Closed form of recurrence relation $a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}$

Question

Given the following sequence:

$$\begin{cases}a_{0}=2\\a_{1}=\frac{5}{2}\\a_{n}=a_{n-1}(a_{n-2}^{2}-a_{0})-a_{1}\end{cases}$$

Prove that

$$a_{n}=2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}$$

For all natural numbers $n$.

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My try: (using induction)

Let $b_{n}=\frac{2^{n}-(-1)^{n}}{3}$

For $n=0$: correct $\checkmark$

Now for $n+1$:

$$ \begin{aligned} a_{n+1} &= a_{n}(a_{n-1}^{2}-a_{0})-a_{1})=(2^{b_{n}}+2^{-b_{n}})(2^{2b_{n}}+2^{-2b_{n}}+2) \\ &=2^{3b_{n}}+2^{-3b_{n}}+2^{b_{n}}+2^{-b_{n}}+2^{b_{n}+1}+2^{-b_{n}+1} \end{aligned} $$

How do I proceed from here?

Answer 1

You want to proof that $a_n = f(n)$ for any $n\inℕ$ with $$ f(n) := 2^{\frac{2^{n}-(-1)^{n}}{3}}+2^{\frac{(-1)^{n}-2^{n}}{3}}.$$

Check of two (!) initial conditions:

• $f(0) = 2 = a_0 \checkmark$
• $f(1) = \frac52 = a_1 \checkmark$

Inductive step: $$\begin{aligned} a_{n+1} &= a_n (a_{n-1}^2 - a_0) - a_1\\ &= f(n) (f(n-1)^2 - f(0)) - f(1)\\ &= (2^{b_n} + 2^{-b_n}) \cdot ((2^{b_{n-1}} + 2^{-b_{n-1}})^2 - 2) - \frac52\\ &= (2^{b_n} + 2^{-b_n}) \cdot (2^{2b_{n-1}} + 2^{-2b_{n-1}}) - \frac52 \\ &= 2^{b_n+2b_{n-1}} + 2^{b_n-2b_{n-1}} + 2^{-b_n+2b_{n-1}} + 2^{-b_n-2b_{n-1}} - \frac52 \end{aligned}$$ Simple computations show that: $$\begin{aligned} b_n+2b_{n-1} &= \frac13(2^{n+1} + (-1)^n) = b_{n+1}\\ b_n-2b_{n-1} &= (-1)^{n+1}. \end{aligned}$$

Thus $$a_{n+1} = f(n+1) + 2^{(-1)^{n+1}} + 2^{(-1)^n} - \frac52$$

With the equality (for $n\in\mathbb N$) $$2^{(-1)^{n+1}} + 2^{(-1)^n} = \frac52$$ we can finish the proof: $$a_{n+1} = f(n+1).$$