Av(1243, 1342, 1423, 2431)
Generating Function
\(\displaystyle \frac{-\left(x^{2}-x +1\right) \left(x -1\right)^{2} \sqrt{1-4 x}-4 x^{5}+11 x^{4}-17 x^{3}+12 x^{2}-5 x +1}{2 x^{2} \left(x^{2}-2 x +2\right) \left(x -1\right)^{2}}\)
Counting Sequence
1, 1, 2, 6, 20, 67, 224, 754, 2569, 8870, 31007, 109578, 390906, 1405887, 5092080, ...
Implicit Equation for the Generating Function
\(\displaystyle x^{2} \left(x^{2}-2 x +2\right) \left(x -1\right)^{4} F \left(x
\right)^{2}+\left(4 x^{5}-11 x^{4}+17 x^{3}-12 x^{2}+5 x -1\right) \left(x -1\right)^{2} F \! \left(x \right)+4 x^{6}-13 x^{5}+24 x^{4}-25 x^{3}+16 x^{2}-6 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 67\)
\(\displaystyle a \! \left(6\right) = 224\)
\(\displaystyle a \! \left(7\right) = 754\)
\(\displaystyle a \! \left(n +5\right) = \frac{\left(2 n +3\right) a \! \left(n \right)}{7+n}-\frac{\left(29+13 n \right) a \! \left(n +1\right)}{2 \left(7+n \right)}+\frac{\left(23 n +73\right) a \! \left(n +2\right)}{14+2 n}-\frac{\left(95+21 n \right) a \! \left(n +3\right)}{2 \left(7+n \right)}+\frac{2 \left(3 n +17\right) a \! \left(n +4\right)}{7+n}+\frac{3 n^{2}+14 n +25}{14+2 n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 67\)
\(\displaystyle a \! \left(6\right) = 224\)
\(\displaystyle a \! \left(7\right) = 754\)
\(\displaystyle a \! \left(n +5\right) = \frac{\left(2 n +3\right) a \! \left(n \right)}{7+n}-\frac{\left(29+13 n \right) a \! \left(n +1\right)}{2 \left(7+n \right)}+\frac{\left(23 n +73\right) a \! \left(n +2\right)}{14+2 n}-\frac{\left(95+21 n \right) a \! \left(n +3\right)}{2 \left(7+n \right)}+\frac{2 \left(3 n +17\right) a \! \left(n +4\right)}{7+n}+\frac{3 n^{2}+14 n +25}{14+2 n}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 19 rules.
Found on July 23, 2021.Finding the specification took 2 seconds.
Copy 19 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{10}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{10}\! \left(x \right) F_{11}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{7} \left(x \right)^{2} F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= x\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)+F_{4}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right) F_{17}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{10}\! \left(x \right) F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{14}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{18}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{14}\! \left(x \right) F_{16}\! \left(x \right)\\
\end{align*}\)