Av(1342, 1423, 1432, 2413)
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Counting Sequence
1, 1, 2, 6, 20, 71, 264, 1015, 4002, 16094, 65758, 272208, 1139182, 4811807, 20487096, ...
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Implicit Equation for the Generating Function
\(\displaystyle x^{2} F \left(x \right)^{3}-x \left(x -1\right) F \left(x \right)^{2}-F \! \left(x \right)+1 = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 71\)
\(\displaystyle a \! \left(n +6\right) = \frac{8 n \left(n +1\right) a \! \left(n \right)}{15 \left(n +7\right) \left(n +6\right)}-\frac{4 \left(7 n +12\right) \left(n +1\right) a \! \left(n +1\right)}{15 \left(n +7\right) \left(n +6\right)}+\frac{2 \left(4 n^{2}+7 n -6\right) a \! \left(n +2\right)}{15 \left(n +7\right) \left(n +6\right)}-\frac{2 \left(8 n^{2}-7 n -138\right) a \! \left(n +3\right)}{15 \left(n +7\right) \left(n +6\right)}+\frac{2 \left(11 n^{2}+77 n +114\right) a \! \left(n +4\right)}{15 \left(n +7\right) \left(n +6\right)}+\frac{\left(67 n +376\right) a \! \left(n +5\right)}{15 n +105}, \quad n \geq 6\)
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This specification was found using the strategy pack "Point Placements" and has 10 rules.

Found on July 23, 2021.

Finding the specification took 3 seconds.

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