Av(1243, 1324, 1342, 1423, 1432, 2143, 2413, 3142)
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Generating Function
\(\displaystyle \frac{-2 x^{3}+1-\sqrt{-4 x +1}}{2 x \left(x^{5}-x^{2}+1\right)}\)
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Counting Sequence
1, 1, 2, 6, 16, 47, 147, 474, 1571, 5320, 18320, 63959, 225858, 805288, 2894978, ...
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Implicit Equation for the Generating Function
\(\displaystyle x \left(x^{5}-x^{2}+1\right) F \left(x \right)^{2}+\left(2 x^{3}-1\right) F \! \left(x \right)+1 = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 16\)
\(\displaystyle a \! \left(5\right) = 47\)
\(\displaystyle a \! \left(1+n \right) = \frac{2 \left(11+2 n \right) a \! \left(n \right)}{7+n}-\frac{2 \left(11+2 n \right) a \! \left(n +3\right)}{7+n}+a \! \left(n +4\right)+\frac{2 \left(11+2 n \right) a \! \left(n +5\right)}{7+n}-a \! \left(n +6\right), \quad n \geq 6\)
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This specification was found using the strategy pack "Point Placements" and has 16 rules.

Found on July 23, 2021.

Finding the specification took 1 seconds.

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\(\begin{align*} F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{1}\! \left(x \right) &= 1\\ F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\ F_{3}\! \left(x \right) &= F_{10}\! \left(x \right) F_{4}\! \left(x \right)\\ F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\ F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\ F_{6}\! \left(x \right) &= F_{0}\! \left(x \right) F_{10}\! \left(x \right) F_{7}\! \left(x \right)\\ F_{7}\! \left(x \right) &= F_{15}\! \left(x \right)+F_{8}\! \left(x \right)\\ F_{8}\! \left(x \right) &= F_{11}\! \left(x \right)+F_{9}\! \left(x \right)\\ F_{9}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{10}\! \left(x \right)\\ F_{10}\! \left(x \right) &= x\\ F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\ F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\ F_{13}\! \left(x \right) &= F_{14} \left(x \right)^{2} F_{10}\! \left(x \right)\\ F_{14}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{12}\! \left(x \right)\\ F_{15}\! \left(x \right) &= F_{11}\! \left(x \right) F_{14}\! \left(x \right)\\ \end{align*}\)