Av(1324, 1342, 1432, 2431)
Generating Function
\(\displaystyle \frac{-4 \left(x -\frac{1}{2}\right)^{2} \sqrt{-4 x +1}-2 x^{3}+10 x^{2}-6 x +1}{2 x^{3}}\)
Counting Sequence
1, 1, 2, 6, 20, 69, 242, 858, 3068, 11050, 40052, 145996, 534888, 1968685, 7276050, ...
Implicit Equation for the Generating Function
\(\displaystyle x^{3} F \left(x
\right)^{2}+\left(2 x^{3}-10 x^{2}+6 x -1\right) F \! \left(x \right)+x^{3}+6 x^{2}-5 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(n +2\right) = -\frac{4 \left(1+2 n \right) a \! \left(n \right)}{5+n}+\frac{6 \left(3+n \right) a \! \left(n +1\right)}{5+n}, \quad n \geq 3\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(n +2\right) = -\frac{4 \left(1+2 n \right) a \! \left(n \right)}{5+n}+\frac{6 \left(3+n \right) a \! \left(n +1\right)}{5+n}, \quad n \geq 3\)
This specification was found using the strategy pack "Point Placements" and has 17 rules.
Found on July 23, 2021.Finding the specification took 4 seconds.
Copy 17 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{5}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{5} \left(x \right)^{2} F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= x\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{5} \left(x \right)^{2} F_{11}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{11}\! \left(x \right) F_{14}\! \left(x \right) F_{5}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{14}\! \left(x \right) F_{8}\! \left(x \right)\\
\end{align*}\)