Permutations avoiding 1243, 1342, 2314, 3124, 3412, 4123

Proof tree for permutations avoiding 1243, 1342, 2314, 3124, 3412, 4123

Legend

A = Av(0132, 0231, 1203, 2013, 2301, 3012)

B = Av(012, 2301)

C = Av(021, 120, 2013, 3012)

D = Av(021, 120, 201)

E = Av(012, 021, 120)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{10}}{ F_{8}}$
${ F_{8}}=x$
${ F_{10}}={ F_{28}}+{ F_{35}}$
${ F_{28}}={\frac {-4{x}^{4}+9{x}^{3}-10{x}^{2}+5x-1}{ \left( 2x-1 \right) \left( x-1 \right) ^{4}}}$
${ F_{35}}={ F_{68}}{ F_{8}}$
${ F_{68}}={ F_{160}}+{ F_{162}}$
${ F_{160}}={\frac {-{x}^{2}-1}{ \left( x-1 \right) ^{3}}}$
${ F_{162}}={\frac {2{x}^{4}+{x}^{3}-x}{ \left( 2x-1 \right) \left( x-1 \right) ^{4}}}$

Coefficients:
$1, 1, 2, 6, 18, 47, 110, 240, 502, 1025,\ldots$

Minimal polynomial:
$ \left( 2x-1 \right) \left( x-1 \right) ^{4}F \left( x \right) -2{x}^{5}+3{x}^{4}-8{x}^{3}+10{x}^{2}-5x+1$

Generating function:
${\frac {2{x}^{5}-3{x}^{4}+8{x}^{3}-10{x}^{2}+5x-1}{ \left( 2x-1 \right) \left( x-1 \right) ^{4}}}$

Recurrence relation:
$12a \left( n \right) -6a \left( n+1 \right) -{n}^{3}+12{n}^{2}-17n+6$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =18$
$a \left( 5 \right) =47$

Closed form:
$\cases{1&$n=0$\cr -2+1/6{n}^{3}-3/2{n}^{2}+n/3+{2}^{n+1}&otherwise\cr}$