$λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.

Question

Determine the value of the constant $λ > 0$ such that $2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$ is true for all positive integers n.

I tried equating it to n = 1 and found that $\lfloor \lambda \rfloor = 0 , 2$

Then I equated it to n = 2 and found out that $\lfloor \lambda\rfloor = 2$

I use the original equation to get (let x = fractional part of lambda)

$2\lfloor{\lambda n}\rfloor = 1 - n + \lfloor\lambda \lfloor\lambda n \rfloor\rfloor$

= $2\lfloor{(2+x) n}\rfloor = 1 - n + \lfloor (2+x) \lfloor(2+x) n \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor (2+x)(2n+ \lfloor nx \rfloor\rfloor$

->$4n+2\lfloor{nx}\rfloor = 1 - n + \lfloor 4n +2\lfloor nx \rfloor + 2nx +x\lfloor nx \rfloor\rfloor$

-> $n-1 = \lfloor 2nx + x\lfloor nx \rfloor \rfloor$.

I do not know how to proceed, I tried doing $n-1<y< n$ but it didn't give me anything

Answer 1

Take $n=1$, we have $2\lfloor{\lambda }\rfloor = \lfloor\lambda \lfloor\lambda \rfloor\rfloor\Rightarrow \lfloor{\lambda }\rfloor=2,\ \lfloor{2\lambda }\rfloor=4$ (another situation is impossible).

Take $n=2$, we have $2\lfloor{2\lambda }\rfloor =-1+ \lfloor\lambda \lfloor2\lambda \rfloor\rfloor\Rightarrow \lfloor{4\lambda }\rfloor=9$.

Let $a_1=2,a_2=4,a_3=9$, and $a_{n+1}=\lfloor{a_n\lambda }\rfloor$. Then the condition becomes $$2a_n=1-a_{n-1}+a_{n+1}\Rightarrow a_n=\frac{1}{2}+\frac{(1-\sqrt2)^{n+1}+(1+\sqrt2)^{n+1}}{4}.$$

Hence, $$\frac{a_{n+1}}{a_n}\le \lambda<\frac{a_{n+1}+1}{a_n},\quad\forall n\in\mathbb{N}.$$ As a consequence, $\lambda$ has the unique value $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1+\sqrt2.$

Answer 2

$$ \lfloor \lambda \lfloor \lambda n \rfloor \rfloor =2\lfloor{\lambda n}\rfloor + n - 1 $$

$$ 2\lfloor{\lambda n}\rfloor + n - 1 \le \lambda \lfloor \lambda n \rfloor \lt 2\lfloor{\lambda n} \rfloor + n$$

$$ n - 1 \le (\lambda - 2) \lfloor \lambda n \rfloor \lt n$$

Since this has to be true for all $n$, we must have $\lambda > 2$.

So $$ (\lambda - 2)(\lambda n - 1) < (\lambda - 2)\lfloor \lambda n \rfloor \le (\lambda - 2)\lambda n$$

Then

$$ n - 1 \le (\lambda - 2) \lfloor \lambda n \rfloor \le (\lambda - 2)\lambda n$$ $$ n - 1 \le (\lambda - 2)\lambda n$$ $$ n - 1 \le \lambda^2 n - 2\lambda n$$ $$ \lambda^2 n - 2\lambda n + n \ge 2n-1 $$ $$ (\lambda - 1)^2 n \ge 2n-1$$ $$\lambda \ge 1 + \sqrt{2 - \frac 1n} $$

Letting $n \to \infty$, we find $$\lambda \ge 1 + \sqrt 2 $$

Also, since

$$ (\lambda - 2)(\lambda n - 1) < (\lambda - 2)\lfloor \lambda n \rfloor < n$$

Then

$$ (\lambda - 2)(\lambda n - 1) < n$$ $$\lambda^2n - (2n+1)\lambda < n - 2$$ $$\lambda^2 - \left(2+\frac 1n \right)\lambda < 1 - \frac 2n$$

Letting $n \to \infty$,

$$\lambda^2 - 2\lambda \le 1$$ $$\lambda^2 - 2\lambda + 1 \le 2$$ $$(\lambda - 1)^2 \le 2$$ $$\lambda \le 1+\sqrt 2$$

So $\lambda = 1 + \sqrt 2$