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.)