Permutations avoiding 1243, 1324, 1342, 1423, 1432, 2143, 2413

Proof tree for permutations avoiding 1243, 1324, 1342, 1423, 1432, 2143, 2413

Legend

A = Av(0132, 0213, 0231, 0312, 0321, 1032, 1302)

B = Av(021)

C = Av(01, 10)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{8}}{ F_{9}}$
${ F_{8}}=x$
${ F_{9}}={ F_{0}}+{ F_{29}}$
${ F_{29}}=1/4{\frac { \left( -2{x}^{2}+\sqrt {-4x+1}-1 \right) \left( \sqrt {-4x+1}-1 \right) { F_{0}}}{x}}$

Coefficients:
$1, 1, 2, 6, 17, 51, 161, 524, 1747, 5939,\ldots$

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

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

Recurrence relation:
$ \left( 10+4n \right) a \left( n \right) + \left( -4-n \right) a \left( 1+n \right) + \left( 46+12n \right) a \left( n+2 \right) + \left( -26-7n \right) a \left( n+3 \right) + \left( 66+13n \right) a \left( n+4 \right) + \left( -46-7n \right) a \left( n+5 \right) + \left( 34+5n \right) a \left( n+6 \right) + \left( -8-n \right) a \left( n+7 \right) $
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =17$
$a \left( 5 \right) =51$
$a \left( 6 \right) =161$

Closed form:
N/A