I do not have access to the above article at this time but as the results are fairly basic I will try to include a proof here, for the sake of completeness and with no claim to originality.
We will use the Polya Enumeration Theorem. By definition we have that $$\Big\langle {n\atop k}\Big\rangle = [z^n] Z(C_k)\left(\frac{z}{1-z}\right)$$ where $Z(C_k)$ is the cycle index of the cyclic group acting on $k$ slots. The notation here is from the first response and should not be confused with Eulerian numbers.
Now we have $$Z(C_k) = \frac{1}{k} \sum_{q|k} \varphi(q) a_q^{k/q}.$$
Substituting $Z(C_k)$ into the above we obtain $$\Big\langle {n\atop k}\Big\rangle = [z^n] \frac{1}{k} \sum_{q|k} \varphi(q) \left(\frac{z^q}{1-z^q}\right)^{k/q} = [z^n] \frac{1}{k} \sum_{q|k} \varphi(q) \frac{z^k}{(1-z^q)^{k/q}} \\ = [z^{n-k}] \frac{1}{k} \sum_{q|k} \varphi(q) \frac{1}{(1-z^q)^{k/q}}.$$
The next step is to expand the rational term in $z$ using the Newton binomial taking care to note that it only contains exponents that are multiples of $q$ in its power series. This yields $$\frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) {\frac{n-k}{q} + \frac{k}{q} - 1 \choose \frac{k}{q} -1} = \frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) {\frac{n}{q} - 1 \choose \frac{k}{q} -1} \\= \frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) \frac{k/q}{n/q} {n/q \choose k/q } = \frac{1}{n} \sum_{q|\gcd(k,n)} \varphi(q) {n/q \choose k/q }.$$ This establishes the first formula.
For the sum we have that $$\sum_{k=1}^n \Big\langle {n\atop k}\Big\rangle = \frac{1}{n} \sum_{k=1}^n \sum_{q|\gcd(k,n)} \varphi(q) {n/q \choose k/q }.$$ Re-indexing this on $q$ and putting $k=pq$ yields $$\frac{1}{n} \sum_{q|n} \varphi(q) \sum_{p=1}^{n/q} {n/q \choose p} = \frac{1}{n} \sum_{q|n} \varphi(q) (2^{n/q} - 1) \\ = - \frac{1}{n} \sum_{q|n} \varphi(q) + \frac{1}{n} \sum_{q|n} \varphi(q) 2^{n/q} = - 1 + \frac{1}{n} \sum_{q|n} \varphi(q) 2^{n/q}.$$ This establishes the second formula.
There are many more Polya Enumeration computations at this MSE Meta link.
Addendum. The count for parts at least equal to two is given by $$[z^n] Z(C_k)\left(\frac{z^2}{1-z}\right)$$
which yields $$[z^{n-2k}] \frac{1}{k} \sum_{q|k} \varphi(q) \frac{1}{(1-z^q)^{k/q}}$$ which produces $$\frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) {\frac{n-2k}{q} + \frac{k}{q} - 1 \choose \frac{k}{q} -1} = \frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) {\frac{n-k}{q} - 1 \choose \frac{k}{q} -1} \\ = \frac{1}{k} \sum_{q|k \wedge q|n} \varphi(q) \frac{k/q}{(n-k)/q} {\frac{n-k}{q} \choose \frac{k}{q}} = \frac{1}{n-k} \sum_{q|\gcd(k,n)} \varphi(q) {(n-k)/q \choose k/q}.$$
This seems to be the right count e.g. for $k=5$ and $n\ge 10$ we get $$1, 1, 3, 7, 14, 26, 42, 66, 99, 143, 201, 273, 364, 476, 612, 776,\ldots$$ which is OEIS A008646.
For $k=6$ and $n \ge 12$ we get $$1, 1, 4, 10, 22, 42, 80, 132, 217, 335, 504, 728, 1038, 1428, 1944,\ldots$$ which is OEIS A032191.
Note that by a trivial argument these values for parts at least two are also given by $$\Big\langle {n-k\atop k}\Big\rangle$$ (put a value one in every slot then add a cyclical composition into $k$ parts of $n-k$, this assures that all parts are at least two and the initial value does not change the symmetry.)