Sounds very nice ! Any ideas?!

I will prove a stronger problem: Problem: A sequence $P=\left \{ a_{n} \right \}$ is called a $ \text{Permutation}$ of natural numbers (positive integers) if for any natural number $m,$ there exists a unique natural number $n$ such that $a_n=m.$ We also define $S_k(P)$ as: $S_k(P)=a_{1}+a_{2}+\cdots +a_{k}$ (the sum of the first $k$ elements of the sequence). Prove that there exists infinitely many distinct $ \text{Permutations}$ of natural numbers like $P_1,P_2, \cdots$ such that: $\left\{\begin{array}{lll} 1)\ \forall k, \forall i<j: S_k(P_i)|S_k(P_j)\\ \\ 2)\ \forall i\ P_i(a_n) \ \text{has infinitely elements like}\ a_j\ \text{such that gcd}(a_1+a_2+\dots +a_{j-1},a_j)=1 \end{array}\right.$ Solution: Define $P_1$ be an arbitrary permutation of natural numbers such that it satisfy the problem's conditions (for condition 2 after defining the first $m$ elements we can choose $a_{m+1}$ be a prime larger than $a_1+a_2+\dots +a_m$ then obviously $\text{gcd}(a_{m+1},a_1+a_2+\dots+a_m)=1$; we can do this for infinitely natural numbers $m$). For any $r\in \mathbb{N}$ we want to build the sequences $P_2,P_3,\dots,P_r$ such that $P_1,P_2,\dots ,P_r$ satisfy the conditions of problem. we build these sequences by induction on $r$; so assume that we have already build the sequences $P_1,P_2,\dots,P_r=\{a_n\}$. we build the sequence $P_{r+1}=\{b_n\}$ in the following way: Define $b_1$ be a large enough number such that $a_1\mid b_1$ so the sequence $P_{r+1}$ must be different from $P_1,P_2,\dots ,P_r$ because of the first element. Now in each step we choose $b_i$ (for $i\geq 2$) such that it is the least positive integer such that $a_1+a_2+\dots +a_i\mid b_1+b_2+\dots +b_i$ or $b_i\equiv -(b_1+b_2+\dots+b_{i-1})\pmod{a_1+a_2+\dots +a_i}$ such that $b_i\notin \{b_1,b_2,\dots ,b_{i-1}\}$ (we call this a good step). From the assumption there are infinitely index $i$ such that $\text{gcd}(a_i,a_1+a_2+\dots +a_{i-1})=1$ let $A=\{i_1,i_2,\dots \}$ be a set of all these indexes. take one of these indexes and call it $n$; so $\text{gcd}(a_n,a_1+a_2+\dots +a_{n-1})=1\ \bigstar$ now assume that that we have chosen $b_1,b_2,\dots,b_{n-2}$. I claim that we can choose $b_{n-1},b_n$ such that $\text{gcd}(b_n,b_1+b_2+\dots +b_{n-1})=1$: $\left\{\begin{array}{lll} a_1+a_2+\dots +a_{n-1}\mid b_1+b_2+\dots +b_{n-1}\Longrightarrow b_{n-1}=(a_1+a_2+\dots +a_{n-1})u-(b_1+b_2+\dots +b_{n-2})\ u\\ \\ a_1+a_2+\dots +a_n\mid b_1+b_2+\dots +b_n\Longrightarrow b_n=(a_1+a_2+\dots +a_n)v-(b_1+b_2+\dots +b_{n-1})=(a_1+a_2+\dots +a_n)v-(a_1+a_2+\dots +a_{n-1})u \end{array}\right.$ $\Longrightarrow \text{gcd}(b_n,b_1+b_2+\dots +b_{n-1})=1\Longleftrightarrow \text{gcd}\left( (a_1+a_2+\dots +a_n)v-(a_1+a_2+\dots +a_{n-1})u,(a_1+a_2+\dots +a_{n-1})u\right)=\text{gcd}\left( (a_1+a_2+\dots +a_n)v,(a_1+a_2+\dots +a_{n-1})u\right)=1\spadesuit$ Note that from $\bigstar\Longrightarrow \text{gcd}(a_1+a_2+\dots +a_n,a_1+a_2+\dots +a_{n-1})=1$ Hence it's enough to choose $u$ such that $\text{gcd}(u,a_1+a_2+\dots +a_n)=1$ and then choose $v$ such that $\text{gcd}\left(v,(a_1+a_2+\dots +a_n)u\right)=1$ for satisfying $\spadesuit$ hence we defined $b_{n-1},b_n$ such that $\text{gcd}(b_n,b_1+b_2+\dots +b_{n-1})=1$; we call this step a nice step. for satisfying the condition 2 of the problem it's enough to do nice steps for infinitely many indexes like $n\in A=\{i_1,i_2,\dots \}$ (not necessary for all indexes in $A$) instead of good steps . assume that after defining $n$ elements $t$ is the least positive integer that is not among $b_1,b_2,\dots ,b_n$. I claim that we can choose the next elements such that $t$ must be chosen in one good step. take one large enough index $N\in A$ we continue defining $b_{n+1},b_{n+2},\dots ,b_{N-2}$ by good steps.(if $t$ defined during these steps our claim proves so assume otherwise). Now we define $b_{N-1}$ such that: $\left\{\begin{array}{ccc} b_{N-1}\equiv -(b_1+b_2+\dots +b_{N-2})\pmod{a_1+a_2+\dots +a_{N-1}}\\ \\ b_{N-1}\equiv -(b_1+b_2+\dots +b_{N-2}+t)\pmod{a_1+a_2+\dots +a_N}\ \clubsuit \end{array}\right.$ $b_{N-1}$ exists from Chinese remainder theorem. (because $\text{gcd}( a_1+a_2+\dots +a_{N-1},a_1+a_2+\dots +a_N)=\text{gcd}( a_1+a_2+\dots +a_{N-1},a_N)=1$) Now doing a nice step $b_N$ must be equal to the $t$ according to $\clubsuit$. we call the step defining the $b_{N-1}$ a big step. (Note that in both nice step and big step we used indexes in $A$) Hence continuing these three kind of steps we can define $P_{r+1}$ satisfying the problem's conditions. Q.E.D

andria wrote: Now doing a nice step $b_N$ must be equal to the $t$ according to $\clubsuit$. we call the step defining the $b_{N-1}$ a big step. (Note that in both nice step and big step we used indexes in $A$) Sorry for reviving, but I'm confused about how to guarantee $gcd(t,b_1+\cdots+b_{N-1})=1$ (which is equivalent to $gcd(t,a_1+\cdots+a_N)=1$ )as nice step required?

Rickyminer wrote: andria wrote: Now doing a nice step $b_N$ must be equal to the $t$ according to $\clubsuit$. we call the step defining the $b_{N-1}$ a big step. (Note that in both nice step and big step we used indexes in $A$) Sorry for reviving, but I'm confused about how to guarantee $gcd(t,b_1+\cdots+b_{N-1})=1$ (which is equivalent to $gcd(t,a_1+\cdots+a_N)=1$ )as nice step required? You don't need that because of (not necessary for all index in A)... You just need infinitly of nice step

The main idea similar to this easier problem : Prove that there exist a permutation of natural numbers like $P=\left \{ a_{n} \right \}$ , such that for each $n \in \mathbb{N}$ we have : $$n|a_1+a_2+...+a_n$$Actually if you solve this one first , then this beautiful problem will be much easier for you !