Determination of the sizes of optimal geometric orthogonal codes with parameters $(n\times m,k,\lambda ,k-1)$

Abstract

The study of $(n\times m,k,\lambda _a,\lambda _c)$-geometric orthogonal codes ($(n\times m,k,\lambda _a,\lambda _c)$-GOCs) is motivated by the application in DNA origami. The central research on GOCs is to determine the value of $\Phi (n\times m,k,\lambda _a,\lambda _c)$, i.e., the largest possible size among all $(n\times m,k,\lambda _a,\lambda _c)$-GOCs. When $\lambda _a=\lambda _c$, the exact values of $\Phi (n\times m,3,1,1)$ and $\Phi (n\times m,k,k-1,k-1)$ were recently determined by Wang and Su et al. In this paper, we research on the cases of $\lambda _c=k-1$ and $\lambda _a \le k-2$. We determine the exact values of $\Phi (n\times m,k,k-2,k-1)$ and $\Phi (n\times m,k,k-3,k-1)$, and give a calculation method of $\Phi (n\times m,k,\lambda _a,k-1)$ with $\lambda _a\le k-4$ for any positive integers n, m and k.

Appendices

Appendix

Proof of Lemma 3.14

For any $B\in \Gamma _{1,2}$,

$$\begin{aligned} \tau (B)&=\{0,\mu ,\dots ,(i-2)\mu ,{\overline{(i-2)\mu }}, {\underline{(i-1)\mu }},\dots ,(k-2)\mu \},\\ \rho (B)&=\{y_1,y_1+\nu ,\dots ,y_1+(i-2)\nu , {\overline{y_i}}, {\underline{y_i+\nu }},\dots ,y_i+(k-i)\nu \}, \end{aligned}$$

where $2\le i\le k$.

Next, we calculate the exact value of $|\Gamma _{1,2}|$ by Lemma 3.11. For $k\ge 5,$ we only need to calculate the number of all parameters $(i,j,\mu ,\nu ,y_1,x_i,y_i)$ corresponding to $B\in \Gamma _{1,2}$. For $k=4$, we need further minus half of the number of these parameters under the case of $(i,j)=(3,2)$. We count the number by classifying the different (i, j)s. Under the case of each fixed (i, j), it is easy to get that the number of parameters $(\mu ,x_i)$ is $\lfloor \frac{n-1}{k-2}\rfloor .$

(i) When $\nu >0$, according to the size of $y_1$ and $y_i$, we divide it into the following two cases:

 (a) If $y_1=0\le y_i$, similar to the case $\Gamma _{1,1}$, there also require $y_i\ne y_{i-1}$ for $B\in \Gamma _{1,2}$. For each fixed $(i,j=i-1)$, the number of parameters $(\nu ,y_1=0,y_i)$ is $\sum _{\nu =1}^{M_1}(m-(k-i)\nu )-\lfloor \frac{m-1}{k-2}\rfloor ,$ where $M_1=\lfloor \frac{m-1}{i-2}\rfloor$ if $\lfloor \frac{k+2}{2}\rfloor +1\le i\le k$ or $M_1=\lfloor \frac{m-1}{k-i}\rfloor$ if $2\le i\le \lfloor \frac{k+2}{2}\rfloor$.

 (b) If $y_1>y_i=0$, note that $y_i\ne y_{i-1}$ always holds in this case. Therefore for each fixed $(i,j=i-1)$, by the results of the case $\Gamma _{1,1}$, the number of parameters $(\nu ,y_1,y_i=0)$ is $\sum _{\nu =1}^{M_2}(m-1-(i-2)\nu ),$ where $M_2=\min \{\lfloor \frac{m-2}{i-2}\rfloor ,\lfloor \frac{m-1}{k-i}\rfloor \}$ if $2<i<k$ or $M_2=\lfloor \frac{m-1}{k-i}\rfloor$ if $i=2$ or $M_2=\lfloor \frac{m-2}{i-2}\rfloor$ if $i=k$.

Combining the above results, when $k\ge 5$ we have

$$\begin{aligned} |\Gamma ^1_{1,2}|=\sum _{2\le i\le k}\lfloor \frac{n-1}{k-2}\rfloor \cdot \bigg \{\sum _{\nu =1}^{M_1}(m-(k-i)\nu )+ \sum _{\nu =1}^{M_2}(m-1-(i-2)\nu )-\lfloor \frac{m-1}{k-2}\rfloor \bigg \}. \end{aligned}$$

(ii) When $\nu =0$, similar to case (i), we get

$$\begin{aligned} |\Gamma ^2_{1,2}|=\lfloor \frac{n-1}{k-2}\rfloor \cdot (k-1)\cdot (2m-2),~~ k\ge 5. \end{aligned}$$

(When $k=4$, if we limit $\nu =\nu '=0$, then there is no intersection sets in $\Gamma _{1,2}$ by Lemma 3.11. Therefore the above formula for $|\Gamma ^2_{1,2}|$ also holds for $k=4$.)

(iii) When $\nu <0$, by Lemma 3.12, $|\Gamma ^3_{1,2}|=|\Gamma ^1_{1,2}|$.

When $k\ge 5$, we get the conclusion immediately by summing the above three cases. When $k=4$, it is not difficult to calculate that the number of intersection sets is $\lfloor \frac{n-1}{2}\rfloor \cdot (m^2-m-\lfloor \frac{m-1}{2}\rfloor )$ by Lemma 3.11. Here, we need further remove these intersection sets to get the conclusion after detailed simplification. $\square$

Proof of Lemma 3.15

For any $B\in \Gamma _{2,1}$,

$$\begin{aligned} \tau (B)= & {} \left\{ 0,\mu ,\dots ,(i-2)\mu , {\overline{x_i}},(i-1)\mu ,x_i+\mu ,\dots ,\frac{j+i-3}{2}\mu ,x_i \right. \\{} & {} \left. +\frac{j+1-i}{2}\mu , {\underline{x_i+\frac{j+3-i}{2}\mu }},\dots ,x_i+\frac{2k-j-i-1}{2}\mu \right\} ,\\ \rho (B)= & {} \left\{ y_1,y_1+\nu ,\dots ,y_1+(i-2)\nu ,{ \overline{y_i}},y_1+(i-1)\nu ,y_i +\nu ,\dots ,y_1+\frac{j+i-3}{2}\nu ,\right. \\{} & {} \left. y_i+\frac{j+1-i}{2}\nu , {\underline{y_i+\frac{j+3-i}{2}\nu }},\dots ,y_i+\frac{2k-j-i-1}{2}\nu \right\} , \end{aligned}$$

where $2\le i\le k, 1\le j\le k-1, j\ge i, j-i\equiv 1\pmod {2}$, $(i-2)\mu<x_i<(i-1)\mu$. In particular, when $k\equiv 0\pmod {2}$ and $(i,j)=(2,k-1)$, $(x_2,y_2)\ne (\frac{\mu }{2},y_1+\frac{\nu }{2})$.

Next, we calculate the exact value of $|\Gamma _{2,1}|$ by Lemma 3.10. For $k\ge 4$, we only need to calculate the number of all parameters $(i,j,\mu ,\nu ,y_1,x_i,y_i)$ corresponding to $B\in \Gamma _{2,1}$. We count the number by classifying the different (i, j)s. Under the case of each fixed (i, j), we first calculate the number of sets satisfying $\lambda _{\Delta B}((\mu ,\nu ))=k-2$, then remove the subcase of $\lambda (\Delta B)=k-1$ by Lemma 3.2.

By Lemma 3.5, the number of parameters $(\mu ,x_i)$ is $\sum _{\mu =1}^{\min \{N'_2,N_2\}}(\mu -1)+\sum _{\mu =N'_2+1}^{N_2}(n-1-\frac{2k-j+i-5}{2}\mu )$, where $N_2=\lfloor \frac{2n-4}{2k-j+i-5}\rfloor$, $N'_2=\lfloor \frac{2n}{2k-j+i-3}\rfloor$.

(i) When $\nu >0$, according to the size of $y_1$ and $y_i$, we divide it into the following two cases:

 (a) If $y_1=0\le y_i$, then the conditions $y_i+\frac{2k-j-i-1}{2}\nu \le m-1$ and $\frac{j+i-3}{2}\nu \le m-1$ yield $1\le \nu \le \min \{\lfloor \frac{2(m-1)}{2k-j-i-1}\rfloor ,\lfloor \frac{2(m-1)}{j+i-3}\rfloor \}\triangleq M_3, 0\le y_i\le m-1-\frac{2k-j-i-1}{2}\nu$. Therefore for each fixed (i, j), the number of parameters $(\nu ,y_1=0,y_i)$ is $\sum _{\nu =1}^{M_3}(m-\frac{2k-j-i-1}{2}\nu ).$

 (b) If $y_1>y_i=0$, then the conditions $y_1\ge 1,y_1+\frac{j+i-3}{2}\nu \le m-1$ and $\frac{2k-j-i-1}{2}\nu \le m-1$ yield $1\le \nu \le \min \{\lfloor \frac{2(m-1)}{2k-j-i-1}\rfloor ,\lfloor \frac{2(m-2)}{j+i-3}\rfloor \}\triangleq M_4$, $1\le y_1\le m-1-\frac{j+i-3}{2}\nu$. Therefore for each fixed (i, j), the number of parameters $(\nu ,y_1,y_i=0)$ is $\sum _{\nu =1}^{M_4}(m-1-\frac{j+i-3}{2}\nu ).$

For $k\equiv 0 ~(\textrm{mod}~2)$, (i, j) can take $(2,k-1)$. We further remove the sets satisfying $\lambda (\Delta B)=k-1$, which holds only when $(i,j)=(2,k-1)$, $(x_2,y_2)=(\frac{\mu }{2},y_1+\frac{\nu }{2})$ and $y_2>y_1=0$. These sets have the following form:

$$\begin{aligned} \{(0,0),(x_2,y_2),(2x_2,2y_2)\dots ,((k-1)x_2,(k-1)y_2)\}. \end{aligned}$$

Therefore, we get

$$\begin{aligned}{} & {} |\Gamma ^1_{2,1}|=\sum _{\begin{array}{c} 2\le i\le k-2,3\le j\le k-1\\ j>i~\textrm{and}~j-i\equiv 1~(\textrm{mod}~2) \end{array} }\bigg \{\sum _{\mu =1}^{\min \{N'_2,N_2\}}(\mu -1)+\sum _{\mu =N'_2+1}^{N_2}(n-1-\frac{2k-j+i-5}{2}\mu )\bigg \}\\{} & {} ~~~~~~~~~~~~~~~~~\cdot \bigg \{\sum _{\nu =1}^{M_3}(m-\frac{2k-j-i-1}{2}\nu )+\sum _{\nu =1}^{M_4}(m-1 -\frac{j+i-3}{2}\nu )\bigg \}-\lfloor \frac{n-1}{k-1}\rfloor \cdot \lfloor \frac{m-1}{k-1}\rfloor \cdot \gamma _k, \end{aligned}$$

where $\gamma _k=1$ for $k\equiv 0 ~(\textrm{mod}~2); \gamma _k=0$ for $k\equiv 1 ~(\textrm{mod}~2)$.

(ii) When $\nu =0$, similar to case (i), we can get:

$$\begin{aligned} |\Gamma ^2_{2,1}|=\sum _{\begin{array}{c} 2\le i\le k-2,3\le j\le k-1\\ j>i~\textrm{and}~j-i\equiv 1~(\textrm{mod}~2) \end{array} }\bigg \{\sum _{\mu =1}^{\min \{N'_2,N_2\}}(\mu -1)+\sum _{\mu =N'_2+1}^{N_2}(n-1-\frac{2k-j+i-5}{2}\mu )\bigg \}\cdot (2m-1) \end{aligned}$$

$~~~~~~~~~~~~~~~~~~-\lfloor \frac{n-1}{k-1}\rfloor \cdot \gamma _k.$

(iii) When $\nu <0$, by Lemma 3.12, $|\Gamma ^3_{2,1}|=|\Gamma ^1_{2,1}|$.

Summing the above three cases, we get the conclusion immediately. $\square$

Proof of Lemma 3.16

For $B\in \Gamma _{2,2}$,

$$\begin{aligned} \tau (B)= & {} \{0,\mu ,\dots ,(i-2)\mu , {\overline{(i-2)\mu }},\dots ,\frac{j+i-3}{2}\mu ,\frac{j+i-3}{2}\mu ,\\{} & {} {\underline{\frac{j+i-1}{2}\mu }},\dots ,\frac{2k-j+i-5}{2}\mu \},\\ \rho (B)= & {} \{y_1,y_1+\nu ,\dots ,y_1+(i-2)\nu ,{ \overline{y_i}},\dots ,y_1+\frac{j+i-3}{2}\nu ,y_i+\frac{j+1-i}{2}\nu ,\\{} & {} {\underline{y_i+\frac{j+3-i}{2}\nu }},\dots ,y_i+\frac{2k-j-i-1}{2}\nu \}, \end{aligned}$$

where $2\le i\le k, 1\le j\le k-1, j\ge i, j-i\equiv 1\pmod {2}$.

Next, we calculate the exact value of $|\Gamma _{2,2}|$ by Lemma 3.11. For $k\ge 4$, we first calculate the number of all parameters $(i,j,\mu ,\nu ,y_1,x_i,y_i)$ corresponding to $B\in \Gamma _{2,2}$, then minus half of the number of parameters under the case of $(i,j)=(2,k-1)$ for $k\equiv 0\pmod {2}$. We count the number by classifying the different (i, j)s. Under the case of each fixed (i, j), it is easy to know that the number of parameters $(\mu ,x_i)$ is $\lfloor \frac{2(n-1)}{2k-j+i-5}\rfloor .$

(i) When $\nu >0$, according to the size of $y_1$ and $y_i$, we divide it into the following two cases:

 (a) If $y_1=0\le y_i$, similar to the case $\Gamma _{2,1}$, there also require $y_i\ne y_{i-1}$ in $\Gamma _{2,2}$. So it’s easy to get for each fixed (i, j), the number of parameters $(\nu ,y_1=0,y_i)$ is $\sum _{\nu =1}^{M_3}(m-\frac{2k-j-i-1}{2}\nu )-\lfloor \frac{2(m-1)}{2k+i-j-5}\rfloor ,$ where $M_3=\min \{\lfloor \frac{2(m-1)}{2k-j-i-1}\rfloor ,\lfloor \frac{2(m-1)}{j+i-3}\rfloor \}$.

 (b) If $y_1>y_i=0$, note that $y_i\ne y_{i-1}$ always holds in this case. Therefore for each fixed (i, j), by the results of the case $\Gamma _{2,1}$, the number of parameters $(\nu ,y_1,y_i=0)$ is $\sum _{\nu =1}^{M_4}(m-1-\frac{j+i-3}{2}\nu ),$ where $M_4=\min \{\lfloor \frac{2(m-1)}{2k-j-i-1}\rfloor ,\lfloor \frac{2(m-2)}{j+i-3}\rfloor \}$.

By Lemma 3.11, when $k\equiv 0 ~(\textrm{mod}~ 2)$ and $(i,j)=(2,k-1)$, we need to remove the following intersection sets:

$$\begin{aligned} \{(0,0),(0,y_2),(\mu ,\nu ),(\mu ,y_2+\nu ),\dots ,(\frac{k-2}{2}\mu ,\frac{k-2}{2}\nu ),(\frac{k-2}{2}\mu ,y_2+\frac{k-2}{2}\nu )\}, \end{aligned}$$

and the number of such sets is $\sum _{\nu =1}^{\lfloor \frac{2(m-2)}{k-2}\rfloor }(m-1-\frac{k-2}{2}\nu )\cdot \lfloor \frac{2(n-1)}{k-2}\rfloor$. Therefore, we get

$$\begin{aligned}{} & {} |\Gamma ^1_{2,2}|=\sum _{\begin{array}{c} 2\le i\le k-2,3\le j\le k-1\\ j>i~\textrm{and}~j-i\equiv 1~(\textrm{mod}~2) \end{array} }\lfloor \frac{2(n-1)}{2k-j+i-5}\rfloor \cdot \bigg \{\sum _{\nu =1}^{M_3}(m-\frac{2k-j-i-1}{2}\nu )-\lfloor \frac{2(m-1)}{2k+i-j-5}\rfloor \\{} & {} ~~~~~~~~~~~~~~~~~+ \sum _{\nu =1}^{M_4}(m-1-\frac{j+i-3}{2}\nu )\bigg \} -\sum _{\nu =1}^{\lfloor \frac{2(m-1)}{k-2}\rfloor }(m-1-\frac{k-2}{2}\nu )\cdot \lfloor \frac{2(n-1)}{k-2}\rfloor \cdot \gamma _k, \end{aligned}$$

where $\gamma _k=1$ for $k\equiv 0 ~(\textrm{mod}~2); \gamma _k=0$ for $k\equiv 1 ~(\textrm{mod}~2)$.

(ii) When $\nu =0$, similar to case (i), we can get

$$\begin{aligned} |\Gamma ^2_{2,2}|=\sum _{\begin{array}{c} 2\le i\le k-2,3\le j\le k-1\\ j>i~\textrm{and}~j-i\equiv 1~(\textrm{mod}~2) \end{array} }\lfloor \frac{2(n-1)}{2k-j+i-5}\rfloor \cdot (2m-2)-\lfloor \frac{2(n-1)}{k-2}\rfloor \cdot (m-1)\cdot \gamma _k. \end{aligned}$$

(iii) When $\nu <0$, $|\Gamma ^3_{2,2}|=|\Gamma ^1_{2,2}|$.

Summing the above three cases, we get the conclusion immediately. $\square$

Proof of Lemma 3.17

For $B\in \Gamma _{3,1}$,

$$\begin{aligned}{} & {} \tau (B)=\left\{ 0,\mu ,\dots ,(i-2)\mu , {\overline{x_i}},(i-1)\mu ,x_i+\mu ,\dots ,\frac{j+i-4}{2}\mu ,x_i\right. \\{} & {} \quad \left. +\frac{j-i}{2}\mu ,\frac{j+i-2}{2}\mu , {\underline{\frac{j+i}{2}\mu }}, \dots ,\frac{2k-j+i-4}{2}\mu \right\} ,\\{} & {} \rho (B)= \left\{ y_1,y_1+\nu ,\dots ,y_1+(i-2)\nu , {\overline{y_i}},y_1+(i-1)\nu ,y_i+\nu ,\dots ,y_1\right. \\{} & {} \quad \left. +\frac{j+i-4}{2}\nu ,y_i+\frac{j-i}{2}\nu , y_1+\frac{j+i-2}{2}\nu , {\underline{y_1+\frac{j+i}{2}\nu }},\dots ,y_1+\frac{2k-j+i-4}{2}\nu \right\} , \end{aligned}$$

where $2\le i\le k, 1\le j\le k-1, j\ge i, j-i\equiv 0\pmod {2}$, $(i-2)\mu<x_i<(i-1)\mu$. In particular, when $k\equiv 1\pmod {2}$ and $(i,j)=(2,k-1)$, $(x_2,y_2)\ne (\frac{\mu }{2},y_1+\frac{\nu }{2})$.

Next, we calculate the exact value of $|\Gamma _{3,1}|$ by Lemma 3.10. For $k\ge 4$, we only need to calculate the number of all parameters $(i,j,\mu ,\nu ,y_1,x_i,y_i)$ corresponding to $B\in \Gamma _{3,1}$. We count the number by classifying the different (i, j)s. Under the case of each fixed (i, j), we first calculate the number of the sets satisfying $\lambda _{\Delta B}((\mu ,\nu ))=k-2$, then remove the subcase of $\lambda (\Delta B)=k-1$ by Lemma 3.2.

By Lemma 3.5, we get the number of parameters $(\mu ,x_i)$ is $\sum _{\mu =1}^{N_3}(\mu -1),~\textrm{where}~ N_3=\lfloor \frac{2n-2}{2k-j+i-4}\rfloor .$

(i) When $\nu >0$, according to the size of $y_1$ and $y_i$, we divide it into the following two cases:

 (a) If $y_1=0\le y_i$, then the conditions $y_i+\frac{j-i}{2}\nu \le m-1$ and $\frac{2k-j+i-4}{2}\nu \le m-1$ yield $1\le \nu \le \lfloor \frac{2(m-1)}{2k-j+i-4}\rfloor \triangleq M_5, 0\le y_i\le m-1-\frac{j-i}{2}\nu$. Under the case of each fixed (i, j), the number of parameters $(\nu ,y_1=0,y_i)$ is $\sum _{\nu =1}^{M_5}(m-\frac{j-i}{2}\nu ).$

 (b) If $y_1>y_i=0$, then the conditions $y_1\ge 1,y_1+\frac{2k-j+i-4}{2}\nu \le m-1$ and $\frac{j-i}{2}\nu \le m-1$ yield $1\le \nu \le M_6$, where $M_6=\min \{\lfloor \frac{2(m-2)}{2k-j+i-4}\rfloor ,\lfloor \frac{2(m-1)}{j-i}\rfloor \}$ if $i\ne j$ or $M_6=\lfloor \frac{m-2}{k-2}\rfloor$ if $i=j$, $1\le y_1\le m-1-\frac{2k-j+i-4}{2}\nu$. Therefore for each fixed (i, j), the number of parameters $(\nu ,y_1,y_i=0)$ is $\sum _{\nu =1}^{M_6}(m-1-\frac{2k-j+i-4}{2}\nu ).$

For $k\equiv 1 ~(\textrm{mod}~2)$, (i, j) can take $(2,k-1)$. Similar to $\Gamma _{2,1}$, we need to remove the following sets:

$$\{(0,0),(x_2,y_2),(2x_2,2y_2)\dots ,((k-1)x_2,(k-1)y_2)\}.$$

Therefore, we get

$$\begin{aligned} |\Gamma ^1_{3,1}|=\sum _{\begin{array}{c} 2\le i\le k-1,2\le j\le k-1\\ j\ge i~\textrm{and}~j-i\equiv 0~(\textrm{mod}~2) \end{array} }\sum _{\mu =1}^{N_3}\bigg \{\sum _{\nu =1}^{M_5}(m-\frac{j-i}{2}\nu )+\sum _{\nu =1}^{M_6}(m-1-\frac{2k-j+i-4}{2}\nu )\bigg \}\cdot (\mu -1) \end{aligned}$$

$~~~~~~~~~~~~~~~~-\lfloor \frac{n-1}{k-1}\rfloor \cdot \lfloor \frac{m-1}{k-1}\rfloor \cdot (1-\gamma _k),$

where $\gamma _k=1$ for $k\equiv 0 ~(\textrm{mod}~2); \gamma _k=0$ for $k\equiv 1 ~(\textrm{mod}~2)$.

(ii) When $\nu =0$, similar to case (1), we can get:

$$|\Gamma ^2_{3,1}|=\sum _{\begin{array}{c} 2\le i\le k-1,2\le j\le k-1\\ j\ge i~\textrm{and}~j-i\equiv 0~(\textrm{mod}~2) \end{array} }\sum _{\mu =1}^{N_3}(\mu -1)\cdot (2\,m-1)-\lfloor \frac{n-1}{k-1}\rfloor \cdot (1-\gamma _k).$$

(iii) When $\nu <0$, by Lemma 3.12, $|\Gamma ^3_{3,1}|=|\Gamma ^1_{3,1}|$.

Summing the above three cases, we get the conclusion immediately. $\square$

Proof of Lemma 3.18

For $B\in \Gamma _{3,2}$,

$$\begin{aligned} \tau (B)= & {} \left\{ 0,\mu ,\dots ,(i-2)\mu , {\overline{(i-2)\mu }},\dots ,\frac{j+i-4}{2}\mu ,\frac{j+i-4}{2}\mu ,\right. \\{} & {} \left. \frac{j+i-2}{2}\mu , {\underline{\frac{j+i}{2}\mu }},\dots ,\frac{2k-j+i-4}{2}\mu \right\} ,\\ \rho (B)= & {} \left\{ y_1,y_1+\nu ,\dots ,y_1+(i-2)\nu , {\overline{y_i}},\dots ,y_1+\frac{j+i-4}{2}\nu ,y_i+\frac{j-i}{2}\nu ,y_1 \right. \\{} & {} \left. +\frac{j+i-2}{2}\nu , {\underline{y_1+\frac{j+i}{2}\nu }}, \dots ,y_1+\frac{2k-j+i-4}{2}\nu \right\} , \end{aligned}$$

where $2\le i\le k, 1\le j\le k-1, j\ge i, j-i\equiv 0\pmod {2}$.

Next, we calculate the exact value of $|\Gamma _{3,2}|$ by Lemma 3.11. For $k\ge 4$, we only need to calculate the number of all parameters $(i,j,\mu ,\nu ,y_1,x_i,y_i)$ corresponding to $B\in \Gamma _{3,2}$. We count the number by classifying the different (i, j)s. Under the case of each fixed (i, j), we first calculate the number of parameters $(\mu ,x_i)$ is $\lfloor \frac{2(n-1)}{2k-j+i-4}\rfloor .$

(i) When $\nu >0$, according to the size of $y_1$ and $y_i$, we divide it into the following two cases:

 (a) If $y_1=0\le y_i$, similar to the case $\Gamma _{3,1}$, there also require $y_i\ne y_{i-1}$ in $\Gamma _{3,2}$. For each fixed (i, j), it’s easy to get that the number of parameters $(\nu ,y_1=0,y_i)$ is $\sum _{\nu =1}^{M_5}(m-1-\frac{j-i}{2}\nu ),~\textrm{where}~ M_5=\lfloor \frac{2(m-1)}{2k-j+i-4}\rfloor .$

 (b) If $y_1>y_i=0$, note that $y_i\ne y_{i-1}$ always holds in this case. Therefore for each fixed (i, j), by the results of the case $\Gamma _{3,1}$, the number of parameters $(\nu ,y_1,y_i=0)$ is $\sum _{\nu =1}^{M_6}(m-1-\frac{2k-j+i-4}{2}\nu )$, where $M_6=\min \{\lfloor \frac{2(m-2)}{2k-j+i-4}\rfloor ,\lfloor \frac{2(m-1)}{j-i}\rfloor \}$ if $i\ne j$ or $M_6=\lfloor \frac{m-2}{k-2}\rfloor$ if $i=j$.

Therefore, we get

$$\begin{aligned} |\Gamma ^1_{3,2}|=&\sum _{\begin{array}{c} 3\le i\le k-1,3\le j\le k-1\\ j\ge i~\textrm{and}~j-i\equiv 0~(\textrm{mod}~2) \end{array} }\lfloor \frac{2(n-1)}{2k-j+i-4}\rfloor \cdot \bigg \{\sum _{\nu =1}^{M_5}(m-1-\frac{j-i}{2}\nu )\\ {}&+\sum _{\nu =1}^{M_6}(m-1-\frac{2k-j+i-4}{2}\nu )\bigg \}. \end{aligned}$$

(ii) When $\nu =0$, similar to case (i), we can get

$$\begin{aligned} |\Gamma ^2_{3,2}|=\sum _{\begin{array}{c} 3\le i\le k-1,3\le j\le k-1\\ j\ge i~\textrm{and}~j-i\equiv 0~(\textrm{mod}~2) \end{array} }\lfloor \frac{2(n-1)}{2k-j+i-4}\rfloor \cdot (2m-2). \end{aligned}$$

(iii) When $\nu <0$, by Lemma 3.12, $|\Gamma ^3_{3,2}|=|\Gamma ^1_{3,2}|$.

Summing the above three cases, we get the conclusion immediately. $\square$