Problem 6

A sequence

(u_n)

is defined by

u_0=2

u_1=\frac52

, and

u_{n+1}=u_n(u_{n-1}^2-2)-u_1

for

n=1,2,\ldots

. Prove that for every positive integer

n

\lfloor u_n\rfloor=2^{\frac{2^n-(-1)^n}{3}}

, where

\lfloor x\rfloor

denotes the greatest integer less than or equal to

x

Step 1 of 6: Record the recurrence

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In plain words

The recurrence is designed for the identity $(2^r+2^{-r})^2-2=2^{2r}+2^{-2r}$ .

u_0=2,\quad u_1=\frac52,\quad u_{n+1}=u_n(u_{n-1}^2-2)-u_1

Detailed analysis

We must prove the floor identity for the sequence with $u_0=2$ , $u_1=\frac52$ , and $u_{n+1}=u_n(u_{n-1}^2-2)-u_1$ for $n\ge1$ . The useful idea is to represent every term as a sum of a power of $2$ and its reciprocal.