Permutations avoiding 1243, 1324, 1342, 1423, 1432, 2143, 2413, 2431, 3142, 4231

Proof tree for permutations avoiding 1243, 1324, 1342, 1423, 1432, 2143, 2413, 2431, 3142, 4231

Legend

A = Av(0132, 0213, 0231, 0312, 0321, 1032, 1302, 1320, 2031, 3120)

B = Av(021, 3120)

C = Av(021, 201)

D = Av(01, 10)

E = Av(021, 120)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{7}}{ F_{8}}$
${ F_{7}}={ F_{12}}+{ F_{14}}$
${ F_{12}}={\frac {{x}^{3}-5{x}^{2}+4x-1}{4{x}^{3}-8{x}^{2}+5x-1}}$
${ F_{14}}={\frac {2{x}^{4}-3{x}^{3}+x}{ \left( 2x-1 \right) ^{2}}}$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 6, 14, 35, 85, 201, 465, 1057,\ldots$

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

Generating function:
${\frac {2{x}^{6}-5{x}^{5}+4{x}^{4}-5{x}^{2}+4x-1}{ \left( x-1 \right) \left( 2x-1 \right) ^{2}}}$

Recurrence relation:
$-4a \left( n \right) +4a \left( n+1 \right) -a \left( n+2 \right) +1$
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =6$
$a \left( 4 \right) =14$
$a \left( 5 \right) =35$
$a \left( 6 \right) =85$

Closed form:
$\cases{1&$n=0$\cr 1&$n=1$\cr 2&$n=2$\cr 6&$n=3$\cr 1+1/16 \left( 4n-3 \right) {2}^{n}&otherwise\cr}$