Av(1342, 1423, 1432)
Counting Sequence
1, 1, 2, 6, 21, 80, 322, 1347, 5798, 25512, 114236, 518848, 2384538, 11068567, 51817118, ...
Implicit Equation for the Generating Function
\(\displaystyle x F \left(x
\right)^{3}-x F \left(x
\right)^{2}+\left(x -1\right) F \! \left(x \right)+1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(n +4\right) = \frac{15 n \left(n +1\right) a \! \left(n \right)}{4 \left(2 n +9\right) \left(n +4\right)}+\frac{\left(13 n +24\right) \left(n +1\right) a \! \left(n +1\right)}{2 \left(2 n +9\right) \left(n +4\right)}+\frac{3 \left(9 n^{2}+37 n +37\right) a \! \left(n +2\right)}{\left(2 n +9\right) \left(n +4\right)}+\frac{\left(10 n^{2}+85 n +174\right) a \! \left(n +3\right)}{2 \left(2 n +9\right) \left(n +4\right)}, \quad n \geq 4\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(n +4\right) = \frac{15 n \left(n +1\right) a \! \left(n \right)}{4 \left(2 n +9\right) \left(n +4\right)}+\frac{\left(13 n +24\right) \left(n +1\right) a \! \left(n +1\right)}{2 \left(2 n +9\right) \left(n +4\right)}+\frac{3 \left(9 n^{2}+37 n +37\right) a \! \left(n +2\right)}{\left(2 n +9\right) \left(n +4\right)}+\frac{\left(10 n^{2}+85 n +174\right) a \! \left(n +3\right)}{2 \left(2 n +9\right) \left(n +4\right)}, \quad n \geq 4\)
This specification was found using the strategy pack "Point Placements" and has 8 rules.
Found on July 23, 2021.Finding the specification took 1 seconds.
Copy 8 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{0} \left(x \right)^{2} F_{4}\! \left(x \right) F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= x\\
\end{align*}\)