Av(1243, 1324, 1342, 1423, 2143, 2413, 4132)
Generating Function
\(\displaystyle \frac{\left(x^{4}-x^{3}+x^{2}-x +1\right) \sqrt{1-4 x}-3 x^{4}+3 x^{3}-x^{2}+x -1}{2 \left(x -1\right) x}\)
Counting Sequence
1, 1, 2, 6, 17, 49, 150, 480, 1585, 5356, 18423, 64274, 226864, 808604, 2906166, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(x -1\right)^{2} x F \left(x
\right)^{2}+\left(x -1\right) \left(3 x^{4}-3 x^{3}+x^{2}-x +1\right) F \! \left(x \right)+x^{8}-x^{6}-x^{5}+3 x^{4}-2 x^{3}+2 x^{2}-2 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 17\)
\(\displaystyle a \! \left(5\right) = 49\)
\(\displaystyle a \! \left(6\right) = 150\)
\(\displaystyle a \! \left(7\right) = 480\)
\(\displaystyle a \! \left(8\right) = 1585\)
\(\displaystyle a \! \left(n +6\right) = -\frac{2 \left(-5+2 n \right) a \! \left(n \right)}{7+n}+\frac{\left(-13+9 n \right) a \! \left(1+n \right)}{7+n}-\frac{2 \left(2+5 n \right) a \! \left(n +2\right)}{7+n}+\frac{2 \left(12+5 n \right) a \! \left(n +3\right)}{7+n}-\frac{2 \left(22+5 n \right) a \! \left(n +4\right)}{7+n}+\frac{2 \left(17+3 n \right) a \! \left(n +5\right)}{7+n}, \quad n \geq 9\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 17\)
\(\displaystyle a \! \left(5\right) = 49\)
\(\displaystyle a \! \left(6\right) = 150\)
\(\displaystyle a \! \left(7\right) = 480\)
\(\displaystyle a \! \left(8\right) = 1585\)
\(\displaystyle a \! \left(n +6\right) = -\frac{2 \left(-5+2 n \right) a \! \left(n \right)}{7+n}+\frac{\left(-13+9 n \right) a \! \left(1+n \right)}{7+n}-\frac{2 \left(2+5 n \right) a \! \left(n +2\right)}{7+n}+\frac{2 \left(12+5 n \right) a \! \left(n +3\right)}{7+n}-\frac{2 \left(22+5 n \right) a \! \left(n +4\right)}{7+n}+\frac{2 \left(17+3 n \right) a \! \left(n +5\right)}{7+n}, \quad n \geq 9\)
This specification was found using the strategy pack "Point Placements" and has 20 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 20 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{12}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{12}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{13}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{9} \left(x \right)^{2} F_{0}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{9} \left(x \right)^{2} F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= x\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{12}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{12}\! \left(x \right) F_{17}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{18}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{12}\! \left(x \right) F_{19}\! \left(x \right)\\
F_{19}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{18}\! \left(x \right)\\
\end{align*}\)