Av(1342, 1432, 2143, 3142)
Counting Sequence
1, 1, 2, 6, 20, 69, 245, 894, 3339, 12708, 49108, 192164, 759908, 3032102, 12192282, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(x -1\right) x^{2} F \left(x
\right)^{3}+x \left(x^{2}-2 x +2\right) F \left(x
\right)^{2}+\left(x -1\right) \left(x +1\right) F \! \left(x \right)-x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 69\)
\(\displaystyle a \! \left(6\right) = 245\)
\(\displaystyle a \! \left(7\right) = 894\)
\(\displaystyle a \! \left(8\right) = 3339\)
\(\displaystyle a \! \left(9\right) = 12708\)
\(\displaystyle a \! \left(10\right) = 49108\)
\(\displaystyle a \! \left(n +11\right) = \frac{n \left(n +1\right) a \! \left(n \right)}{\left(n +12\right) \left(n +11\right)}-\frac{\left(75 n +134\right) \left(n +1\right) a \! \left(n +1\right)}{5 \left(n +12\right) \left(n +11\right)}+\frac{\left(2059 n^{2}+9747 n +11308\right) a \! \left(n +2\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{\left(6127 n^{2}+41355 n +69202\right) a \! \left(n +3\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{2 \left(5878 n^{2}+52109 n +114923\right) a \! \left(n +4\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{2 \left(7944 n^{2}+87762 n +241471\right) a \! \left(n +5\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{2 \left(7800 n^{2}+103044 n +339197\right) a \! \left(n +6\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{2 \left(5520 n^{2}+84417 n +321847\right) a \! \left(n +7\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{\left(5451 n^{2}+94221 n +406270\right) a \! \left(n +8\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{\left(1761 n^{2}+33831 n +162200\right) a \! \left(n +9\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{\left(65 n +658\right) a \! \left(n +10\right)}{5 n +60}, \quad n \geq 11\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(4\right) = 20\)
\(\displaystyle a \! \left(5\right) = 69\)
\(\displaystyle a \! \left(6\right) = 245\)
\(\displaystyle a \! \left(7\right) = 894\)
\(\displaystyle a \! \left(8\right) = 3339\)
\(\displaystyle a \! \left(9\right) = 12708\)
\(\displaystyle a \! \left(10\right) = 49108\)
\(\displaystyle a \! \left(n +11\right) = \frac{n \left(n +1\right) a \! \left(n \right)}{\left(n +12\right) \left(n +11\right)}-\frac{\left(75 n +134\right) \left(n +1\right) a \! \left(n +1\right)}{5 \left(n +12\right) \left(n +11\right)}+\frac{\left(2059 n^{2}+9747 n +11308\right) a \! \left(n +2\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{\left(6127 n^{2}+41355 n +69202\right) a \! \left(n +3\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{2 \left(5878 n^{2}+52109 n +114923\right) a \! \left(n +4\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{2 \left(7944 n^{2}+87762 n +241471\right) a \! \left(n +5\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{2 \left(7800 n^{2}+103044 n +339197\right) a \! \left(n +6\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{2 \left(5520 n^{2}+84417 n +321847\right) a \! \left(n +7\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{\left(5451 n^{2}+94221 n +406270\right) a \! \left(n +8\right)}{25 \left(n +12\right) \left(n +11\right)}-\frac{\left(1761 n^{2}+33831 n +162200\right) a \! \left(n +9\right)}{25 \left(n +12\right) \left(n +11\right)}+\frac{\left(65 n +658\right) a \! \left(n +10\right)}{5 n +60}, \quad n \geq 11\)
This specification was found using the strategy pack "Point Placements" and has 17 rules.
Found on July 23, 2021.Finding the specification took 6 seconds.
Copy 17 equations to clipboard:
\(\begin{align*}
F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\
F_{1}\! \left(x \right) &= 1\\
F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\
F_{3}\! \left(x \right) &= F_{4}\! \left(x \right) F_{5}\! \left(x \right)\\
F_{4}\! \left(x \right) &= x\\
F_{5}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{6}\! \left(x \right)\\
F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\
F_{7}\! \left(x \right) &= F_{0}\! \left(x \right) F_{4}\! \left(x \right) F_{8}\! \left(x \right)\\
F_{8}\! \left(x \right) &= F_{5}\! \left(x \right)+F_{9}\! \left(x \right)\\
F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\
F_{10}\! \left(x \right) &= F_{11}\! \left(x \right) F_{14}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{11}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{12}\! \left(x \right)\\
F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)\\
F_{13}\! \left(x \right) &= F_{11}\! \left(x \right) F_{4}\! \left(x \right)\\
F_{14}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{15}\! \left(x \right)\\
F_{15}\! \left(x \right) &= F_{16}\! \left(x \right)\\
F_{16}\! \left(x \right) &= F_{0}\! \left(x \right) F_{14}\! \left(x \right) F_{4}\! \left(x \right)\\
\end{align*}\)