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Suppose $a_{1}=a_{2}=\frac{1}{3}$, for $n = 3, 4, 5, ...$ , we have $$a_{n}=\frac{a_{n-1}^2(1−2a_{n-2})}{2a_{n-1}^2 − 4a_{n-2}a_{n-1}^2+a_{n-2}}$$

(1) Prove that $\frac{1}{a_{n}} −2$ are square numbers

(2) Find out the general term formula of ${a_{n}}$

I have done the first part using induction by assuming $x_n=\frac{1}{a_n} -2$, we get $$x_n=\frac{1}{x_{n-2}a_{n-1}^2}$$ on simplifying the relation but not able to solve to get general term, Can anyone suggest, how to proceed?

Anne Bauval
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Anirudh Kumar
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1 Answers1

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It seems that $\big(\frac{1}{a_n}-2\big)^{1/2}$ is sequence A001835 . Can you prove this by induction?

Find everything you need to know there. Many references. And in particular $$ \big(\frac{1}{a_n}-2\big)^{1/2} = \frac{(3 + \sqrt3)^{2n - 1} + (3 - \sqrt3)^{2n - 1}}{6^n} $$

GEdgar
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  • yeah, this is sequence A001835, but how you reached here, want to know? – Anirudh Kumar Jun 25 '23 at 12:31
  • I did what I suggested in my comment. Computed the first 10 terms, used your hint and computed $\big(\frac{1}{a_n}-2\big)^{1/2}$, then searched for that at OEIS. – GEdgar Jun 25 '23 at 17:29