Av(1342, 1423, 2413, 2431, 4132)
Generating Function
\(\displaystyle \frac{\left(2 x^{3}-2 x^{2}-x +1\right) \sqrt{1-4 x}+2 x^{4}-4 x^{3}+4 x^{2}+x -1}{4 x^{2}-2 x}\)
Counting Sequence
1, 1, 2, 6, 19, 60, 192, 625, 2071, 6980, 23892, 82910, 291178, 1033264, 3699616, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(2 x -1\right)^{2} x F \left(x
\right)^{2}-\left(2 x -1\right) \left(2 x^{4}-4 x^{3}+4 x^{2}+x -1\right) F \! \left(x \right)+x^{7}-x^{5}-5 x^{4}+9 x^{3}-x^{2}-3 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 60\)
\(\displaystyle a \! \left(6\right) = 192\)
\(\displaystyle a \! \left(7\right) = 625\)
\(\displaystyle a \! \left(n +5\right) = -\frac{8 \left(-3+2 n \right) a \! \left(n \right)}{6+n}+\frac{4 \left(-1+7 n \right) a \! \left(1+n \right)}{6+n}-\frac{2 \left(-4+3 n \right) a \! \left(n +2\right)}{6+n}-\frac{2 \left(25+6 n \right) a \! \left(n +3\right)}{6+n}+\frac{\left(34+7 n \right) a \! \left(n +4\right)}{6+n}, \quad n \geq 8\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 19\)
\(\displaystyle a \! \left(5\right) = 60\)
\(\displaystyle a \! \left(6\right) = 192\)
\(\displaystyle a \! \left(7\right) = 625\)
\(\displaystyle a \! \left(n +5\right) = -\frac{8 \left(-3+2 n \right) a \! \left(n \right)}{6+n}+\frac{4 \left(-1+7 n \right) a \! \left(1+n \right)}{6+n}-\frac{2 \left(-4+3 n \right) a \! \left(n +2\right)}{6+n}-\frac{2 \left(25+6 n \right) a \! \left(n +3\right)}{6+n}+\frac{\left(34+7 n \right) a \! \left(n +4\right)}{6+n}, \quad n \geq 8\)
This specification was found using the strategy pack "Point Placements" and has 18 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 18 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{12}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{12}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{13}\! \left(x \right)+F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{9} \left(x \right)^{3}\\
F_{9}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{9} \left(x \right)^{2} F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= x\\
F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{17}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{15}\! \left(x \right) F_{2}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{12}\! \left(x \right) F_{15}\! \left(x \right)\\
F_{17}\! \left(x \right) &= F_{10} \left(x \right)^{2}\\
\end{align*}\)