Permutations avoiding 132, 2314, 2341, 3124, 3214, 4312, 4321

Proof tree for permutations avoiding 132, 2314, 2341, 3124, 3214, 4312, 4321

Legend

A = Av(021, 1203, 1230, 2013, 2103, 3201, 3210)

B = Av(012, 021, 210)

C = Av(021, 120, 201, 210)

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

Coefficients:
$1, 1, 2, 5, 8, 7, 6, 6, 6, 6,\ldots$

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

Generating function:
${\frac {{x}^{6}+{x}^{5}-3{x}^{4}-3{x}^{3}-{x}^{2}-1}{x-1}}$

Recurrence relation:
$-a \left( n \right) +6$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =8$
$a \left( 5 \right) =7$
$a \left( 6 \right) =6$

Closed form:
$\cases{1&$n=0$\cr 1&$n=1$\cr 2&$n=2$\cr 5&$n=3$\cr 8&$n=4$\cr 7&$n=5$\cr 6&otherwise\cr}$