Safal wrote:

GMMeowChand wrote:

Can anyone solve that problem without the 2? Let me rewrite

Consider the sequence defined by
$a_1 = 2$ , $a_2 = 3$ , and
$a_{2k+1} = 2 + a_k, \quad a_{2k+2} = 2 + a_k + a_{k+1},$ for all integers $k \geq 1$ . Determine all positive integers $n$ such that
$\frac{a_n}{n}$ is an integer.

Edit: Actually I misread the problem statement and tried that problem instead for some times and couldn’t get how to prove that 2k+2 doesn’t divide (a_(2k+2)) and I even dont know if this claim is true. And I got a lot of bounding for other cases

I am hacking this idea from other post above of this problem or this particular thread.

(No Offense Lol!!!)

Show that $\frac{a_n}{n}\leq 2$ for all $n$ , then only possible integer solution for $a_n$ is $n$ or $2n$ .

I think this is the shorter way to kill your problem and also the original problem.

BTW the bound I wrote for $\frac{a_n}{n}$ is obvious , due to the solution present by others, also values of new sequence you give will be smaller than or equal to the original sequence.

Actually i have got that like i got $a_{2k+1}<2k+1$ for some bigger k and the k is not that much big and $a_{2k+2}<4k+4$ so the only thing that I have to care is that $2k+2=a_{2k+2}$ and yeah that's the place where I stuck how to prove if that is true or false. If it is false then how can I get to the contradiction