Av(1243, 1342, 1423, 2143, 2413, 3142)
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Generating Function
\(\displaystyle \frac{x^{3}+x -1+\sqrt{x^{6}+2 x^{4}-6 x^{3}+9 x^{2}-6 x +1}}{2 \left(x -1\right) x}\)
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Counting Sequence
1, 1, 2, 6, 18, 56, 182, 612, 2112, 7438, 26626, 96598, 354388, 1312486, 4900384, ...
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Implicit Equation for the Generating Function
\(\displaystyle -\left(x -1\right) x F \left(x \right)^{2}+\left(x^{3}+x -1\right) F \! \left(x \right)-x +1 = 0\)
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Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 18\)
\(\displaystyle a \! \left(5\right) = 56\)
\(\displaystyle a \! \left(6\right) = 182\)
\(\displaystyle a \! \left(n +7\right) = \frac{\left(n -1\right) a \! \left(n \right)}{n +8}-\frac{\left(n -1\right) a \! \left(1+n \right)}{n +8}+\frac{2 \left(n +2\right) a \! \left(n +2\right)}{n +8}-\frac{\left(25+8 n \right) a \! \left(n +3\right)}{n +8}+\frac{3 \left(22+5 n \right) a \! \left(n +4\right)}{n +8}-\frac{3 \left(28+5 n \right) a \! \left(n +5\right)}{n +8}+\frac{\left(47+7 n \right) a \! \left(n +6\right)}{n +8}, \quad n \geq 7\)
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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 2 seconds.

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