BARRON SQUARES
I submitted to FiveThirtyEight's Riddler Express a puzzle where every row and every column had to obey the same rule: the leftmost digit times the rightmost digit equals the two-digit number in the middle — e.g. 6 · 7 = 42. This would read off as 6, 4, 2, 7 Here's a 4×4 example:
6 │ 4 2 │ 7
──┼─────┼──
4 │ 2 0 │ 5
8 │ 4 8 │ 6
──┼─────┼──
8 │ 6 4 │ 8
Rows: 6·7=42, 4·5=20, 8·6=48, 8·8=64. Columns: 6·8=48, 4·6=24, 2·4=08, 7·8=56. I call these Barron Squares. There is a generalization: a Barron Square of order k is a 4k × 4k matrix where every row and column satisfies
the k-digit number on the left, times the k-digit number on the right, equals the 2k-digit number in the middle.
The question "how many exist at each order" ends up being a mildly fun exercise.
What we know
| Object | Count | How |
|---|---|---|
| Order-1 (4×4) | 118 | Exhaustive — all 9⁴ = 6,561 corner quadruples, under a second in Python |
| Order-2 (8×8) | 1,248 | Parallel C search, ~50 hours on 12 cores. Independently re-verified by a second, structurally different C program (~161 CPU-hours) and spot-checked with the Z3 SMT solver |
| 4×4×4 Barron Cube | 0 | Exhaustive over 9⁸ ≈ 43 million vertex octuplets |
| Order-4 (16×16) | 0 in 500,000 random trials | Random probing |
Two structural theorems fell out of staring at the 1,248 order-2 squares: an endpoint pairing dichotomy between inner rows and columns, and, as a corollary, the fact that the inner 4×4 center of every 8×8 Barron Square is a symmetric matrix. Both hold for all 1,248 squares. Proving the dichotomy algebraically (rather than by computer enumeration) is the top open question.
Read more
- Paper: 18-page PDF — full proofs, enumeration details, and open questions
- Code and data: github.com/trbarron/barron-squares — C/Python search, all 118 and 1,248 squares as JSON
- The sequence
(118, 1248, …)will be submitted to OEIS (I'm excited)
If you find a third term I'd love to hear about it.