I would like to know if there is a way to get the recurrence relation $$a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}},\qquad (a_1=1,a_2=2)$$ in closed form, or if there is no such way, how one could proceed to find the limit of $(a_n)$ to by some sort of estimation (or some other method I don't know about).
Solving the recurrence relation $a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}}$
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sequences-and-series
recurrence-relations
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2A simple note: let $a_n=\frac{p_n}{q_n}$. Then $\displaystyle a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}} = \frac{\frac{p_{n-1}^2}{q_{n-1}^2}+\frac{p_{n-2}^2}{q_{n-2}^2}}{\frac{p_{n-1}}{q_{n-1}}+\frac{p_{n-2}}{q_{n-2}}} = \frac{p_{n-1}^2q_{n-2}^2+p_{n-2}^2q_{n-1}^2}{p_{n-1}q_{n-1}q_{n-2}^2+p_{n-2}q_{n-2}q_{n-1}^2}$, so the original recurrence can be expressed as a pair of recurrences $p_n = p_{n-1}^2q_{n-2}^2+p_{n-2}^2q_{n-1}^2$ and $q_n=p_{n-1}q_{n-1}q_{n-2}^2+p_{n-2}q_{n-2}q_{n-1}^2$ for the numerators and denominators of the sequence. – 2012-11-03
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0The limit $1.7919405510508976479$ is not identified by the ISC http://isc.carma.newcastle.edu.au – 2012-11-03
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0Limit varies with $a_2$ as follows: (1, 1)(2, 1.7919)(3, 2.6879)(4,3.6226)(5,4.5775)(100, 99.3487) – 2012-11-03