Av(1432, 2143, 2413)
Counting Sequence
1, 1, 2, 6, 21, 77, 289, 1107, 4322, 17162, 69137, 281917, 1161404, 4826652, 20211146, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(x -1\right) x^{2} F \left(x
\right)^{3}+x \left(2 x^{2}-3 x +2\right) F \left(x
\right)^{2}+\left(x -1\right) F \! \left(x \right)+\left(x -1\right)^{2} = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 21\)
\(\displaystyle a \! \left(5\right) = 77\)
\(\displaystyle a \! \left(6\right) = 289\)
\(\displaystyle a \! \left(7\right) = 1107\)
\(\displaystyle a \! \left(8\right) = 4322\)
\(\displaystyle a \! \left(9\right) = 17162\)
\(\displaystyle a \! \left(10\right) = 69137\)
\(\displaystyle a \! \left(n +11\right) = \frac{8 n \left(2 n +1\right) a \! \left(n \right)}{\left(n +12\right) \left(n +11\right)}-\frac{\left(293 n^{2}+754 n +456\right) a \! \left(n +1\right)}{2 \left(n +12\right) \left(n +11\right)}+\frac{\left(2359 n^{2}+10829 n +12132\right) a \! \left(n +2\right)}{4 \left(n +12\right) \left(n +11\right)}-\frac{\left(5579 n^{2}+36901 n +60192\right) a \! \left(n +3\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{3 \left(5797 n^{2}+50283 n +108048\right) a \! \left(n +4\right)}{8 \left(n +12\right) \left(n +11\right)}-\frac{\left(18922 n^{2}+203843 n +545328\right) a \! \left(n +5\right)}{8 \left(n +12\right) \left(n +11\right)}+\frac{\left(14753 n^{2}+190168 n +609708\right) a \! \left(n +6\right)}{8 \left(n +12\right) \left(n +11\right)}-\frac{\left(4123 n^{2}+61820 n +230826\right) a \! \left(n +7\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{\left(805 n^{2}+13733 n +58395\right) a \! \left(n +8\right)}{2 \left(n +12\right) \left(n +11\right)}-\frac{3 \left(n +10\right) \left(138 n +1253\right) a \! \left(n +9\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{\left(31 n +312\right) a \! \left(n +10\right)}{2 n +24}, \quad n \geq 11\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 21\)
\(\displaystyle a \! \left(5\right) = 77\)
\(\displaystyle a \! \left(6\right) = 289\)
\(\displaystyle a \! \left(7\right) = 1107\)
\(\displaystyle a \! \left(8\right) = 4322\)
\(\displaystyle a \! \left(9\right) = 17162\)
\(\displaystyle a \! \left(10\right) = 69137\)
\(\displaystyle a \! \left(n +11\right) = \frac{8 n \left(2 n +1\right) a \! \left(n \right)}{\left(n +12\right) \left(n +11\right)}-\frac{\left(293 n^{2}+754 n +456\right) a \! \left(n +1\right)}{2 \left(n +12\right) \left(n +11\right)}+\frac{\left(2359 n^{2}+10829 n +12132\right) a \! \left(n +2\right)}{4 \left(n +12\right) \left(n +11\right)}-\frac{\left(5579 n^{2}+36901 n +60192\right) a \! \left(n +3\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{3 \left(5797 n^{2}+50283 n +108048\right) a \! \left(n +4\right)}{8 \left(n +12\right) \left(n +11\right)}-\frac{\left(18922 n^{2}+203843 n +545328\right) a \! \left(n +5\right)}{8 \left(n +12\right) \left(n +11\right)}+\frac{\left(14753 n^{2}+190168 n +609708\right) a \! \left(n +6\right)}{8 \left(n +12\right) \left(n +11\right)}-\frac{\left(4123 n^{2}+61820 n +230826\right) a \! \left(n +7\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{\left(805 n^{2}+13733 n +58395\right) a \! \left(n +8\right)}{2 \left(n +12\right) \left(n +11\right)}-\frac{3 \left(n +10\right) \left(138 n +1253\right) a \! \left(n +9\right)}{4 \left(n +12\right) \left(n +11\right)}+\frac{\left(31 n +312\right) a \! \left(n +10\right)}{2 n +24}, \quad n \geq 11\)
This specification was found using the strategy pack "Row And Col Placements" and has 19 rules.
Found on July 23, 2021.Finding the specification took 11 seconds.
Copy 19 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{4}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{5}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{3}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{0}\! \left(x \right) F_{5}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= x\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)+F_{3}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right) F_{17}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{14}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{13}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{0}\! \left(x \right) F_{13}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{18}\! \left(x \right)+F_{4}\! \left(x \right)\\
F_{18}\! \left(x \right) &= F_{17}\! \left(x \right) F_{8}\! \left(x \right)\\
\end{align*}\)