Av(1243, 1342, 1423, 2143, 2431)
Generating Function
\(\displaystyle \frac{2 x^{5}-\sqrt{-4 x +1}\, x^{3}+5 x^{3}-4 x^{2}+\sqrt{-4 x +1}+2 x -1}{2 x^{2} \left(x^{2}+3\right) \left(x -1\right)}\)
Counting Sequence
1, 1, 2, 6, 19, 60, 194, 643, 2175, 7481, 26089, 92037, 327869, 1177771, 4261438, ...
Implicit Equation for the Generating Function
\(\displaystyle x^{2} \left(x^{2}+3\right) \left(x -1\right)^{2} F \left(x
\right)^{2}-\left(x -1\right) \left(2 x^{5}+5 x^{3}-4 x^{2}+2 x -1\right) F \! \left(x \right)+x^{6}+2 x^{4}-3 x^{3}+2 x^{2}-2 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 60\)
\(\displaystyle a \! \left(6\right) = 194\)
\(\displaystyle a \! \left(7\right) = 643\)
\(\displaystyle a \! \left(n +5\right) = \frac{2 \left(3+2 n \right) a \! \left(n \right)}{3 \left(7+n \right)}+\frac{5 \left(7+3 n \right) a \! \left(2+n \right)}{3 \left(7+n \right)}+\frac{\left(11+3 n \right) a \! \left(n +1\right)}{21+3 n}+\frac{2 \left(13+4 n \right) a \! \left(n +3\right)}{3 \left(7+n \right)}+\frac{\left(17+3 n \right) a \! \left(n +4\right)}{7+n}+\frac{6 n +18}{7+n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 60\)
\(\displaystyle a \! \left(6\right) = 194\)
\(\displaystyle a \! \left(7\right) = 643\)
\(\displaystyle a \! \left(n +5\right) = \frac{2 \left(3+2 n \right) a \! \left(n \right)}{3 \left(7+n \right)}+\frac{5 \left(7+3 n \right) a \! \left(2+n \right)}{3 \left(7+n \right)}+\frac{\left(11+3 n \right) a \! \left(n +1\right)}{21+3 n}+\frac{2 \left(13+4 n \right) a \! \left(n +3\right)}{3 \left(7+n \right)}+\frac{\left(17+3 n \right) a \! \left(n +4\right)}{7+n}+\frac{6 n +18}{7+n}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 17 rules.
Found on July 23, 2021.Finding the specification took 2 seconds.
Copy 17 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{10}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{10}\! \left(x \right) F_{11}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{7} \left(x \right)^{2} F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= x\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)+F_{4}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{10}\! \left(x \right) F_{14}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{10}\! \left(x \right) F_{15}\! \left(x \right)\\
\end{align*}\)