Next, apply the following way of combining codes. Suppose we have an \((n_1,M_1,d_1)\) code \(C_1\) and an \((n_2,M_2,d_2)\) code \(C_2\). The combined \((an_1+bn_2,\min(M_1,M_2),ad_1+bd_2)\) code \(a C_1\oplus b C_2\) may be constructed by pasting \(a\) copies of \(C_1\) and \(b\) copies of \(C_2\) together and omitting the last \(|M_1-M_2|\) rows. Applying this to construct a Levenshtein code with length \(n\) and distance \(d\), define \(k=\lfloor d/(2d-n)\rfloor\), \(a=d(2k+1)-n(k+1)\), and \(b=kn-d(2k-1)\). If \(n\) is even, construct \(\frac{a}{2}A_{4k}'\oplus\frac{b}{2}A_{4k+4}'\). If \(n\) is odd and \(k\) is even, construct \(aA_{2k}\oplus\frac{b}{2}A_{4k+4}'\). If \(n\) is odd and \(k\) is odd, construct \(\frac{a}{2}A_{4k}'\oplus b A_{2k+2}\).
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