If you have two types of objects, $a$, and $b$. You have $n_1$ of the first object, $n_2$ of the second object. How many distinct ways can you rearrange these objects on a ring. Arrangements are only unique if the ring can't be rotated or flipped to form another arrangement.
1 Answers
Consulting the following fact sheet on necklaces and bracelets we get for the cycle index of the cyclic group
$$Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}.$$
We will apply PET so we need with $k$ objects of type $A$ and $m$ objects of type $B$ for the cyclic component:
$$[A^k B^m] Z(C_{k+m}; A+B) = [A^k B^m] \frac{1}{k+m} \sum_{d|k+m} \varphi(d) (A^d + B^d)^{(k+m)/d} \\ = [A^k B^m] \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) (A^d + B^d)^{(k+m)/d} \\ = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) [A^k B^m] (A^d + B^d)^{(k+m)/d} \\ = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) [A^{k/d} B^{m/d}] (A + B)^{(k+m)/d}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{P_{k,m} = \frac{1}{k+m} \sum_{d|\gcd(k,m)} \varphi(d) {(k+m)/d\choose k/d}.}$$
Now for reflectional i.e. dihedral symmetry we get for $k+m$ odd the extra contributiom
$$[A^k B^m] \frac{1}{2} (A+B) (A^2 + B^2)^{(k+m-1)/2}.$$
If $k$ is the odd one this is
$$[A^{k-1} B^m] \frac{1}{2} (A^2 + B^2)^{(k+m-1)/2} \\ = [A^{(k-1)/2} B^{m/2}] \frac{1}{2} (A + B)^{(k+m-1)/2}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{ Q_{1,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m-1)/2 \choose m/2}.}$$
On the other hand if $m$ is the odd one we get
$$\bbox[5px,border:2px solid #00A000]{ Q_{2,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m-1)/2 \choose k/2}.}$$
With $k+m$ even they can both be odd or both be even. The contribution is
$$[A^k B^m] \frac{1}{4} (A+B)^2 (A^2 + B^2)^{(k+m)/2-1} + [A^k B^m] \frac{1}{4} (A^2 + B^2)^{(k+m)/2}.$$
If they are both odd we are left with just
$$[A^k B^m] \frac{1}{2} AB (A^2 + B^2)^{(k+m)/2-1} \\ = [A^{k-1} B^{m-1}] \frac{1}{2} (A^2 + B^2)^{(k+m)/2-1} \\ = [A^{(k-1)/2} B^{(m-1)/2}] \frac{1}{2} (A + B)^{(k+m)/2-1}.$$
We obtain
$$\bbox[5px,border:2px solid #00A000]{ Q_{3,k,m} = \frac{1}{2} P_{k,m} + \frac{1}{2} {(k+m)/2-1 \choose (k-1)/2}.}$$
The last one is when both are even and we obtain
$$[A^k B^m] \frac{1}{4} (A^2 + B^2) (A^2 + B^2)^{(k+m)/2-1} + [A^{k/2} B^{m/2}] \frac{1}{4} (A + B)^{(k+m)/2} \\ = [A^{k-2} B^m] \frac{1}{4} (A^2 + B^2)^{(k+m)/2-1} + [A^k B^{m-2}] \frac{1}{4} (A^2 + B^2)^{(k+m)/2-1} \\ + \frac{1}{4} {(k+m)/2\choose k/2} \\ = [A^{k/2-1} B^{m/2}] \frac{1}{4} (A + B)^{(k+m)/2-1} + [A^{k/2} B^{m/2-1}] \frac{1}{4} (A + B)^{(k+m)/2-1} \\ + \frac{1}{4} {(k+m)/2\choose k/2}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{ \begin{array}{l} Q_{4,k,m} & = \frac{1}{2} P_{k,m} + \frac{1}{4} {(k+m)/2-1\choose m/2} + \frac{1}{4} {(k+m)/2-1\choose k/2} \\ & + \frac{1}{4} {(k+m)/2\choose k/2}. \end{array}}$$
There is some Maple code to verify this which computes the desired statistic from PET and from the closed form.
with(numtheory); with(combinat);pet_varinto_cind := proc(poly, ind) local subs1, subs2, polyvars, indvars, v, pot, res, k;
res := ind; polyvars := indets(poly); indvars := indets(ind); for v in indvars do pot := op(1, v); subs1 := [seq(polyvars[k]=polyvars[k]^pot, k=1..nops(polyvars))]; subs2 := [v=subs(subs1, poly)]; res := subs(subs2, res); od; res;end;
pet_cycleind_cyclic := proc(n) local d;
add(phi(d)*a[d]^(n/d), d in divisors(n))/n;end;
pet_cycleind_dihedral := proc(n) local s;
s := 1/2*pet_cycleind_cyclic(n); if(type(n, odd)) then s := s + 1/2*a[1]*a[2]^((n-1)/2); else s := s + 1/4*(a[1]^2*a[2]^((n-2)/2) + a[2]^(n/2)); fi; s;end;
CYCLIC_CIND := proc(k,m) option remember; local cind, sind;
cind := pet_cycleind_cyclic(k+m); sind := pet_varinto_cind(A+B, cind); coeff(coeff(expand(sind), A, k), B, m);end;
CYCLIC_X := proc(k, m) local d;
1/(k+m)*add(phi(d)*binomial((k+m)/d, k/d), d in divisors(gcd(k,m)));end;
DIHEDRAL_CIND := proc(k,m) option remember; local cind, sind;
cind := pet_cycleind_dihedral(k+m); sind := pet_varinto_cind(A+B, cind); coeff(coeff(expand(sind), A, k), B, m);end;
DIHEDRAL_X := proc(k, m) local d; if type(k+m, odd) then if type(k, odd) then 1/2CYCLIC_X(k,m) + 1/2binomial((k+m-1)/2, m/2) else 1/2CYCLIC_X(k,m) + 1/2binomial((k+m-1)/2, k/2) fi; else if type(k, odd) then 1/2CYCLIC_X(k,m) + 1/2binomial((k+m)/2-1, (k-1)/2); else 1/2CYCLIC_X(k,m) + 1/4binomial((k+m)/2-1, m/2) + 1/4binomial((k+m)/2-1, k/2) + 1/4binomial((k+m)/2, k/2); fi; fi; end;
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