Erratum to: Des. Codes Cryptogr. (2014) 72:529–537 DOI 10.1007/s10623-012-9782-3

Several errors in the original publication of this article are noted. It has been corrected in this erratum.

Theorem 4.2

In the proof of Theorem 4.2, the computation of \(\frac{F(63, 4\cdot 63+4;[0,2,4,6,8,10,12,14])}{10321920}\) is incorrect. We exchange “Let \(D^{\prime \prime }\) be a self-orthogonal \({\ldots }\) (page 535, line 5 up)” to

Let \(D^{\prime \prime }\) be a self-orthogonal 8-\((24m,4m+4,\lambda _8 )\) design, where \(\lambda _8 = \left( {\begin{array}{c}5m-2\\ m-1\end{array}}\right) \frac{(4m-1)(4m-2)(4m-3)}{(24m-5)(24m-6)(24m-7)}\). We set \(A_s^u =\sum _{i=0}^{4m+4} (i)_s n_i^u =(u)_s \lambda _s \) for \(0\le \hbox {s}\le 8\). For the design \(D^{\prime \prime }\), we have

$$\begin{aligned} F(m,u; [x_1 ,x_2 ,x_3 ,x_4 ,x_5 ,x_6 ,x_7 ,x_8 ])&= \sum \limits _{i=0}^{4m+4} (i-x_1 )(i-x_2 )\ldots (i-x_8 )n_i^u \\&= \sum \limits _{\theta =0}^8 (-1)^{\theta }\sigma _{\theta ,8} \left( \sum \limits _{h=0}^{8-\theta } S(8-\theta ,h)A_h^u \right) . \end{aligned}$$

Then, we have \(n_{16}^u =\frac{F(m, u;[0,2,4,6,8,10,12,14])}{10321920 }-9n_{18}^u -45n_{20}^u -\cdots -\left( {\begin{array}{c}2m+2\\ 8\end{array}}\right) n_{4m+4}^u \). Put \(u=4m+8\). In the case \(m=63\), by a computation using Magma and Mathematica, we have

$$\begin{aligned}&\frac{F(63,4{\cdot }63+8;[0,2,4,6,8,10,12,14])}{10321920} \\&\quad =43477008963170791885401824066553255650102446561069494920895005670086011251615/4. \end{aligned}$$

Hence \(n_{16}^{4\cdot 63+8}\) is not an integer. Therefore, if \(m=63\), there is no self-orthogonal 8-\((24m,4m+4,\lambda _8 )\) design.

Thus Theorem 4.2 is correct.

Theorem 4.3

For Theorem 4.3, we examined again by using Magma and Mathematica. Then we found some errors.

In Theorem 4.3 (1), in the set \(\{58,90,113\}\) should be 58.

In Theorem 4.3 (2), the set \(\left\{ {10, 79, 93, 118, 120, 123,125, 142} \right\} \) should be \(\{10, 23, 79, 93, 118, 120, 123, 125, 142\}\). The set \(\{79, 93, 118, 120, 123, 125, 142\}\) should be \(\{23, 79, 93, 118, 120, 123, 125, 142\}\).