Permutations avoiding 1243, 1342, 1423, 2134, 2341

Proof tree for permutations avoiding 1243, 1342, 1423, 2134, 2341

Legend

A = Av(0132, 0231, 0312, 1023, 1230)

B = Av(012)

C = Av(102, 0132, 0231, 0312, 1230)

╱ = Av(10)

╲ = Av(01)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{438}}+{ F_{6370}}$
${ F_{438}}={ F_{3142}}{ F_{6370}}{ F_{992}}$
${ F_{992}}=x$
${ F_{6370}}=1/2{\frac {-2x-\sqrt {-4x+1}+1}{x}}$
${ F_{3142}}={\frac { \left( {x}^{2}-x-1 \right) x}{2x-1}}$

Coefficients:
$1, 1, 2, 6, 19, 58, 181, 578, 1888, 6297,\ldots$

Minimal polynomial:
$x \left( 2x-1 \right) ^{2} \left( F \left( x \right) \right) ^{2}+ \left( x+1 \right) \left( {x}^{3}-2{x}^{2}+x-1 \right) \left( 2x-1 \right) ^{2}F \left( x \right) +{x}^{9}-2{x}^{8}-{x}^{7}+2{x}^{6}+3{x}^{5}-3{x}^{4}-{x}^{3}+5{x}^{2}-4x+1$

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

Recurrence relation:
$ \left( -20+8n \right) a \left( n \right) + \left( 16-14n \right) a \left( 1+n \right) + \left( -13-n \right) a \left( n+2 \right) + \left( 67+21n \right) a \left( n+3 \right) + \left( -95-21n \right) a \left( n+4 \right) + \left( 46+8n \right) a \left( n+5 \right) + \left( -7-n \right) a \left( n+6 \right) $
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =19$
$a \left( 5 \right) =58$
$a \left( 6 \right) =181$
$a \left( 7 \right) =578$
$a \left( 8 \right) =1888$
$a \left( 9 \right) =6297$

Closed form:
N/A