Permutations avoiding 123, 2413, 2431, 3421, 4231, 4321

Proof tree for permutations avoiding 123, 2413, 2431, 3421, 4231, 4321

Legend

A = Av(012, 1302, 1320, 2310, 3120, 3210)

B = Av(01, 3210)

C = Av(012, 120, 210)

D = Av(01, 210)

E = Av(10, 012)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{8}}{ F_{9}}$
${ F_{8}}=x$
${ F_{9}}={ F_{12}}+{ F_{23}}$
${ F_{12}}={x}^{3}+{x}^{2}+x+1$
${ F_{23}}=x \left( x+1 \right) \left( {x}^{3}+5{x}^{2}+3x+1 \right) $

Coefficients:
$1, 1, 2, 5, 9, 6, 1, 0, 0, 0,\ldots$

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

Generating function:
${x}^{6}+6{x}^{5}+9{x}^{4}+5{x}^{3}+2{x}^{2}+x+1$

Recurrence relation:
$a \left( n \right) $
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =9$
$a \left( 5 \right) =6$
$a \left( 6 \right) =1$

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