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Let us say that we have the order $\{1,2,3,4,5,...,k\}$ of $k$ elements. Now, I want to determine in how many different ways can we pick a family of increasing substrings from $\{1,...,k\}$? The rules are the following:

  1. We do not consider the trivial case of picking the entire $\{1,...,k\}$ as a whole.
  2. We want to pick at least one increasing substring, but we can pick multiple different ones (such as if $k=5$ we can pick just $\{1,2\}$, but we may also pick $\{1,2\}$ and $\{3,4,5\}$, using the entire sequence and it would also be possible to pick something like $\{1,2\}$ and $\{3,4\}$).
  3. The substrings may not overlap with one another (i.e. if for example $k=5$ we cannot have $\{1,2,3\}$ and $\{3,4\}$).
  4. The increasing substrings must only consist of consecutive numbers (i.e. we cannot have anything like $\{1,3,4\}$).
  5. A substring cannot be just one single number i.e. $\{1\}$.

So basically, we are curious about how many different combinations of these substrings we can have in total as families. An example would be, for the case of $k=4$ the following:

$\Big\{ \{1,2\}, \{1,2,3\}, \{2,3\}, \{3,4\}, \{2,3,4\}, \{\{1,2\}, \{3,4\}\} \Big\}$

so the number of families in this case would be $6$.

I have thought about it this way: if I have $\{1,...,k\}$ and enumerate them starting from only having one increasing substring such as $\{1,2\}$ or anything $\{1,...,j\}$ where $j \le k-1$. There are the total of $k-1$ of these.

Then we would enumerate any substrings of the length two on their own (minus 1 from the previous part), length three (minus 1 from the previous part), length $k-1$ (minus one from the previous part).

I believe how this is how we would do it for the case of considering only single substrings, but we also want to consider different combinations of multiple substrings, and I am not sure how to go about doing this.

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    This is not clear. You say you want to count the number of increasing substrings, but it appears that you really want something else. Like the number of families of increasing substrings with some conditions. I don't think these rules are explained very well. I suggest editing your post to include some examples. For $k=3$, for instance, why not work out the answer you want completely, showing all the "good" choices. If $k=3$ is trivial for some reason, then do $k=4$ whatever it takes to illustrate the question properly. – lulu Jan 08 '21 at 11:22
  • Yes! I am looking for the number of families of increasing substrings with these conditions. I edited the post. For $k=3$ the case would be ${{1,2},{2,3}}$, meaning that the number of families of increasing substrings would be $2$. –  Jan 08 '21 at 11:43
  • So, the question appears to be about choosing pairs of integers, representing the first and last terms of your substrings.I think you just want to choose evenly many elements of your set. – lulu Jan 08 '21 at 11:51
  • Yes, the question could be viewed like that, but with the conditions of (1)-(4) i.e. the integers cannot 'overlap' and such. –  Jan 08 '21 at 11:54
  • Those conditions are automatic. write your subset in increasing order $a_1<a_2<\cdots<a_{2n}$ and now look at the substring connecting $a_1$ to $a_2$ then $a_3$ to $a_4$ and so on. I'll post something below. – lulu Jan 08 '21 at 12:05

1 Answers1

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This is the same as asking for the number of subsets of $\{1, \cdots, k\}$ with evenly many elements (not including the empty set or the set $\{1,k\}$). Indeed, we just pair the elements in the subset sequentially to get the first and last elements of your substrings. Thus, with $k=11$ the subset $\{3,6,7,11\}$ corresponds to the family $\{\{3,4,5,6\}, \{7,8,9,10,11\}\}$.

In general there are $2^{k-1}$ subsets with evenly many elements (see, e.g., this question). As we wish to exclude the empty set and the set $\{1,k\}$ the answer is $$\boxed {S_k=2^{k-1}-2}$$

Note that $S_3=4-2=2$ and $S_4=8-2=6$ in harmony with your examples.

lulu
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