1
$\begingroup$

Let $\zeta$ be the positive root of $x^2-2023x-1=0$

Define a sequence $\psi_i$ such that $\psi_0=1$ and $$\psi_{n+1}=\left \lfloor{\psi_n \zeta}\right \rfloor , n\geq0$$ $( \left \lfloor{x}\right \rfloor$ denotes Greatest Integer Function)

Find remainder when $\psi_{2023}$ is divided by $\sqrt \psi_2$


$\zeta=\frac{2023+\sqrt{4092533}}{2}$

I was not able to solve the problem analytically but numerically.

$\psi_2=2023^2$

Using Mathematica, I got the remainder as 1012.

But the answer of the problem claims the answer to be 1.

I do not have the solution, and it is my first time dealing with sequences involving the floor function.

$\endgroup$

1 Answer 1

3
$\begingroup$

Hint: Prove that $\psi_{n + 2} = (2023^2 + 2)\psi_n - \psi_{n - 2} - 2023$. Note that $\lfloor \zeta^2 \rfloor = 2023^2 + 1$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .