Permutations avoiding 132, 563247819

Proof tree for permutations avoiding 132, 563247819

Legend

A = Av(120, 865104237)

B = Av(120, 65104237)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{1}}=1$
${ F_{3}}={ F_{0}}{ F_{12}}{ F_{8}}$
${ F_{12}}=-{\frac {4{x}^{3}-10{x}^{2}+6x-1}{{x}^{4}-10{x}^{3}+15{x}^{2}-7x+1}}$
${ F_{8}}=x$

Coefficients:
$1, 1, 2, 5, 14, 42, 132, 429, 1430, 4861,\ldots$

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

Generating function:
${\frac { \left( x-1 \right) \left( {x}^{3}-9{x}^{2}+6x-1 \right) }{ \left( {x}^{2}-3x+1 \right) \left( 5{x}^{2}-5x+1 \right) }}$

Recurrence relation:
$5a \left( n \right) -20a \left( n+1 \right) +21a \left( n+2 \right) -8a \left( n+3 \right) +a \left( n+4 \right) $
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =14$

Closed form:
$\cases{1&$n=0$\cr {\frac { \left( 5+\sqrt {5} \right) \left( \left( 1/2+1/10\sqrt {5} \right) ^{-n}\sqrt {5}-5 \left( 3/2-1/2\sqrt {5} \right) ^{-n}\sqrt {5}-4 \left( 1/2-1/10\sqrt {5} \right) ^{-n}\sqrt {5}+10 \left( 3/2+1/2\sqrt {5} \right) ^{-n}+5 \left( 1/2+1/10\sqrt {5} \right) ^{-n}+15 \left( 3/2-1/2\sqrt {5} \right) ^{-n}+10 \left( 1/2-1/10\sqrt {5} \right) ^{-n} \right) }{200}}&otherwise\cr}$