Third, start with the same six-piece dissection as before. Now further dissect each into n congruent pieces using arcs of unit circles from the center. Since n of these make a shape with reflective symmetry, patches of n can be reflected (they need not have started in the same piece of the six-piece dissection). The n = 2 case, with all six pairs of pieces flipped is the second figure that Michael sent earlier; perhaps he had intended this more general situation.
Let's enumerate solutions of this form. Firstly, assume without loss of generality that the spirals are clockwise instead of anticlockwise. We can 'flip' between 1 and 6 patches of n pieces. For want of a better word, this will be the 'depth' of the solution. To enumerate them, it would be best to consider each depth separately: Depth = 1: solutions = 1. (Trivially one up to rotation.) Depth = 2: solutions = 2n + 1. (There can be between 0 and 2n pieces between the two flipped patches.) Depth = 6: solutions = 1. (Trivially one up to rotation.) For depth = 3, 4 and 5, it's probably best to apply Burnside's Lemma. In each case, we are effectively enumerating cyclic partitions of (6-d)n objects into d (possibly empty) sets. We have a preliminary value of (6-d)n+d-1 choose d-1, but we need to take into consideration that some of these are identical. Depth = 3: The partition is (a,b,c), and the cyclic group of order 3 has the following three elements: Identity -- fixes all (3n+2 choose 2) partitions; Rotation by 2pi/3 -- fixes only (n,n,n); Rotation by -2pi/3 -- fixes only (n,n,n); So, we have a total of [(3n+2 choose 2) + 2]/3 distinct solutions. Depth = 5: The cyclic group of order 5 has the following five elements: Identity -- fixes all (n+4 choose 4) partitions; 4 rotations -- fixes only (n/5,n/5,n/5,n/5,n/5), if it exists; So, we have a total of [(n+4 choose 4) + 4]/5 solutions if 5 divides n, and [n+4 choose 4]/5 otherwise. Depth = 4: This is the tricky one. We have four elements: Identity -- fixes all (2n+3 choose 3) partitions; Rotation by pi -- fixes n+1 partitions of the form (a,n-a,a,n-a); Rotation by pi/2 -- fixes only (n/2,n/2,n/2,n/2), if it exists; Rotation by pi/2 -- fixes only (n/2,n/2,n/2,n/2), if it exists; For this depth, there are [(2n+3 choose 3) + n + 3]/4 solutions if n is even, and [(2n+3 choose 3) + n + 1]/4 otherwise. To summarise, we have the following: Depth = 1, solutions = 1; Depth = 2, solutions = 2n+1; Depth = 3, solutions = [3n^2 + 3n + 2]/2; Depth = 4, solutions = [2n^3 + 6n^2 + 7n + 3]/6 (for n odd); or solutions = [2n^3 + 6n^2 + 7n + 6]/6 (for n even); Depth = 5, solutions = [n^4 + 10n^3 + 35n^2 + 50n + 24]/120 (for n not divisible by 5); or solutions = [n^4 + 10n^3 + 35n^2 + 50n + 120]/120 (for n divisible by 5). Of course, for the sequence of 'known partitions of the unit disk into p congruent pieces, at least one of which doesn't touch the centre', we have to add the curvilinear solution for p = 12. So, we get 0,0,0,0,0,0,0,0,0,0,0,30,0,0,0,0,0,56,0,0,0,0,0,98, ... The sequence for p = 6n is: 0, 30, 56, 98, 157, 236, 339, ... Mathematica code (for n >= 3): f[7] /. {f -> Function[{n}, 1 + 2 n + 1 + (3 n^2 + 3 n + 2)/2 + Floor[(2 n^3 + 6 n^2 + 7 n + 6)/6] + Floor[(n^4 + 10 n^3 + 35 n^2 + 50 n + 120)/120]]} New OEIS entry, perhaps? Sincerely, Adam P. Goucher