OEIS A348567 · A348566

Symmetric Sandpiles

Put grains of sand on the cells of an n × n grid. A cell holding four or more grains topples: it sends one grain to each neighbour, and grains pushed off the edge are lost. Configurations that come back again and again when you keep dropping sand at random are called recurrent. They form a group, and the identity of that group draws the famous fractal mandalas below.

A348567 counts the recurrent configurations that look the same under all eight symmetries of the square. Eleven terms were known, found by search, along with a conjecture linking them to a neighbouring sequence. The count turns out to be a product of eigenvalues, which gives every term exactly and proves the conjecture:

a(n) = ∏h ≤ k, both odd ( 4 − 2 cos hπ⁄(n+1) − 2 cos kπ⁄(n+1) )

The identity, at every size

Each grid's identity element is symmetric under all eight symmetries of the square, so it is one of the configurations that a(n) counts. Drag the slider: the picture is recomputed by toppling sand in your browser.

  • 0 grains
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A symmetric configuration is fixed by its values on the wedge, a triangle holding about one eighth of the cells. Toppling all copies of a wedge cell together is the same as toppling on a folded grid, so the problem lives entirely on the triangle.

All of them, for small n

Here is every fully symmetric recurrent configuration for one grid size, sorted by number of grains. The outlined tiles are the identity and the all-3s pile. Their count is a(n).

Why it is a product

The count of symmetric recurrents equals the determinant of the folded Laplacian (Florescu, Morar, Perkinson, Salter and Xu, 2015). That determinant is the product of the Laplacian's eigenvalues on symmetric patterns. The symmetric eigenpatterns are the waves shown here, one for each pair of odd frequencies h ≤ k, and each is labelled with its eigenvalue. They fill a triangle, just like the wedge.

Dropping the diagonal symmetry lets every ordered pair (h, k) through, which gives the horizontally and vertically symmetric count b(n) = A348566(n, n). Each off-diagonal factor then appears twice, so a(n)² = b(n) · ∏h (4 − 4cos hπ⁄(n+1)), and the last product is exactly 2A030451(n). That is the conjecture in the entry.

The terms

na(n)digitssource

How it was checked

  1. Brute force. Every symmetric configuration for n ≤ 7, tested with Dhar's burning algorithm, matches the product.
  2. Exact determinant. The folded Laplacian's integer determinant, computed with no floating point, agrees with the product for every n ≤ 40.
  3. High precision. The product is evaluated with 60 more digits than the answer has, and it lands within 10−30 of an integer for every n ≤ 100.
  4. The conjecture. a(n)² = A348566(n, n) · 2A030451(n) holds for every n ≤ 100, as the proof says it must.
  5. Domino tilings. For even sizes, a(2m) = 2m · A065072(m), a square root of the number of domino tilings of the 2m × 2m square. That holds against the independently computed A065072 table for every m ≤ 50, which covers every even term up to a(100). Laura Florescu's 2011 Reed thesis had already linked these symmetric sandpiles to weighted tilings; the product formula covers odd sizes too.

Growth: log a(n) ≈ G n² ⁄ (2π), where G is Catalan's constant: one eighth of the spanning-tree entropy of the square lattice. Full proof, programs and the table to n = 100: oeis-results/A348567. Computed by James Guillochon with Claude, November 2026.