Permutations avoiding 1234, 1243, 2314, 3124, 3412

Proof tree for permutations avoiding 1234, 1243, 2314, 3124, 3412

Legend

A = Av(0123, 0132, 1203, 2013, 2301)

B = Av(012, 021, 2301)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{10}}{ F_{8}}$
${ F_{8}}=x$
${ F_{10}}={ F_{0}}+{ F_{34}}$
${ F_{34}}={ F_{69}}{ F_{8}}$
${ F_{69}}={ F_{467}}+{ F_{469}}$
${ F_{467}}={\frac {{x}^{2}-x+1}{ \left( x-1 \right) ^{4}}}$
${ F_{469}}={\frac {x \left( {x}^{2}-x+1 \right) }{ \left( x-1 \right) ^{5} \left( 2x-1 \right) }}$

Coefficients:
$1, 1, 2, 6, 19, 56, 152, 385, 923, 2119,\ldots$

Minimal polynomial:
$ \left( 2x-1 \right) \left( x-1 \right) ^{6}F \left( x \right) +4{x}^{6}-15{x}^{5}+30{x}^{4}-33{x}^{3}+21{x}^{2}-7x+1$

Generating function:
$-{\frac {4{x}^{6}-15{x}^{5}+30{x}^{4}-33{x}^{3}+21{x}^{2}-7x+1}{ \left( 2x-1 \right) \left( x-1 \right) ^{6}}}$

Recurrence relation:
$240a \left( n \right) -120a \left( n+1 \right) +{n}^{5}-5{n}^{4}+45{n}^{3}-55{n}^{2}+134n-120$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =19$
$a \left( 5 \right) =56$
$a \left( 6 \right) =152$

Closed form:
$-5+6{2}^{n}-{\frac {11{n}^{3}}{24}}-{n}^{2}-{\frac {68n}{15}}-{\frac {{n}^{5}}{120}}$