Av(1243, 1324, 1432, 2143, 2413)
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Generating Function
\(\displaystyle \frac{\left(-2 x^{2}+2 x -1+\sqrt{-4 x +1}\right) \left(x -1\right)}{2 x^{2} \left(x^{2}-2 x +2\right)}\)
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Counting Sequence
1, 1, 2, 6, 19, 61, 200, 670, 2286, 7918, 27770, 98424, 351983, 1268541, 4602752, ...
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Implicit Equation for the Generating Function
\(\displaystyle x^{2} \left(x^{2}-2 x +2\right) F \left(x \right)^{2}+\left(x -1\right) \left(2 x^{2}-2 x +1\right) F \! \left(x \right)+\left(x -1\right)^{2} = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(n +4\right) = -\frac{\left(5+2 n \right) a \! \left(n \right)}{6+n}-\frac{\left(70+19 n \right) a \! \left(2+n \right)}{2 \left(6+n \right)}+\frac{\left(42+13 n \right) a \! \left(n +1\right)}{12+2 n}+\frac{2 \left(14+3 n \right) a \! \left(n +3\right)}{6+n}, \quad n \geq 4\)
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This specification was found using the strategy pack "Point Placements" and has 15 rules.

Found on July 23, 2021.

Finding the specification took 3 seconds.

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