Permutations avoiding 1324, 1342, 2143, 2314, 2413, 3142, 3412

Proof tree for permutations avoiding 1324, 1342, 2143, 2314, 2413, 3142, 3412

Legend

A = Av(0213, 0231, 1032, 1203, 1302, 2031, 2301)

B = Av(021, 120)

╱ = Av(10)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{0}}+{ F_{19}}$
${ F_{19}}=-{\frac {x}{ \left( 2x-1 \right) \left( x-1 \right) ^{2}}}$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 6, 17, 43, 100, 220, 467, 969,\ldots$

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

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

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

Closed form:
$-1/2{n}^{2}-3/2n-1+{2}^{n+1}$