Permutations avoiding 132, 1234, 2314, 3214, 4123, 4231

Proof tree for permutations avoiding 132, 1234, 2314, 3214, 4123, 4231

Legend

A = Av(021, 0123, 1203, 2103, 3012, 3120)

B = Av(021, 201, 0123, 1203, 2103)

C = Av(012, 021, 120, 210)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}=-{\frac {x \left( {x}^{6}+3{x}^{5}+2{x}^{4}+4{x}^{3}+3{x}^{2}+x+1 \right) }{x-1}}$

Coefficients:
$1, 1, 2, 5, 9, 11, 14, 15, 15, 15,\ldots$

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

Generating function:
$-{\frac {{x}^{7}+3{x}^{6}+2{x}^{5}+4{x}^{4}+3{x}^{3}+{x}^{2}+1}{x-1}}$

Recurrence relation:
$-a \left( n \right) +15$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =9$
$a \left( 5 \right) =11$
$a \left( 6 \right) =14$
$a \left( 7 \right) =15$

Closed form:
$\cases{1&$n=0$\cr 1&$n=1$\cr 2&$n=2$\cr 5&$n=3$\cr 9&$n=4$\cr 11&$n=5$\cr 14&$n=6$\cr 15&otherwise\cr}$