Av(1243, 1324, 1342, 1423, 1432)
Generating Function
\(\displaystyle \frac{-\left(x -1\right)^{2} \sqrt{-4 x +1}+5 x^{2}-4 x +1}{2 x^{2}}\)
Counting Sequence
1, 1, 2, 6, 19, 62, 207, 704, 2431, 8502, 30056, 107236, 385662, 1396652, 5088865, ...
Implicit Equation for the Generating Function
\(\displaystyle x^{2} F \left(x
\right)^{2}+\left(-5 x^{2}+4 x -1\right) F \! \left(x \right)+x^{3}+2 x^{2}-3 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(2+n \right) = -\frac{2 \left(-1+2 n \right) a \! \left(n \right)}{4+n}+\frac{\left(11+5 n \right) a \! \left(n +1\right)}{4+n}, \quad n \geq 3\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(2+n \right) = -\frac{2 \left(-1+2 n \right) a \! \left(n \right)}{4+n}+\frac{\left(11+5 n \right) a \! \left(n +1\right)}{4+n}, \quad n \geq 3\)
This specification was found using the strategy pack "Point Placements" and has 14 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 14 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{5}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{5} \left(x \right)^{2} F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= x\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right) F_{4}\! \left(x \right) F_{5}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{11}\! \left(x \right) F_{8}\! \left(x \right)\\
\end{align*}\)