In this paper, the covering radius of codes over \(\mathbb R ={\mathbb Z_2}{R^{*}},\) where \(R^{*}={\mathbb Z_2}+v{\mathbb Z_2},v^{2}=v\) with different weight are discussed. The block repetition codes over \(\mathbb R\) is defined and the covering radius for block repetition codes, simplex code of \(\alpha\)-type and \(\beta\)-type in \(\mathbb R\) are obtained.
Codes over finite commutative rings have been studied for almost 50 years. The main motivation of studying codes over rings is that they can be associated with codes over finite fields through the Gray map. Recently, coding theory over finite commutative non-chain rings is a hot research topic. Recently, there has been substantial interest in the class of additive codes. In [1, 2], Delsarte contributes to the algebraic theory of association scheme where the main idea is to characterize the subgroups of the underlying abelian group in a given association scheme.
The covering radius is an important geometric parameter of codes. It not only indicates the maximum error correcting capability of codes, but also relates to some practical problems such as the data compression and transmission. Studying of the covering radius of codes has attracted many coding scientists for almost 30 years. The covering radius of linear codes over binary finite fields was studied in [3].
Additive codes over \(\mathbb Z_{2}\mathbb Z_{4}\) have been extensively studied in [4, 5, 6, 7].Enormous results were made available on the simplex codes over finite fields and finite rings. A few of them are [8, 9, 10]. In [11], the authors, in particular, gave lower and upper bounds on the covering radius of codes over the ring \(R={\mathbb Z_2}+u{\mathbb Z_2}\) where \(u^{2}=0\) with respect to different distance, and they explained the covering radius of various repetition codes, Simplex Codes (Type \(\alpha\) and \(\beta\) over R.) The above results motivate us to work in this area.
In this paper, the simplex codes of types \(\alpha\) and \(\beta\) over \(\mathbb R\) are obtained. At this juncture, the meaning of constructing new codes is to concatenate binary and quaternary simplex codes of types \(\alpha\) and \(\beta.\) These results in the simplex codes over \(\mathbb R,\) which contain the corresponding binary and \(R^{*}\) codes as subclasses. The rest of this paper is organized as follows
In Section 2, some properties related to additive codes over \(\mathbb R={\mathbb Z_2}{R^{*}},\) where \(R^*=\{0,1,v,1+v\}, v^2=v\) are given. The covering radius of codes over \(\mathbb R\) is considered in Section 3. In Section 4, the block repetition codes over \(\mathbb R\) is defined and find their covering radius. The construction of simplex codes over \(\mathbb R\) of types \(\alpha\) and \(\beta\) and the covering radius of these codes is also considered in Sections 5.
Throughout this paper, \(\mathbb R={\mathbb Z_2}{R^{*}},\) where \(R^*=\{0,1,v,1+v\}, v^2=v.\) In this section, some preliminary results are given based on [5] and [7]. A non empty set \(C\) is a \(\mathbb R\)-additive code if it is a subgroup of \({\mathbb Z_2^{\gamma}\times R^{{*}^\delta}}\). In this case, \(C\) is also isomorphic to an abelian structure \({\mathbb Z_2^{\lambda}\times R^{{*}^\mu}}\) for some \(\lambda\) and \(\mu.\) \(C\) is of type \({{2^\lambda}{{R^{*}}^\mu}}\) as a group. It follows that it has \(\mid C \mid=2^{\lambda+2\mu}\) codewords, and the number of order two codewords in \(C\) is \(\mid C \mid=2^{\lambda+\mu}.\)
Consider the following extension of the Gray map \begin{equation*} \phi : {\mathbb Z_2^{\gamma}\times R^{{*}^\delta}}\rightarrow \mathbb Z_{2}^n, \;\; \text{with}\;\; n=\gamma+2\delta, \end{equation*} given by \begin{equation*} \phi\left(u,w\right)=\left(u, \phi\left(w_1\right),\cdots \phi\left(w_\delta\right)\right), \forall~ u~\in ~\mathbb Z_2^{\gamma}, \forall ~\left(w_1,\cdots, w_{\delta}\right)~\in ~R^{{*}^\delta}, \end{equation*} where \begin{equation*} \phi : R^{*}\rightarrow \mathbb Z_2^{2}, \end{equation*} is the Gray map given by \(\phi\left(0\right)=\left(0,~0\right), \phi\left(1\right)=\left(0,~1\right), \phi\left(v\right)=\left(1,~1\right),\) and \(\phi\left(1+v\right)=\left(1,~0\right).\) Then the binary image of a \({\mathbb R}\)-additive code under the extended Gray map is called a \({\mathbb R}\)-linear code of length \(n=\gamma+2\delta.\) The Hamming weight of \(u,\) denoted by \(wt_{H}\left(u\right)\) and \(wt_{L}\left(w\right)\) and \(wt_{E}\left(w\right)\) the Lee and Euclidean weights of \(w\) respectively, where \(u \in \mathbb Z_2^{\gamma}\) and \( w\in R^{{*}^\delta}\) are defined as \begin{equation*} wt_{E}\left(v_{i}\right) = \begin{cases} 0 & \quad \text{if } v_{i}=0\\ 1 & \quad \text{if } v_{i}=1 ~\text{or} ~\left(1+v\right)\\ 4 & \quad \text{if } v_{i}=v \end{cases} \end{equation*} and \begin{equation} wt_{L}\left(v_{i}\right) = \begin{cases} 0 & \quad \text{if } v_{i}=0\\ 1 & \quad \text{if } v_{i}=1 ~\text{or} ~\left(1+v\right)\\ 2 & \quad \text{if } v_{i}=v \end{cases} \end{equation} The Lee weight of \(x\) is defined as \(wt_{L}\left(x\right)=wt_{H}\left(u\right)+wt_{L}\left(w\right),\) where \(x=\left(u, ~w\right)~\in ~\mathbb Z_2^{\gamma}\times R^{{*}^\delta},\) and \(u=\left(u_1, ~\cdots, ~u_\gamma\right)~\epsilon~ \mathbb Z_2^{\gamma}\) and \(w=\left(w_1, ~\cdots, ~w_\delta\right)~\epsilon ~R^{{*}^\delta}\) and the Euclidean weight of \(x\) is defined as \(wt_{E}\left(x\right)=wt_{H}\left(u\right)+wt_{E}\left(w\right).\) The Gray map defined above is an isometry which transforms the Lee distance defined over \(\mathbb Z_2^{\gamma}\times R^{{*}^\delta}\) to the Hamming distance defined over \(\mathbb Z_2^{n},\) with \(n=\gamma+2\delta.\)Proposition 1. Let \(C\) be a code over \(\mathbb Z_2^{\gamma}\times R^{{*}^\delta}\) and \(\phi \left(C\right)\) be the Gray map image of \(C.\) Then \(r_{L}\left(C\right)=r_{H}\left(\phi \left(C\right)\right).\)
The following result is useful for determining the covering radius of codes over rings. This is a generalization of the result in [13] for codes over finite fields.Proposition 2. If \(C_{0}\) and \(C_{1}\) are codes over \({\mathbb R},\) of lengths \(n_{0}\) and \(n_{1},\) minimum distance \(d_{0}\) and \(d_{1},\) and generated by matrices \(G_{0}\) and \(G_{1},\) respectively, and if \(C\) is the code generated by \begin{equation*} G=\left[\begin{array}{c|c} 0&G_{1} \\\hline G_{0}&A \end{array}\right], \end{equation*} then \(r_{d}\left(C\right)\leq {r_d}\left(C_{0}\right)+{r_d}\left(C_{1}\right),\) and the covering radius of the concatenation of \(C_{0}\) and \(C_{1},\) denoted \(C_{c},\) satisfies $$r_{d}\left(C_c\right)\geq {r_d}\left(C_{0}\right)+{r_d}\left(C_{1}\right)$$ for all distances \(d\) over \({\mathbb R}.\)
In order to determine the covering radius of simplex codes of types \(\alpha\) and \(\beta\) over \({\mathbb R},\) the some classes of block codes over \({\mathbb R}\) is defind and the approach in [14] is used to obtain the covering radius.
The block repetition code \(C^{n}\) over \({\mathbb R}\) is a \({\mathbb R}\)-additive code of length \(n=\sum\limits_{j=1}^{7}n_j\) with generator matrix \(G=\left(\overbrace{0101\cdots 01}^{n_1}\overbrace{0v0v\cdots 0v}^{n_2}\overbrace{01+v01+v\cdots 01+v}^{n_3}\\ \overbrace{1010\cdots 10}^{n_4}\overbrace{1111\cdots 11}^{n_5} \overbrace{1v1v\cdots 1v}^{n_6}\overbrace{11+v11+v\cdots 11+v}^{n_7}\right).\)
If, for a fixed \(1\leq i\leq 7.\) For all \(1\leq j\neq i\leq 7, n_j=0,\) then the code \(C^{n}=C^{n_{i}}\) is denoted by \(C_{i}.\) Therefore, the seven basic repetition codes are given below. That,
Theorem 3. The covering radius of \(C_{j}, 1\leq j\leq 7,\) with respect to the Euclidean weight is given by
Proof.
For \(c\in C_{j}, 1\leq j\leq 7,\) let \(t_{i}\left(c\right), 0\leq i\leq 7\) denote the number of occurrences of symbol \(i\) in the codeword \(c.\)
Considering \(1 ~\text{to}~ 5,\) that
$$
r_{E}\left(C_{j}\right)=\max\limits_{x\in \mathbb R^{n}}\{d_{E}\left(x, C_{j}\right); 1\leq j\leq 7\}.
$$
Let \(x\in {\mathbb R^n}.\) If \(x\) is given by \(\left(t_{0}, t_{1}, t_{2}, t_{3}, t_{4}, t_{5}, t_{6}, t_{7}\right),\) where \(\sum\limits_{j=0}^{7}t_{j}=n,\)
then
\({d_{E}\left(x, \overline{00}\right)=n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7}}, {d_{E}\left(x, \overline{01}\right)=n-t_{1}+3t_{3}+t_{4}+t_{6}+4t_{7}},\)
\({d_{E}\left(x, \overline{0v}\right)=n-t_{2}+3t_{0}+4t_{4}+t_{5}+t_{7}},{d_{E}\left(x, \overline{01+v}\right)=n-t_{3}+3t_{1}+t_{4}+4t_{5}+t_{6}},\)
\({d_{E}\left(x, \overline{10}\right)=n-t_{4}+t_{1}+4t_{2}+t_{3}+3t_{6}},{d_{E}\left(x, \overline{11}\right)=n-t_{5}+t_{0}+t_{2}+4t_{3}+3t_{7}},\)
\({d_{E}\left(x, \overline{1v}\right)=n-t_{6}+4t_{0}+t_{1}+t_{3}+3t_{4}}, {d_{E}\left(x, \overline{11+v}\right)=n-t_{7}+t_{0}+4t_{1}+t_{2}+3t_{5}}.\)
Therefore, \(d_{E}(x, C_{1})=d_{E}(x, C_{3})=\min\{(n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7}), (n-t_{1}+3t_{3}+t_{4}+t_{6}+4t_{7}), (n-t_{2}+3t_{0}+4t_{4}+t_{5}+t_{7}),
(n-t_{3}+3t_{1}+t_{4}+4t_{5}+t_{6})\}\)
\(\leq \frac{4n+2(t_{0}+t_{1}+t_{2}+t_{3}+t_{4}+t_{5}+t_{6}+t_{7})+2(t_{4}+t_{5}+t_{6}+t_{7})}{4}
\leq 2n\) and hence
$$
r_{E}\left(C_{1}\right)=r_{E}\left(C_{3}\right)\leq 2n.
$$
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{4}}\overbrace{01\cdots 01}^{\frac{n}{4}}\overbrace{0v\cdots 0v}^{\frac{n}{4}}
\overbrace{01+v\cdots 01+v}^{\frac{n}{4}}\right)~\in {\mathbb R}^n,\)
then $$d_{E}\left(x, \overline{00}\right)=d_{E}\left(x, \overline{01}\right)=d_{E}\left(x, \overline{0u}\right)=d_{E}\left(x, \overline{01+u}\right)=\frac{n}{8}+4\left(\frac{n}{8}\right)+\frac{n}{8}
=\frac{3n}{4}$$ and \(r_{E}\left(C_{1}\right)=r_{E}\left(C_{3}\right)\geq \frac{3n}{4}.\) Thus, \(\frac{3n}{4}\leq r_{E}\left(C_{1}\right)=r_{E}\left(C_{3}\right)\leq 2n.\)
\begin{eqnarray*}d_{E}\left(x, C_{2}\right)&=&\min\left\{\left(n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7}\right), \left(n-t_{2}+3t_{0}+4t_{4}+t_{5}+t_{7}\right)\right\}\\ &\leq&
\frac{2n+2\left(t_{0}+t_{2}+t_{5}+t_{7}\right)+4\left(t_{4}+t_{6}\right)}{2}\leq 3n.\end{eqnarray*} Thus \(r_{E}\left(C_{2}\right)\leq 3n.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{2}}\overbrace{0v\cdots 0v}^{\frac{n}{2}}\right)~\in ~{\mathbb R}^n,\)
then $$d_{E}\left(x, \overline{00}\right)=d_{E}\left(x, \overline{0v}\right)=4\left(\frac{n}{4}\right)=n.$$
Thus \(r_{E}\left(C_{2}\right)\geq n,\) and so, \(n\leq r_{E}\left(C_{2}\right)\leq 3n.\)
\begin{eqnarray*}d_{E}\left(x, C_{4}\right)&=&\min\left\{\left(n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7}\right), \left(n-t_{4}+t_{1}+4t_{2}+t_{3}+3t_{6}\right)\right\}\\& \leq&
\frac{2n-t_{0}+t_{1}+7t_{2}+t_{3}-t_{4}+t_{5}+7t_{6}+t_{7}}{2}\leq 4n\end{eqnarray*} and hence \(r_{E}\left(C_{4}\right)\leq 4n.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{2}}\overbrace{10\cdots 10}^{\frac{n}{2}}\right)~\in ~{\mathbb R}^n,\)
then $$d_{E}\left(x, \overline{00}\right)=d_{E}\left(x, \overline{10}\right)=\frac{n}{4}.$$
Thus \(r_{E}\left(C_{4}\right)\geq \frac{n}{4},\) and so \(\frac{n}{4}\leq r_{E}\left(C_{4}\right)\leq 4n.\)
\begin{equation}
\begin{aligned}
d_{E}(x, C_{5})=d_{E}(x, C_{7})=\min\{(n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7}), (n-t_{1}+3t_{3}+t_{4}+t_{6}+4t_{7})\\,
(n-t_{2}+3t_{0}+4t_{4}+t_{5}+t_{7}), (n-t_{3}+3t_{1}+t_{4}+4t_{5}+t_{6}), (n-t_{4}+t_{1}+4t_{2}+t_{3}+3t_{6})\\,
(n-t_{5}+t_{0}+t_{2}+4t_{3}+3t_{7}), (n-t_{6}+4t_{0}+t_{1}+t_{3}+3t_{4}), (n-t_{7}+t_{0}+4t_{1}+t_{2}+3t_{5})\}\leq 2n,
\end{aligned}
\end{equation}
then \(r_{E}(C_{5})=r_{E}(C_{7})\leq 2n.\)
If
\(x=\left(\overbrace{00\cdots 00}^{\frac{n}{8}}\overbrace{01\cdots 01}^{\frac{n}{8}}\overbrace{0v\cdots 0v}^{\frac{n}{8}}
\overbrace{01+v\cdots 01+v}^{\frac{n}{8}}\overbrace{10\cdots 10}^{\frac{n}{8}}\overbrace{11\cdots 11}^{\frac{n}{8}}
\overbrace{1v\cdots 1v}^{\frac{n}{8}}\\ \overbrace{11+v\cdots 11+v}^{\frac{n}{8}}\right)~\in ~{\mathbb R}^n,\)
then
\(d_{E}\left(x, \overline{00}\right)=d_{E}\left(x, \overline{01}\right)=d_{E}\left(x, \overline{0v}\right)=d_{E}\left(x, \overline{01+v}\right)=d_{E}\left(x, \overline{10}\right)=
d_{E}\left(x, \overline{11}\right) =d_{E}\left(x, \overline{1v}\right)=d_{E}\left(x, \overline{11+v}\right)=\frac{n}{16}+4\left(\frac{n}{16}\right)+\frac{n}{16}+\frac{n}{16}+\frac{n}{8}+\frac{n}{16}+4\left(\frac{n}{16}\right)+\frac{n}{8}=n.\)
Thus \(r_{E}\left(C_{5}\right)=r_{E}\left(C_{7}\right)\geq n,\) and so \(n\leq r_{E}\left(C_{5}\right)=r_{E}\left(C_{7}\right)\leq 2n.\)
\(d_{E}(x, C_{6})=\min\{(n-t_{0}+3t_{2}+t_{5}+4t_{6}+t_{7})\),
\((n-t_{2}+3t_{0}+4t_{4}+t_{5}+t_{7}),
(n-t_{4}+t_{1}+4t_{2}+t_{3}+3t_{6})\),
\((n-t_{6}+4t_{0}+t_{1}+t_{3}+3t_{4})\}\leq \frac{4n+2n+4n}{4}\),
so then \(r_{E}\left(C_{6}\right)\leq \frac{5n}{2}.\)
If
\(x=\left(\overbrace{00\cdots 00}^{\frac{n}{4}}\overbrace{0v\cdots 0v}^{\frac{n}{4}}\overbrace{10\cdots 10}^{\frac{n}{4}}
\overbrace{1v\cdots 1v}^{\frac{n}{4}}\right)~\epsilon ~{\mathbb R}^n,\)
then
\(d_{E}\left(x, \overline{00}\right)=d_{E}\left(x, \overline{0v}\right)=d_{E}\left(x, \overline{10}\right)=d_{E}\left(x, \overline{1v}\right)=4\left(\frac{n}{8}\right)+\frac{n}{8}+\frac{n}{8}+4\left(\frac{n}{8}\right)=\frac{5n}{4}.\)
Thus \(r_{E}\left(C_{6}\right)\geq \frac{5n}{4},\) and so \(\frac{5n}{4}\leq r_{E}\left(C_{6}\right)\leq \frac{5n}{2}.\)
Theorem 4. The Covering radius of \(C_{j}, 1\leq j\leq 7,\) with respect to the Lee weight is given by
Proof.
For \(c\in C_{j}, 1\leq j\leq 7,\) let \(t_{i}\left(c\right), 0\leq i\leq 7\) denote the number of occurrences of symbol \(i\) in the codeword \(c.\)
Considering \(1 ~\text{to}~ 5,\) that
$$
r_{L}\left(C_{j}\right)=\max\limits_{x\in {\mathbb R}^{n}} \{d_{L}\left(x, C_{j}\right); 1\leq j\leq 7\}.
$$
Let \(x\in {\mathbb R}^{n}.\) If \(x\) is given by \(\left(t_{0}, t_{1}, t_{2}, t_{3}, t_{4}, t_{5}, t_{6}, t_{7}\right),\) where
\(\sum\limits_{j=0}^{7}t_{j}=n,\) then
\({d_{L}\left(x, \overline{00}\right)=n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}}, {d_{L}\left(x, \overline{01}\right)=n-t_{1}+t_{3}+t_{4}+t_{6}+2t_{7}},\)
\({d_{L}\left(x, \overline{0v}\right)=n-t_{2}+t_{0}+2t_{4}+t_{5}+t_{7}},{d_{L}\left(x, \overline{01+v}\right)=n-t_{3}+t_{1}+t_{4}+2t_{5}+t_{6}},\)
\({d_{L}\left(x, \overline{10}\right)=n-t_{4}+t_{1}+2t_{2}+t_{3}+t_{6}},{d_{L}\left(x, \overline{11}\right)=n-t_{5}+t_{0}+t_{2}+2t_{3}+t_{7}},\)
\({d_{L}\left(x, \overline{1v}\right)=n-t_{6}+2t_{0}+t_{1}+t_{3}+t_{4}},{d_{L}\left(x, \overline{11+v}\right)=n-t_{7}+t_{0}+2t_{1}+t_{2}+t_{5}}.\)
Therefore, \(d_{L}(x, C_{1})=d_{L}(x, C_{3})=\min\{(n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}), (n-t_{1}+t_{3}+t_{4}+t_{6}+2t_{7}),
(n-t_{2}+t_{0}+2t_{4}+t_{5}+t_{7}), (n-t_{3}+t_{1}+t_{4}+2t_{5}+t_{6})\}\leq \frac{4n+4(t_{4}+t_{5}+t_{6}+t_{7})}{4}\leq 2n,\) then
\(r_{L}\left(C_{1}\right)=r_{L}\left(C_{3}\right)\leq 2n.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{4}}\overbrace{01\cdots 01}^{\frac{n}{4}}\overbrace{0v\cdots 0v}^{\frac{n}{4}}
\overbrace{01+v\cdots 01+v}^{\frac{n}{4}}\right)\in{\mathbb R}^n,\)
then \(d_{L}\left(x, \overline{00}\right)=d_{L}\left(x, \overline{01}\right)=d_{L}\left(x, \overline{0v}\right)=d_{L}\left(x, \overline{01+v}\right)=\frac{n}{8}+2\left(\frac{n}{8}\right)+\frac{n}{8}
=\frac{n}{2}.\) Thus \(r_{L}\left(C_{1}\right)=r_{L}\left(C_{3}\right)\geq \frac{n}{2},\) and so \(\frac{n}{2}\leq r_{L}\left(C_{1}\right)=r_{L}\left(C_{3}\right)\leq 2n.\)
\(d_{L}\left(x, C_{2}\right)=\min\{\left(n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}\right), \left(n-t_{2}+t_{0}+2t_{4}+t_{5}+t_{7}\right)\} \leq \frac{2n+2\left(t_{4}+t_{5}+t_{6}+t_{7}\right)}{2}\leq
2n.\) Then \(r_{L}\left(C_{2}\right)\leq 2n.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{2}}\overbrace{0v\cdots 0v}^{\frac{n}{2}}\right) \in {\mathbb R}^n,\) then
\(d_{L}\left(x, \overline{00}\right)=d_{L}\left(x, \overline{0v}\right)=2\left(\frac{n}{4}\right)=\frac{n}{2}.\)
Thus \(r_{L}\left(C_{2}\right)\geq \frac{n}{2}\) and so \(\frac{n}{2}\leq r_{L}\left(C_{2}\right)\leq 2n.\)
\(d_{L}\left(x, C_{4}\right)=\min\{\left(n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}\right), \left(n-t_{4}+t_{1}+2t_{2}+t_{3}+t_{6}\right)\} \leq \frac{2n+2n}{2}\leq 2n,\) so then \(r_{L}\left(C_{4}\right)\leq 2n.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{2}}\overbrace{10\cdots 10}^{\frac{n}{2}}\right) \in {\mathbb R}^n,\)
then \(d_{L}\left(x, \overline{00}\right)=d_{L}\left(x, \overline{10}\right)=\left(\frac{n}{4}\right)=\frac{n}{4}.\) Thus \(r_{L}\left(C_{4}\right)\geq \frac{n}{4}\) and hence
\(\frac{n}{4}\leq r_{L}\left(C_{4}\right)\leq 2n.\)
\begin{equation}
\begin{aligned}
d_{L}(x, C_{5})=d_{L}(x, C_{7})=\min\{(n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}), (n-t_{1}+t_{3}+t_{4}+t_{6}+2t_{7}), (n-t_{2}+t_{0}+2t_{4}+t_{5}+t_{7}),\\
(n-t_{3}+t_{1}+t_{4}+2t_{5}+t_{6}), (n-t_{4}+t_{1}+2t_{2}+t_{3}+t_{6}), (n-t_{5}+t_{0}+t_{2}+2t_{3}+t_{7}),\\ (n-t_{6}+2t_{0}+t_{1}+t_{3}+t_{4}),
(n-t_{7}+t_{0}+2t_{1}+t_{2}+t_{5})\}\leq \frac{8n-n+5n}{8}\leq \frac{3n}{2},
\end{aligned}
\end{equation}
so then \(r_{L}\left({C_{5}}\right)=r_{L}\left({C_{7}}\right)\leq \frac{3n}{2}.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{8}}\overbrace{01\cdots 01}^{\frac{n}{8}}\overbrace{0v\cdots 0v}^{\frac{n}{8}}\overbrace{01+v\cdots 01+v}^{\frac{n}{8}}
\overbrace{10\cdots 10}^{\frac{n}{8}}\overbrace{11\cdots 11}^{\frac{n}{8}}\overbrace{1v\cdots 1v}^{\frac{n}{8}}
\overbrace{11+v\cdots 11+v}^{\frac{n}{8}}\right)~\in ~{\mathbb R}^n,\)
then \(d_{L}\left(x, \overline{00}\right)=d_{L}\left(x, \overline{01}\right)=d_{L}\left(x, \overline{0v}\right)=d_{L}\left(x, \overline{01+v}\right)=d_{L}\left(x, \overline{10}\right)=
d_{L}\left(x, \overline{11}\right)=d_{L}\left(x, \overline{1v}\right)=d_{L}\left(x, \overline{11+v}\right)=\frac{n}{16}+2\left(\frac{n}{16}\right)+\frac{n}{16}+\frac{n}{16}+\frac{n}{16}+\frac{n}{16}+\frac{n}{16}+2\left(\frac{n}{16}\right)+\frac{n}{16}+\frac{n}{16}=\frac{3n}{4}.\)
Thus \(r_{L}\left(C_{5}\right)=r_{L}\left(C_{7}\right)\geq \frac{3n}{4},\) and so \(\frac{3n}{4}\leq r_{L}\left(C_{5}\right)=r_{L}\left(C_{7}\right)\leq \frac{3n}{2}.\)
\(d_{L}(x, C_{6})=\min\{(n-t_{0}+t_{2}+t_{5}+2t_{6}+t_{7}), (n-t_{2}+t_{0}+2t_{4}+t_{5}+t_{7}),
(n-t_{4}+t_{1}+2t_{2}+t_{3}+t_{6}), (n-t_{6}+2t_{0}+t_{1}+t_{3}+t_{4})\}\leq \frac{4n+2n}{4}\leq \frac{3n}{2},\) so then \(r_{L}\left(C_{6}\right)\leq \frac{3n}{2}.\)
If \(x=\left(\overbrace{00\cdots 00}^{\frac{n}{4}}\overbrace{0v\cdots 0v}^{\frac{n}{4}}\overbrace{10\cdots 10}^{\frac{n}{4}}
\overbrace{1v\cdots 1v}^{\frac{n}{4}}\right)\in {\mathbb R}^n,\)
then \(d_{L}\left(x, \overline{00}\right)=d_{L}\left(x, \overline{0v}\right)=d_{L}\left(x, \overline{10}\right)=d_{L}\left(x, \overline{1v}\right)=2\left(\frac{n}{8}\right)+\frac{n}{8}+\frac{n}{8}+2\left(\frac{n}{8}\right)=\frac{3n}{4}.\)
Thus \(r_{L}\left(C_{6}\right)\geq \frac{3n}{4},\) and hence \(\frac{3n}{4}\leq r_{L}\left(C_{6}\right)\leq \frac{3n}{2}.\)
Theorem 5. The covering radius of the block repetition code \(C^{n}\) have the following properties; $$ {r_{E}\left(C^{n}\right)\leq \frac{1}{2}\left[5\left(n_{1}+n_{3}+n_{6}\right)+3n_{2}+9n_{4}\right]+2\left(n_{5}+n_{7}\right),} $$ and if \(n_{1}=\cdots=n_{7}=n\) $$ r_{L}\left(C^{7n}\right)=2n. $$
Proof.
By Proposition 2, Theorem 3 and Theorem 4, let \(x=x_{1}x_{2}x_{3}x_{4}x_{5}x_{6}x_{7} \in {\mathbb R}^n\)
with \(x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{6}, x_{7}\) is
\(\left(a_{0}, a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}\right), \left(b_{0}, b_{1}, b_{2}, b_{3}, b_{4}, b_{5}, b_{6}, b_{7}\right),\)
\(\left(c_{0}, c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}\right), \left(d_{0}, d_{1}, d_{2}, d_{3}, d_{4}, d_{5}, d_{6}, d_{7}\right),\)
\(\left(e_{0}, e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}, e_{7}\right), \left(f_{0}, f_{1}, f_{2}, f_{3}, f_{4}, f_{5}, f_{6}, f_{7}\right),\)
\(\left(g_{0}, g_{1}, g_{2}, g_{3}, g_{4}, g_{5}, g_{6}, g_{7}\right),\) respectively such that
\(n_{1}=\sum\limits_{j=0}^{7}a_{j}, ~n_{2}=\sum\limits_{j=0}^{7}b_{j}, ~n_{3}=\sum\limits_{j=0}^{7}c_{j}, ~n_{4}=\sum\limits_{j=0}^{7}d_{j},\)
\(~n_{5}=\sum\limits_{j=0}^{7}e_{j}, ~n_{6}=\sum\limits_{j=0}^{7}f_{j},~n_{7}=\sum\limits_{j=0}^{7}g_{j}.\)
Then
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{00}\right)=n_{1}-a_{0}+3a_{2}+a_{5}+4a_{6}+a_{7}+n_{2}-b_{0}+3b_{2}+b_{5}+4b_{6}+b_{7}+n_{3}-c_{0}+3c_{2}+c_{5}+4c_{6}\\+c_{7}+n_{4}-d_{0}+3d_{2}+d_{5}+
4d_{6}+d_{7}+n_{5}-e_{0}+3e_{2}+e_{5}+4e_{6}+e_{7}+n_{6}-f_{0}+3f_{2}+f_{5}+4f_{6}+f_{7}+n_{7}-g_{0}+3g_{2}+g_{5}+4g_{6}+g_{7},
\end{aligned}
\end{equation}
where
\(\overline{00}=\overbrace{00\cdots00}^{n_{1}}\overbrace{00\cdots00}^{n_{2}}\overbrace{00\cdots00}^{n_{3}}\overbrace{00\cdots00}^{n_{4}}\overbrace{00\cdots00}^{n_{5}}\overbrace{00\cdots00}^{n_{6}}\overbrace{00\cdots00}^{n_{7}},\)
is the first vector of \(C^{n},\) where \(n=n_1.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_{1}}\right)=n_{1}-a_{1}+3a_{3}+a_{4}+a_{6}+4a_{7}+n_{2}-b_{2}+3b_{0}+4b_{4}+b_{5}+b_{7}+n_{3}-c_{3}+3c_{1}+c_{4}+4c_{5}+c_{6}+n_{4}-d_{4}\\+d_{1}+4d_{2}+
d_{3}+3d_{6}+n_{5}-e_{5}+e_{0}+e_{2}+4e_{3}+3e_{7}+n_{6}-f_{6}+4f_{0}+f_{1}+f_{3}+3f_{4}+n_{7}-g_{7}+g_{0}+4g_{1}+g_{2}+3g_{5},
\end{aligned}
\end{equation}
where
\(\overline{y_1}=\overbrace{01\cdots01}^{n_{1}}\overbrace{0v\cdots0v}^{n_{2}}\overbrace{01+v\cdots01+v}^{n_{3}}\overbrace{10\cdots10}^{n_{4}}
\overbrace{11\cdots11}^{n_{5}}\overbrace{1v\cdots1v}^{n_{6}}\overbrace{11+v\cdots11+v}^{n_{7}},\) is the second vector of \(C^{n},\) where \(n=n_2.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_2}\right)=n_{1}-a_{1}+3a_{3}+a_{4}+a_{6}+4a_{7}+n_{2}-b_{2}+3b_{0}+4b_{4}+b_{5}+b_{7}+n_{3}-c_{3}+3c_{1}+c_{4}+4c_{5}+\\c_{6}+n_{4}-d_{0}+3d_{2}+d_{5}+
4d_{6}+d_{7}+n_{5}-e_{1}+3e_{3}+e_{4}+e_{6}+4e_{7}+n_{6}-f_{2}+3f_{0}+4f_{4}+f_{5}+f_{7}+n_{7}-g_{3}+3g_{1}+g_{4}+4g_{5}+g_{6},
\end{aligned}
\end{equation}
where
\(\overline{y_2}=\overbrace{01\cdots01}^{n_{1}}\overbrace{0v\cdots0v}^{n_{2}}\overbrace{01+v\cdots01+v}^{n_{3}}\overbrace{00\cdots00}^{n_{4}}
\overbrace{01\cdots01}^{n_{5}}\overbrace{0v\cdots0v}^{n_{6}}\overbrace{01+v\cdots01+v}^{n_{7}},\)
is the third vector of \(C^{n},\) where \(n=n_3.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_3}\right)=n_{1}-a_{2}+3a_{0}+4a_{4}+a_{5}+a_{7}+n_{2}-b_{0}+3b_{2}+b_{5}+4b_{6}+b_{7}+n_{3}-c_{2}+3c_{0}+4c_{4}+c_{5}+c_{7}+n_{4}-d_{0}\\+3d_{2}+d_{5}+
4d_{6}+d_{7}+n_{5}-e_{2}+3e_{0}+4e_{4}+e_{5}+e_{7}+n_{6}-f_{0}+3f_{2}+f_{5}+4f_{6}+f_{7}+n_{7}-g_{2}+3g_{0}+4g_{4}+g_{5}+g_{7},
\end{aligned}
\end{equation}
where
\(\overline{y_3}=\overbrace{0v\cdots0v}^{n_{1}}\overbrace{00\cdots00}^{n_{2}}\overbrace{0v\cdots0v}^{n_{3}}\overbrace{00\cdots00}^{n_{4}}
\overbrace{0v\cdots0v}^{n_{5}}\overbrace{00\cdots00}^{n_{6}}\overbrace{0v\cdots0v}^{n_{7}},\) is the fourth vector of \(C^{n},\) where \(n=n_4.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_4}\right)=n_{1}-a_{3}+3a_{1}+a_{4}+4a_{5}+a_{6}+n_{2}-b_{2}+3b_{0}+4b_{4}+b_{5}+b_{7}+n_{3}-c_{1}+3c_{3}+c_{4}+c_{6}+4c_{7}+n_{4}\\-d_{0}+3d_{2}+d_{5}+
4d_{6}+d_{7}+n_{5}-e_{3}+3e_{1}+e_{4}+4e_{5}+e_{6}+n_{6}-f_{2}+3f_{0}+4f_{4}+f_{5}+f_{7}+n_{7}-g_{1}+3g_{3}+g_{4}+g_{6}+4g_{7},
\end{aligned}
\end{equation}
where
\(\overline{y_4}=\overbrace{01+v\cdots01+v}^{n_{1}}\overbrace{0v\cdots0v}^{n_{2}}\overbrace{01\cdots01}^{n_{3}}\overbrace{00\cdots00}^{n_{4}}
\overbrace{01+v\cdots01+v}^{n_{5}}\overbrace{0v\cdots0v}^{n_{6}}\\\overbrace{01+v\cdots01+v}^{n_{7}},\)
is the fifth vector of \(C^{n},\) where \(n=n_5.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_5}\right)=n_{1}-a_{0}+3a_{2}+a_{5}+4a_{6}+a_{7}+n_{2}-b_{0}+3b_{2}+b_{5}+4b_{6}+b_{7}+n_{3}-c_{0}+3c_{2}+c_{5}+4c_{6}+c_{7}+n_{4}\\-d_{4}+d_{1}+4d_{2}+
d_{3}+3d_{6}+n_{5}-e_{4}+e_{1}+4e_{2}+e_{3}+3e_{6}+n_{6}-f_{4}+f_{1}+4f_{2}+f_{3}+3f_{6}+n_{7}-g_{4}+g_{1}+4g_{2}+g_{3}+3g_{6},
\end{aligned}
\end{equation}
where
\begin{equation}
\begin{aligned} \overline{y_5}=\overbrace{00\cdots00}^{n_{1}}\overbrace{00\cdots00}^{n_{2}}\overbrace{00\cdots00}^{n_{3}}\overbrace{10\cdots10}^{n_{4}}\overbrace{10\cdots10}^{n_{5}}\overbrace{10\cdots10}^{n_{6}}\overbrace{10\cdots10}^{n_{7}},
\end{aligned}
\end{equation}
is the sixth vector of \(C^{n},\) where \(n=n_6.\)
\begin{equation}
\begin{aligned} d_{E}\left(x,\overline{y_6}\right)=n_{1}-a_{2}+3a_{0}+4a_{4}+a_{5}+a_{7}+n_{2}-b_{0}+3b_{2}+b_{5}+4b_{6}+b_{7}+n_{3}-c_{2}+3c_{0}+4c_{4}+c_{5}\\+c_{7}+n_{4}-d_{4}+d_{1}+4d_{2}+d_{3}+3d_{6}+n_{5}-e_{6}+4e_{0}+e_{1}+e_{3}+3e_{4}+n_{6}-f_{4}+f_{1}+4f_{2}+f_{3}+3f_{6}+n_{7}-g_{6}+4g_{0}+g_{1}+g_{3}+3g_{4},
\end{aligned}
\end{equation}
where
\(\overline{y_6}=\overbrace{0v\cdots0v}^{n_{1}}\overbrace{00\cdots00}^{n_{2}}\overbrace{0v\cdots0v}^{n_{3}}
\overbrace{10\cdots10}^{n_{4}}\overbrace{1v\cdots1v}^{n_{5}}\overbrace{10\cdots10}^{n_{6}}\overbrace{1v\cdots1v}^{n_{7}},\)
is the seventh vector of \(C^{n},\) where \(n=n_7.\)
\begin{equation}
\begin{aligned}
d_{E}\left(x, \overline{y_7}\right)=n_{1}-a_{3}+3a_{1}+a_{4}+4a_{5}+a_{6}+n_{2}-b_{2}+3b_{0}+4b_{4}+b_{5}+b_{7}+n_{3}-c_{1}+3c_{3}+c_{4}+c_{6}+4c_{7}+n_{4}-d_{4}\\+d_{1}+4d_{2}+ d_{3}+3d_{6}+n_{5}-e_{7}+e_{0}+4e_{1}+e_{2}+3e_{5}+n_{6}-f_{6}+4f_{0}+f_{1}+f_{3}+3f_{4}+n_{7}-g_{5}+g_{0}+g_{2}+4g_{3}+3g_{7},
\end{aligned}
\end{equation}
where
\(\overline{y_7}=\overbrace{01+v\cdots01+v}^{n_{1}}\overbrace{0v\cdots0v}^{n_{2}}\overbrace{01\cdots01}^{n_{3}}
\overbrace{10\cdots10}^{n_{4}}\overbrace{11+v\cdots11+v}^{n_{5}}\overbrace{1v\cdots1v}^{n_{6}}\overbrace{11\cdots11}^{n_{7}},\)
is the eighth vector of \(C^{n},\) where \(n=n_8.\)
Thus
\begin{eqnarray*}&&r_{E}\left(C^{n}\right)\leq \frac{8n_{1}+4\left(a_{0}+a_{1}+a_{2}+a_{3}\right)+12\left(a_{4}+a_{5}+a_{6}+a_{7}\right)+8n_{2}+8\left(b_{0}+b_{2}+b_{5}+b_{7}\right)+16\left(b_{4}+b_{6}\right)}{8}\\
&&+\frac{8n_{3}+4\left(c_{0}+c_{1}+c_{2}+c_{3}\right)+12\left(c_{4}+c_{5}+c_{6}+c_{7}\right)+8n_{4}-4\left(d_{0}+d_{4}\right)+4\left(d_{1}+d_{3}+d_{5}+d_{7}\right)+28\left(d_{2}+d_{6}\right)}{8}\\
&&+\frac{8n_{5}+8\left(e_{0}+e_{1}+e_{2}+e_{3}+e_{4}+e_{5}+e_{6}+e_{7}\right)+8n_{6}+12\left(f_{0}+f_{2}+f_{4}+f_{6}\right)+4\left(f_{1}+f_{3}+f_{5}+f_{7}\right)}{8}\\
&&+\frac{8n_{7}+8\left(g_{0}+g_{1}+g_{2}+g_{3}+g_{4}+g_{5}+g_{6}+g_{7}\right)}{8}.\end{eqnarray*}
Hence
$$
{r_{E}\left(C^{n}\right)\leq \frac{1}{2}\left[5\left(n_{1}+n_{3}+n_{6}\right)+3n_{2}+9n_{4}\right]+2\left(n_{5}+n_{7}\right).}
$$
For the second part, that \(\phi\left(C^{7n}\right)\) is the set given by
\({\{000\cdots000000\cdots000000\cdots000000\cdots000000\cdots000000\cdots000000\cdots000,}\)
\({001\cdots001011\cdots011010\cdots010100\cdots100101\cdots101111\cdots111110\cdots110,}\)
\({001\cdots001011\cdots011010\cdots010000\cdots000001\cdots001011\cdots011010\cdots010,}\)
\({011\cdots011000\cdots000011\cdots011000\cdots000011\cdots011000\cdots000011\cdots011,}\)
\({010\cdots010011\cdots011001\cdots001000\cdots000010\cdots010011\cdots011001\cdots001,}\)
\({000\cdots000000\cdots000000\cdots000100\cdots100100\cdots100100\cdots100100\cdots100,}\)
\({011\cdots011000\cdots000011\cdots011100\cdots100111\cdots111100\cdots100111\cdots111,}\)
\({010\cdots010011\cdots011001\cdots001100\cdots100110\cdots110111\cdots111101\cdots101\}.}\)
Then \(r_{L}\left(c^{7n}\right)=r\left(\phi\left(C^{7n}\right)\right)=2n.\)
Theorem 6. The covering radius of the \(\mathbb R\)-simplex codes of type \(\alpha\) are upper bounded as follows: \(r_{L}\left(S_{k}^{\alpha}\right)\leq 3. 2^{3k-1}\) and \(r_{E}\left(S_{k}^{\alpha}\right)\leq 2^{k}\left(\frac{7.2^{2k}-1}{3}\right).\)
Proof. In \(\mathbb R\)-Simplex codes of type \(\alpha\) have a Lee weight equal to \(2^{3k}\) or \(3.2^{3k-1}.\) Hence, from (1), Proposition 2 and Theorem 5, we have \begin{eqnarray*} r_{L}\left(S_{k}^{\alpha}\right)&\leq& r_{L}\left(2^{2k}S_{2,~k}^{\alpha}\right)+r_{L}\left(2^{k}S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}r_{L}\left(S_{2,~k}^{\alpha}\right)+2^{k}r_{L}\left(S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}r_{H}\left(S_{2,~k}^{\alpha}\right)+2^{k}r_{L}\left(S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}\left(2^{k-1}\right)+2^{k}\left[\left(3.2^{2\left(k-1\right)}+3.2^{2\left(k-2\right)}+\cdots +3.2^{2.1}\right)+r_{L}\left(S_{R^{*},~1}^{\alpha}\right)\right]\\ &\leq& 2^{3k-1}+2^{k}\left[\left(2^{2k}-1\right)+1\right]\\ &\leq& 2^{3k-1}+2^{k}.2^{2k}\\ &\leq& 2^{3k-1}+2^{3k}\leq 3.2^{3k-1}. \end{eqnarray*} Thus \(r_{L}\left(S_{k}^{\alpha}\right)\leq 3. 2^{3k-1}.\) Similar arguments using (1), Proposition 2 and Theorem 5 give that \begin{eqnarray*} r_{E}\left(S_{k}^{\alpha}\right)&\leq& r_{E}\left(2^{2k}S_{2,~k}^{\alpha}\right)+r_{E}\left(2^{k}S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}r_{E}\left(S_{2,~k}^{\alpha}\right)+2^{k}r_{E}\left(S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}r_{H}\left(S_{2,~k}^{\alpha}\right)+2^{k}r_{E}\left(S_{R^{*},~k}^{\alpha}\right)\\ &\leq& 2^{2k}.2^{k-1}+2^{k}\left(\frac{11\left(2^{2k}-1\right)+9}{6}\right)\\ &\leq& 2^{k}\left[2^{2k-1}+\left(\frac{11\left(2^{2k}-1\right)+9}{6}\right)\right]\\ &\leq& 2^{k}\left(\frac{7.2^{2k}-1}{3}\right). \end{eqnarray*}
Theorem 7.
The covering radius of the \({\mathbb R}\)-Simplex codes of type \(\beta\) are given by
\(\left(i\right) r_{L}\left(S_{k}^{\beta}\right)\leq 2^{k-1}\left[\left(2^{k-1}+1\right)\left(2^{k}-1\right)-2\right],\)
\(\left(ii\right) r_{E}\left(S_{k}^{\beta}\right)\leq 2^{k-1}\left(\frac{14.2^{2k}-449}{6}\right).\)
Proof. From (1), (2), Proposition 2 and Theorem 5, we have \begin{eqnarray*} r_{L}\left(S_{k}^{\beta}\right)&\leq& r_{L}\left(2^{k}S_{2,~k}^{\beta}\right)+r_{L}\left(2^{k-1}S_{R^{*},~k}^{\beta}\right)\\ &\leq& 2^{k}r_{L}\left(S_{2,~k}^{\beta}\right)+2^{k-1}r_{L}\left(S_{R^{*},~k}^{\beta}\right)\\ &\leq &2^{k}r_{H}\left(S_{2,~k}^{\beta}\right)+2^{k-1}r_{L}\left(S_{R^{*},~k}^{\beta}\right)\\ &\leq &2^{k}(\frac{2^{k}-1}){2}+2^{k-1}\left[2^{k-1}\left(2^{k}-1\right)-2\right]\\ &\leq &2^{k-1}\left(2^{k}-1\right)+2^{k-1}\left[2^{k-1}\left(2^{k}-1\right)-2\right]\\ &\leq &2^{k-1}\left[\left(2^{k-1}+1\right)\left(2^{k}-1\right)-2\right]. \end{eqnarray*} Similar arguments using (1), Proposition 2 and Theorem 5 give that \begin{eqnarray*} r_{E}\left(S_{k}^{\beta}\right)&\leq& r_{E}\left(2^{k}S_{2,~k}^{\beta}\right)+r_{E}\left(2^{k-1}S_{R^{*},~k}^{\beta}\right)\\ &\leq& 2^{k}r_{E}\left(S_{2,~k}^{\beta}\right)+2^{k-1}r_{E}\left(S_{R^{*},~k}^{\beta}\right)\\ &\leq& 2^{k}r_{H}\left(S_{2,~k}^{\beta}\right)+2^{k-1}r_{E}\left(S_{R^{*},~k}^{\beta}\right)\\ &\leq& 2^{k}\left(\frac{2^{k}-1}{2}\right)+2^{k-1}\left[2^{k}\left(2^{k+1}-1\right)+\frac{1}{3}\left(2^{2k}-1\right)-\frac{147}{2}\right] \left(\text{since} r_{E}\left(S_{2}^{\beta}\right)\leq 25\right)\\ &\leq& 2^{k-1}\left(2^{k}-1\right)+2^{k-1}\left[2^{k}\left(2^{k+1}-1\right)+\frac{1}{3}\left(2^{2k}-1\right)-\frac{147}{2}\right]\\ &\leq& 2^{k-1}\left[2^{k}\left(2^{k+1}-1\right)+\frac{1}{3}\left(2^{2k}-1\right)+\left(2^{k}-1\right)-\frac{147}{2}\right]\\ &\leq& 2^{k-1}\left(\frac{14.2^{2k}-449}{6}\right). \end{eqnarray*}
