Av(1342, 1423, 1432, 3412)
View Raw Data
Generating Function
\(\displaystyle \frac{x^{3}-x^{2}+3 x -1}{2 x^{3}-3 x^{2}+4 x -1}\)
Counting Sequence
1, 1, 2, 6, 20, 66, 216, 706, 2308, 7546, 24672, 80666, 263740, 862306, 2819336, ...
Implicit Equation for the Generating Function
\(\displaystyle \left(2 x^{3}-3 x^{2}+4 x -1\right) F \! \left(x \right)-x^{3}+x^{2}-3 x +1 = 0\)
Recurrence
\(\displaystyle a \! \left(0\right) = 1\)
\(\displaystyle a \! \left(1\right) = 1\)
\(\displaystyle a \! \left(2\right) = 2\)
\(\displaystyle a \! \left(3\right) = 6\)
\(\displaystyle a \! \left(n +3\right) = 2 a \! \left(n \right)-3 a \! \left(n +1\right)+4 a \! \left(n +2\right), \quad n \geq 4\)
Explicit Closed Form
\(\displaystyle \left\{\begin{array}{cc}1 & n =0 \\ \frac{\left(\left(\left(15 \,\mathrm{I} \,3^{\frac{1}{3}}-5 \,3^{\frac{5}{6}}\right) \sqrt{38}-285 \,\mathrm{I} \,3^{\frac{5}{6}}+285 \,3^{\frac{1}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}+2850+\left(\left(-21 \,\mathrm{I} \,3^{\frac{2}{3}}-21 \,3^{\frac{1}{6}}\right) \sqrt{38}+57 \,\mathrm{I} \,3^{\frac{1}{6}}+19 \,3^{\frac{2}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}\right) \left(\frac{\left(27+6 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}}{12}-\frac{\mathrm{I} \,3^{\frac{5}{6}} \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}}{12}+\frac{1}{2}+\frac{\left(\left(9 \,\mathrm{I}-2 \sqrt{38}\right) 3^{\frac{1}{6}}+\left(-2 \,\mathrm{I} \sqrt{38}+3\right) 3^{\frac{2}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}}{300}\right)^{-n}}{17100}\\+\\\frac{\left(\left(\left(-15 \,\mathrm{I} \,3^{\frac{1}{3}}-5 \,3^{\frac{5}{6}}\right) \sqrt{38}+285 \,\mathrm{I} \,3^{\frac{5}{6}}+285 \,3^{\frac{1}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}+2850+\left(\left(21 \,\mathrm{I} \,3^{\frac{2}{3}}-21 \,3^{\frac{1}{6}}\right) \sqrt{38}-57 \,\mathrm{I} \,3^{\frac{1}{6}}+19 \,3^{\frac{2}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}\right) \left(\frac{\left(27+6 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}}{12}+\frac{\mathrm{I} \,3^{\frac{5}{6}} \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}}{12}+\frac{1}{2}+\frac{\left(\left(-9 \,\mathrm{I}-2 \sqrt{38}\right) 3^{\frac{1}{6}}+\left(2 \,\mathrm{I} \sqrt{38}+3\right) 3^{\frac{2}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}}{300}\right)^{-n}}{17100}\\-\\\frac{\left(\frac{\left(2 \sqrt{38}\, 3^{\frac{1}{6}}-3 \,3^{\frac{2}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}}{150}-\frac{\left(27+6 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}}{6}+\frac{1}{2}\right)^{-n} \left(\left(-\frac{3^{\frac{5}{6}} \sqrt{38}}{57}+3^{\frac{1}{3}}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{1}{3}}-5+\left(-\frac{7 \sqrt{38}\, 3^{\frac{1}{6}}}{95}+\frac{3^{\frac{2}{3}}}{15}\right) \left(9+2 \sqrt{38}\, \sqrt{3}\right)^{\frac{2}{3}}\right)}{30} & \text{otherwise} \end{array}\right.\)

This specification was found using the strategy pack "Insertion Point Placements Expand Verified" and has 16 rules.

Found on January 21, 2022.

Finding the specification took 3 seconds.

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\(\begin{align*} F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{1}\! \left(x \right) &= 1\\ F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\ F_{3}\! \left(x \right) &= F_{11}\! \left(x \right) F_{4}\! \left(x \right)\\ F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\ F_{5}\! \left(x \right) &= F_{2}\! \left(x \right)+F_{6}\! \left(x \right)\\ F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\ F_{7}\! \left(x \right) &= F_{11}\! \left(x \right) F_{12}\! \left(x \right) F_{8}\! \left(x \right)\\ F_{8}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{9}\! \left(x \right)\\ F_{9}\! \left(x \right) &= F_{10}\! \left(x \right)\\ F_{10}\! \left(x \right) &= F_{11}\! \left(x \right) F_{8}\! \left(x \right)\\ F_{11}\! \left(x \right) &= x\\ F_{12}\! \left(x \right) &= F_{13}\! \left(x \right)+F_{14}\! \left(x \right)\\ F_{13}\! \left(x \right) &= F_{2}\! \left(x \right) F_{8}\! \left(x \right)\\ F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{15}\! \left(x \right) &= F_{2}\! \left(x \right) F_{9}\! \left(x \right)\\ \end{align*}\)

This specification was found using the strategy pack "Insertion Row And Col Placements Expand Verified" and has 23 rules.

Found on January 21, 2022.

Finding the specification took 5 seconds.

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\(\begin{align*} F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{1}\! \left(x \right) &= 1\\ F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\ F_{3}\! \left(x \right) &= F_{12}\! \left(x \right) F_{4}\! \left(x \right)\\ F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\ F_{5}\! \left(x \right) &= F_{2}\! \left(x \right)+F_{6}\! \left(x \right)\\ F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\ F_{7}\! \left(x \right) &= F_{12}\! \left(x \right) F_{8}\! \left(x \right)\\ F_{8}\! \left(x \right) &= F_{22}\! \left(x \right)+F_{9}\! \left(x \right)\\ F_{9}\! \left(x \right) &= F_{10}\! \left(x \right) F_{2}\! \left(x \right)\\ F_{10}\! \left(x \right) &= F_{11}\! \left(x \right)+F_{13}\! \left(x \right)\\ F_{11}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{12}\! \left(x \right)\\ F_{12}\! \left(x \right) &= x\\ F_{13}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{17}\! \left(x \right)\\ F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)\\ F_{15}\! \left(x \right) &= F_{12}\! \left(x \right) F_{16}\! \left(x \right)\\ F_{16}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{14}\! \left(x \right)\\ F_{17}\! \left(x \right) &= F_{18}\! \left(x \right)+F_{19}\! \left(x \right)+F_{21}\! \left(x \right)\\ F_{18}\! \left(x \right) &= 0\\ F_{19}\! \left(x \right) &= F_{12}\! \left(x \right) F_{20}\! \left(x \right)\\ F_{20}\! \left(x \right) &= F_{12}\! \left(x \right)+F_{17}\! \left(x \right)\\ F_{21}\! \left(x \right) &= F_{12}\! \left(x \right) F_{14}\! \left(x \right)\\ F_{22}\! \left(x \right) &= F_{16} \left(x \right)^{2} F_{2}\! \left(x \right)\\ \end{align*}\)

This specification was found using the strategy pack "Insertion Point Row Placements Expand Verified" and has 16 rules.

Found on January 21, 2022.

Finding the specification took 5 seconds.

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\(\begin{align*} F_{0}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{1}\! \left(x \right) &= 1\\ F_{2}\! \left(x \right) &= F_{3}\! \left(x \right)\\ F_{3}\! \left(x \right) &= F_{13}\! \left(x \right) F_{4}\! \left(x \right)\\ F_{4}\! \left(x \right) &= F_{0}\! \left(x \right)+F_{5}\! \left(x \right)\\ F_{5}\! \left(x \right) &= F_{2}\! \left(x \right)+F_{6}\! \left(x \right)\\ F_{6}\! \left(x \right) &= F_{7}\! \left(x \right)\\ F_{7}\! \left(x \right) &= F_{10}\! \left(x \right) F_{13}\! \left(x \right) F_{8}\! \left(x \right)\\ F_{8}\! \left(x \right) &= F_{14}\! \left(x \right)+F_{9}\! \left(x \right)\\ F_{9}\! \left(x \right) &= F_{10}\! \left(x \right) F_{2}\! \left(x \right)\\ F_{10}\! \left(x \right) &= F_{1}\! \left(x \right)+F_{11}\! \left(x \right)\\ F_{11}\! \left(x \right) &= F_{12}\! \left(x \right)\\ F_{12}\! \left(x \right) &= F_{10}\! \left(x \right) F_{13}\! \left(x \right)\\ F_{13}\! \left(x \right) &= x\\ F_{14}\! \left(x \right) &= F_{15}\! \left(x \right)+F_{2}\! \left(x \right)\\ F_{15}\! \left(x \right) &= F_{11}\! \left(x \right) F_{2}\! \left(x \right)\\ \end{align*}\)