Av(1342, 2143, 2413, 3142)
Generating Function
\(\displaystyle \frac{-x^{3}+x^{2}-3 x +1-\sqrt{x^{6}-6 x^{5}+31 x^{4}-52 x^{3}+35 x^{2}-10 x +1}}{2 x \left(x^{2}-3 x +1\right)}\)
Counting Sequence
1, 1, 2, 6, 20, 69, 243, 870, 3160, 11627, 43283, 162833, 618424, 2368800, 9142922, ...
Implicit Equation for the Generating Function
\(\displaystyle x \left(x^{2}-3 x +1\right) F \left(x
\right)^{2}+\left(x^{3}-x^{2}+3 x -1\right) F \! \left(x \right)+x^{2}-3 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 69\)
\(\displaystyle a \! \left(6\right) = 243\)
\(\displaystyle a \! \left(7\right) = 870\)
\(\displaystyle a \! \left(n +8\right) = -\frac{n a \! \left(n \right)}{9+n}+\frac{9 \left(1+n \right) a \! \left(1+n \right)}{9+n}-\frac{10 \left(12+5 n \right) a \! \left(n +2\right)}{9+n}+\frac{\left(522+151 n \right) a \! \left(n +3\right)}{9+n}-\frac{3 \left(335+74 n \right) a \! \left(n +4\right)}{9+n}+\frac{\left(939+167 n \right) a \! \left(n +5\right)}{9+n}-\frac{6 \left(74+11 n \right) a \! \left(n +6\right)}{9+n}+\frac{\left(102+13 n \right) a \! \left(n +7\right)}{9+n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 69\)
\(\displaystyle a \! \left(6\right) = 243\)
\(\displaystyle a \! \left(7\right) = 870\)
\(\displaystyle a \! \left(n +8\right) = -\frac{n a \! \left(n \right)}{9+n}+\frac{9 \left(1+n \right) a \! \left(1+n \right)}{9+n}-\frac{10 \left(12+5 n \right) a \! \left(n +2\right)}{9+n}+\frac{\left(522+151 n \right) a \! \left(n +3\right)}{9+n}-\frac{3 \left(335+74 n \right) a \! \left(n +4\right)}{9+n}+\frac{\left(939+167 n \right) a \! \left(n +5\right)}{9+n}-\frac{6 \left(74+11 n \right) a \! \left(n +6\right)}{9+n}+\frac{\left(102+13 n \right) a \! \left(n +7\right)}{9+n}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 17 rules.
Found on July 23, 2021.Finding the specification took 3 seconds.
Copy 17 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{11}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{0}\! \left(x \right) F_{11}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{4}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{12}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{11}\! \left(x \right) &= x\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{11}\! \left(x \right) F_{14}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{12}\! \left(x \right)\\
\end{align*}\)