IV\] (1) is completed.

Case for $n=6m+2$ {#sec: case for n=6m+2}
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In this section, we consider the case of extremal Type IV $[6m+2,3m+1,2m+2]$ codes ($m \leq 19$). Let $C$ be an extremal Type IV $[6m+2,3m+1,2m+2]$ code, and $D_{2m+2}^{6m+2}$ be the support (with duplicates omitted) design of the minimum weight of $C$. By [@MOSW1978 Theorem 14], the number of codewords of minimum nonzero weight of $C$ is equal to $$\frac{3(6m+1)}{m+1}\binom{3m+1}{m}.$$ Therefore, by the Assmus–Mattson theorem, $D_{2m+2}^{6m+2}$ is a $3$-design with parameters $$\left(6m+2,2m+2,\frac{1}{3}\binom{3m}{m} \right).$$

\[prop:type4 6m+2 not 8-design\] Let $D_{2m+2}^{6m+2}$ be the support $t$-design of the minimum weight of an extremal Type IV code of length $n=6m+2$.

1.  If $t \geq 4$, then $D_{2m+2}^{6m+2}$ is a $5$-design and $m$ must be $11$.

2.  $D_{2m+2}^{6m+2}$ is never an $8$-design.

\(1) By Theorem \[thm:[Calderbank-Delsarte]{}\] (2), $D_{2m+2}^{6m+2}$ is a $5$-design if $t \geq 4$. If $D_{2m+2}^{6m+2}$ is a $5$-design, by Lemma \[lem: divisible\], $$\lambda_{4}=\frac{2m-1}{6m-1}\frac{1}{3}\binom{3m}{m} \mbox{ and } 
\lambda_{5}=\frac{(2m-1)(2m-2)}{(6m-1)(6m-2)}\frac{1}{3}\binom{3m}{m}$$ are positive integers. By a computation for $m \leq 19$, if $\lambda_{4}$ and $\lambda_{5}$ are positive integers, then we have $m=11$.

\(2) For $m=11$, we have checked that $$\begin{aligned}
\lambda_{6}&=\frac{(2m-1)(2m-2)(2m-3)}{(6m-1)(6m-2)(6m-3)}\frac{1}{3}\binom{3m}{m},\mbox{ and }\\
\lambda_{7}&=\frac{(2m-1)(2m-2)(2m-3)(2m-4)}{(6m-1)(6m-2)(6m-3)(6m-4)}\frac{1}{3}\binom{3m}{m}\end{aligned}$$ are positive integers and $$\lambda_{8}=\frac{(2m-1)(2m-2)(2m-3)(2m-4)(2m-5)}{(6m-1)(6m-2)(6m-3)(6m-4)(6m-5)}\frac{1}{3}\binom{3m}{m}$$ is not a positive integer. Therefore, by Lemma 