Abstract
A cyclic Latin square of order 2n, which has no orthogonal Latin square mate, is shown to have (2n-1)(2n-2) mutually orthogonal F(2n;2n-1,2n-1)-squares. This is a complete set of F-squares for the cyclic Latin square. Row and column operations areused to construct this complete set of F-squares from a Hadamard matrix and 2n-1 OF(2n;2n-1,2n-1)-squares into which the Latin square is decomposed. Tables of complete sets of mutually orthogonal F(2n;2n-1,2n-1)-squares are given for n=2 and 3, i.e., for cyclic Latin squares of orders 4 and 8.
| Original language | English |
|---|---|
| Pages (from-to) | 207-218 |
| Number of pages | 12 |
| Journal | Journal of Statistical Planning and Inference |
| Volume | 10 |
| Issue number | 2 |
| DOIs | |
| State | Published - Aug 1984 |
| Externally published | Yes |
Keywords
- F-square design
- Factorial design
- Hadamard matrix
- Hadamard product
- Latin square
- Orthogonal F-squares
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