Av(1243, 1342, 1423, 2143, 2413)
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Generating Function
\(\displaystyle \frac{-x +1-\sqrt{4 x^{4}-8 x^{3}+9 x^{2}-6 x +1}}{2 x \left(x^{2}-x +1\right)}\)
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Counting Sequence
1, 1, 2, 6, 19, 62, 210, 734, 2628, 9589, 35530, 133335, 505753, 1935925, 7468725, ...
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Implicit Equation for the Generating Function
\(\displaystyle x \left(x^{2}-x +1\right) F \left(x \right)^{2}+\left(x -1\right) F \! \left(x \right)-x +1 = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 62\)
\(\displaystyle a \! \left(n +6\right) = -\frac{4 \left(1+n \right) a \! \left(n \right)}{7+n}+\frac{12 \left(n +2\right) a \! \left(1+n \right)}{7+n}-\frac{3 \left(20+7 n \right) a \! \left(n +2\right)}{7+n}+\frac{\left(89+23 n \right) a \! \left(n +3\right)}{7+n}-\frac{4 \left(19+4 n \right) a \! \left(n +4\right)}{7+n}+\frac{\left(40+7 n \right) a \! \left(n +5\right)}{7+n}, \quad n \geq 6\)
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This specification was found using the strategy pack "Point Placements" and has 12 rules.

Found on July 23, 2021.

Finding the specification took 3 seconds.

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