Av(1243, 1342, 1432, 2143, 3142)
Counting Sequence
1, 1, 2, 6, 19, 63, 218, 779, 2852, 10642, 40325, 154752, 600236, 2349320, 9267240, ...
Implicit Equation for the Generating Function
\(\displaystyle -x^{2} F \left(x
\right)^{3}+x \left(x^{2}+2\right) F \left(x
\right)^{2}+\left(-x -1\right) F \! \left(x \right)+1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 63\)
\(\displaystyle a \! \left(6\right) = 218\)
\(\displaystyle a \! \left(7\right) = 779\)
\(\displaystyle a \! \left(n +8\right) = -\frac{4 \left(2 n +3\right) \left(n -1\right) a \! \left(n \right)}{5 \left(n +9\right) \left(n +8\right)}+\frac{2 n \left(11 n +31\right) a \! \left(n +1\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{\left(259 n^{2}+1019 n +840\right) a \! \left(n +2\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(325 n^{2}+1841 n +2364\right) a \! \left(n +3\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{2 \left(266 n^{2}+2089 n +3930\right) a \! \left(n +4\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(445 n^{2}+4793 n +12696\right) a \! \left(n +5\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{4 \left(91 n^{2}+1175 n +3771\right) a \! \left(n +6\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(35 n +249\right) a \! \left(n +7\right)}{5 n +45}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 63\)
\(\displaystyle a \! \left(6\right) = 218\)
\(\displaystyle a \! \left(7\right) = 779\)
\(\displaystyle a \! \left(n +8\right) = -\frac{4 \left(2 n +3\right) \left(n -1\right) a \! \left(n \right)}{5 \left(n +9\right) \left(n +8\right)}+\frac{2 n \left(11 n +31\right) a \! \left(n +1\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{\left(259 n^{2}+1019 n +840\right) a \! \left(n +2\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(325 n^{2}+1841 n +2364\right) a \! \left(n +3\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{2 \left(266 n^{2}+2089 n +3930\right) a \! \left(n +4\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(445 n^{2}+4793 n +12696\right) a \! \left(n +5\right)}{25 \left(n +9\right) \left(n +8\right)}-\frac{4 \left(91 n^{2}+1175 n +3771\right) a \! \left(n +6\right)}{25 \left(n +9\right) \left(n +8\right)}+\frac{\left(35 n +249\right) a \! \left(n +7\right)}{5 n +45}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 13 rules.
Found on July 23, 2021.Finding the specification took 4 seconds.
Copy 13 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{5}\! \left(x \right)\\
F_{4}\! \left(x \right) &= x\\
F_{5}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{10}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{0}\! \left(x \right) F_{5}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{0}\! \left(x \right) F_{12}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{10}\! \left(x \right)\\
\end{align*}\)