Av(1324, 1342, 1423, 2143, 2413, 2431, 4132)
Generating Function
\(\displaystyle \frac{\left(x -1\right) \left(x^{3}-x +1\right) \sqrt{1-4 x}+2 x^{5}-x^{4}+x^{3}+x^{2}-2 x +1}{2 x \left(x -1\right)^{2}}\)
Counting Sequence
1, 1, 2, 6, 17, 48, 144, 456, 1500, 5065, 17429, 60850, 214939, 766624, 2756951, ...
Implicit Equation for the Generating Function
\(\displaystyle x \left(x -1\right)^{4} F \left(x
\right)^{2}-\left(x +1\right) \left(2 x^{4}-3 x^{3}+4 x^{2}-3 x +1\right) \left(x -1\right)^{2} F \! \left(x \right)+x^{9}-x^{7}+4 x^{5}-4 x^{4}-2 x^{3}+6 x^{2}-4 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 17\)
\(\displaystyle a \! \left(5\right) = 48\)
\(\displaystyle a \! \left(6\right) = 144\)
\(\displaystyle a \! \left(7\right) = 456\)
\(\displaystyle a \! \left(8\right) = 1500\)
\(\displaystyle a \! \left(9\right) = 5065\)
\(\displaystyle a \! \left(n +5\right) = -\frac{2 \left(-3+2 n \right) a \! \left(n \right)}{6+n}+\frac{\left(-6+5 n \right) a \! \left(1+n \right)}{6+n}+\frac{\left(10+3 n \right) a \! \left(n +2\right)}{6+n}-\frac{\left(32+9 n \right) a \! \left(n +3\right)}{6+n}+\frac{2 \left(14+3 n \right) a \! \left(n +4\right)}{6+n}-\frac{3 n}{6+n}, \quad n \geq 10\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 17\)
\(\displaystyle a \! \left(5\right) = 48\)
\(\displaystyle a \! \left(6\right) = 144\)
\(\displaystyle a \! \left(7\right) = 456\)
\(\displaystyle a \! \left(8\right) = 1500\)
\(\displaystyle a \! \left(9\right) = 5065\)
\(\displaystyle a \! \left(n +5\right) = -\frac{2 \left(-3+2 n \right) a \! \left(n \right)}{6+n}+\frac{\left(-6+5 n \right) a \! \left(1+n \right)}{6+n}+\frac{\left(10+3 n \right) a \! \left(n +2\right)}{6+n}-\frac{\left(32+9 n \right) a \! \left(n +3\right)}{6+n}+\frac{2 \left(14+3 n \right) a \! \left(n +4\right)}{6+n}-\frac{3 n}{6+n}, \quad n \geq 10\)
This specification was found using the strategy pack "Point Placements" and has 23 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 23 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{5}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{5} \left(x \right)^{2} F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= x\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)+F_{17}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right) F_{5}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{5} \left(x \right)^{2}\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{16}\! \left(x \right) F_{18}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{19}\! \left(x \right)\\
F_{19}\! \left(x \right) &= F_{20}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{20}\! \left(x \right) &= F_{21}\! \left(x \right)\\
F_{21}\! \left(x \right) &= F_{22}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{22}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{21}\! \left(x \right)\\
\end{align*}\)