# find a disjoint set of non-trivial arithmetic sequences such that no two sequences have the same jump and $N = \bigcup_{m}\{(a_m,d_m)\}$

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.

$$a_m=2^{m-1},d_m=2^m$$ 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,

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

$$(2,2),(3,3),(1,4),(1,6),(11,12)$$.

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.