Av(12453, 12534, 12543, 21453, 21534, 21543)
The requested set is not in the database, but a symmetry of it is.
Counting Sequence
1, 1, 2, 6, 24, 114, 598, 3336, 19402, 116302, 713368, 4455650, 28240942, 181180912, 1174280146, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(-x +1\right) F \left(x
\right)^{3}+\left(x +3\right) \left(x -1\right) F \left(x
\right)^{2}+\left(-2 x +3\right) F \! \left(x \right)+x -1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 24\)
\(\displaystyle a \! \left(5\right) = 114\)
\(\displaystyle a \! \left(n +6\right) = -\frac{3 \left(2 n +1\right) \left(n -1\right) a \! \left(n \right)}{\left(2 n +11\right) \left(n +5\right)}-\frac{\left(16 n^{2}+10 n +9\right) a \! \left(n +1\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(112 n^{2}+312 n +227\right) a \! \left(n +2\right)}{\left(2 n +11\right) \left(n +5\right)}-\frac{6 \left(25 n^{2}+107 n +112\right) a \! \left(n +3\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(52 n^{2}+256 n +261\right) a \! \left(n +4\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(10 n^{2}+104 n +265\right) a \! \left(n +5\right)}{\left(2 n +11\right) \left(n +5\right)}, \quad n \geq 6\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 24\)
\(\displaystyle a \! \left(5\right) = 114\)
\(\displaystyle a \! \left(n +6\right) = -\frac{3 \left(2 n +1\right) \left(n -1\right) a \! \left(n \right)}{\left(2 n +11\right) \left(n +5\right)}-\frac{\left(16 n^{2}+10 n +9\right) a \! \left(n +1\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(112 n^{2}+312 n +227\right) a \! \left(n +2\right)}{\left(2 n +11\right) \left(n +5\right)}-\frac{6 \left(25 n^{2}+107 n +112\right) a \! \left(n +3\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(52 n^{2}+256 n +261\right) a \! \left(n +4\right)}{\left(2 n +11\right) \left(n +5\right)}+\frac{\left(10 n^{2}+104 n +265\right) a \! \left(n +5\right)}{\left(2 n +11\right) \left(n +5\right)}, \quad n \geq 6\)
This specification was found using the strategy pack "Point Placements Tracked Fusion" and has 20 rules.
Found on January 22, 2022.Finding the specification took 156 seconds.
Copy 20 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{17}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{2}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{17}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{10}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{2}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right) F_{17}\! \left(x \right) F_{18}\! \left(x \right)\\
F_{12}\! \left(x \right) &= \frac{F_{13}\! \left(x \right)}{F_{17}\! \left(x \right)}\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right) F_{17}\! \left(x \right) F_{18}\! \left(x \right)\\
F_{15}\! \left(x \right) &= \frac{F_{16}\! \left(x \right)}{F_{17}\! \left(x \right)}\\
F_{16}\! \left(x \right) &= F_{5}\! \left(x \right)\\
F_{17}\! \left(x \right) &= x\\
F_{18}\! \left(x \right) &= \frac{F_{19}\! \left(x \right)}{F_{17}\! \left(x \right)}\\
F_{19}\! \left(x \right) &= F_{2}\! \left(x \right)\\
\end{align*}\)