OBIBD construction for designs with r = (v-1)/2 and odd k > 3.

Let v be a prime number or power of the form 4kt+2k+1, and x be a primitive root of the Galois Field of order v. If there exists an integer m such that (x^m-1),(x^4t+m+2-1),..., (x^{(k-1)(4t+2)+m}-1) are all odd, or all even, powers of x, then an OBIBD(v) can be constructed from the initial blocks:

(x²ⁱ, x^2i+m; x^4t+2i+2, x^4t+2i+m+2;...; x^{(k-1)(4t+2)+2i}, x^{(k-1)(4t+2)+2i+m}), for i = 0,1,...,2t

(The full design is obtained by developing over the elements of the Galois Field.)

Proof.The difference square formed by the typical block is exactly anologous to that formed for the case with k = 3. The rest of the proof therefore follows.

There is also a need to show that the designs for first and second treatments are BIBDs. Given any initial block, the orbit of that block under the transformations induced by the doubly homogeneous group of degree v and order v(v-1)/2 will comprise a BIBD. The initial block given for the first set of treatments is fixed by a subgroup (of order k) of the multiplicative subgroup: so the length of the orbit (the number of blocks of the BIBD) will be v(v-1)/2k, the order of the group divided by the order of the subgroup. So we have a BIBD with parameters (4kt+2k+1, (2t+1)(4kt+2k+1),k(2t+1),k,( k-1)/2) so that k must be odd. The same argument applies for the second set of treatments, or else we can use the same argument as was used for k = 3.