Permutations avoiding 123, 1432, 2413, 2431, 3241, 3421, 4132, 4231

Proof tree for permutations avoiding 123, 1432, 2413, 2431, 3241, 3421, 4132, 4231

Legend

A = Av(012, 0321, 1302, 1320, 2130, 2310, 3021, 3120)

B = Av(012, 021, 102, 120, 201)

C = Av(01, 10)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{18}}+{ F_{20}}$
${ F_{18}}={\frac {{x}^{3}-{x}^{2}-1}{x-1}}$
${ F_{20}}={\frac {x \left( {x}^{4}-3{x}^{3}+{x}^{2}+x+1 \right) }{ \left( x-1 \right) ^{2}}}$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 5, 7, 7, 8, 9, 10, 11,\ldots$

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

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

Recurrence relation:
$a \left( n \right) -n-2$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =7$
$a \left( 5 \right) =7$
$a \left( 6 \right) =8$

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