## Concatenation of Two OOAs

Given two (linear) ordered orthogonal arrays A1, A2 with parameters OOA(M1, s1, SM2, T1, k) and OOA(M2, s2, Sb, T2, k), a new (linear) OOA(M1, s1s2, Sb, T1T2, k) can be constructed [1, Theorem 4.1]. Using duality, a linear [(s1s2, T1T2), s1s2T1T2m1m2, k + 1]-NRT-code over Fb can be constructed based on a linear [(s1, T1), s1T1m1, k + 1]-NRT-code over Fbm2 and a linear [(s2, T2), s2T2m2, k + 1]-NRT-code over Fb.

### Construction

Let φ : SM2A2Sb(s2, T2) denote an arbitrary bijection. If A1 and A2 are linear, let φ : Fbm2A2 be Fb-linear. Then the resulting OOA A is defined as

A := {(φ(xi))(i, j) ∈ {1,…, s1}×{1,…, T1}  :  xA1}

with the resulting indices for depth ordered first by j and secondly by the depth-index from A2. More formally and assuming that the columns of Ah are indexed by {0,…, sh – 1}×{0,…, Th – 1} for h = 1, 2, A is defined as

A := {((φ(x⌊i/s2⌋,⌊j/T2))i mod s2, j mod T2)(i, j) ∈ {0,…, s1s2−1}×{0,…, T1T2−1}  :  xA1}.

### Application to Nets

If either A1 or A2 has depth 1 (i.e., it is an orthogonal array) and the other one has depth k, A has also depth equal to strength. Since OOAs with depth equal to strength are equivalent to (mk, m, s)-nets, this procedure can be used for obtaining a new net based on an orthogonal array and a net. In particular, we have the following two corollaries:

1. Given a (digital) (mnetk, mnet, snet)-net in base MOA and a (linear) OA(MOA, sOA, Sb, k), a (digital) (mnetmOAk, mnetmOA, snetsOA)-net in base b can be constructed (established for the digital case in [2, Threorem 11]).

2. Given a (linear) OA(MOA, sOA, Sbmnet, k) and a (digital) (mnetk, mnet, snet)-net in base b, a (digital) (mnetmOAk, mnetmOA, snetsOA)-net in base b can be constructed (established for the digital case in [2, Threorem 12]).