Permutations avoiding 1243, 1324, 1432, 2143, 2413, 4132

Proof tree for permutations avoiding 1243, 1324, 1432, 2143, 2413, 4132

Legend

A = Av(0132, 0213, 0321, 1032, 1302, 3021)

B = Av(021)

╱ = Av(10)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{0}}+{ F_{19}}$
${ F_{19}}={ F_{59}}{ F_{8}}$
${ F_{8}}=x$
${ F_{59}}={ F_{293}}+{ F_{295}}$
${ F_{293}}={ F_{312}}+{ F_{7}}$
${ F_{312}}={\frac {\sqrt {-4x+1}-1}{2x-2}}$
${ F_{295}}=-1/8{\frac { \left( \sqrt {-4x+1}-1 \right) ^{3}{ F_{0}}}{{x}^{2}}}$

Coefficients:
$1, 1, 2, 6, 18, 54, 166, 527, 1724, 5781,\ldots$

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

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

Recurrence relation:
$ \left( -4-8n \right) a \left( n \right) + \left( 20+18n \right) a \left( 1+n \right) + \left( -40-16n \right) a \left( n+2 \right) + \left( 26+7n \right) a \left( n+3 \right) + \left( -5-n \right) a \left( n+4 \right) +2$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =18$
$a \left( 5 \right) =54$

Closed form:
N/A