INMO 2025, Question 1

Question

Question:

Consider the sequence defined by

$a_1 = 2 $

$a_2 = 3$,

and $a_{2k+2} = 2+a_k +a_{k+1}$ and $a_{2k+1} =2+2a_k$ for all integers $ k \ge 1$.

Determine all positive integers n such that $\frac{a_n}{n}$ is an integer.

P.S.: This question is already on Mathstackexchange here, but I solved it in a different way and am not able to understand where I went wrong. Can you please help me find where my solution is wrong?

My Attempt:

Let the first equation be $a_{2k+2} = 2+a_k +a_{k+1}$ ----eq 1

Second equation $a_{2k+1} =2+2a_k$ -----eq 2

from eq 1, $a_{2k} = 2+a_k +a_{k-1}$ for $k \ge 2$

$a_{2k} = (1+a_k) +(1+a_{k-1})$

From eq 2, $\frac{a_{2k+1}}{2} =1+a_k$ . Similarly, $\frac{a_{2k-1}}{2} =1+a_{k-1} $ for $k \ge 2$

Therefore , $a_{2k} = \frac{a_{2k+1}}{2} +\frac{a_{2k-1}}{2}$

$2a_{2k} = a_{2k+1} +a_{2k-1}$ ----- eq3

$2a_{2k+1} = a_{2k} - a_{2k-1}$ for $k \ge 2$

Now, I got a recursive formula for $a_k$ when $k$ is odd. Next, I tried to find the recursive formula when $k$ is even.

Let's assume an integer m which is odd.

Therefore, $2a_{m} = a_{m-1} - a_{m-2}$ for $m \ge 5$

using eq 1, $a_{2m} = 2+a_m +a_{m-1}$ for $m \ge 2$ . Substituting $a_m$,

$a_{2m} = 2+ 2a_{m-1} - a_{m-2} +a_{m-1}$

$a_{2m} = 2+ 3a_{m-1} - a_{m-2}$

$a_{2m} = 3(1+a_{m-1}) -(1+ a_{m-2})$ for $m \ge 5$

$a_{2m} = 3(\frac{a_{2m-1}}{2}) -(\frac{a_{2m-3}}{2})$

$2a_{2m} = 3a_{2m-1} - a_{2m-3}$

using eq3 , $2a_{2m-2} = a_{2m-1} +a_{2m-3}$ for $ m \ge 4$

$a_{2m-1} - 2a_{2m-2} = - a_{2m-3}$

Substituting,

$2a_{2m} = 3a_{2m-1} + a_{2m-1} - 2a_{2m-2} $

$a_{2m} = 2a_{2m-1} - a_{2m-2} $ for $ m \ge 5$

We proved this formula works for $a_{2m}$ where m is odd. Now let $n=2m$ be even.

Using the relation $a_{2m} = 2a_{2m-1} - a_{2m-2} $ for $ m \ge 5$ we got right now,

$a_{n} = 2a_{n-1} - a_{n-2} $ for $ n \ge 10$ and n is even

this is the same formula we got while deriving the relation for m when it is odd. So, it's the same calculation and we get to the conclusion that

$a_{2n} = 2a_{2n-1} - a_{2n-2} $ for $ n \ge 10$

Now, we have proved this formula works for every number even or odd.

so in general, $a_{r} = 2a_{r-1} - a_{r-2} $ for $ r \ge 20$

But this formula is holds incorrect for r = 24

Why??