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MacMahon's count of 3x3 semi-magic squares by line sum

Proved
MagicSquares.semi_magic_count_three

by Yuxuan Xu · Sep 16, 2026 · Mathlib 0df444a (Lean v4.33.1)

combinatoricsmagic-squares

MacMahon's 1915 formula for the number of 3×33 \times 33×3 semi-magic squares of a given line sum.

Write H3(t)H_{3}(t)H3​(t) for the number of 3×33 \times 33×3 arrays of nonnegative integers whose three rows and three columns all sum to ttt (the diagonals are unconstrained, and entries need not be distinct). Then

H3(t)=3(t+34)+(t+22).H_{3}(t) = 3\binom{t+3}{4} + \binom{t+2}{2}.H3​(t)=3(4t+3​)+(2t+2​).

Unlike the magic count M3(t)M_{3}(t)M3​(t), this is an honest polynomial in ttt of degree (3−1)2=4(3-1)^{2} = 4(3−1)2=4: Ehrhart and Stanley proved that Hn(t)H_{n}(t)Hn​(t) is a polynomial of degree (n−1)2(n-1)^{2}(n−1)2 for every nnn, satisfying the reciprocity law Hn(−n−t)=(−1)n−1Hn(t)H_{n}(-n-t) = (-1)^{n-1}H_{n}(t)Hn​(−n−t)=(−1)n−1Hn​(t).

Formalization Note semiMagicCount n t counts arrays with entries in Fin (t+1) whose row and column sums are ttt after coercion to ℕ; the bound on entries makes the finite search space exact.

Preamble
import Mathlib
import Definitions.Def_MagicSquares
open MagicSquares
Formal statement
namespace MagicSquares

theorem semi_magic_count_three (t : ℕ) :
    semiMagicCount 3 t = 3 * ((t + 3).choose 4) + ((t + 2).choose 2) := by sorry

end MagicSquares
Source
Beck, Cohen, Cuomo & Gribelyuk, The number of ``magic'' squares, cubes and hypercubes, Amer. Math. Monthly 110 (2003), 707-717; arXiv:math/0201013v3, Section 2, Theorem 1 (MacMahon's formula for H3(t)H_{3}(t)H3​(t)).
Human review
  • Endorsed by Shuze Chen · Sep 17, 2026

  • Endorsed by Yuxuan Xu · Sep 17, 2026

    Confirmed by the mission captain (proposal self-audit).

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