Bump this. Can i get a hint?

A Counter proof: $\gcd (a_{n+1} a_n + 1 , a_i)=1 \Rightarrow \gcd (a_{n+1},a_i ) =1$ for each $i=1,2, \cdots n$. $\Rightarrow $ $\forall i,j \geq 3$ : $ \gcd ( a_i , a_j ) =1$. So if $\{a_n\}$ contains all positive integers, then there exists $k,l > 3$ such that $a_k = p^b, a_l=p^q$ $\Rightarrow p \mid \gcd(a_k,a_l)$, a contradiction.

chystudent1-_- wrote: A Counter proof: $\gcd (a_{n+1} a_n + 1 , a_i)=1 \Rightarrow \gcd (a_{n+1},a_i ) =1$ for each $i=1,2, \cdots n$. $\Rightarrow $ $\forall i,j \geq 3$ : $ \gcd ( a_i , a_j ) =1$. So if $\{a_n\}$ contains all positive integers, then there exists $k,l > 3$ such that $a_k = p^b, a_l=p^q$ $\Rightarrow p \mid \gcd(a_k,a_l)$, a contradiction. How does $\gcd(a_{n + 1} a_n + 1, a_i) = 1$ imply $\gcd(a_{n + 1}, a_i) = 1$?

Bump. Does anyone have a solution?

Bump this elegant and hard problem. Does anybody have any ideas?

Does anyone have a solution？ I have thought about it for so long that i am starved for a solution

I suspect it needs a very clever representation of natural numbers , I found that the increasing sequence of primes is placed on the triangular numbers , Thus I think we need some kind of representation like this : $$\begin{matrix} 1 & 2 & a_2 & \cdots \\ 3 & b_1 & b_2 & \cdots \\ 5 & c_1 & c_2 & \cdots \\ 7 & d_1 & d_2 & \cdots\\ 11 & \cdots \end{matrix}$$where we move to and fro on the diagonals : $2 \rightarrow 2 \rightarrow 3 \rightarrow a_2 \rightarrow b_1 \rightarrow 5 \rightarrow \cdots$

Hedra wrote: I suspect it needs a very clever representation of natural numbers , I found that the increasing sequence of primes is placed on the triangular numbers , Thus I think we need some kind of representation like this : $$\begin{matrix} 1 & 2 & a_2 & \cdots \\ 3 & b_1 & b_2 & \cdots \\ 5 & c_1 & c_2 & \cdots \\ 7 & d_1 & d_2 & \cdots\\ 11 & \cdots \end{matrix}$$where we move to and fro on the diagonals : $2 \rightarrow 2 \rightarrow 3 \rightarrow a_2 \rightarrow b_1 \rightarrow 5 \rightarrow \cdots$ I didn't understand. Could you please elaborate more?

I did a OEIS search and found that this sequence is already known and prior to my suspicion , It does have a clever representation of the natural numbers and @above you will find out in this link what I meant. Though proving that this sequence is indeed completely equivalent to the sequence asked is kinda difficult , I have been working on this and will post a solution when (and if) I get the complete solution.

Hedra wrote: I did a OEIS search and found that this sequence is already known and prior to my suspicion , It does have a clever representation of the natural numbers and @above you will find out in this link what I meant. Though proving that this sequence is indeed completely equivalent to the sequence asked is kinda difficult , I have been working on this and will post a solution when (and if) I get the complete solution. I think it's not the same sequence. $12\times 13+1=157$, which is prime. So $a_{13}=13$, not $15$.

I am afraid you are correct.

I'm longing for a solution.

Open Problem?