We can answer this quite easily using the Polya Enumeration Theorem
and the cycle index $Z(D_{2N})$ of the dihedral group. We will not
derive this cycle index here but it can be worked out from first
principles without too much difficulty. We then have for the answer
$$[A_1^2 A_2^2 \cdots A_N^2] Z(D_{2N})
\left(A_1+A_2+\cdots+A_N\right).$$
The cycle index is
$$Z(D_{2N}) = \frac{1}{4N} \sum_{d|2N} \varphi(d) a_d^{2N/d}
+ \frac{1}{4} a_1^2 a_2^{N-1} + \frac{1}{4} a_2^N.$$
where we have contributions from the cycle index of the cyclic group
and from two types of reflections (axis passes through opposite slots
or edges).
As we only have two instances of each color and we substitute $a_d
= A_1^d + A_2^d + \cdots + A_N^d$ the contribution reduces to
$$\frac{1}{4N} \varphi(1) a_1^{2N}
+ \frac{1}{4N} \varphi(2) a_2^{N}
+ \frac{1}{4} a_1^2 a_2^{N-1} + \frac{1}{4} a_2^N$$
which works out to
$$\frac{1}{4N} a_1^{2N}
+ \frac{N+1}{4N} a_2^{N}
+ \frac{1}{4} a_1^2 a_2^{N-1}.$$
Do the substitution to get
$$\frac{1}{4N} (A_1+A_2+\cdots+A_N)^{2N}
+ \frac{N+1}{4N} (A_1^2+A_2^2+\cdots+A_N^2)^{N}
\\ + \frac{1}{4} (A_1+A_2+\cdots+A_N)^2
(A_1^2+A_2^2+\cdots+A_N^2)^{N-1}.$$
Extracting coefficients now yields
$$\frac{1}{4N} {2N\choose 2,2,\ldots, 2}
+ \frac{N+1}{4N} {N\choose 1,1,\ldots, 1}
+ \frac{1}{4} {N\choose 1} {N-1\choose 1,1,\ldots, 1}.$$
This simplifies to
$$\frac{1}{4N} \frac{(2N)!}{2^N}
+ \frac{N+1}{4} (N-1)! + \frac{1}{4} N!$$
and we obtain the closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{4N} \frac{(2N)!}{2^N} + \frac{2N+1}{4} (N-1)!.}$$
This yields the sequence
$$1, 2, 11, 171, 5736, 312240, 24327000, 2554072920,
\\ 347351195520, 59397023589120, \ldots $$
which points us to OEIS A120445 where
additional material awaits (and the above result is
confirmed). (Observe that they use the term necklace when we actually
have a bracelet.)
The following memory efficient Perl script can be used to compute
the first six values of the sequence (output matches result from PET).
#! /usr/bin/perl -w
#
sub recurse {
my ($n, $slots, $sofar, $orbref) = @_;
my $rest = scalar(@$slots);
if($rest == 0){
my @perm = (-1) x (2*$n);
my $color = 1;
map {
$perm[$sofar->[2*$_]-1] = $color;
$perm[$sofar->[1+2*$_]-1] = $color;
$color++;
} (0..$n-1);
my (%orbit, $key);
for(my $rot = 0; $rot < 2*$n; $rot++){
my @rotact = ();
push @rotact,
@perm[$rot..(2*$n-1)],
@perm[0..($rot)-1];
$key = join('-', @rotact);
$orbit{$key} = 1;
if($rot % 2 == 0){
my @reflact1 = reverse @rotact;
$key = join('-', @reflact1);
$orbit{$key} = 1;
my @reflact2;
push @reflact2,
@reflact1[1..(2*$n-1)],
$reflact1[0];
$key = join('-', @reflact2);
$orbit{$key} = 1;
}
}
$key = join('|', sort(keys(%orbit)));
$orbref->{$key} = 1;
return;
}
for(my $p=0; $p < $rest; $p++){
for(my $q = $p+1; $q < $rest; $q++){
push @$sofar, $slots->[$p], $slots->[$q];
my @data;
push @data,
@$slots[0..$p-1],
@$slots[$p+1..$q-1],
@$slots[$q+1..$rest-1];
recurse($n, \@data, $sofar, $orbref);
splice @$sofar, -2, 2;
}
}
1;
}
MAIN: {
my $n = shift || 5;
my %orbits;
recurse($n, [1..2*$n], [], \%orbits);
print scalar(keys(%orbits));
print "\n";
1;
}
Addendum. We can also do this from first principles without
using PET. Introduce the total count of bracelets before applying
symmetry and call it $D_N.$ Let $R_N$ be the bracelets that have
rotational symmetry, $F_N$ the ones that have reflectional symmetry
with an axis passing through opposite slots, $G_N$ the ones that have
reflectional symmetry with an axis passing through opposite edges. Put
$U_N$ for the ones with no symmetry. We are thus interested in the
quantity
$$E_N = R_N + F_N + G_N + U_N.$$
It is not difficult to see that with $N$ sufficiently large the
bracelets with rotational symmetry do not have reflectional symmetries
which would otherwise require inclusion-exclusion.
Now we have $R_N = \frac{1}{2} \frac{1}{N} N!$ and $F_N = \frac{1}{2}
N \times (N-1)!$ and $G_N = \frac{1}{2} N!.$
Observe also that when computing the sizes of the orbits we find
$$D_N = {2N\choose 2,2,\ldots,2} =
2 N R_N + 2 N F_N + 2 N G_N
+ 4 N U_N$$
which implies
$${2N\choose 2,2,\ldots,2} =
2N \times \frac{1}{2} (1 + 2N) (N-1)!
+ 4 N U_N \\ = (1+2N) \times N! + 4 N U_N $$
so that
$$E_N = \frac{1}{2} (1+2N) (N-1)!
+ \frac{1}{4N} {2N\choose 2,2,\ldots,2}
- \frac{1+2N}{4} (N-1)!
\\ = \frac{1}{4N} \frac{(2N)!}{2^N}
+ \frac{1+2N}{4} (N-1)!$$
as before.
A reference for this technique may be found at the following MSE
link.