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In a generalized arithmetic progression there is a set of constant differences you can choose from at each step. So a generalized arithmetic progression starting at 0 with possible constant differences 2, 3, and 5 would contain at least every multiple of 2, every multiple of 3, and every multiple of 5. I'm interested in a more limited case, where the constant differences have a fixed order (not necessarily sorted), and each step we round robin the differences -- we can no longer make an arbitrary choice of constant difference at each step. Is there a name for that I can Google for to learn about?

For example, for such a 'circular arithmetic progression' (what I'm calling it for lack of name) if the constant differences were (2, 3, 5), the progression would be:

0 2 5 10 12 15 20 22 ...

Which would be distinct from if the constant differences were (3, 5, 2):

0 3 8 10 13 18 20 23 ...
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Looks like one gets a collection of interlaced ordinary arithmetic progressions. For example in 0,2,5,10,12,15... there are three interlaced: 0,10,20,... and 2,12,22,... and 5,15,25... each being a usual arithmetic progression. – coffeemath Nov 28 '12 at 13:26
@coffeemath: Yep, I noticed that too. Maybe they're not specifically studied and that's why? It definitely would make it easier to prove properties about them. – Joseph Garvin Nov 28 '12 at 14:57
For example for the three difference case $(a,b,c)$, if $a+b+c=k$, then subtraction of the sequence $0,0,0,k,k,k,2k,2k,2k...$ from the sequence gives a periodic sequence $a,b,c,a,b,c,...$. So it might be viewed as adding a step function to a periodic sequence. – coffeemath Nov 28 '12 at 15:32
@coffeemath: Whoa, that's a cool trick. What prompted you to think of that? – Joseph Garvin Nov 28 '12 at 16:13
up vote 2 down vote accepted

Your sequences are unions of arithmetic progressions which are the open sets in the topology below:

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