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Assume you have a set of $n$ distinct elements. A single element can be placed in a group, and groups can be combined to form more groups. A group must contain at least one element to be valid, and cannot contain any element more than once.

Valid Groups

$(1)$

$(1\to2)$

$(3\to(1\to2))$

Invalid Groups

$(1\to1)$

$()$

When combining groups, the order of the elements does not matter.

$(1\to2) = (2\to1)$

However, each group is it's own distinct entity so the order of combinations does matter

$((1\to2)\to3) \ne ((1\to3)\to2)$

Under these rules, how many valid groups can be created from a set of $n$ elements? I know, calculating by hand that when $n = 1$ there is $1$ valid group, when $n = 2$ there are $3$ valid groups, when $n = 3$ there are $9$ valid groups, and when $n = 4$ there are $37$ valid groups.

Apologies if I'm using the notation and terminology incorrectly. I'm ignorant to whatever branch of mathematics this is.

Mike Earnest
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Medynsky
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  • This isn't group theory, P,ease read the tag description before ising it, and if you don't understand it, then don't use it. – Arturo Magidin Jan 17 '22 at 02:33
  • The Online Encyclopedia of Integer Sequences knows a lot of sequences containing $1,3,9,37$; https://oeis.org/search?q=1%2C3%2C9%2C37&language=english&go=Search – Gerry Myerson Jan 17 '22 at 03:55
  • I think when $n=3$ there are only 7 valid groups and when $n=4$ there are only 15 valid groups. – Selrach Dunbar Jan 17 '22 at 05:28
  • The Integer Sequence Encyclopedia is a nice resource! However I did a bit more work and found n=5 for my specific sequence is 195, which doesn't match any of the sequences. That's 5 size 1 groups, 10 size two groups, 30 size three groups, 75 size 4 groups, and 75 size 5 groups. – Medynsky Jan 17 '22 at 05:45
  • The following are all 9 of the valid groups I could find for $n=3$: $(1),(2),(3),(1\to2),(1\to3),(2\to3),((1\to2)\to3),((1\to3)\to2),((2\to3)\to1)$. The last three are all different groups. You can't have a group like $(1\to2\to3)$, you have to combine two elements first before combining the third. – Medynsky Jan 17 '22 at 05:53
  • For the simpler problem where the elements are all equal and thus interchangeable you would get OEIS A173404 which are the partial sums of OEIS A002572: the number of binary rooted trees. – Fabius Wiesner Jan 17 '22 at 13:34
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    @Medynsky My answer gives $225$ groupings for $n=5$, contradicting your comment. You said there were 75 size 5 groups; I think there should be $105=3\cdot 5\cdot 7$ size 5 groups. I think you may have missed the $30$ groupings of the form (a -> b) -> (c -> (d -> e)). – Mike Earnest Jan 18 '22 at 01:16

1 Answers1

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Answer: The number of ways to form groups is $$ \binom n1+\binom n2 1!!+\binom n3 3!! + \binom n4 5!!+ \dots + \binom nn (2n-3)!! $$ Here, $(2k-1)!!=(2k-1)\times (2k-3)\times \cdots \times 3\times 1$ is the product of the first $k$ odd numbers. The above formula can be written succinctly as $$\sum_{k=1}^n \binom nk (2k-3)!!$$ as long as we adopt the funny-looking convention that $(-1)!!=1$.

Proof: First, let us count the number of ways to form this hierarchical group structure which uses all of the numbers in $\{1,\dots,n\}$. This is exactly answered by this other MSE question:

How many ways to write a commutative non-associative product of $n$ terms?

Two answers there prove that the number of ways is $(2n-3)!!$ for all $n\ge 1$. For your question, you first need choose the number of elements $k$ your group structure will use, then choose the particular elements in $\binom{n}k$ ways, then finally put a group structure on them in $(2k-3)!!$ ways. This leads exactly to the advertised sum.

For example, when $n=4$, my formula gives $$ \binom 41 + \binom 42 + \binom 43\cdot 3+\binom 44\cdot 3\cdot 5=4+6+12+15=37\;\;\color{green}{\checkmark} $$

Mike Earnest
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