Av(1342, 1423, 1432, 2413, 2431, 3142, 4132)
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Generating Function
\(\displaystyle \frac{2 x^{3}-\sqrt{-4 x +1}\, x +\sqrt{-4 x +1}+x -1}{2 \left(x^{2}+x -1\right) x}\)
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Counting Sequence
1, 1, 2, 6, 17, 51, 158, 506, 1665, 5603, 19202, 66795, 235223, 836906, 3003669, ...
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Implicit Equation for the Generating Function
\(\displaystyle \left(x^{2}+x -1\right)^{2} x F \left(x \right)^{2}-\left(x^{2}+x -1\right) \left(2 x^{3}+x -1\right) F \! \left(x \right)+x^{5}+x^{3}-2 x +1 = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 17\)
\(\displaystyle a \! \left(n +4\right) = \frac{2 \left(3+2 n \right) a \! \left(n \right)}{n +5}-\frac{\left(11+n \right) a \! \left(1+n \right)}{n +5}-\frac{2 \left(9+4 n \right) a \! \left(n +2\right)}{n +5}+\frac{2 \left(11+3 n \right) a \! \left(n +3\right)}{n +5}, \quad n \geq 5\)
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This specification was found using the strategy pack "Point Placements" and has 13 rules.

Found on July 23, 2021.

Finding the specification took 1 seconds.

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