Av(1243, 1324, 1342, 2143)
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Generating Function
\(\displaystyle \frac{\left(x -1\right) \sqrt{-4 x +1}-2 x^{2}+5 x -1}{2 x^{3}-8 x^{2}+10 x -2}\)
Counting Sequence
1, 1, 2, 6, 20, 69, 243, 869, 3145, 11491, 42312, 156807, 584288, 2187298, 8221257, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(x^{3}-4 x^{2}+5 x -1\right) F \left(x \right)^{2}+\left(2 x^{2}-5 x +1\right) F \! \left(x \right)+x = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(n +5\right) = \frac{2 \left(3+2 n \right) a \! \left(n \right)}{n +5}-\frac{\left(21 n +41\right) a \! \left(n +1\right)}{n +5}+\frac{\left(103+41 n \right) a \! \left(n +2\right)}{n +5}-\frac{3 \left(11 n +37\right) a \! \left(n +3\right)}{n +5}+\frac{2 \left(5 n +21\right) a \! \left(n +4\right)}{n +5}, \quad n \geq 5\)

This specification was found using the strategy pack "Point Placements" and has 14 rules.

Found on July 23, 2021.

Finding the specification took 4 seconds.

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