###### Av(123, 132, 321)
Generating Function
$$\displaystyle \left(1+x \right) \left(x^{3}+2 x^{2}+1\right)$$
Counting Sequence
1, 1, 2, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...
Implicit Equation for the Generating Function
$$\displaystyle -F \! \left(x \right)+\left(1+x \right) \left(x^{3}+2 x^{2}+1\right) = 0$$
Recurrence
$$\displaystyle a \! \left(0\right) = 1$$
$$\displaystyle a \! \left(1\right) = 1$$
$$\displaystyle a \! \left(2\right) = 2$$
$$\displaystyle a \! \left(3\right) = 3$$
$$\displaystyle a \! \left(4\right) = 1$$
$$\displaystyle a \! \left(n \right) = 0, \quad n \geq 5$$
Explicit Closed Form
$$\displaystyle \left\{\begin{array}{cc}1 & n =0\text{ or } n =1 \\ 2 & n =2 \\ 3 & n =3 \\ 1 & n =4 \\ 0 & \text{otherwise} \end{array}\right.$$

### This specification was found using the strategy pack "Point Placements" and has 15 rules.

Found on January 17, 2022.

Finding the specification took 1 seconds.

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\begin{align*} F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{1}\! \left(x \right) &= 1\\ F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\ F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{5}\! \left(x \right)\\ F_{4}\! \left(x \right) &= x\\ F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)+F_{7}\! \left(x \right)\\ F_{6}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{4}\! \left(x \right)\\ F_{7}\! \left(x \right) &= F_{10}\! \left(x \right)+F_{8}\! \left(x \right)\\ F_{8}\! \left(x \right) &= F_{9}\! \left(x \right)\\ F_{9}\! \left(x \right) &= F_{4}\! \left(x \right) F_{6}\! \left(x \right)\\ F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)+F_{12}\! \left(x \right)+F_{14}\! \left(x \right)\\ F_{11}\! \left(x \right) &= 0\\ F_{12}\! \left(x \right) &= F_{13}\! \left(x \right) F_{4}\! \left(x \right)\\ F_{13}\! \left(x \right) &= F_{4}\! \left(x \right)\\ F_{14}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\ \end{align*}