It may be of interest to relate this problem to cycle indices and the
Polya Enumeration Theorem (PET). Suppose we distribute the eight red
beads on the necklace first. Next we distribute two blue beads into
every space between two red beads, leaving $32-2\times 8 = 16$ blue
beads. The problem now becomes equivalent to distributing the
remaining $16$ blue beads into the eight slots under rotation or
rotation and reflection.
To compute the count under rotations we need the cycle index
$Z(C_8)$ of the cyclic group on eight elements. We now enumerate the
permutations in this cycle index. There is the identity, which
contributes $a_1^8$. A rotation by a distance of four maps opposite
slots to each other and creates two-cycles, giving $a_2^4.$ A rotation
by a distance of two or six creates four-cycles, giving $2\times
a_4^2.$ A rotation by a distance of $1,3,5$ or $7$ creates
eight-cycles giving $4\times a_8.$
This gives the cycle index
$$Z(C_8) = \frac{1}{8} (a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8).$$
The desired value is given by
$$[z^{16}] Z(C_8)\left(\frac{1}{1-z}\right).$$
This is
$$\frac{1}{8}
\left([z^{16}] \left(\frac{1}{1-z}\right)^8 +
[z^{16}] \left(\frac{1}{1-z^2}\right)^4 +
2 [z^{16}] \left(\frac{1}{1-z^4}\right)^2 +
4 [z^{16}] \left(\frac{1}{1-z^8}\right)\right).$$
which yields
$$\frac{1}{8}
\left({16+7\choose 7}
+ {8+3\choose 3}
+ 2{4+1\choose 1}
+ 4{2+0\choose 0}\right) = 30667.$$
When reflections are included we have dihedral symmetry and we need
the cycle index $Z(D_8)$ of the dihedral group on eight elements. The
additional permutations are four reflections about an axis passing
through the midpoint of opposite edges connecting two slots, giving
$4\times a_2^4$ and four reflections about an axis passing through two
opposite slots giving $4\times a_1^2 a_2^3.$ Hence the cycle index is
$$Z(D_8) = \frac{1}{2} Z(C_8) +
\frac{1}{16} (4 a_2^4 + 4 a_1^2 a_2^3).$$
This yields for sixteen blue beads the count
$$\frac{1}{2} 30667
+ \frac{1}{16}
\left(4 [z^{16}] \left(\frac{1}{1-z^2}\right)^4 +
4 [z^{16}] \left(\frac{1}{1-z}\right)^2
\left(\frac{1}{1-z^2}\right)^3
\right)$$
which is
$$\frac{1}{2} 30667
+ \frac{1}{16}
\left(4 {8+3\choose 3} +
4 \sum_{q=0}^8 {q+2\choose 2} {16-2q+1\choose´1}
\right) = 15581.$$
The sequences OEIS A032193 and
OEIS A005514 are relevant here.
Remark. We can also apply Burnside directly to the cycle indices
and bypass PET. E.g. for a permutation of cycle type $a_2^4$ to fix an
assignment to the four cycles it is necessary that we
choose a number $q$ of blue beads for each the four two-cycles
and then place twice that number $q$ of beads on the four two-cycles
with $q$ beads in each of the two slots on a two cycle.
Choosing a pair has generating function
$$\frac{1}{1-z^2}$$
and the total contribution is
$$[z^{16}]
\frac{1}{1-z^2}
\frac{1}{1-z^2}
\frac{1}{1-z^2}
\frac{1}{1-z^2}.$$
Similarly an assignment for a permutation of cycle type $a_4^2$ that
is fixed by this shape of permutation means that we have to choose a
number $q$ of blue beads for each of the two cycles and then place
four times that number on each cycle with $q$ beads in each of the
slots on the four-cycle. This gives
$$[z^{16}]
\frac{1}{1-z^4}
\frac{1}{1-z^4}.$$
This can be continued and it is essentially the mechanism by which PET is proved starting from Burnside.
There are many more related links at
MSE Meta on Burnside/Polya.