I was trying to find finite partitions of integer inverses that would add up to one, after a few experiments I struck upon this beautiful partition:
$$ 1 = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{2 × 3 × 7} $$
I tried to determine if there was a general rule to finding more such partitions adding more terms but the problem quickly gets unwieldly
Is there a general principle for obtaining this family of partitions? The numerand on the first set of elements seems to be a permanent of the partition elements, and the Diophantic condition for said partitions seems to be that the permanent must be one less than the product of all the elements
$$ perm(a_0,a_1 \dots, a_n) = ( \prod_i a_i) - 1$$
But of course this only applies on this "special" family, I haven't been able to figure out if there can be partitions of 1 that don't abide to this amusing pattern
NB: The terminology of the title is accurate to best of my knowledge, which isn't much, so please correct me if it's the case
If you just want $\sum_{k\in S} \frac{1}{k} = 1$, then note that for any unit fraction we have $\frac{1}{k} = \frac{1}{k + 1} + \frac{1}{k(k + 1)}$. Your example is a special case of this:
$$ 1 = \frac{1}{2} + \frac{1}{2} = \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{6}\right) = \frac{1}{2} + \frac{1}{3} + \frac{1}{7} + \frac{1}{42} $$
In particular, you can define a sequence where $a_1 = 2$ and $a_{n + 1} = 1 + \prod_{k=1}^n a_k$. Then, $\frac{1}{a_N - 1} + \sum_{k=1}^{N - 1} \frac{1}{a_k} = 1$ for all $N$.
Verification code on tio.run.
Another interesting class of solutions can be obtained from perfect numbers. Take $N = 2^{k-1}(2^k-1)$ where $2^k - 1$ is a prime. We know that this is a perfect number. Now consider its divisors $S = \{d \in \mathbb{N} : d \mid N\}$. Summing the reciprocal of divisors, we have
$$ \sum_{d \in S} \frac{1}{d} = \sum_{d} \frac{1}{d} = \left(1 + \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{k - 1}}\right)\left(1 + \frac{1}{2^k - 1}\right) = 2 $$
Subtracting the $1$ gives another unit fraction decomposition of $1$, say $1 = \frac{1}{2} + \frac{1}{4} + \frac{1}{7} + \frac{1}{14} + \frac{1}{28}$.