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 6 of 6: Take the floor

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In plain words

An integer plus a positive fraction less than $1$ has that integer as its floor; the closed form for $x_n$ is exactly the exponent in the target formula.

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

Detailed analysis

For $n\ge1$ , $x_n$ is a positive integer, so $2^{x_n}$ is an integer and $0<2^{-x_n}<1$ . Since $u_n=a_n=2^{x_n}+2^{-x_n}$ , we get $\lfloor u_n\rfloor=2^{x_n}$ . Substituting the closed form from Step 2 yields $\boxed{\lfloor u_n\rfloor=2^{\frac{2^n-(-1)^n}{3}}}$ , as required.