Proof. By definition, the integer \(a\) is the coset leader of \(C_{n}(a)\) if and only if \[\label{fan} a \bmod n \leq aq^i \bmod n \quad for any integeri\geq 0.\tag{3}\] Noticing that \(a \bmod n = a- n\cdot \lfloor \frac{ a}{ n } \rfloor\) and \(\lambda a \bmod \lambda n=\lambda a - \lambda n \cdot \lfloor \frac{\lambda a}{\lambda n}\rfloor\), we can assert that \[\notag \lambda \cdot (a \bmod n) = \lambda a - \lambda n \cdot \lfloor \frac{ a}{ n }\rfloor = \lambda a - \lambda n \cdot \lfloor \frac{ \lambda a}{\lambda n }\rfloor = \lambda a \bmod \lambda n.\] Similarly, we also have \[\label{qq1} \lambda \cdot (aq^i \bmod n) = \lambda a q^i \bmod \lambda n.\tag{4}\] Therefore, the inequality in (3 ) is equivalent to \[\notag \lambda a \bmod \lambda n \leq \lambda a q^i \bmod \lambda n \quad for any integeri\geq 0.\] By definition, this holds if and only if \(\lambda a\) is the coset leader of \(C_{\lambda n}(\lambda a)\). Therefore, the first statment follows.

In addition, the equality in (4 ) also implies that \[\notag aq^{i} \bmod n =a \quad if and only if \quad \lambda a q^i \bmod \lambda n =\lambda a.\] It follows that the smallest integer \(\ell_a\) such that \(aq^{\ell_a} \bmod n = a\) is equal to the smallest integer \(\ell_{\lambda a}\) such that \(\lambda aq^{\ell_{\lambda a}} \bmod \lambda n=\lambda a\). Therefore, we have \(|C_{n}(a)|=|C_{ \lambda n}(\lambda a)|.\) This completes the proof. ◻

Proof. If \(2\leq \delta \leq \frac{q^h-1}{\lambda}+1\), then we have \(\widetilde{f}(\delta)=0\) and \(g(\delta)=0\). If \(\delta>\frac{q^h-1}{\lambda}+1\), then we distinguish the following two cases.

Case 1. Suppose that \(\delta_h\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\). Then we have \(\widetilde{\mu}(\delta)=\delta_h\) and \(w_{\delta}=\delta_hq^h\). By substituting them into equation (6 ), we obtain \[\notag \begin{align} \widetilde{f}(\delta)&=\lfloor \frac{\delta_h+\delta_{h}q^h}{\lambda}\rfloor-\lfloor \frac{\delta_{h}q^h}{\lambda}\rfloor + \sum\limits_{t=1}^{\delta_h-1}\left[ \lfloor \frac{2t}{\lambda} \rfloor -\lfloor \frac{t}{\lambda} \rfloor\right] \\ &=\sum\limits_{t=1}^{\delta_h}\left[ \lfloor \frac{2t}{\lambda} \rfloor -\lfloor \frac{t}{\lambda} \rfloor\right]. \end{align}\] In addition, it is straightforward to verify that \[\notag \tau(\delta)= \begin{cases} 1 &if \lambda \mid 2\delta_h,\\ 0 & if \lambda \nmid 2\delta_h. \end{cases}\] Recalling equation (8 ), it follows that \[\notag g(\delta)=\begin{cases} \lfloor \frac{\left(3+(-1)^{\lambda}\right)(\delta_h-1)}{2\lambda}\rfloor+ 1 &if \lambda \mid 2\delta_h,\\ \lfloor \frac{\left(3+(-1)^{\lambda}\right)(\delta_h-1)}{2\lambda}\rfloor & if \lambda \nmid 2\delta_h. \end{cases}\]

Case 2. Suppose that \(\delta_h> \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\). Then \(\widetilde{\mu}(\delta)=\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}=\delta_0\), \(w_{\delta}=\delta_hq^h\) and \(\tau(\delta)=0\). By substituting these values into equations (6 ) and (8 ), we obtain \[\notag \widetilde{f}(\delta)=\lfloor \frac{\delta_0+\delta_h}{\lambda}\rfloor-\lfloor \frac{2\delta_h}{\lambda}\rfloor+\sum\limits_{t=1}^{\delta_h}\left[ \lfloor \frac{2t}{\lambda} \rfloor -\lfloor \frac{t}{\lambda} \rfloor\right]\] and \[\notag g(\delta)= \lfloor \frac{\left(3+(-1)^{\lambda}\right)(\delta_h-1)}{2\lambda}\rfloor.\]

Finally, substituting the values of \(g(\delta)\) and \(\widetilde{f}(\delta)\) for corresponding cases into equation (?? ), we derive the desired equation (?? ). ◻

Proof. If \(\delta \leq \frac{q^{h+1}-1}{\lambda}+1,\) we can directly have \(f(\delta)=0\). If \(\delta >\frac{q^{h+1}-1}{\lambda}+1,\) then we have \(k_{\delta}=1\). We distinguish following cases.

Case 1. Suppose that \(\delta_h>0\). Then we have \(s_{\delta}=0\). This leads to

• \(\mathcal{T}_0(\delta)=\left\{t\in \mathbb{Z}: q\leq t\leq \delta_{h+1}q+ \delta_h-1 and q\nmid t\right\}\);

• \(\mathcal{T}_1(\delta)=\left\{t\in \mathbb{Z}:1 \leq t \leq \delta_{h+1}\right\}\);

• \(w_{\delta}=\delta_{h+1}q^{h+1}+\delta_hq^h\);

• \(\mu(\delta)=\min\left\{\delta_{h+1},\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\right\}\).

Recall equation (5 ). It follows that \[\label{cor3e1} \begin{align} f(\delta)=&\lfloor \frac{\mu(\delta)+w_{\delta}}{\lambda}\rfloor - \lfloor \frac{{\lfloor \mu(\delta})/q\rfloor+w_{\delta}}{\lambda}\rfloor + \sum\limits_{t=q, q\nmid t}^{\delta_{h+1}q+\delta_{h}-1}\left[ \lfloor\frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{t}{\lambda}\rfloor\right]+ \sum\limits_{t=1}^{\delta_{h+1}}\left[ \lfloor\frac{ tq+t}{\lambda}\rfloor - \lfloor\frac{2t}{\lambda}\rfloor\right]. \end{align}\tag{24}\] Note that \(\mu(\delta)=\delta_{h+1}\) if \(\delta_{h+1}\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\), and \(\mu(\delta)=\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}=\delta_0\) if \(\delta_{h+1}> \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\). Thus, \[\label{588} \lfloor \frac{\mu(\delta)+w_{\delta}}{\lambda}\rfloor - \lfloor \frac{{\lfloor \mu(\delta})/q\rfloor+w_{\delta}}{\lambda}\rfloor =\begin{cases} \lfloor\frac{\delta_0+\delta_{h+1}+\delta_h}{\lambda} \rfloor - \lfloor\frac{\delta_{h+1}+\delta_h}{\lambda} \rfloor& if\delta_{h+1}> \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell},\\ \lfloor\frac{2\delta_{h+1}+\delta_h}{\lambda} \rfloor - \lfloor\frac{\delta_{h+1}+\delta_h}{\lambda} \rfloor & if\delta_{h+1}\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}. \end{cases}\tag{25}\] By applying Lemma 6, we derive \[\notag \sum\limits_{t=q,q\nmid t}^{\delta_{h+1}q}\left[ \lfloor\frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{t}{\lambda}\rfloor\right] = \frac{\delta_{h+1}(\delta_{h+1}-1)(q-1)}{2\lambda}.\] Since each integer \(t\in \left[\delta_{h+1}q+1,\delta_{h+1}q+\delta_h-1\right]\) can be uniquely expressed as \(t=\delta_{h+1}q+i\) with \(i\in \left[1,\delta_h-1\right]\), we have \[\notag \sum\limits_{t=\delta_{h+1}q+1}^{\delta_{h+1}q+\delta_h-1}\left[ \lfloor\frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{t}{\lambda}\rfloor\right] = \sum\limits_{i=1}^{\delta_h-1}\left[\lfloor \frac{2\delta_{h+1}+i}{\lambda}\rfloor - \lfloor \frac{\delta_{h+1}+i}{\lambda}\rfloor \right].\] Adding above two sums, we get \[\label{59} \sum\limits_{t=q, q\nmid t}^{\delta_{h+1}q+\delta_{h}-1}\left[ \lfloor\frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{t}{\lambda}\rfloor\right]= \frac{\delta_{h+1}(\delta_{h+1}-1)(q-1)}{2\lambda} + \sum\limits_{i=1}^{\delta_h-1}\left[\lfloor \frac{2\delta_{h+1}+i}{\lambda}\rfloor - \lfloor \frac{\delta_{h+1}+i}{\lambda}\rfloor \right].\tag{26}\] Furthermore, it is straightforward to verify that \[\label{60} \begin{align} \sum\limits_{t=1}^{\delta_{h+1}}\left[ \lfloor\frac{ tq+t}{\lambda}\rfloor - \lfloor\frac{2t}{\lambda}\rfloor\right]&= \sum\limits_{t=1}^{\delta_{h+1}}\frac{t(q-1)}{\lambda}\\ &= \frac{\delta_{h+1}(\delta_{h+1}+1)(q-1)}{2\lambda}. \end{align}\tag{27}\] We can now conclude from (24 ) – (27 ) that \[\notag f(\delta)= \begin{cases} \lfloor\frac{\delta_0+\delta_{h+1}+\delta_h}{\lambda} \rfloor - \lfloor\frac{\delta_{h+1}+\delta_h}{\lambda} \rfloor + \frac{\delta_{h+1}^2(q-1)}{\lambda}+\sum\limits_{i=1}^{\delta_h-1}\left[\lfloor \frac{2\delta_{h+1}+i}{\lambda}\rfloor - \lfloor \frac{\delta_{h+1}+i}{\lambda}\rfloor \right]& if\delta_h>0 and \delta_{h+1}> \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell},\\ \frac{\delta_{h+1}^2(q-1)}{\lambda}+\sum\limits_{i=1}^{\delta_h}\left[\lfloor \frac{2\delta_{h+1}+i}{\lambda}\rfloor - \lfloor \frac{\delta_{h+1}+i}{\lambda}\rfloor \right]& if\delta_h>0 and \delta_{h+1}\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}. \end{cases}\]

Case 2. Suppose that \(\delta_h=0\). Then we have \(s_{\delta}=1\). This leads to

• \(\mathcal{T}_0(\delta)=\left\{t\in \mathbb{Z}: q\leq t\leq \delta_{h+1}q-1 and q\nmid t\right\}\);

• \(\mathcal{T}_1(\delta)=\{t\in \mathbb{Z}:1\leq t\leq \delta_{h+1}-1\}\);

• \(w_{\delta}=\delta_{h+1}q^{h+1}\);

• \(\mu(\delta)=\min\left\{ \delta_{h+1}q, \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell} \right\}.\)

Note that \(\mu(\delta)=\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}=\delta_1q+\delta_0\) if \(\delta_{h+1}q>\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\), and \(\mu(\delta)=\delta_{h+1}q\) if \(\delta_{h+1}q\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\). Thus, by substituting these terms into equation (5 ), we can use an analogous argument as in Case 1 to obtain \[\notag f(\delta)= \begin{cases} \frac{(\delta_{h+1}^2-\delta_{h+1}+\delta_1)(q-1)}{\lambda}+\lfloor \frac{\delta_1+\delta_0+\delta_{h+1}}{\lambda}\rfloor -\lfloor \frac{\delta_1+\delta_{h+1}}{\lambda}\rfloor& if\delta_h=0 and \delta_{h+1}q>\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell},\\ \frac{\delta_{h+1}^2(q-1)}{\lambda} &if \delta_h=0 and \delta_{h+1}q\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}. \end{cases}\]

Finally, substituting the value of \(f(\delta)\) into equation (?? ) for each corresponding case, we can conclude that (?? ) holds. This completes the proof. ◻

Proof. Note that \(\lambda \mid q-1\) implies that \(\mathrm{gcd}(q,\lambda)=1\). Since \(q\nmid \delta\), it follows that \(q\nmid \lambda\delta\). Additionally, it is clear that \(2\leq \lambda\delta<q^{m-h}\). By [40], it follows that \(\lambda \delta\) is a coset leader modulo \(\lambda n\). Then applying Lemma 1, we conclude that \(\delta\) is a coset leader modulo \(n\). Consequently, \(d_B(\mathcal{C}_{(q,n,\delta)})=\delta.\) ◻

Proof. We show that equation (?? ) holds through the following cases.

Case 1. Suppose that \(\sum\limits_{\ell=0}^{h-j_{\delta}} \delta_{\ell} q^{\ell} > \sum\limits_{\ell=h - r_{\delta} + 1}^{h + j_{\delta}} \delta_{\ell} q^{\ell - (h - r_{\delta} + 1)}\). By Remarks 4 and 5, the inequality implies that \(\lambda \delta\not\in \mathcal{A}_{j_{\delta}}\) for every integer \(i\in [-j_{\delta}+1,j_{\delta}]\). Therefore, by Theorem 1, we conclude \(\lambda\delta\not \in \mathcal{S}\). Note that the assumptions \(q\nmid\delta\) and \(\lambda \mid q-1\) imply \(q\nmid \lambda \delta\). It follows that \(\lambda \delta\) is a coset leader modulo \(\lambda n\). Applying Lemma 1, we further obtain that \(\delta\) is a coset leader modulo \(n\). Consequently, we have \(d_B(\mathcal{C}_{(q,n,\delta)})=\delta\).

Case 2. Suppose that \(\sum\limits_{\ell=0}^{h-j_{\delta}} \delta_{\ell} q^{\ell} \leq \sum\limits_{\ell=h - r_{\delta} + 1}^{h + j_{\delta}} \delta_{\ell} q^{\ell - (h - r_{\delta} + 1)}\). With an argument similar to that in the proof of [28], we can show that each integer in \([\lambda\delta, \hat{\delta}]\) is not a coset leader modulo \(\lambda n\). Applying Lemma 1, it follows that each integer in \([\delta, \lfloor \frac{\hat{\delta}}{\lambda}\rfloor ]\) is not a coset leader modulo \(n\). Notice that when \(m\) is odd, \[\hat{\delta}=\lambda \lfloor\frac{\hat{\delta}}{\lambda} \rfloor+\hat{\delta}\bmod \lambda= \sum\limits_{\ell = h + r_{\delta}}^{h + j_{\delta}} \delta_{\ell} q^{\ell} + \sum\limits_{\ell =h - r_{\delta}+1}^{h + j_{\delta}} \delta_{\ell} q^{\ell - ( h- r_{\delta}+1)}.\] Thus, we have \[\lambda \lfloor\frac{\hat{\delta}}{\lambda} \rfloor+ \lambda = \sum\limits_{\ell = h + r_{\delta}}^{h + j_{\delta}} \delta_{\ell} q^{\ell} + \sum\limits_{\ell = h - r_{\delta}+2}^{h + j_{\delta}} \delta_{\ell} q^{\ell - (h - r_{\delta}+1)} +\delta_{h-r_{\delta}+1}-\hat{\delta}\bmod \lambda+\lambda.\] This implies that \[q\mid \lambda \lfloor\frac{\hat{\delta}}{\lambda} \rfloor+ \lambda\quad if and only if\quad \delta_{h-r_{\delta}+1}+\lambda-q=\hat{\delta} \bmod \lambda.\] Then we distinguish the following two subcases:

Subcase 1. If \(\delta_{h-r_{\delta}+1}+\lambda-q\neq \hat{\delta} \bmod \lambda\), then \(q\nmid \lambda\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+\lambda\). Applying a similar argument as in Case 1, we can show that \(\lambda \lfloor \frac{\hat{\delta}}{\lambda}\rfloor+\lambda\) is a coset leader modulo \(\lambda n=q^m-1\). By applying lemma 1, it follows that \(\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+1\) is a coset leader modulo \(n\). Consequently, \(d_B(\mathcal{C}_{(q,n,\delta)}) =\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+1.\)

Subcase 2. If \(\delta_{h-r_{\delta}+1}+\lambda -q= \hat{\delta} \bmod \lambda\), then \(q\mid \lambda \lfloor \frac{\hat{\delta}}{\lambda}\rfloor+\lambda\) and \(q\nmid \lambda \lfloor \frac{\hat{\delta}}{\lambda}\rfloor+2\lambda\). This implies that \(\lambda \lfloor \frac{\hat{\delta}}{\lambda}\rfloor+\lambda\) is not a coset leader of \(\lambda n\). Furthermore, using a similar argument as in Case 1 again, we can conclude that while \(\lambda\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+2\lambda\) is a coset leader modulo \(\lambda n\). By applying Lemma 1, it follows that \(\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+1\) is not a coset leader modulo \(n\), while \(\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+2\) is a coset leader modulo \(n\). Therefore, we have \(d_B(\mathcal{C}_{(q,n,\delta)}) =\lfloor \frac{\hat{\delta}}{\lambda}\rfloor+2.\) ◻

Proof. We can use an argument analous to that in the proof of Theorem 5 to conclude that ?? ) hlods if \(r_{\delta}\neq 0\). Thus, we now only demonstrate equation (?? ) by considering the following cases.

Case 1. Suppose that \(\sum\limits_{\ell=0}^{h-j_{\delta}-1}\delta_\ell q^\ell \ge \sum\limits_{\ell=h}^{h+j_{\delta}}\delta_\ell q^{\ell-h}\). If the inequality is strict, by Theorem 1 together and Remarks 4–5, we have \(\lambda\delta\notin\mathcal{S}\cup\mathcal{H}\); If equality holds, by Lemma 2, we \(\lambda\delta\in\mathcal{H}\). Note that the assumptions \(q\nmid\delta\) and \(\lambda \mid q-1\) imply \(q\nmid \lambda \delta\). Therefore, in either case, \(\lambda\delta\) is a coset leader modulo \(\lambda n\). By Lemma 1, it follows that \(\delta\) is a coset leader of modulo \(n\). Therefore, \(d_B(\mathcal{C}_{(q,n,\delta)})=\delta\).

Case 2. Suppose that \(\sum\limits_{\ell=0}^{h-j_{\delta}-1}\delta_\ell q^\ell< \sum\limits_{\ell=h}^{h+j_{\delta}}\delta_\ell q^{\ell-h}\). We can use an analogous argument as in the proof of [28] to show that each integer in \([\lambda\delta, \hat{\delta})\) is not a coset leader modulo \(\lambda n\). By Lemma 1, it follows that each integer \(a\in \left[\delta, \lfloor \frac{\hat{\delta}}{\lambda}\rfloor \right)\) is not a coset leader modulo \(n\). Then we distinguish the following two subcases:

Subcase 1. Assume that \(\lambda\mid \hat{\delta}\). Using Lemma 2, it is straightforward to verify that \(\hat{\delta}\in \mathcal{H}\), and hence \(\hat{\delta}\) is a coset leader modulo \(\lambda n.\) It follows that \(\frac{\hat{\delta}}{\lambda}\) is a coset leader modulo \(n\). Therefore, we can obtain \(d_B(\mathcal{C}_{(q,n,\delta)})=\frac{\hat{\delta}}{{\lambda}}\).

Subcase 2. Assume that \(\lambda \nmid \hat{\delta}\). Then we have \(\lambda\lfloor \frac{\hat{\delta}}{\lambda}\rfloor<\hat{\delta}\). Recalling that each integer in \([\lambda \delta, \hat{\delta} )\) is not a coset leader modulo \(\lambda n\), this implies that \(\lambda \lfloor \frac{\hat{\delta}}{\lambda}\rfloor\) is not a coset leader modulo \(\lambda n\). By Lemma 1, it follows that \(\lfloor \frac{\hat{\delta}}{\lambda}\rfloor\) is not a coset leader modulo \(n\). Then we can use an analogous argument as in the proof of Case 2 of Theorem 5 to conclude that \(d_B(\mathcal{C}_{(q,n,\delta)})=\lfloor \frac{\hat{\delta}}{\lambda}\rfloor + 1\) \(None\) if \(\delta_h+\lambda-q\neq \hat{\delta} \bmod \lambda\), and \(d_B(\mathcal{C}_{(q,n,\delta)})=\lfloor \frac{\hat{\delta}}{\lambda}\rfloor + 2\) if \(\delta_h+\lambda-q= \hat{\delta} \bmod \lambda\). ◻

Proof. Note that \(q\nmid b\) implies that \(q\nmid \lambda \delta\). Since \(\lambda\) is a factor \(q-1\), it follows that \(q\nmid \delta\). Then by applying Theorem 5, we can directly derive equation (?? ).

We now demonstrate that (?? ) holds. It is clear that \(V(\lambda(\delta-1))\) has the form \[\notag (\mathbf{0}_{h-k}, a, \mathbf{0}_{2k_{}-1},\delta_{h-k},\ldots,\delta_0)\] with \(\sum\limits_{\ell=0}^{h-k}\delta_{\ell}q^{\ell}=b-\lambda\). By definition, it follows that

• \(s_{\delta}=k_{\delta}=k\);

• \(w_{\delta}=aq^{h+k}\);

• \(\mu(\delta)=\min\left\{b-\lambda, a q^{2k-1}\right\}\);

• \(\mathcal{T}_i(\delta)=\left[q^{k-i},a q^{k-i}\right]\cap \{t\in \mathbb{Z}: q\nmid t\}\) for each integer \(i\in [-k+1,k]\).

Then applying Lemma 10, we can obtain \[\notag \sum\limits_{t\in \mathcal{T}_i(\delta)} \left[\lfloor \frac{\lfloor tq^{2i}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor\right] = \begin{cases} \frac{1}{2\lambda}(a^2-1)(q-1)^2q^{2k-3}& if -k+2 \leq i\leq k-1,\\ \frac{1}{2\lambda} a(a-1)(q-1)q^{2k-2}& if i=kor -k+1. \end{cases}\] It follows that \[\notag \begin{align} \sum\limits_{i=-k_{\delta}+1}^{k_{\delta}}\sum\limits_{t\in \mathcal{T}_i(\delta)}N(tq^{2i-1}+1)&= \sum\limits_{i=-k+1}^{k}\sum\limits_{t\in \mathcal{T}_i(\delta)}N(tq^{2i-1}+1) \\ &=\frac{1}{\lambda}\left[a (a-1)(q-1)q^{2k-2}+(a^2-1)(k-1)(q-1)^2q^{2k-3}\right]. \end{align}\] By substituting the values of \(k_{\delta}\) and \(\mu(\delta)\), and the above expression into (5 ), we obtain \[\notag f(\delta)= \begin{cases} \frac{q-1}{\lambda}a^2q^{2k-3}\left[k(q-1)+1\right]& if b\geq a q^{2k-1}+\lambda,\\ \frac{q-1}{\lambda} a q^{2k-3}\left[ ak(q-1)+a-q \right]+\frac{b+a-\lambda}{\lambda} - \lfloor\frac{\lfloor (b-\lambda+aq)/q\rfloor}{\lambda} \rfloor &ifb< a q^{2k-1}+\lambda. \end{cases}\] Furthermore, notice that \[\notag\begin{align} N(\delta)- \left[\frac{b+a-\lambda}{\lambda} - \lfloor\frac{\lfloor (b-\lambda+aq)/q\rfloor}{\lambda} \rfloor\right]& = \frac{aq^{h+k}+b-\lambda}{\lambda} -\lfloor \frac{aq^{h+k}+b-\lambda}{\lambda q}\rfloor -\frac{b+a-\lambda}{\lambda}+\lfloor\frac{\lfloor (b-\lambda+aq)/q\rfloor}{\lambda} \rfloor\\ &= \frac{aq^{h+k-1}(q-1)}{\lambda} -\lfloor \frac{aq+b-\lambda}{\lambda q}\rfloor + \lfloor\frac{\lfloor (b-\lambda+aq)/q\rfloor}{\lambda} \rfloor\\ &= \frac{aq^{h+k-1}(q-1)}{\lambda}. \end{align}\] Therefore, we can apply Theorem 3 to derive the desired equality in (?? ). ◻

Proof. By Theorem 6, it is easy to conclude that (?? ) holds if \(k=0\), and (?? ) holds if \(k\geq 1\). Therefore, we only demonstrate that (?? ) holds if \(k=0\), and (?? ) holds if \(k\geq 1\).

If \(k=0\), then we have \(\lambda(\delta-1) = a q^h+b-\lambda\) with \(0\leq b-\lambda<q^{h-1}\). It follows that

• \(\frac{q^h-1}{\lambda}+1<\delta\leq \frac{q^{h+1}-1}{\lambda}+1\);

• \(\delta_h=a\);

• \(\widetilde{\mu}( \delta)=\min\{b-\lambda, a\}\);

• \(w_{\delta}=aq^h\);

• \(\tau(\delta)=1\) if \(a \leq b-\lambda\) and \(\lambda\mid 2a\), and \(\tau(\delta)=0\) if otherwise.

By substituting the above values for the corresponding case into equations (6 ) and (8 ), we derive \[\notag \widetilde{f}(\delta)=\begin{cases} \frac{a+b-\lambda}{\lambda}-\lfloor\frac{2a}{\lambda}\rfloor+ \sum\limits_{t=1}^a\left[ \lfloor \frac{2t}{\lambda}\rfloor - \lfloor \frac{t}{\lambda}\rfloor \right]& ifb<a+\lambda,\\ \sum\limits_{t=1}^a\left[ \lfloor \frac{2t}{\lambda}\rfloor - \lfloor \frac{t}{\lambda}\rfloor \right]& ifb\geq a+\lambda, \end{cases}\] and \[\notag g(\delta)=\begin{cases} \frac{(a-1)(3+(-1)^{\lambda})}{2\lambda}+1& if b\geq a+\lambdaand\lambda\mid 2a,\\ \frac{(a-1)(3+(-1)^{\lambda})}{2\lambda}& otherwise. \end{cases}\] Then by Theorem 3, it follows that (?? ) holds.

If \(k\geq 1\), then we have \(k_{\delta}=k\) and \(V(\lambda(\delta -1))\) has the form \[\notag (\mathbf{0}_{h-k-1}, a, \mathbf{0}_{2k}, \delta_{h-k-1}, \ldots,\delta_0)\]with \(\sum\limits_{\ell=0}^{h-k}\delta_{\ell}q^{\ell}=b-\lambda\). By definition, it follows that

• \(w_{\delta}=aq^h\);

• \(s_{\delta}=k_{\delta}=k\);

• \(\tau(\delta)=0\);

• \(\widetilde{\mu}(\delta)=\min\left\{b-\lambda, a q^{2k}\right\}\);

• \(\phi(\delta)=aq^k-1\);

• \(\mathcal{T}_i(\delta)=\left[q^{k-i},a q^{k-i}-1\right]\cap \{t\in \mathbb{Z}: q\nmid t\}\) for each integer \(i\in [-k,k]\).

Then applying Lemmas 8 and 10, we obtain \[\notag \sum\limits_{t\in \mathcal{T}_i(\delta)} \left[ \lfloor\frac{\lfloor tq^{2i}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor\right]= \begin{cases} \frac{1}{2\lambda}(a^2-1)(q-1)^2q^{2k-2} &if-k+1\leq i \leq k-1andi\neq 0,\\ \frac{1}{2\lambda}a(a-1)(q-1)q^{2k-1} &ifi=kor-k,\\ {\left(\begin{align} &\textstyle{\frac{1}{2\lambda}(a^2-1)(q-1)^2q^{2k-2}}\\&+\textstyle{\frac{(3+(-1)^{\lambda})}{4\lambda}(a-1)(q-1)q^{k-1}} \end{align}\right)} & if i=0. \end{cases}\] It follows that \[\notag \begin{align} \sum\limits_{i=-k_{\delta}}^{k_{\delta}}\sum\limits_{t\in \mathcal{T}_i(\delta)}\left[ \lfloor\frac{\lfloor tq^{2i}\rfloor+t}{\lambda}\rfloor - \lfloor\frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor\right]=& \frac{1}{\lambda}(k-\frac{1}{2})(a^2-1)(q-1)^2q^{2k-2}+\frac{1}{\lambda}a(a-1)(q-1)q^{2k-1}\\ & + \frac{(3+(-1)^{\lambda})(a-1)q^{k-1}(q-1)}{4\lambda}. \end{align}\] By substituting \(k_{\delta}\), \(\tau(\delta)\) \(w_{\delta}\), \(\widetilde{\mu}(\delta)\), \(\phi(\delta)\) and the above expression into (6 ) and (8 ), we obtain \[\notag \widetilde{f}(\delta)=\begin{cases} \frac{q-1}{\lambda}a^2q^{2k-2}\left[\left(k-\frac{1}{2}\right)(q-1)+q\right]+\frac{q-1}{2\lambda}aq^{k-1}& ifb> aq^{2k} + \lambdaand \lambdais odd,\\ \frac{q-1}{\lambda}a^2q^{2k-2}\left[\left(k-\frac{1}{2}\right)(q-1)+q\right]+\frac{q-1}{2\lambda}\left[(2a-1)q^{k-1}+1\right]& ifb> aq^{2k} + \lambdaand \lambdais even,\\ {\left(\begin{align} &\textstyle\frac{q-1}{\lambda}aq^{2k-2} \left[a(k-\textstyle\frac{1}{2})(q-1)+(a-1)q\right] \\\ &+ \textstyle\frac{q-1}{2\lambda}aq^{k-1}+\frac{a+b-\lambda}{\lambda}- \lfloor\textstyle \frac{\lfloor(a+bq-\lambda)/q\rfloor}{\lambda} \rfloor\end{align}\right)} &ifb\leq aq^{2k}+\lambdaand\lambdais odd, \\ {\left(\begin{align} &\textstyle\frac{q-1}{\lambda}aq^{2k-2} \left[a(k-\textstyle\frac{1}{2})(q-1)+(a-1)q\right]+\textstyle\frac{a+b-\lambda}{\lambda}\\ &+\textstyle\frac{q-1}{2\lambda}\left[(2a-1)q^{k-1}+1\right]- \lfloor \textstyle\frac{\lfloor(a+bq-\lambda)/q\rfloor}{\lambda} \rfloor\end{align}\right)} &ifb\leq aq^{2k}+\lambdaand\lambdais even, \end{cases}\] and \[\notag g(\delta)=\begin{cases} \frac{a(q-1)q^{k-1}}{\lambda},& if \lambda is odd,\\ \frac{2a(q-1)q^{k-1}}{\lambda},& if \lambda is even. \end{cases}\] By Theorem 3, it follows that (?? ) holds. ◻

Proof. We demonstrate the proof only for the case where \(m\) is even, as the case \(m\) is odd follows similarly. First, observe that \[\begin{align} V\left(\sum\limits_{\ell=h+r_b}^{h+j_b}b_{\ell}q^{\ell}+q^{h-j_{b}}-1\right) = (\mathbf{0}_{h-1-j_b}, b_{h+j_b},\ldots,b_{h+r_b}, \mathbf{0}_{2r_b}, q-1,\ldots,q-1), \end{align}\] and \[V(\lambda b)\leq V(\lambda a)\leq V\left(\sum\limits_{\ell=h+r_b}^{h+j_b}b_{\ell}q^{\ell}+q^{h-j_{b}}-1\right)\] for all integers \(a\in \left[b,b+ \frac{q^{h-j_b}-1}{\lambda}-\lceil \frac{\sum\limits_{\ell=0}^{h-j_{b}-1}b_{\ell}q^{\ell}}{\lambda}\rceil \right]\). Since \(\sum\limits_{\ell=0}^{h-j_b-1}b_{\ell}q^{\ell}>\sum\limits_{\ell=h-r_{b}}^{h+j_{b}}b_{\ell}q^{\ell-(h-r_{b})}\), it follows that \(V(\lambda a)\) must have the form \[\label{eee} V(\lambda a)=(\mathbf{0}_{h-1-j_b}, b_{h+j_b}, \ldots, b_{h+r_b}, \mathbf{0}_{2r_{b}},a_{h-j_b-1},\ldots,a_0)\tag{28}\] with \(\sum\limits_{\ell=0}^{h-j_b-1}a_{\ell}q^{\ell}>\sum\limits_{\ell=h-r_{b}}^{h+j_{b}}b_{\ell}q^{\ell-(h-r_{b})}\) for all integers \(a\in \left[b,b+ \frac{q^{h-j_b}-1}{\lambda}-\lfloor \frac{\sum\limits_{\ell=0}^{h-j_{b}-1}b_{\ell}q^{\ell}}{\lambda}\rfloor \right]\).

Next, we define a function \(f\) for the integers in \(\left[b,b+ \frac{q^{h-j_b}-1}{\lambda}-\lfloor \frac{\sum\limits_{\ell=0}^{h-j_{b}-1}b_{\ell}q^{\ell}}{\lambda}\rfloor \right]\) as \[f(a)= \frac{a}{q^{t_a}},\] where \(t_a\) is largest integer such that \(q^{t_a}\mid a\). We now show that \[\label{HAHA} f(a)\in \mathcal{L}^n_m \quad for alla\in \left[b,b+ \frac{q^{h-j_b}-1}{\lambda}-\lfloor \frac{\sum\limits_{\ell=0}^{h-j_{b}-1}b_{\ell}q^{\ell}}{\lambda}\rfloor \right]\tag{29}\] by considering the following cases.

Case 1. Suppose that \(t_a=0\), i.e., \(q\nmid a\). Then \(f(a)= a\). It follows that \(V(\lambda f(a))\) has the form given in (28 ). By Remarks 4 and 5, this implies that \(\lambda f(a)\not \in \mathcal{B}_{j_b}(i)\) for any \(i\in [-j_b,j_b]\). Applying Theorem 1, we conclude that \(\lambda f(a) \not \in \mathcal{H}\cup \mathcal{S}\). Therefore, \(\lambda f(a) \in \mathcal{L}_m^{\lambda n}\). By Lemma 1, this implies that \(f(a)\in \mathcal{L}^n_m\).

Case 2. Suppose that \(1\leq t_a\leq j_b\). In this case, we first have \(f(a)\leq b-1\). Since \(V(\lambda a )\) has the form given in (28 ), we derive \[\notag V\left( \lambda f(a) \right) = (\mathbf{0}_{h-1-j_b+t_a}, b_{h+j_b}, \ldots, b_{h+r_b}, \mathbf{0}_{2r_{b}},a_{h-j_b-1},\ldots,a_{t_a}).\] Furthermore, the inequality \(\sum\limits_{\ell=0}^{h-j_b-1}a_{\ell}q^{\ell}>\sum\limits_{\ell=h-r_{b}}^{h+j_{b}}b_{\ell}q^{\ell-(h-r_{b})}\) implies that \(a_{j_b+r_b}>0\). It follows that \(\lambda f(a) \not \in \mathcal{B}_{j_b-t_a}(i)\) for any \(i\in [t_a-j_b, j_b-t_a]\), and hence \(\lambda f(a) \not \in \mathcal{H}\cup \mathcal{S}\). This implies that \(\lambda f(a) \in \mathcal{L}^{\lambda n}_m\), and hence \(f(a)\in \mathcal{L}^n_m\).

Case 3. Suppose \(t_a\geq j_b+1\). In this case, we have \(\lambda f(a)<q^h\) and \(\lambda f(a) q^h \bmod \lambda n \neq \lambda f(a)\). By Lemma 2, we have \(|C_{\lambda n}(\lambda f(a))|=m\). Moreover, by applying [40], we conclude that \(\lambda f(a)\) is a coset leader modulo \(\lambda n\). It follows that \(f(a)\in \mathcal{L}^n_m\).

By now we have demonstrated that (29 ) holds. Noticing that \(C_n(a)=C_n(f(a))\) and \(f\) is injective, it follows that \[\begin{align} \left|\bigcup\limits_{a=b}^{b+\delta-2}C_n(a) \right|= \left|\bigcup\limits_{a=b}^{b+\delta-2}C_n(f(a))\right|=\sum\limits_{a=b}^{b+\delta-2} |C_n(f(a))|=m(\delta-1) \end{align}\] and \[\begin{align} \bigcup\limits_{a=b}^{b+\delta-2}C_n(a) = \bigcup\limits_{a=b}^{b+\delta-2}C_n(f(a)) \neq \bigcup\limits_{a=b}^{b+\delta-1}C_n(f(a)) = \bigcup\limits_{a=b}^{b+\delta-1}C_n(a) \end{align}\] for any integer \(\delta\) such that \(2\leq \delta \leq \frac{q^{h-j_b}-1}{\lambda}-\lfloor \frac{\sum\limits_{\ell=0}^{h-j_{b}-1}b_{\ell}q^{\ell}}{\lambda}\rfloor +1\). Therefore, both (?? ) and (?? ) hold. ◻

Proof. It is clear that \[\label{l70} \{a\in [1,x]: q\nmid aand \lambda\mid a+y \}= \{a\in [1,x]: \lambda\mid a+y \}-\{a\in [1,x]: q\mid aand \lambda\mid a+y \}.\tag{30}\] Next, we show that \[\label{l71} \left| \{a\in [1,x]: q\mid aand \lambda\mid a+y\}\right|=\left |\left\{a\in \left[1,\lfloor {x}/{q}\rfloor\right]: \lambda\mid a+y\right\}\right|.\tag{31}\] Notice that there exists a bijective \(a\mapsto{a}/{q}\) between the integers in \([1,x]\) that are divisible by \(q\) and the integers in \(\left[1,\lfloor x/q \rfloor\right]\). Furthermore, for any integer \(a\) such that \(q\mid a\), we have \[\notag a+y=\frac{a}{q}(q-1)+\frac{a}{q}+y.\] Since \(\lambda\mid q-1\), it follows that \(\lambda\mid a+y\) if and only if \(\lambda\mid \frac{a}{q}+y\) for any integer \(a\) such that \(q\mid a\). Therefore, we can conclude that there exists a one-to-one correspondence between the two sets in (31 ) by mapping \(a\) to \({a}/{q}\), and hence the equality in (31 ) holds. With (30 ), it follows that \[\notag \begin{align} \left| \{a\in [1,x]: q\nmid aand \lambda\mid a+y \}\right|&= \left| \{a\in \left[\lfloor x/q\rfloor+1,x\right]: \lambda\mid a+y \}\right| \\ &=\lfloor \frac{x+y}{\lambda}\rfloor -\lfloor\frac{\lfloor {x}/{q}\rfloor+y}{\lambda} \rfloor. \end{align}\] This completes the proof. ◻

Proof. First, we can apply a similar argument as utilized in the proof of Lemma 3 to conclude that \[\notag \left| \{\alpha\in [x,y]:\lambda\mid 2\alphaand q\mid \alpha \}\right|=\left| \left\{\alpha\in \big[\lceil {x}/{q}\rceil,\lfloor {y}/{q}\rfloor\big]: \lambda\mid 2\alpha\right\}\right|.\] It follows that \[\label{35} \left| \{\alpha\in [x,y]:\lambda\mid 2\alphaand q\nmid \alpha \}\right| =\left| \{\alpha\in [x,y]:\lambda\mid 2\alpha \}\right|-\left| \left\{\alpha\in \big[\lceil {x}/{q}\rceil,\lfloor {y}/{q}\rfloor\big]: \lambda\mid 2\alpha\right\}\right|.\tag{32}\] Then we distinguish the following two cases:

Case 1. Suppose that \(\lambda\) is odd. Then \(\lambda\mid 2\alpha\) holds if and only if \(\lambda \mid \alpha.\) Therefore, \[\notag \left| \{\alpha\in [x,y]: \lambda\mid 2\alpha\} \right| = \left| \{\alpha\in [x,y]: \lambda\mid \alpha\} \right|=\lfloor\frac{y}{\lambda}\rfloor -\lfloor\frac{x-1}{\lambda} \rfloor.\] Similarly, we also have \[\notag \left| \{\alpha\in [\lceil {x}/{q} \rceil, \lfloor {y}/{q} \rfloor]: \lambda\mid 2\alpha\} \right| = \lfloor \frac{\lfloor y/q\rfloor}{\lambda}\rfloor-\lfloor \frac{\lceil {x}/{q} \rceil-1}{\lambda} \rfloor.\]

Case 2. Suppose that \(\lambda\) is even. Then \(\lambda\mid 2\alpha\) holds if and only if \(\frac{\lambda}{2}\mid \alpha\). Consequently, \[\notag \left| \{\alpha\in [x,y]: \lambda\mid 2\alpha\} \right| = \left| \left\{\alpha\in [x,y]: \frac{\lambda}{2} \mid \alpha\right\} \right|=\lfloor\frac{2y}{\lambda}\rfloor -\lfloor\frac{2x-2}{\lambda} \rfloor.\] Similarly, we also have \[\notag \left| \{\alpha \in \big[\lceil {x}/{q} \rceil, \lfloor {y}/{q} \rfloor\big]: \lambda\mid 2\alpha \} \right| = \lfloor \frac{2\lfloor y/q\rfloor}{\lambda}\rfloor-\lfloor \frac{2\lceil {x}/{q} \rceil-2}{\lambda} \rfloor.\]

We now can derive the desired equation from equation (32 ) and the discussion for the above two cases. This completes the proof. ◻

Proof. For simplicity, we set \(\beta_1= x - \lambda\cdot \lfloor \frac{x}{\lambda} \rfloor\) and \(\beta_2= y - \lambda\cdot \lfloor \frac{y}{\lambda} \rfloor\). Then \[\label{9} \sum\limits_{t=1}^{q-1} \left [\lfloor \frac{t+x}{\lambda}\rfloor- \lfloor \frac{t+y}{\lambda}\rfloor \right]= \sum\limits_{t=1}^{q-1} \left[\lfloor \frac{x}{\lambda} \rfloor -\lfloor \frac{y}{\lambda}\rfloor+ \lfloor \frac{t+\beta_1}{\lambda}\rfloor-\lfloor \frac{t+\beta_2}{\lambda}\rfloor\right].\tag{33}\] It is straightforward to verify that \[\notag \lfloor\frac{t+\beta_1}{\lambda}\rfloor =\begin{cases} 0, &fort\in [1, \lambda-\beta_1-1], \\ i,&fort\in [i\lambda-\beta_1,(i+1)\lambda-1-\beta_1],\\ \frac{q-1}{\lambda},&fort\in [q-1-\beta_1,q-1], \end{cases}\] where \(i\) can be any integer in \([1,\frac{q-1}{\lambda}-1].\) Therefore, we can obtain \[\notag \sum\limits_{t=1}^{q-1}\lfloor \frac{t+\beta_1}{\lambda}\rfloor=\frac{(q-1)(\beta_1+1)}{\lambda}+\sum\limits_{i=1}^{\frac{q-1}{\lambda}-1}i\lambda.\] Similarly, we can also derive \[\notag \sum\limits_{t=1}^{q-1}\lfloor \frac{t+\beta_2}{\lambda}\rfloor=\frac{(q-1)(\beta_2+1)}{\lambda}+\sum\limits_{i=1}^{\frac{q-1}{\lambda}-1}i\lambda.\] Combining the above two equalities, we obtain \[\notag \sum\limits_{t=1}^{q-1} \left[\lfloor \frac{t+\beta_1}{\lambda}\rfloor-\lfloor \frac{t+\beta_2}{\lambda}\rfloor\right]=\frac{(q-1)(\beta_1-\beta_2)}{\lambda}.\] With the equality in (33 ), it follows that \[\notag \sum\limits_{t=1}^{q-1}\left[\lfloor \frac{t+x}{\lambda}\rfloor- \lfloor \frac{t+y}{\lambda}\rfloor \right] = \frac{q-1}{\lambda}\left(\lambda \cdot \lfloor \frac{x}{\lambda} \rfloor +\beta_1\right)- \frac{q-1}{\lambda}\left(\lambda \cdot \lfloor \frac{y}{\lambda} \rfloor +\beta_2\right)=\frac{(q-1)(x-y)}{\lambda}.\]This completes the proof. ◻

Proof. Notice that each integer \(t\in [q,aq]\) such that \(q\nmid t\) admits a unique decomposition \[t=iq+j \quad withi\in [1,a-1]andj\in [1,q-1].\] Substituting this decomposition, we have \[\notag \begin{align} \lfloor \frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor- \lfloor \frac{t}{\lambda}\rfloor &= \lfloor \frac{i+iq+j}{\lambda}\rfloor-\lfloor \frac{iq+j}{\lambda}\rfloor\\ &= \lfloor \frac{j+2i}{\lambda}\rfloor-\lfloor \frac{j+i}{\lambda}\rfloor. \end{align}\] Then by applying Lemma 5, we obtain \[\notag \begin{align} \sum\limits_{t=q, q\nmid t }^{aq}\left [\lfloor \frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor- \lfloor \frac{t}{\lambda}\rfloor \right] &=\sum\limits_{i=1}^{a-1}\sum\limits_{j=1}^{q-1}\left[ \lfloor \frac{j+2i}{\lambda}\rfloor-\lfloor \frac{j+i}{\lambda}\rfloor \right]\\ &=\sum\limits_{i=1}^{a-1}\frac{(q-1)i}{\lambda}\\ &=\frac{a(a-1)(q-1)}{2\lambda}. \end{align}\] This completes the proof. ◻

Proof. Notice that each integer \(t\in [0,q-2]\) admits a unique decomposition \[t=i\lambda+j \quad withi\in \left[0,\frac{q-1}{\lambda}-1\right]and j\in [0,\lambda-1].\] Therefore, \[\label{le8e1} \begin{align} \sum\limits_{t=1}^{q-1}\left[\lfloor \frac{2t+x}{\lambda}\rfloor- \lfloor \frac{t+x}{\lambda}\rfloor\right]&=\sum\limits_{t=0}^{q-1}\left[\lfloor \frac{2t+x}{\lambda}\rfloor- \lfloor \frac{t+x}{\lambda}\rfloor\right]\\ &=\sum\limits_{i=0}^{\frac{q-1}{\lambda}-1}\sum\limits_{j=0}^{\lambda-1}\left[ \lfloor \frac{2i\lambda+2j+x}{\lambda}\rfloor-\lfloor \frac{i\lambda+j+x}{\lambda}\rfloor \right]+\lfloor \frac{2(q-1)+x}{\lambda}\rfloor-\lfloor \frac{q-1+x}{\lambda}\rfloor\\ &=\sum\limits_{i=0}^{\frac{q-1}{\lambda}-1}\sum\limits_{j=0}^{\lambda-1}\left[ i+ \lfloor \frac{2j+x}{\lambda} \rfloor-\lfloor \frac{j+x}{\lambda}\rfloor \right]+\frac{q-1}{\lambda}\\ &=\frac{(q-1)(q+1)}{2\lambda}-\frac{q-1}{2}+\frac{q-1}{\lambda}\sum\limits_{j=0}^{\lambda-1}\left[ \lfloor \frac{2j+x}{\lambda} \rfloor-\lfloor \frac{j+x}{\lambda}\rfloor \right]. \end{align}\tag{34}\] Let \(y=x-\lambda\cdot\lfloor \frac{x}{\lambda} \rfloor\). Then we have \(0\leq y\leq \lambda-1\) and \[\label{le8e2} \lfloor \frac{2j+x}{\lambda} \rfloor-\lfloor \frac{j+x}{\lambda}\rfloor = \lfloor \frac{2j+y}{\lambda} \rfloor-\lfloor \frac{j+y}{\lambda}\rfloor.\tag{35}\] Noting that \(\lfloor \frac{j+y}{\lambda} \rfloor =0\) for each integer \(j\in [0,\lambda-y-1]\), and \(\lfloor \frac{j+y}{\lambda} \rfloor =1\) for each integer \(j\in [\lambda-y, \lambda-1]\), we can derive \[\label{le8e3} \sum\limits_{j=0}^{\lambda-1}\lfloor \frac{j+y}{\lambda}\rfloor = y.\tag{36}\] Next, we determine the value of \(\sum\limits_{j=0}^{\lambda-1}\lfloor \frac{2j+y}{\lambda} \rfloor\) through the following two cases:

Case 1. Suppose that \(\lambda\) is even. Since \(x\) is even, it follows that \(y\) is also even. We can obtain \[\notag \lfloor \frac{2j+y}{\lambda} \rfloor =\begin{cases} 0&if 0\leq j\leq \frac{\lambda-y}{2}-1, \\ 1&if \frac{\lambda-y}{2}\leq j \leq \frac{2\lambda-y}{2}-1, \\ 2 &if\frac{2\lambda-y}{2} \leq j \leq \lambda -1. \end{cases}\] It follows that \[\notag \begin{align} \sum\limits_{j=0}^{\lambda-1}\lfloor \frac{2j+y}{\lambda} \rfloor &= \left(\frac{2\lambda-y}{2}- \frac{\lambda-y}{2}\right)+2\left(\lambda - \frac{2\lambda-y}{2}\right) \\ &=\frac{\lambda}{2}+y. \end{align}\]

Case 2. Suppose that \(\lambda\) is odd. Then we can obtain \[\notag \lfloor \frac{2j+y}{\lambda} \rfloor =\begin{cases} 0&if 0\leq j\leq \frac{\lambda-y}{2}-1, \\ 1&if \frac{\lambda+1}{2}-\lfloor \frac{y+1}{2}\rfloor\leq j \leq \lambda-\lceil \frac{y+1}{2}\rceil, \\ 2 &if\lambda -\lceil \frac{y+1}{2} \rceil +1 \leq j \leq \lambda -1. \end{cases}\] Consequently, we have \[\notag \begin{align} \sum\limits_{j=0}^{\lambda-1}\lfloor \frac{2j+y}{\lambda} \rfloor &= \left( \lambda-\lceil \frac{y+1}{2}\rceil -\frac{\lambda+1}{2}+ \lfloor \frac{y+1}{2}\rfloor +1 \right) + 2 \left(\lceil \frac{y+1}{2}\rceil - 1 \right) \\ &= \frac{\lambda-1}{2}-1+\lceil \frac{y+1}{2}\rceil+\lfloor \frac{y+1}{2}\rfloor +1\\ &= \frac{\lambda-1}{2}+y. \end{align}\] From equations (35 ) and (36 ), and the discussion for the above two cases, we can conclude that \[\notag \sum\limits_{j=0}^{\lambda-1}\left[\lfloor \frac{2j+x}{\lambda} \rfloor-\lfloor \frac{j+x}{\lambda}\rfloor \right] =\begin{cases} \frac{\lambda}{2},& if\lambdais even, \\ \frac{\lambda-1}{2},& if\lambdais odd. \end{cases}\] Then by applying (34 ), we obtain the desired equation. This completes the proof. ◻

Proof. Note that each integer \(t\in \left[q^k,aq^{k}\right]\) such that \(q\nmid t\) can be uniquely decomposed as \[t=iq+j \quad withi\in [q^{k-1}, aq^{k-1}-1]and j\in [1,q-1].\] Therefore, by substituting this decomposition, we obtain \[\notag \begin{align} \sum\limits_{t=q^{k},q\nmid t}^{aq^{k}-1}\left[\lfloor \frac{2t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{-1}\rfloor+t}{\lambda}\rfloor\right] &= \sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\sum\limits_{j=1}^{q-1} \left[\lfloor \frac{2iq+2j}{\lambda}\rfloor -\lfloor \frac{i+iq+j}{\lambda}\rfloor\right]\\ &= \sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\sum\limits_{j=1}^{q-1} \left[\frac{i(q-1)}{\lambda}+\lfloor \frac{2j+2i}{\lambda}\rfloor -\lfloor \frac{j+2i}{\lambda}\rfloor \right]\\ &= \sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\frac{i(q-1)^2}{\lambda}+\sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\sum\limits_{j=1}^{q-1}\left[\lfloor \frac{2j+2i}{\lambda}\rfloor-\lfloor\frac{j+2i}{\lambda} \rfloor\right]. \end{align}\] It is straightforward to obtain \[\notag \sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\frac{i(q-1)^2}{\lambda} = \frac{q^{k-1}(q-1)^2(aq^{k-1}+q^{k-1}-1)}{2\lambda}.\] In addition, by applying Lemma 7, we have \[\notag \sum\limits_{i=q^{k-1}}^{aq^{k-1}-1}\sum\limits_{j=1}^{q-1}\left[\lfloor \frac{2j+2i}{\lambda}\rfloor-\lfloor\frac{j+2i}{\lambda} \rfloor\right]=\begin{cases} \frac{q^k(q-1)(a-1)}{2\lambda}& if \lambdais odd, \\ \frac{q^{k-1}(q-1)(q+1)(a-1)}{2\lambda}& if \lambdais even. \end{cases}\] Combining the above three equations, we obtain the desired equation. This completes the proof. ◻

Proof. By definition, we first have \[\label{l6ee1} \sum\limits_{t=q^{k}}^{aq^{k}-1}N(t+1)=\sum\limits_{t=q^{k}}^{aq^{k}-1}t-\sum\limits_{t=q^{k}}^{aq^{k}-1}\lfloor t/q\rfloor.\tag{37}\] Through direct computation, we obtain \[\label{l6ee2} \sum\limits_{t=q^{k}}^{aq^{k}-1}t= \begin{cases} \frac{1}{2}a(a-1)& ifk=0,\\ \frac{1}{2}(a^2-1)q^{2k}-\frac{1}{2}(a-1)q^k& ifk\geq 1. \end{cases}\tag{38}\]

We now determine the value of \(\sum\limits_{t=q^{k}}^{aq^{k}-1}\lfloor t/q\rfloor.\) First, note that \(\lfloor t/q\rfloor=0\) for all \(t\in \left[q^k,aq^{k}-1\right]\) when \(k=0\). This implies that \(\sum\limits_{t=q^k}^{aq^{k}-1}\lfloor t/q\rfloor =0\) if \(k=0\). When \(k\geq 1,\) the interval \(\left[q^{k}, q^{k+1}-1\right]\) can be partitioned as \[\left[q^{k}, q^{k+1}-1\right]=\bigsqcup\limits_{i=0}^{q^{k-1}(a-1)-1} \left[q^{k}+i q, q^{k}+(i+1)q-1\right].\] Moreover, for each integer \(i\in \left[0,q^{k-1}(a-1)-1\right]\) and each integer \(t\in \left[q^{k}+i q, q^{k}+(i+1)q-1\right]\), we have \(\lfloor t/q\rfloor =q^{k-1}+i\). Therefore, \[\notag \begin{align}\sum\limits_{t=q^{k}}^{aq^{k}-1}\lfloor t/q\rfloor&=\sum\limits_{i=0}^{q^{k-1}(a-1)-1}\sum\limits_{t=q^{k}+i q}^{q^k+(i+1)q-1}(q^{k-1}+i)\\ &=\frac{1}{2}(a^2-1)q^{2k}-\frac{1}{2}(a-1)q^k \end{align}\] if \(k\geq 1.\) Combining (37 ), (38 ) and the value of \(\sum\limits_{t=q^k}^{q^{k+1}-1}\lfloor t/q\rfloor\) given as above, we obtain the desired equality. ◻

Proof. The arguments used to derive (?? ) and (?? ) are analogus, so we only demonstrate equation (?? ) holds through the following two cases.

Case 1. Suppose that \(1\leq i\leq k\). In this case, we first have \[\notag \lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor= \frac{t(q^{2i-1}-1)}{\lambda}+\lfloor \frac{2t}{\lambda}\rfloor \quad and\quad \lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor= \frac{t(q^{2i-2}-1)}{\lambda}+\lfloor \frac{2t}{\lambda}\rfloor.\] It follows that \[\notag \lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor =\frac{tq^{2i-2}(q-1)}{\lambda}.\] This leads to \[\label{3} \sum\limits_{t=q^{k-i},q\nmid t}^{aq^{k-i}-1}\left[\lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor\right]= \sum\limits_{t=q^{k-i}}^{aq^{k-i}-1} \frac{tq^{2i-2}(q-1)}{\lambda} - \sum\limits_{t\in \mathcal{W}_i}\frac{tq^{2i-2}(q-1)}{\lambda}\tag{39}\] with \(\mathcal{W}_i=\left[q^{k-i},aq^{k-i}-1 \right]\cap \{t\in \mathbb{Z}: q\mid t\}\).

Note that \(\mathcal{W}_i=\varnothing\) if \(i=k\). Thus, \[\label{4} \sum\limits_{t\in \mathcal{W}_i}\frac{tq^{2i-2}(q-1)}{\lambda}=0 \quad ifi=k.\tag{40}\] On the other hand, it can be easily verified that \(\mathcal{W}_i =\{q^{k-i}+ j q \mid j =0,\ldots, q^{k-i-1}(a-1)-1\}\) if \(1\leq i \leq k-1.\) Therefore,

if \(1\leq i\leq k-1\). Additionally, by straightforward computation, we can obtain \[\label{6} \begin{align} \sum\limits_{t=q^{k-i}}^{aq^{k-i}-1}\frac{tq^{2i-2}(q-1)}{\lambda} =\begin{cases}\frac{1}{2\lambda}a(a-1)(q-1)^2q^{2k-2} \quadifi=k,\\ \frac{1}{2\lambda}(q-1)(a-1)\left[(a+1)q^{2k-2}-q^{k+i-2}\right] \\ \quad \quad if 1\leq i\leq k-1. \end{cases}\end{align}\tag{41}\] By substituting equations (40 )-(41 ) into equation (39 ), we conclude that equation (?? ) holds for \(1\leq i\leq k\).

Case 2. Suppose that \(-k+1\leq i\leq 0\). Then for any integer \(t\in \left[q^{k-i}, aq^{k-i}-1\right]\cap \{t\in \mathbb{Z}: q\nmid t\}\) with \(q\)-adic expansion \(\sum\limits_{\ell=0}^{k-i}t_{\ell}q^{\ell}\), we have \[\label{lhx} \lfloor tq^{2i-1} \rfloor = \lfloor tq^{2i-2} \rfloor \cdot q+t_{-2i+1}.\tag{42}\] This leads to \[\notag \begin{align} \lfloor \frac{t+\lfloor tq^{2i-1}\rfloor}{\lambda}\rfloor = \frac{ \sum\limits_{\ell=0}^{k-i}t_{\ell} (q^{\ell}-1)}{\lambda} + \frac{\lfloor tq^{2i-2}\rfloor(q-1)}{\lambda} + \lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor+t_{-2i+1}}{\lambda} \rfloor. \end{align}\] We also have \[\notag \begin{align} \lfloor \frac{t+\lfloor tq^{2i-2}\rfloor}{\lambda}\rfloor = \frac{ \sum\limits_{\ell=0}^{k-i}t_{\ell} (q^{\ell}-1)}{\lambda} + \lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor}{\lambda} \rfloor. \end{align}\] Combining the above two equalities, we can obtain \[\notag \lfloor \frac{t+\lfloor tq^{2i-1} \rfloor}{\lambda}\rfloor-\lfloor \frac{t+\lfloor tq^{2i-2}\rfloor}{\lambda}\rfloor= \lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor+t_{-2i+1}}{\lambda} \rfloor -\lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor}{\lambda} \rfloor +\frac{\lfloor tq^{2i-2}\rfloor(q-1)}{\lambda}.\] By applying Lemma 5 with \(x=\sum\limits_{\ell=1}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor+t_{-2i+1}\) and \(y= \sum\limits_{\ell=1}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor\), we have \[\notag \sum\limits_{t_0=1}^{q-1} \left[ \lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor+t_{-2i+1}}{\lambda} \rfloor -\lfloor \frac{\sum\limits_{\ell=0}^{k-i}t_{\ell}+\lfloor tq^{2i-2}\rfloor}{\lambda} \rfloor \right]= \frac{(q-1)t_{-2i+1}}{\lambda}.\] Noticing that the value of \(\lfloor tq^{2i-2}\rfloor\) is independent of \(t_0\), we get \[\notag \sum\limits_{t_0=1}^{q-1}\frac{\lfloor tq^{2i-2}\rfloor(q-1)}{\lambda} = \frac{\lfloor tq^{2i-2}\rfloor(q-1)^2}{\lambda}.\] Noting that \(\lfloor tq^{2i-2}\rfloor=\lfloor\frac{\lfloor tq^{2i-1}\rfloor}{q}\rfloor\) and recalling equation (42 ), we can add the above two sums to obtain \[\label{whlx} \begin{align} \sum\limits_{t_0=1}^{q-1}\left[ \lfloor \frac{t+\lfloor tq^{2i-1} \rfloor}{\lambda}\rfloor-\lfloor \frac{t+\lfloor tq^{2i-2}\rfloor}{\lambda}\rfloor \right]&= \frac{\lfloor tq^{2i-2}\rfloor(q-1)^2}{\lambda} + \frac{(q-1)t_{-2i+1}}{\lambda}\\ &=\frac{q-1}{\lambda}N(\lfloor tq^{2i-1}\rfloor+1). \end{align}\tag{43}\]

In addition, each integer \(t\in \left[q^{k-i}, aq^{k-i+1}-1\right]\cap \{t\in \mathbb{Z}: q\nmid t\}\) with \(q\)-adic expansion \(\sum\limits_{\ell=0}^{k-i}t_{\ell}q^{\ell}\) can be uniquely decomposed as \[\notag t=\lfloor tq^{2i-1} \rfloor\cdot q^{-2i+1}+\sum\limits_{\ell=1}^{-2i}t_{\ell}q^{\ell}+t_0.\] Furthermore, as \(t\) ranges over integers from \(q^{k-i}\) to \(q^{k-i+1} - 1\) that are not divisible by \(q\), the value of \(\lfloor tq^{2i-1} \rfloor\) ranges from \(q^{k+i-1}\) to \(aq^{k+i-1}-1\). Additionally, for each fixed value of \(\lfloor tq^{2i-1} \rfloor\), the sum \(\sum\limits_{\ell=1}^{-2i} t_{\ell} q^{\ell}\) ranges over integers from \(0\) to \(q^{-2i}-1\), while \(t_0\) varies from \(1\) to \(q - 1\). Therefore, we can conclude that \[\notag \begin{align} \sum\limits_{\substack{t=q^{k-i}, q\nmid t}}^{ q^{k-i+1}-1}\left[\lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor\right]&= \sum\limits_{\lfloor tq^{2i-1} \rfloor=q^{k+i-1}}^{aq^{k+i-1}-1 } \sum\limits_{\sum_{\ell=1}^{-2i}t_{\ell}q^{\ell}=0}^{q^{-2i}-1}\sum\limits_{t_0=1}^{q-1}\left[ \lfloor \frac{t+\lfloor tq^{2i-1} \rfloor}{\lambda}\rfloor-\lfloor \frac{t+\lfloor tq^{2i-2}\rfloor}{\lambda}\rfloor \right]\\ &=\sum\limits_{\lfloor tq^{2i-1} \rfloor=q^{k+i-1}}^{aq^{k+i-1}-1 } \frac{q^{-2i}(q-1)}{\lambda}N(\lfloor tq^{2i-1}\rfloor+1), \end{align}\] where the second equality follows from equation (43 ). We can now apply Lemma 9 to conclude that (?? ) holds for \(-k+1\leq i\leq 0\).

By now, we have established equation (?? ) ◻

Proof. Suppose that \(m\) is odd. We first aim to show that \[\label{ha2} \left[\sum\limits_{\ell=h+s_{\delta}}^{h+k_{\delta}}\delta_{\ell}q^{\ell}, \lambda(\delta-1)\right]\cap \mathcal{S}\cap \mathcal{D}_{\lambda}=\left[\sum\limits_{\ell=h+s_{\delta}}^{h+k_{\delta}}\delta_{\ell}q^{\ell}, \lambda(\delta-1)\right]\cap \mathcal{A}_{k_{\delta}}(s_{\delta}) \cap \mathcal{D}_{\lambda} .\tag{44}\] For any integer \(a\in \left[\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell},\lambda(\delta-1)\right]\) with \(q\)-adic expansion \(\sum\limits_{\ell=0}^{h+k_{\delta}}a_{\ell}q^{\ell}\), we have \[V(\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell})\leq V(a)\leq V(\lambda(\delta-1)).\] Noting that \[\notag V(\lambda(\delta-1))=(\mathbf{0}_{h-k_{\delta}},\delta_{h+k_{\delta}},\ldots,\delta_{0})\] and \[\notag V(\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell})=(\mathbf{0}_{h-k_{\delta}},\delta_{h+k_{\delta}},\ldots,\delta_{h-k_{\delta}+1}, \mathbf{0}_{h-k_{\delta}+1}),\] it follows that \[\label{vvv} V(a)=(\mathbf{0}_{h-k_{\delta}},a_{h+k_{\delta}},\ldots,a_0)=(\mathbf{0}_{h-k_{\delta}},\delta_{h+k_{\delta}},\ldots,\delta_{h-k_{\delta}+1} , a_{h-k_{\delta}},\ldots,a_0).\tag{45}\] Recalling the definition of \(s_{\delta}\), it follows that \(s_{\delta}\) is the smallest integer in \([-k_{\delta}+1,k_{\delta}]\) such that \(a_{h+s_{\delta}}>0\). By Remark 5, this implies \(a\not \in \mathcal{A}_{k_{\delta}}(i)\) for any integer \(i\neq s_{\delta}\). Therefore, \(\left[\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell},\delta-1\right]\cap \mathcal{A}_{k_{\delta}}(i)=\varnothing\) for any integer \(i\neq s_{\delta}.\) Then applying Theorem 1, we can conclude that (44 ) holds.

Now, let us count the number of integers in the set \(\left[\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell},\delta-1\right]\cap \mathcal{A}_{k_{\delta}}(s_{\delta})\cap \mathcal{D}_{\lambda}.\) Recall that \(w_{\delta}= \sum\limits_{\ell=h+s_{\delta}}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\), \(\mu(\delta)=\min\left \{\sum\limits_{\ell=0}^{h-k_{\delta}}\delta_{\ell}q^{\ell}, \sum\limits_{\ell=h-s_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell-(h-s_{\delta}+1)} \right\}\), and \(\alpha(a)=\sum\limits_{\ell=0}^{k_{\delta}+s_{\delta}-1}a_{\ell}q^{\ell}\) for each integer \(a\in \mathcal{A}_{k_{\delta}}(s_{\delta})\) with \(q\)-adic expansion \(\sum\limits_{\ell=0}^{h+k_{\delta}}a_{\ell}q^{\ell}\). We conclude from equation (45 ) and the definition of \(\mathcal{A}_{k_{\delta}}(s_{\delta})\) that an integer \(a\in \left[\sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell},\delta-1\right]\cap \mathcal{A}_{k_{\delta}}(s_{\delta})\cap \mathcal{D}_{\lambda}\) if and only if \(a\) admits the decomposition \[a=w_{\delta}+\alpha(a)\] with \(\alpha(a) \in \{ \alpha\in \left[1, \mu(\delta)\right]: q\nmid \alpha and \lambda\mid \alpha+w_{\delta}\}.\) Consequently, we have \[\notag \left|\left[\sum\limits_{\ell=h+s_{\delta}}^{h+k_{\delta}}\delta_{\ell}q^{\ell}, \lambda(\delta-1)\right]\cap \mathcal{A}_{k_{\delta}}\cap \mathcal{D}_{\lambda}\right| = \left| \{\alpha\in \left[1,\mu(\delta)\right]: q\nmid \alphaand\lambda \mid \alpha+w_{\delta} \}\right|.\] Then by applying Lemma 3, equation (?? ) follows.

Suppose that \(m\) is even. The equality in (?? ) can be obtained by emplying a similar argument as above. It remains to establish equation (?? ). By applying Lemma 2, we can conclude that \(a\in \left[\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}, \lambda(\delta-1)\right]\cap \mathcal{H}\cap \mathcal{D}_{\lambda}\) if and only if \(V(a)\) has the form \[\notag (\mathbf{0}_{h-k_{\delta}-1},\delta_{h+k_{\delta}},\ldots,\delta_h, \mathbf{0}_{h-k_{\delta}-1},\delta_{h+k_{\delta}},\ldots,\delta_h)\] with \(\delta_h>0\), \(\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}\leq\sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\) and \(\lambda \mid \sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}+\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\). Since \(\lambda\mid q-1\), the condition \(\lambda \mid \sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}+\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\) is equivalent \(\lambda \mid 2\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}.\) Therefore, an integer satisfying the above condition exists and is unique if and only if \(\delta_h>0\), \(\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}\leq \sum\limits_{\ell=0}^{h-1}\delta_{\ell}q^{\ell}\) and \(\lambda \mid 2\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\). It follows that equation (?? ) holds. ◻

Proof. Suppose that \(m\) is odd. By definition, an integer \(a \in \left[q^{h+k_{\delta}}, \; \sum_{\ell=h-k_{\delta}+1}^{h+k_{\delta}} \delta_{\ell} q^{\ell}\right) \cap \mathcal{A}_{k_{\delta}}(i)\) if and only if the following conditions are satisfied:

• \(V(a)\) is of the form given in ([p33]) and satisfies ([p31]) and ([p32]) for \(k = k_{\delta}\);

• \(V(q^{h+k_{\delta}}) \leq V(a) < V\left(\sum_{\ell=h-k_{\delta}+1}^{h+k_{\delta}} \delta_{\ell} q^{\ell}\right)\).

Notice that \[V(q^{h+k_{\delta}}) = (\mathbf{0}_{h-k_{\delta}}, 1, \mathbf{0}_{h+k_{\delta}}) \quad \text{and} \quad V\left(\sum_{\ell=h-k_{\delta}+1}^{h+k_{\delta}} \delta_{\ell} q^{\ell}\right) = (\mathbf{0}_{h-k_{\delta}}, \delta_{h+k_{\delta}}, \ldots, \delta_{h-k_{\delta}+1}, \mathbf{0}_{h-k_{\delta}+1}).\] Therefore, the form of \(V(a)\) in ([p33]) together with the inequality \(a_0 > 0\) in ([p32]) imply that condition () is satisfied if and only if \[(1, \mathbf{0}_{2k_{\delta}-1}) \leq (a_{h+k_{\delta}}, \ldots, a_{h+i}, \mathbf{0}_{k_{\delta}+i-1}) < (\delta_{h+k_{\delta}}, \ldots, \delta_{h-k_{\delta}+1}).\] Since \(s_{\delta}\) is the smallest integer in \([-k_{\delta}+1, k_{\delta}]\) such that \(\delta_{h+s_{\delta}} > 0\), this is further equivalent to \[q^{k_{\delta}-i} \leq \sum_{\ell=h+i}^{h+k_{\delta}} a_{\ell} q^{\ell-h-i} < \sum_{\ell=h+s_{\delta}}^{h+k_{\delta}} \delta_{\ell} q^{\ell-h-i}.\]

Recall the definition of the set \(\mathcal{T}_i(\delta)\) and Remarks 1 – 3. We now can conclude that for each integer \(i\in \left[-k_{\delta}+1, k_{\delta}\right]\), an integer \(a\in \left[q^{h+k_{\delta}}, \sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\right)\cap \mathcal{A}_{k_{\delta}}(i)\cap \mathcal{D}_{\lambda}\) if and only if \(a\) admits the decomposition \[a=t(a)\cdot q^{h+i}+\alpha(a)\] with \[\notag \begin{cases} \lambda \mid t(a)+\alpha(a),\\ t(a)\in \mathcal{T}_i(\delta),\\ 1\leq \alpha(a)\leq \lfloor t(a)\cdot q^{2i-1}\rfloorandq\nmid \alpha(a). \end{cases}\] Consequently, we can apply Lemma 3 to conclude that \[\notag \begin{align} \left| \left[q^{h+k_{\delta}}, \sum\limits_{\ell=h-k_{\delta}+1}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\right)\cap \mathcal{A}_{k_{\delta}}(i)\cap \mathcal{D}_{\lambda}\right|& =\sum\limits_{t\in \mathcal{T}_i(\delta)}\left| \left\{ \alpha\in [1, \lfloor tq^{2i-1} \rfloor]: q\nmid \alphaand \lambda\mid \alpha + t \right\}\right|\\ &=\sum\limits_{t\in \mathcal{T}_i(\delta)}\left[\lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor\right]. \end{align}\] Then applying Theorem 1, it follows that (?? ) holds.

Suppose that \(m\) is even. We can first utilize a similar argument as above to obtain (?? ). Next, we demonstrate that equation (?? ) holds. By applying Lemma 2, we can conclude that an integer \(a\in \left[q^{h+k_{\delta}}, \sum\limits_{\ell=h}^{h+k_{\delta}} \delta_{\ell}q^{\ell}\right)\cap \mathcal{H} \cap \mathcal{D}_{\lambda}\) if and only if \(a\) can be decomposed as decomposition \[\label{de} a=\sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\cdot q^{h}+\sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\tag{46}\] with \(q\nmid \sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\), \(q^{h+k_{\delta}}\leq a<\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\) and \(\lambda\mid a\). It is easy to see that \(q^{h+k_{\delta}}\leq a <\sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell}\) holds if and only if \(q^{k_{\delta}}\leq \sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\leq \sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}-1\). Furthermore, since \(\lambda\mid q-1\), the condition \(\lambda\mid a\) is satisfied if and only if \(\lambda\mid 2\sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\). Therefore, an integer \(a\in \left[q^{h+k_{\delta}}, \sum\limits_{\ell=h}^{h+k_{\delta}} \delta_{\ell}q^{\ell}\right)\cap \mathcal{H} \cap \mathcal{D}_{\lambda}\) if and only if \(a\) admits the decomposition given in (46 ) with \[\notag \begin{cases} q^{k_{\delta}}\leq \sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}\leq \sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}-1;\\ q\nmid \sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell};\\ \lambda\mid 2\sum\limits_{\ell=0}^{k_{\delta}}a_{\ell}q^{\ell}. \end{cases}\] Consequently, \[\notag \begin{align} \left|\left[q^{h+k_{\delta}}, \sum\limits_{\ell=h}^{h+k_{\delta}} \delta_{\ell}q^{\ell}\right)\cap \mathcal{H} \cap \mathcal{D}_{\lambda}\right| = \left|\left \{\alpha \in \left[q^{k_{\delta}}, \sum\limits_{\ell=h}^{h+k_{\delta}}\delta_{\ell}q^{\ell-h}-1\right]: \lambda \mid 2\alphaandq \nmid \alpha \right\}\right| \end{align}\] By applying Lemma 4, it follows that equation (?? ) holds. ◻

Proof. The arguments used to establish (?? ) and (?? ) are analogous. Therefore, we only demonstrate that (?? ) holds when \(m\) is odd. Using reasoning similar to that in the proof of Assertion 2, we can conclude that an integer \(a \in \mathcal{A}_{k_{\delta}}(i) \cap \mathcal{D}_{\lambda}\) if and only if \(a\) admits the decomposition \[a = t(a) \cdot q^{h+i} + \alpha(a)\] with \[\begin{cases} \lambda \mid t(a) + \alpha(a); \\[1ex] t(a) \in \left[q^{k-i},\, q^{k-i+1} - 1 \right] \cap \{ t \in \mathbb{Z} : q \nmid t \}; \\[1ex] 1 \leq \alpha(a) \leq \left\lfloor t(a) \cdot q^{2i-1} \right\rfloor \text{ and } q \nmid \alpha(a). \end{cases}\] Consequently, by applying Lemma 3, we derive \[\notag \begin{align} \left| \mathcal{A}_{k_{\delta}}(i)\cap \mathcal{D}_{\lambda}\right|& =\sum\limits_{t=q^{k-i},q\nmid t}^{q^{k-i+1}-1}\left| \left\{ \alpha\in \left[1, \lfloor tq^{2i-1} \rfloor\right]: q\nmid \alphaand \lambda\mid \alpha + t \right\}\right|\\ &=\sum\limits_{t=q^{k-i},q\nmid t}^{q^{k-i+1}-1} \left[\lfloor \frac{\lfloor tq^{2i-1}\rfloor+t}{\lambda}\rfloor -\lfloor \frac{\lfloor tq^{2i-2}\rfloor+t}{\lambda}\rfloor\right]. \end{align}\] Then applying Lemma 10, we obtain equation (?? ). ◻

Proof. Following a similar approach to the proof of Assertion 3, we can obtain \[\notag |B_k(0) \cap \mathcal{D}_{\lambda}| = \sum_{t=q^{k}, q \nmid t}^{q^{k+1}-1} \left[ \lfloor \frac{2t}{\lambda} \rfloor - \lfloor \frac{\left\lfloor {t}{q}^{-1} \right\rfloor + t}{\lambda}\rfloor \right].\] Then applying Lemmas 7 and 8, the assertion follows. ◻

Proof. By applying Lemma 2, we can conclude that an integer \(a\in \mathcal{H}\cap [q^{h+k},q^{h+k+1})\cap \mathcal{D}_{\lambda}\) if and only if \(a\) adimits the decomposition \[\label{as5eh} a=\sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}\cdot q^{h}+\sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}.\tag{47}\] with \(a_0>0\), \(a_{k}>0\), and \(\lambda\mid a\). Since \(\lambda\mid q-1\), it follows that \(\lambda\mid a\) is equivalent to \(\lambda \mid 2\sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}\). In addition, it is straightforward to verify that the conditions \(a_{0}>0\) and \(a_{k}>0\) are satisfied if and only if \(q^k\leq \sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell} \leq q^{k+1}-1\) and \(q\nmid \sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}\). Therefore, an integer \(a\in \mathcal{H}\cap [q^{h+k},q^{h+k+1})\cap \mathcal{D}_{\lambda}\) if and only if \(a\) has the decomposition as given in (47 ) with \[\notag \begin{cases} q\nmid \sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell};\\ q^k\leq \sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}\leq q^{k+1}-1;\\ \lambda \mid 2\sum\limits_{\ell=0}^{k}a_{\ell}q^{\ell}. \end{cases}\] Consequently, \[\notag \left|\mathcal{H}\cap \left[q^{h+k},q^{h+k+1}\right)\cap \mathcal{D}_{\lambda}\right| =\left| \left\{ \alpha \in \left[q^k, q^{k+1}-1\right]: \lambda\mid 2\alphaand q \nmid \alpha\right\}\right|.\] Then by applying Lemma 4, the assertion follows. ◻