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Say we have an arithmetic progression in $Z_n$ like $3, 6, 9, 12, ...$ etc. If you move a sliding window of at least 3 values over the progression the 'density' in that subset compared to if the window 'had no holes', is always the same, $\frac{1}{3}$ (I think the term I'm looking for is Natural Density, but this is a new concept for me so I'm not sure if I'm using it 100% correctly).

Arithmetic progressions also have the nice property that they can be described with a fixed amount of space, e.g. for an arithmetic progression you only need to know the starting number and the constant difference to derive the rest of the numbers in the progression.

What types of monotonically increasing progressions can be described with fixed space are known that have "varying" density? By "fixed space" I mean to avoid a hard coded list, instead focusing on sequences that can be generated from some finite set of starting numbers. By "varying" I mean:

  1. A sliding window over the values should not see the same density every time, and that the density seen should increase and decrease, so the density should not be strictly monotonically increasing/decreasing, even though the progression is. So for example a sequence of triangular numbers would not work, because they only ever become further apart.
  2. I should be able to choose a window size vastly greater than the amount of space used to describe the progression and still have it vary (so no cycling through a fixed list of differences).

Apologies if I have not formalized this enough :) I think I'm stumbling on something that must be studied already but I don't know the terminology.

Edit: I was trying to keep things succinct but more constraints are needed. My ultimate goal is to use these sequences on a computer, so sequences that are more easily computable have preference. I'd also like to be able to easily compute the Nth element, without calculating every prior element.

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Let $A\subset\mathbb N$ be the set of all numbers with a leading 8 or 9. –  Mario Carneiro Dec 16 '12 at 21:31
    
Note that the only way for a oscillating pattern to be discernible and not be dampened out regardless of the window size is if the pattern gradually slows down (picture $\cos(\log x)$) as in my example. –  Mario Carneiro Dec 16 '12 at 21:34
    
@MarioCarneiro: Couldn't you get away with just saying all numbers with a leading 8? –  Joseph Garvin Dec 16 '12 at 21:35
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That is fine, the process works in any base. How about "the second-leading digit is 1" for numbers greater than 2? (By leading digit in this context I do mean the one which is always 1 in base 2.) Also, you should edit the OP to reflect the constraints, so others don't have to read the comments. –  Mario Carneiro Dec 16 '12 at 21:47
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Sorry, but no. Any constraint on the last few digits is equivalent to $n\bmod 2^k\in A$ for some $k$ and some $A\subseteq\{0,1,2,\dots,2^k-1\}$, and so it will always have the same "density" with window size $m\cdot2^k$ for $m\geq 1$. The thing which makes my trick work is that as the number gets bigger, the "period of oscillation" grows along with it, so it will eventually be longer than any fixed window. –  Mario Carneiro Dec 16 '12 at 21:57
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If you put any constraint on any set of fixed bits (as in fixed with respect to the radix point), the constraint can be converted to the form "the least-significant $m$ digits form a number in this list...". Let me back up this claim with a proof. Say the constraint is $P(n):=[n_{x_1}=b_1\wedge n_{x_2}=b_2\wedge\dots\wedge n_{x_k}=b_k]$, where $n_i$ is the $i$-th least-significant bit, the $\{x_i\}_1^k$ form an increasing sequence, and $b_i\in\{0,1\}$ are fixed constants for each $i$.

Then the most-significant constraint is for the $x_k$-th bit, so letting $m=x_k+1$, if I let $A=\{n\in\{0,1,\dots,2^m-1\}:P(n)\}$, then if there is some $n\in A$, adding $2^m$ to $n$ will not change any bits below the $m$-th position, so $P(n+2^m)$ is also true, and indeed $P(n+2^ma)$ is true for $a\in\mathbb Z$ as well. Moreover, the same is true for numbers that do not satisfy the constraint, so that $P(n+2^ma)\iff P(n)$ for all $a\in\mathbb Z$. Therefore, $$P(n)\iff n\bmod2^m\in A.$$

Thus if the window is of size $2^m$, the mapping $n\mapsto n\bmod2^m$ takes any set $\{c,\dots,c+2^m-1\}$ to $\{0,\dots,2^m-1\}$, so the "window density" is $|A|/2^m$ regardless of $c$. You get this same result if the window size is a multiple of $2^m$, so it fails your "cycling through a fixed set of differences" test.

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I'm probably being dense, but I don't see how your first sentence could be true. If the number is 32-bits for example, fixing bits 3, 5, and 7 to particular values doesn't put any constraint in the "leading digits", that is, the leading bits, 8-32. Unless by first digits you mean the least significant bits, but it doesn't impose any constraint on bits 0-2 either, so I'm still not sure what you mean. Is k indexing the total number of bits, or is it indexing only the bits that we've locked? –  Joseph Garvin Dec 16 '12 at 22:52
    
Sorry, by "first" I meant it in little-endian sense, i.e. "least significant". I've edited it to reflect that. –  Mario Carneiro Dec 16 '12 at 23:15
    
Your edit helps a bit, thanks. So it sounds like you're saying that if you take the most significant locked bit, $x_k$, you can keep adding $x_k + 1$ and get numbers satisfying the constraint that are evenly spaced. But isn't the neglecting the unfixed bits that are below $x_k$? Why don't we get variable density from those? I'm actually unsure if your conclusion is answering the question or saying it can't be answered -- my test is to eliminate possibilities, so failing the test would mean a proof that fixing bits works, but in the comments on the question you said fixing them doesn't? –  Joseph Garvin Dec 16 '12 at 23:24
    
To take a more manageable example, if I say the 2nd bit is 1 and leave the other two free, this is equivalent to $n\bmod8\in\{4,5,6,7\}$. The way the "unconstrained" numbers are accounted for is that I am listing all solutions, ranging over all unspecified bits. You do get variable density, but my argument says that for large enough windows, you lose all of that variation, since it is a long but repeating pattern, and you are averaging over the whole thing. (And my proof is answering the question, in the negative.) –  Mario Carneiro Dec 16 '12 at 23:27
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If I say "the 2nd bit is 1", then $P(n):=[n_2=1]$, so $k=1$, $x_1=2$, and $b_1=1$. If I said "3rd is 0 and 5th is 1", then $k=2$, $x_1=3$, $x_2=5$, $b_1=0$, and $b_2=1$. Does this help? –  Mario Carneiro Dec 16 '12 at 23:39
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