Permutations avoiding 1234, 1243, 1342, 2314, 2341, 4123

Proof tree for permutations avoiding 1234, 1243, 1342, 2314, 2341, 4123

Legend

A = Av(0123, 0132, 0231, 1203, 1230, 3012)

B = Av(012)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{402}}+{ F_{405}}$
${ F_{402}}=1/2{\frac {-\sqrt {-4x+1}+1}{x}}$
${ F_{405}}={\frac {{x}^{3}}{ \left( x-1 \right) ^{4}}}$

Coefficients:
$1, 1, 2, 6, 18, 52, 152, 464, 1486, 4946,\ldots$

Minimal polynomial:
$x \left( x-1 \right) ^{8} \left( F \left( x \right) \right) ^{2}- \left( 3{x}^{4}-4{x}^{3}+6{x}^{2}-4x+1 \right) \left( x-1 \right) ^{4}F \left( x \right) +{x}^{8}-6{x}^{7}+24{x}^{6}-50{x}^{5}+66{x}^{4}-55{x}^{3}+28{x}^{2}-8x+1$

Generating function:
$1/2{\frac {3{x}^{4}-4{x}^{3}+6{x}^{2}-4x-\sqrt {- \left( 4x-1 \right) \left( x-1 \right) ^{8}}+1}{x \left( x-1 \right) ^{4}}}$

Recurrence relation:
$ \left( -8n-4 \right) a \left( n \right) + \left( 2n+4 \right) a \left( 1+n \right) +{n}^{4}-4{n}^{3}+{n}^{2}+2n$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =18$
$a \left( 5 \right) =52$

Closed form:
N/A