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## 2.Some factors at more than three levels

Let us have g factors, at m1, ..., mg levels. Strength 3 means that each of the mimjmk combinations of symbols in columns i, j and k occurs equally often, that is, N/mimjmk times. This quotient must be an integer, and our assumption is that N is not more than 100. This restricts the possible values of the mi.

If g=3 then the only possibility is the trivial design: for some constant m we take each combination of the symbols m times. So assume g > 3.

Let us say that we have a factors at two levels, b factors at three levels, and c factors at four levels. Then we are left with the following possibilities for the mi:

 N m1...mg existence 8 2a a <= 4 (H) 16 4 2a a <= 3 (M) 16 2a a <= 8 (H) 24 6 2a a <= 3 (M) 24 3 2a a <= 4 24 2a a <= 12 (H) 27 3b b <= 4 32 8 2a a <= 3 (M) 32 4 4 2a a <= 4 (*) 32 4 2a a <= 7 (M) 32 2a a <= 16 (H) 36 3 3 2 2 (T) 40 10 2a a <= 3 (M) 40 5 2a a <= 6 (X) 40 2a a <= 20 (H) 48 12 2a a <= 3 (M) 48 6 4 2a a <= 2 (O) 48 6 2a a <= 7 (M) 48 4 3 2a a <= 4 48 4 2a a <= 11 (M) 48 3 2a a <= 9 48 2a a <= 24 (H) 54 6 3b b <= 3 54 3b b <= 5 54 3b 2 b <= 5 56 14 2a a <= 3 (M) 56 7 2a a <= 6 (X) 56 2a a <= 28 (H) 60 5 3 2 2 (T) 64 16 2a a <= 3 (M) 64 8 4 2a a <= 4 64 8 2a a <= 7 (M) 64 4c c <= 6 (L) 64 4 4 4 4 4 2a a <= 2 (R) 64 4 4 4 4 2a a <= 6 64 4 4 4 2a a <= 8 (R) 64 4 4 2a a <= 12 (**) 64 4 2a a <= 15 (M) 64 2a a <= 32 (H) 72 18 2a a <= 3 (M) 72 9 2a a <= 6 (X) 72 6 6 2a a <= 2 (O) 72 6 3 2a a <= 4 (O') 72 6 2a a <= 11 (M) 72 4 3 3 2 (T) 72 3 3 2a ? 72 3 2a ? 72 2a a <= 36 (H) 80 20 2a a <= 3 (M) 80 10 4 2a a <= 2 (O) 80 10 2a a <= 7 (M) 80 5 4 2a ? 80 5 2a ? 80 4 2a a <= 19 (M) 80 2a a <= 40 (H) 81 9 3b b <= 4 (***) 81 3b b <= 10 84 7 3 2 2 (T) 88 22 2a a <= 3 (M) 88 11 2a a <= 6 (X) 88 2a a <= 44 (H) 90 5 3 3 2 (T) 96 24 2a a <= 3 (M) 96 12 4 2a a <= 4 96 12 2a a <= 7 (M) 96 8 6 2a a <= 2 (O) 96 8 3 2a a <= 4 (O') 96 8 2a a <= 11 (M) 96 6 4 4 2a ? 96 6 4 2a ? 96 6 2a a <= 15 (M) 96 4 4 3 2a ? 96 4 4 2a a <= 20 96 4 3 2a ? 96 4 2a a <= 23 (M) 96 3 2a ? 96 2a a <= 48 (H) 100 5 5 2 2 (T)

Let us first worry about existence for each of these 78 lines. Assume that the mi are sorted in nonincreasing order.

If all mi are 2, then g is at most N/2, and in case of equality we have a Hadamard matrix, and conversely Hadamard matrices do give orthogonal arrays of strength 3 whenever N is a multiple of 8. The corresponding rows have been labeled (H). That settles existence for 12 cases, leaving 66.

If g=4 and N is a multiple of the product of the mi, then at least the trivial design exists; there may be others as well. The corresponding rows have been labeled (T). This settles existence in 6 more cases, leaving 60.

Given a design with g factors at m1, ..., mg levels and N runs, we can form a design at e.m1, ..., mg levels and eN runs, for any positive integer e. Let us call this construction (M), with M for multiply. It settles existence in 23 more cases, leaving 37.

When m1 = 3 (so that all mi are 2 or 3) then no label is given, and the result is quoted from the previous section. This settles 5 more cases, leaving 32.

(*) The case N=32 with groupsizes 4,4,2,2,2,2 follows by taking a Latin square of order 4 (that is, an orthogonal array of strength 2 and N=16 of type 4,4,4) and a Hadamard matrix H of order 4, and replacing the symbol i in the third column of the OA by the i-th row of H and by its complement.

Alternatively, take the binary code spanned by

```00 1111
01 0011
02 0101
10 0011
20 0101
```
where the first two positions are quaternary (with alphabet Z2xZ2) and the remaining four positions are binary.

(**) The case N=64 with groupsizes 42 212 follows by taking the binary code spanned by

```00 111111110000
00 000011111111
01 001100110011
02 010101010101
10 001100111100
20 010101011010
```
where the first two positions are quaternary (with alphabet Z2xZ2) and the remaining twelve positions are binary.

(***) The case N=81 with groupsizes 9,3,3,3,3 follows by taking the ternary code spanned by

```0 1011
0 0112
1 0010
3 0001
```
where the first position is 9-ary (with alphabet Z3xZ3) and the remaining four positions are ternary.

(X) Applying the Rao bound to the derived designs of N=8s with groups s 2a, we find a <= 7. For even s we already found these with (M), but for odd s there is no system with a = 7. Indeed, all s derived designs are essentially Fano planes, and have 7 triples of columns where half of the combinations occur twice and half of the combinations occur not at all. We need a matching, where each time 000 occurs twice in one Fano plane, it occurs not at all in some other Fano plane. But 7s is odd, and no such matching exists.

On the other hand, such designs exist with a = 6 when s is at least 5. Indeed, juxtaposing a design with N=8s for s 26 and one with N=16 for 27 we get one with N=8(s+2) for (s+2) 26, so it suffices to do the case s=5, N=40. We need five designs on 8 rows and six columns. Make them by fixing the first column and cyclically permuting the remaining five, starting from

```000001
111001
100010
011010
001100
110100
101111
010111
```

(R) One can always replace a 4-level factor by two 2-level factors. Thus, from 46 one gets 45 22.

This is best possible for N=64, since no 45 23 exists: The strength 2 derived structure with N=16 and groups 44 23 is unique (the affine plane of order 4 together with the three ways of partitioning its four parallel classes into two sets of two). Parallel lines have the same binary tail. But that means that for a given run 00000 000 there are 15 runs (0xxxx) 000 with nonzero x, and the tail 000 occurs 16 times (instead of 8).

(L) stands for linear code. The quaternary [6,3,4] code gives N=64, 46.

(O) is the label given for a <= 2 in case m.n.2a where N = 2mn and m and n are even, and n/2 is odd. Indeed "even sum" gives a solution for m.n.22, and no OA(2n, n.23, 2) exists.

(O') is the label given for a <= 4 in case m.3.2a where N = 12m and m is even. Indeed, no OA(12,3.25,2) exists. On the other hand, juxtaposing solutions for m=4 and m=6 one finds solutions for all m > 2.

Corrections and improvements are welcome.

Andries Brouwer - aeb@cwi.nl

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