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Is it possible to find a disjoint set of non-trivial arithmetic sequences $\{(a_m,d_m)\} := \{a_m,a_m+d_m,a_m+2d_m,...\}$, such that no two sequences have the same jump $(m \neq n \rightarrow d_m \neq d_n)$ and $N = \bigcup_{m}\{(a_m,d_m)\}$, where $N$ denotes the natural numbers.

Well, I'm kinda confused of how to approach this, any help would be highly appreciated.

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$a_m=2^{m-1},d_m=2^m$ for $m=1,2,\cdots .$

$m=1$ gives odd numbers. Then $m=2$ gives evens not divisible by $4.$ Then $m=3$ gives multiples of $4$ not divisible by $8.$

Since every natural is uniquely a power of $2$ times an odd, we get all naturals in the union, and the sequences are pairwise disjoint.

Note for clarity: Given $m,$ the $k$-th term ($k=0,1,2,\cdots$) of that sequence is $$2^{m-1}+k \cdot 2^m=2^{m-1}(1+2k),$$ of the form a power of $2$ (including $2^0=1,$) times an odd positive integer. This brings out both the disjointness of the sequences and that they cover all positive integers,

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    $\begingroup$ I see Gerry's way does it using only $5$ sequenes, my way needs infinitely many. [Might be called trivial...] $\endgroup$ – coffeemath May 7 '19 at 0:15
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    $\begingroup$ Yes, but my way isn't disjoint, yours is. $\endgroup$ – Gerry Myerson May 7 '19 at 0:16
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    $\begingroup$ Thank you very much @coffeemath. $\endgroup$ – Ilan Aizelman WS May 7 '19 at 8:24
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    $\begingroup$ @IlanAizelmanWS It was good question. Curious-- was there some way to apply the existence of these sequences? $\endgroup$ – coffeemath May 7 '19 at 11:00
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$(2,2),(3,3),(1,4),(1,6),(11,12)$.

But you asked, how to approach it. Well, $(2,2),(1,2)$ gets everything, but uses jump $2$ twice.

So split $(1,2)$ into $(1,4),(3,4)$. But now we're using jump $4$ twice.

So split $(3,4)$ into $(3,12),(7,12),(11,12)$. And notice that $(3,12)$ is contained in $(3,3)$, and $(7,12)$ is contained in $(1,6)$.

Oops – you wanted the sequences to be disjoint. That can only be done if you allow infinitely many sequences, as in the other answer that has been posted.

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