Permutations avoiding 132, 567438219

Proof tree for permutations avoiding 132, 567438219

Legend

A = Av(120, 852103467)

B = Av(120, 52103467)

Equations:
${ F_{0}}={ F_{1}}+{ F_{3}}$
${ F_{3}}={ F_{0}}{ F_{12}}{ F_{8}}$
${ F_{12}}={\frac {-5{x}^{12}+66{x}^{11}-396{x}^{10}+1384{x}^{9}-3123{x}^{8}+4771{x}^{7}-5049{x}^{6}+3719{x}^{5}-1890{x}^{4}+647{x}^{3}-142{x}^{2}+18x-1}{ \left( x-1 \right) \left( {x}^{6}-11{x}^{5}+33{x}^{4}-46{x}^{3}+30{x}^{2}-9x+1 \right) ^{2}}}$
${ F_{8}}=x$
${ F_{1}}=1$

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

Minimal polynomial:
$ \left( 2x-1 \right) \left( 3{x}^{12}-43{x}^{11}+281{x}^{10}-1054{x}^{9}+2524{x}^{8}-4061{x}^{7}+4499{x}^{6}-3446{x}^{5}+1807{x}^{4}-633{x}^{3}+141{x}^{2}-18x+1 \right) F \left( x \right) - \left( x-1 \right) \left( {x}^{6}-11{x}^{5}+33{x}^{4}-46{x}^{3}+30{x}^{2}-9x+1 \right) ^{2}$

Generating function:
${\frac { \left( x-1 \right) \left( {x}^{6}-11{x}^{5}+33{x}^{4}-46{x}^{3}+30{x}^{2}-9x+1 \right) ^{2}}{ \left( 2x-1 \right) \left( 3{x}^{12}-43{x}^{11}+281{x}^{10}-1054{x}^{9}+2524{x}^{8}-4061{x}^{7}+4499{x}^{6}-3446{x}^{5}+1807{x}^{4}-633{x}^{3}+141{x}^{2}-18x+1 \right) }}$

Recurrence relation:
$6a \left( n \right) -89a \left( n+1 \right) +605a \left( n+2 \right) -2389a \left( n+3 \right) +6102a \left( n+4 \right) -10646a \left( n+5 \right) +13059a \left( n+6 \right) -11391a \left( n+7 \right) +7060a \left( n+8 \right) -3073a \left( n+9 \right) +915a \left( n+10 \right) -177a \left( n+11 \right) +20a \left( n+12 \right) -a \left( n+13 \right) $
$a \left( 0 \right) =1$
$a \left( 1 \right) =1$
$a \left( 2 \right) =2$
$a \left( 3 \right) =5$
$a \left( 4 \right) =14$
$a \left( 5 \right) =42$
$a \left( 6 \right) =132$
$a \left( 7 \right) =429$
$a \left( 8 \right) =1430$
$a \left( 9 \right) =4861$
$a \left( 10 \right) =16776$
$a \left( 11 \right) =58561$
$a \left( 12 \right) =206126$
$a \left( 13 \right) =729736$

Closed form:
N/A