Av(132, 4231)
Generating Function
\(\displaystyle \frac{x^{3}-5 x^{2}+4 x -1}{\left(-1+x \right) \left(2 x -1\right)^{2}}\)
Counting Sequence
1, 1, 2, 5, 13, 33, 81, 193, 449, 1025, 2305, 5121, 11265, 24577, 53249, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(-1+x \right) \left(2 x -1\right)^{2} F \! \left(x \right)-x^{3}+5 x^{2}-4 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 5\)
\(\displaystyle a \! \left(n +2\right) = -4 a \! \left(n \right)+4 a \! \left(n +1\right)+1, \quad n \geq 4\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 5\)
\(\displaystyle a \! \left(n +2\right) = -4 a \! \left(n \right)+4 a \! \left(n +1\right)+1, \quad n \geq 4\)
Explicit Closed Form
\(\displaystyle \left\{\begin{array}{cc}1 & n =0 \\ 1+\frac{\left(n -1\right) 2^{n}}{4} & \text{otherwise} \end{array}\right.\)
This specification was found using the strategy pack "Point Placements" and has 14 rules.
Found on January 17, 2022.Finding the specification took 2 seconds.
Copy 14 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\
F_{5}\! \left(x \right) &= F_{10}\! \left(x \right) F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{8}\! \left(x \right) F_{9}\! \left(x \right)\\
F_{8}\! \left(x \right) &= x\\
F_{9}\! \left(x \right) &= 2 F_{7}\! \left(x \right)+F_{1}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{13}\! \left(x \right) &= 2 F_{12}\! \left(x \right)+F_{1}\! \left(x \right)\\
\end{align*}\)