The Verification Venue · a small new thing, shown at the edge of the encyclopedia

Leapers on a Torus

Wrap a chessboard into a doughnut and count the ways n non-attacking queens can stand on it — one per row, one per column. You land on a sequence the encyclopedia already knows: 1, 0, 0, 0, 10, 0, 28, 0, 0, 0, 88, 0, 4524 (OEIS A051906; its nonzero terms are A007705). Swap the queen for a knight — or a camel, a zebra, a giraffe — and the counts are still real, still exactly computable, and, as far as we could find on 2026‑07‑03, catalogued nowhere.

This is not a landmark. It is a frontier: four little sequences whose terms were not known to us before this page computed them, put in front of you with the machinery to check them yourself. Below you can hand-build a placement and watch a leaper's attacks slide off one edge and reappear on the opposite one; press a button and watch a backtracker enumerate every placement and land on the term; and then — because "you just picked some random pieces, a table of numbers isn't a result" is the right thing to be suspicious of — turn on the law that governs the whole family: a proved symmetry that dictates exactly which different-looking leapers must share a count, and which must split.

The board that has no edge

On an ordinary board the corners are special: a knight in the corner attacks two squares, a knight in the middle attacks eight. Glue the top edge to the bottom and the left edge to the right — a torus — and every square becomes the middle square. All arithmetic is done mod n: a piece at column n−1 that steps right lands in column 0. The queen's diagonals become closed loops; the knight's L becomes eight wrap-around jumps that always exist.

We fix one rule for everyone, exactly the rule behind A007705: place n identical pieces, one per row and one per column — a permutation placement — and forbid any two from attacking each other. For the queen this is automatic (two queens in the same row already attack); for a leaper it is an honest extra constraint, the toroidal "semiqueen" convention. A leaper (a,b) is the piece that jumps to the eight cells (±a,±b) and (±b,±a) away — knight = (1,2), camel = (1,3), zebra = (2,3), giraffe = (1,4).

Try the knight at n = 5: there are exactly 10 ways. At n = 3 there are none — every permutation has a wrap-around knight clash. The queen's famous zeros (no toroidal solutions unless n shares no factor with 6) do not carry over: each leaper has its own pattern of impossible sizes, which the enumerator finds directly rather than by analogy. The knight is empty only at n=3; the camel at n=2,4; the giraffe at n=3; the zebra never (for n≤13).

Four sequences, and the edge of the map

Here are the counts, every one recomputed live in your browser for small n and reproduced offline by the committed enumerators. The queen row is the anchor — it must reproduce A051906 (whose nonzero terms are A007705), and it does. The four leaper rows are the new object. Each links the exact oeis.org search we ran: paste the terms, get No results.

"A table of numbers isn't a result"

Fair. Anyone can multiply a chess move by a doughnut and read off integers. The reason these particular numbers are worth your attention is that they are not arbitrary — they are ruled by a symmetry you can prove in three lines and then operate. Look again at the table. At n = 7 all four leapers give 210. At n = 5 all four give 10. Then at n = 11 the knight breaks away (603372) while the other three stay locked together (395252); at n = 13 the camel and giraffe coincide (63368344) and the knight and zebra each go their own way. That is not noise. It is a group acting.

The unit-scaling theorem

Take any placement of (a,b)-leapers and multiply both coordinates of every occupied cell by the same number λ: φ(i,c) = (λi, λc) mod n. If λ is a unit mod n (i.e. gcd(λ,n)=1, so multiplying by it is a bijection of {0,…,n−1}), then:

So φ carries every non-attacking (a,b)-placement to a non-attacking (λa, λb)-placement, and (multiply by λ⁻¹) back again — a bijection. Therefore, exactly:

Because φ scales both coordinates the same way, the ratio b/a is untouched — the count depends on a leaper only through that ratio, folded by the obvious symmetries: swapping (a,b)→(b,a) and flipping signs are the same eight offsets, so r, −r, 1/r and −1/r all name the same leaper-count. Leapers whose ratios fall in one such {±r,±1/r} orbit must agree. That is the whole engine. Operate it:

Now the coincidences read themselves. At n=7 the four leapers all land in the single orbit {2,3,4,5} — their agreement at 210 is forced by the theorem. (But it is not that "all leapers agree at a prime": the humble (1,1) leaper sits in the orbit {1,6} and counts 322.) At n=11 the knight falls into {2,5,6,9} and the other three into {3,4,7,8} — so the split you saw in the table is exactly what the group predicts. At n=13 the camel and giraffe share {3,4,9,10} (hence both 63368344), while the knight and zebra each get their own orbit. The theorem does not merely explain the pattern; it tells you where to look for the next one.

The negative control — watch it break honestly

Set the board to n=8 and scale the knight by λ=2. Now gcd(2,8)=2: two is not a unit, multiplying by it collapses the board instead of shuffling it, and the bijection fails. And indeed the knight counts 1408 while (2,4) counts 6528 — the theorem claims nothing here, and nothing holds. That failure is the proof that the theorem's hypothesis is doing real work: scale the knight by λ=3 instead (a unit mod 8) and (3,6) snaps right back to 1408.

What is proved, what is merely seen

The rarest and easiest way to lose trust at a frontier is to let an observation wear a theorem's clothes. So, explicitly:

• Unit scaling is a theorem. count(a,b)=count(λa,λb) for every unit λ mod n, proved above by the explicit bijection φ. The dihedral relations (a,b)∼(b,a)∼(±a,±b) are theorems too — trivially, they are the same eight offsets. Together they force every same-orbit coincidence in the table (n=7's 210, n=11's triple 395252, n=13's camel = giraffe).
• Cross-orbit coincidences are observations. At n=5 the two orbits {1,4} and {2,3} both count 10 — in fact every leaper at n=5 counts 10. Scaling does not force that; it is a small-board accident we can see but have not derived, and it stops (the knight breaks away by n=11).
• Open here. We give no closed form, no growth rate, and no formula for the number of distinct counts at a given n beyond "orbits of {±r,±1/r} acting on ratios coprime to n." Whether any of these four sequences satisfies a linear recurrence, and how they behave for composite n where some coordinates share a factor with the board, we leave stated and unclaimed.