Theorem. For every odd n >= 5 there are at least 2^((n-5)/2) pairwise inequivalent placements of 2n - 2 queens on an n x n board in which every queen attacks exactly 2 others (attacks blocked by the nearest queen). None of them is fixed by any symmetry of the square.
If 2n - 2 is the maximum, this gives A051758(n) >= 2^((n-5)/2) for odd n >= 5. The maximum is
verified by exhaustive search for n <= 17 (data/k2_no_2n_minus_1.tsv). For n = 5..13 this
family is exactly the set of 6-turn classes the search finds (checked as sets with
tools/cmpall.py).
Let n = 2m + 1 with m >= 2. Coordinates are (row, column), 0-based. Let T = {2, …, n-4}. For r = 2, …, m-1 the pairs {r, n-2-r} partition T into m - 2 pairs. When m = 2, T is empty. Choose S, a subset of T, holding exactly one element of each pair. Let S’ = T \ S = {n-2-s : s in S}, the complementary choice. Place queens in four groups:
| group | line | positions | count |
|---|---|---|---|
| A | column 0 | rows {0, 1, n-1} u S | m + 1 |
| B | row n-2 | columns {1, n-1} u S | m |
| C | column n-2 | rows {0, n-3} u S’ | m |
| D | row 0 | columns {0, n-3, n-2} u S’ | m + 1 |
A and D share (0, 0); C and D share (0, n-2). Every other pair of groups is disjoint, so there
are 4m = 2n - 2 queens. (tools/gen6.py builds exactly these boards from their cycle words.)
Throughout, S and S’ are disjoint subsets of T = [2, n-4], and n >= 5.
Rows. Row 0 holds group D. Row n-2 holds group B. Every other row holds at most one queen: row 1 holds only (1, 0); row n-3 only (n-3, n-2); row n-1 only (n-1, 0); row r in S only (r, 0); row r in S’ only (r, n-2). The groups’ row sets meet only in the shared corners, because S and S’ are disjoint and 1, n-3, n-1 lie outside T.
Columns. Similarly, column 0 holds A and column n-2 holds C. Every other column holds at most one queen: column 1 holds (n-2, 1), column n-1 holds (n-2, n-1), column n-3 holds (0, n-3), column c in S holds (n-2, c), and column c in S’ holds (0, c).
So the attacking pairs along rows and columns are exactly the consecutive queens of four chains: column 0 (A), row 0 (D), row n-2 (B) and column n-2 (C). A queen inside a chain attacks its two chain neighbours. The chain ends are:
Diagonals. Index the anti-diagonals by r + c and the main diagonals by r - c.
| group | r + c | r - c |
|---|---|---|
| A | {0, 1, n-1} u S | {0, 1, n-1} u S |
| B | {n-1, 2n-3} u (n-2+S) | {n-3, -1} u S’ |
| C | {n-2, 2n-5} u (n-2+S’) | {-(n-2), -1} u (-S) |
| D | {0, n-3, n-2} u S’ | {0, -(n-3), -(n-2)} u (-S’) |
(For B: (n-2) - c with c in S is n-2-S = S’. For C: r - (n-2) with r in S’ is -(n-2-S’) = -S.)
Within one group the values are distinct. Across groups, the only common values are:
Every other coincidence is excluded by the ranges. S, S’ lie in [2, n-4], so n-2+S and n-2+S’ lie in [n, 2n-6]. Also 2n-5 >= n and 2n-3 > 2n-6 for n >= 5, and S n S’ is empty. So exactly two diagonal attacks exist, each between adjacent cells. Each gives the second attack to the two chain ends that had only one. Every queen attacks exactly 2.
Lines map to lines under the symmetries of the square. Row 0 and column 0 are the only lines holding m + 1 queens: rows n-2 and column n-2 hold m, other rows and columns at most 1, and diagonals at most 2. So a symmetry g taking board(S1) to board(S2) maps {row 0, column 0} to itself, and fixes their common cell (0, 0). That leaves the identity and the transpose (r, c) -> (c, r).
The transpose would map row 0 of board(S1), with columns {0, n-3, n-2} u S1’, onto column 0 of board(S2). But column 0 always contains row 1, and 1 is not in {0, n-3, n-2} u S1’ for n >= 5. So g is the identity and S1 = S2. The 2^(m-2) boards are pairwise inequivalent, and (taking S1 = S2) none is fixed by any non-trivial symmetry. Each class contributes 8 placements.
python tools/gen6.py N > f.txt then python tools/checkboards.py f.txt N 2: 0 bad
boards for N = 5..31 (8192 classes at N = 31).