Av(1243, 1324, 1342, 4132)
Generating Function
\(\displaystyle \frac{\left(-x^{4}+6 x^{3}-7 x^{2}+4 x -1\right) \sqrt{1-4 x}+7 x^{4}-18 x^{3}+15 x^{2}-6 x +1}{2 x^{2} \left(x -1\right)^{2}}\)
Counting Sequence
1, 1, 2, 6, 20, 68, 232, 796, 2755, 9630, 33990, 121048, 434546, 1571048, 5715649, ...
Implicit Equation for the Generating Function
\(\displaystyle x^{2} \left(x -1\right)^{4} F \left(x
\right)^{2}-\left(7 x^{4}-18 x^{3}+15 x^{2}-6 x +1\right) \left(x -1\right)^{2} F \! \left(x \right)+x^{7}-10 x^{5}+29 x^{4}-34 x^{3}+21 x^{2}-7 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 68\)
\(\displaystyle a \! \left(6\right) = 232\)
\(\displaystyle a \! \left(7\right) = 796\)
\(\displaystyle a \! \left(n +6\right) = -\frac{2 \left(2 n -1\right) a \! \left(n \right)}{8+n}-\frac{\left(59 n +128\right) a \! \left(2+n \right)}{8+n}+\frac{\left(31+29 n \right) a \! \left(n +1\right)}{8+n}+\frac{\left(205+57 n \right) a \! \left(n +3\right)}{8+n}-\frac{\left(31 n +158\right) a \! \left(n +4\right)}{8+n}+\frac{\left(59+9 n \right) a \! \left(n +5\right)}{8+n}+\frac{4}{8+n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 68\)
\(\displaystyle a \! \left(6\right) = 232\)
\(\displaystyle a \! \left(7\right) = 796\)
\(\displaystyle a \! \left(n +6\right) = -\frac{2 \left(2 n -1\right) a \! \left(n \right)}{8+n}-\frac{\left(59 n +128\right) a \! \left(2+n \right)}{8+n}+\frac{\left(31+29 n \right) a \! \left(n +1\right)}{8+n}+\frac{\left(205+57 n \right) a \! \left(n +3\right)}{8+n}-\frac{\left(31 n +158\right) a \! \left(n +4\right)}{8+n}+\frac{\left(59+9 n \right) a \! \left(n +5\right)}{8+n}+\frac{4}{8+n}, \quad n \geq 8\)
This specification was found using the strategy pack "Row Placements" and has 21 rules.
Found on July 23, 2021.Finding the specification took 3 seconds.
Copy 21 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{15}\! \left(x \right)+F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{10}\! \left(x \right) F_{6}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{6} \left(x \right)^{2} F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= x\\
F_{10}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)+F_{4}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right) F_{6}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)+F_{14}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{13}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{15}\! \left(x \right)+F_{17}\! \left(x \right)+F_{19}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{18}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)+F_{17}\! \left(x \right)+F_{19}\! \left(x \right)\\
F_{19}\! \left(x \right) &= F_{20}\! \left(x \right)\\
F_{20}\! \left(x \right) &= F_{6} \left(x \right)^{2} F_{13}\! \left(x \right) F_{9}\! \left(x \right)\\
\end{align*}\)