It is well-known that any rational number can be represented using a sum of distinct Egyptian fractions (that is, rational fractions of the form $1/n$ with $n\in\mathbb{N}$). This may be proven by establishing a greedy algorithm that constructs a sequence of such decompositions. For instance, 
\begin{align}
1
&= \frac12+\frac13+\frac16 \\
&= \frac12+\frac13+\frac17+\frac1{42}\\
&= \frac12+\frac13+\frac17+\frac1{43}+\frac1{1806}\\
&= \cdots\\&=\sum\_{k=1}^\infty \frac{1}{x\_k} \tag{1}
\end{align}
where $x\_1=2$ and $x\_k = 1+\prod\_{j=1}^{k-1}x\_j$ for $k>1$ (i.e. Sylvester's sequence [A000058](https://oeis.org/A000058])).


As noted by Bergner & Walker [1], such Egyptian fractions are relevant in the context of groupoid cardinality. To briefly review: Given a groupoid $G$, the groupoid cardinality is defined [2] by $|G|=\sum\_{[\bullet]\in G}\frac{1}{\#\text{Aut}(\bullet)}$ where $[\bullet]$ is a component of $G$ and $\#\text{Aut}(\bullet)$ is the order of the automorphism group of $\bullet$. If $G$ is a group, then this reduces to $|G| = \frac{1}{\# G}$, and if we form the groupoid $G\coprod H$ as a direct union of groups $G,H$ we have $\textstyle |G\coprod H| =\frac{1}{\#G}+\frac{1}{\#H}$.


Here the link to Egyptian fractions arises: Since every rational number can be represented in terms of Egyptian fractions, so too can every rational fraction be obtained as the cardinality of some groupoid. As a key example, since $|\mathbb{Z}/n|=1/n$ the representations in $(1)$ imply $$1 = |\mathbb{Z}/2\coprod \mathbb{Z}/2|=|\mathbb{Z}/2\coprod \mathbb{Z}/3\coprod \mathbb{Z}/6|=\cdots.$$


But this construction, while valid, is quite artificial. In particular, it is far from obvious to me why each successive groupoid has unit cardinality. This suggests the following question: **Is there a natural sequence of groupoids $\{G\_k\}$, each of unit cardinality, which generate $(1)$?**


References:


[1] John Baez, James Dolan, "From Finite Sets to Feynman Diagrams" ([arXiv](https://arxiv.org/abs/math/0004133))


[2] Julia Bergner, Christopher Walker, "Groupoid Cardinality and Egyptian Fractions" ([JSTOR](https://www.jstor.org/stable/10.4169/college.math.j.46.2.122))



Groupoid cardinality and Egyptian fraction representations of 1
First a notational issue: you shouldn't write $G$ for the one-object groupoid corresponding to $G$. A much better name for this groupoid is $BG$, or $\text{pt} / G$. 


A natural way to write down a groupoid whose groupoid cardinality is $1$ is to write down a groupoid / homotopy quotient $X/G$ where $|X| = |G|$. In turn, a natural class of such quotients are the adjoint quotients $G/G$ where $G$ acts on itself by conjugation. This is equivalent to the groupoid 


$$\coprod\_{[g]} BC\_G([g])$$ 


where $C\_G(-)$ denotes the centralizer and the coproduct runs over the conjugacy classes of $G$. We get a potential Egyptian fraction representation


$$1 = \sum\_{[g]} \frac{1}{|C\_G([g])|}$$


if the sizes of the centralizers are distinct. The largest $n$ such that $\frac{1}{n}$ appears in this sum will always come from the conjugacy class of the identity, where $C\_G(e) = G$, so $n = |G|$, and everything else will divide $|G|$. This identity is just the usual class equation divided by $|G|$. 


To get the first sum this way we want a group of order $6$ with $3$ conjugacy classes having centralizers of orders $2, 3, 6$, or equivalently $3$ conjugacy classes of sizes $3, 2, 1$. There is a unique such group, namely $D\_3$. 


To get the second sum this way we want a group of order $42$ with $4$ conjugacy classes having centralizers of orders $2, 3, 7, 42$, or equivalently $4$ conjugacy classes of sizes $21, 14, 6, 1$. Such a group necessarily has elements of orders $2, 3, 7$, so these must be the non-identity conjugacy classes (in the same order as before, since everything centralizes itself). There are $6$ elements of order $7$, so the Sylow $7$-subgroup $C\_7$ is normal, and so by Schur-Zassenhaus our group $G$ must be a semidirect product $C\_7 \rtimes H$ where $|H| = 6$. We must have $H = D\_3$ or else $G$ has an element of order $6$, so $G$ must be the semidirect product $C\_7 \rtimes D\_3$. There is a unique nontrivial such semidirect product because the only nontrivial action of $S\_3$ on $C\_7$ is the one where $\sigma \in S\_3$ acts by $\text{sgn}(\sigma)$; unfortunately, this group has an element of order $21$. So in fact no group with the desired condition on conjugacy classes exists. 


Nevertheless, the second sum is still related to some geometry, specifically to the geometry of the [(2,3,7) triangle group](https://en.wikipedia.org/wiki/(2,3,7)_triangle_group), or rather to the orbifold quotient of the upper half plane by this group. I don't know about the third sum, though. 


