Degree $d$ non-isomorphic graph count How many non-isomorphic regular graphs are there are $n$-vertices with degree $d$?
 A: To give an idea of the difficulty  of this problem here is the case of
$d=2,$ which means we have a  multiset of cycles. As there is just one
cycle on  $n$ nodes and  the smallest cycle  is a triangle  these have
generating function $$C(z) = \frac{z^3}{1-z}.$$  We get for the set of
non-isomorphic  $2$-regular  graphs  the  generating  function  (Polya
Enumeration Theorem)
$$R_2(z) = 
\exp\left(\sum_{l\ge 1} \frac{1}{l} \frac{z^{3l}}{1-z^l}\right).$$
Differentiate to get
$$R'_2(z) = R_2(z)
\left(3\sum_{l\ge 1} \frac{z^{3l-1}}{1-z^l}
+ \sum_{l\ge 1} \frac{z^{3l}}{(1-z^l)^2} z^{l-1}
\right)$$
or
$$z R'_2(z) = R_2(z)
\left(3\sum_{l\ge 1} \frac{z^{3l}}{1-z^l}
+ \sum_{l\ge 1} \frac{z^{4l}}{(1-z^l)^2}
\right)
\\ = R_2(z)
\left(3\sum_{l\ge 1} \sum_{k\ge 0} z^{(k+3)l}
+ \sum_{l\ge 1} \sum_{k\ge 0} (k+1) z^{(k+4)l}
\right).$$
Introducing $r_{n,2}  = [z^n]  R_2(z)$ and extracting  coefficients we
get
$$n r_{n,2} = \sum_{p=0}^n r_{n-p, 2}
\left(3 \sum_{k+3|p\wedge k\ge 0} 1 
+ \sum_{k+4|p\wedge k\ge 0} (k+1)\right).$$
Now the multiplier on $R_2(z)$ on  the RHS does not include a constant
term in fact it starts at $[z^3]$ so this is
$$r_{n,2} = \frac{1}{n} \sum_{p=3}^{n} r_{n-p, 2}
\left(3 \sum_{k+3|p\wedge k\ge 0} 1 
+ \sum_{k+4|p\wedge k\ge 0} (k+1)\right).$$
for $n\ge  3$ with base cases $r_{0,2}  = 1$ and $r_{1,2}  = r_{2,2} =
0.$ (There are some zero terms in the sum that could be omitted.)
This recurrence produces the sequence
$$1, 0, 0, 1, 1, 1, 2, 2, 3, 4, 5, 6, 9, 10, 13, 17, 21, 25, 
\\ 33, 39, 49, 60, 73, 88,\ldots$$
which  is  OEIS  A008483, partitions  into
parts $\ge 3$, which perfectly matches the problem definition.
