Av(1342, 2314, 2413, 3241)
Generating Function
\(\displaystyle \frac{\left(2 x^{3}+x^{2}-3 x +1\right) \sqrt{1-4 x}+2 x^{4}-6 x^{3}+x^{2}+3 x -1}{2 x \left(x^{2}-3 x +1\right) \left(x -1\right)}\)
Counting Sequence
1, 1, 2, 6, 20, 67, 224, 751, 2534, 8621, 29597, 102558, 358639, 1265163, 4499978, ...
Implicit Equation for the Generating Function
\(\displaystyle x \left(x -1\right)^{2} \left(x^{2}-3 x +1\right)^{2} F \left(x
\right)^{2}-\left(x -1\right) \left(2 x^{2}-1\right) \left(x^{2}-3 x +1\right)^{2} F \! \left(x \right)+x^{7}-2 x^{6}+13 x^{5}-12 x^{4}-9 x^{3}+16 x^{2}-7 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 67\)
\(\displaystyle a \! \left(6\right) = 224\)
\(\displaystyle a \! \left(7\right) = 751\)
\(\displaystyle a \! \left(n +7\right) = -\frac{4 \left(1+2 n \right) a \! \left(n \right)}{8+n}+\frac{10 \left(1+3 n \right) a \! \left(1+n \right)}{8+n}-\frac{11 \left(n -6\right) a \! \left(n +2\right)}{8+n}-\frac{\left(296+59 n \right) a \! \left(n +3\right)}{8+n}+\frac{\left(422+83 n \right) a \! \left(n +4\right)}{8+n}-\frac{\left(266+45 n \right) a \! \left(n +5\right)}{8+n}+\frac{\left(76+11 n \right) a \! \left(n +6\right)}{8+n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 67\)
\(\displaystyle a \! \left(6\right) = 224\)
\(\displaystyle a \! \left(7\right) = 751\)
\(\displaystyle a \! \left(n +7\right) = -\frac{4 \left(1+2 n \right) a \! \left(n \right)}{8+n}+\frac{10 \left(1+3 n \right) a \! \left(1+n \right)}{8+n}-\frac{11 \left(n -6\right) a \! \left(n +2\right)}{8+n}-\frac{\left(296+59 n \right) a \! \left(n +3\right)}{8+n}+\frac{\left(422+83 n \right) a \! \left(n +4\right)}{8+n}-\frac{\left(266+45 n \right) a \! \left(n +5\right)}{8+n}+\frac{\left(76+11 n \right) a \! \left(n +6\right)}{8+n}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 24 rules.
Found on July 23, 2021.Finding the specification took 2 seconds.
Copy 24 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{7}\! \left(x \right) &= x\\
F_{8}\! \left(x \right) &= F_{13}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{10} \left(x \right)^{3}\\
F_{10}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{10} \left(x \right)^{2} F_{7}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{22}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right) F_{17}\! \left(x \right) F_{2}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{15}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{18}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{19}\! \left(x \right)\\
F_{19}\! \left(x \right) &= F_{17}\! \left(x \right) F_{20}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{20}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{21}\! \left(x \right)\\
F_{21}\! \left(x \right) &= F_{20}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{22}\! \left(x \right) &= F_{23} \left(x \right)^{2} F_{20}\! \left(x \right)\\
F_{23}\! \left(x \right) &= F_{11}\! \left(x \right)\\
\end{align*}\)