{2}+x+1$
${ F_{14}}={\frac {x \left( 2{x}^{3}-{x}^{2}-3x-1 \right) }{x-1}}$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 5, 5, 3, 3, 3, 3, 3,\ldots$

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

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

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

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