On the correlation distribution of the generalized maximal length ℤ₄-sequences

Abstract

A family of maximal length ℤ₄-sequences was proposed by Tang, Udaya and Fan in 2005 by using binary sequences based on quadratic forms over finite fields, and was shown to possess low correlation property. However, its correlation distribution still remains open. In this paper, the maximal length ℤ₄-sequences constructed by Tang, Udaya and Fan are equivalently expressed as another form via ℤ₄-valued quadratic forms and then the correlation distribution is completely determined from the approach of ℤ₄-valued quadratic forms.

Appendices

Appendix A: Proof of Proposition 2

For odd m, one has ${\text {tr}_{e}^{n}}(1) = 1$ and then ${\text {tr}_{e}^{n}}(\frac {v^{2}}{1+v^{2}}) = {\text {tr}_{e}^{n}}(\frac {1}{1+v^{2}}+1) = X^{2}+1$. Plugging (10) into ${\text {tr}_{e}^{n}}(z) = a$ and ${\text {tr}_{e}^{n}}(vz) = b$, one then gets

$$\begin{array}{@{}rcl@{}} [a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b)]X^{2}+[b+{\text{tr}_{1}^{e}}(b)]X=a \end{array} $$

(11)

and

$$\begin{array}{@{}rcl@{}} [a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b)]X^{2}+[a+{\text{tr}_{1}^{e}}(a)]X={\text{tr}_{1}^{e}}(b), \end{array} $$

(12)

which means the number of solutions to (9) is the number of pairs $(a,b)\in \mathbb {F}_{2^{e}}\times \mathbb {F}_{2^{e}}$ satisfying (11) and (12). We then discuss the solutions to (9) according to the value of $X={\text {tr}_{e}^{n}}\left (\frac {1}{1+v}\right )$ as follows:

• Case 1: X = 0. If X = 0, then by (11) and (12) one has a = 0 and ${\text {tr}_{1}^{e}}(b) = 0$. According to the balance property of the trace function, the number of pairs $(a,b)\in \mathbb {F}_{2^{e}}\times \mathbb {F}_{2^{e}}$ satisfying a = 0 and ${\text {tr}_{1}^{e}}(b) = 0$ is 2^{e − 1}, i.e., the number of solutions to (9) is 2^{e − 1}. Thus, for this case the rank of F(x) is n − e + 1.

• Case 2: X = 1. Similar to Case 1, from (11) and (12) one has b = 0 and ${\text {tr}_{1}^{e}}(a) = 0$, then the rank of F(x) is also n − e + 1.

• Case 3: $X\in \mathbb {F}_{2^{e}}\backslash \{0,1\}$. Again by (11) and (12) one can deduce

  $$\begin{array}{@{}rcl@{}} a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b) = \frac{a+{\text{tr}_{1}^{e}}(b)}{X}. \end{array} $$

Replacing $a+{\text {tr}_{1}^{e}}(a)+b+{\text {tr}_{1}^{e}}(b)$ in (11) and (12) by $\frac {a+{\text {tr}_{1}^{e}}(b)}{X}$, one gets

$$\begin{array}{@{}rcl@{}}\left\{\begin{array}{ll} (a+b)X=a, \\({\text{tr}_{1}^{e}}(a)+{\text{tr}_{1}^{e}}(b))X={\text{tr}_{1}^{e}}(b). \end{array}\right. \end{array} $$

(13)

Notice that X ≠ 0,1 and ${\text {tr}_{1}^{e}}(a),{\text {tr}_{1}^{e}}(b)\in \{0,1\}$. Thus, according to the second equation in (13), one can claim that ${\text {tr}_{1}^{e}}(a)+{\text {tr}_{1}^{e}}(b) = 0$ and ${\text {tr}_{1}^{e}}(b) = 0$, i.e., ${\text {tr}_{1}^{e}}(a) = {\text {tr}_{1}^{e}}(b) = 0$. On the other hand, from (a + b)X = a, one has $b=\frac {(X+1)a}{X}$. Therefore, (13) holds if and only if $(a,b)\in \mathbb {F}_{2^{e}}\times \mathbb {F}_{2^{e}}$ satisfying ${\text {tr}_{1}^{e}}(a) = 0$ and ${\text {tr}_{1}^{e}}(\frac {(X+1)a}{X}) = 0$. For any fixed 𝜖 ≠ 0,1, by the two-tuple-balanced property of the trace function, i.e., the pair $({\text {tr}_{1}^{e}}(x),{\text {tr}_{1}^{e}}(\epsilon x))$ takes each pair over {0,1} exactly 2^{e − 2} times when x runs through $\mathbb {F}_{2^{e}}$, one then can conclude that the number of pairs (a, b) satisfying (13) is 2^{e − 2}, so is the number of roots of (9). Then, in this case the rank of F(x) is n − e + 2.

Therefore, the rank of F(x) is determined for any δ ≠ 1. Specially, for the case of δ = 0, one has $v=\overline {\delta }=0$ and $X={\text {tr}_{e}^{n}}(\frac {1}{1+v}) = {\text {tr}_{e}^{n}}(1) = m\pmod {2}$. Then, Case 1 and Case 2 imply that the rank of F(x) is n − e + 1 if δ = 0. Further, for the case of $\overline {\delta }\ne 0,1$, the rank distribution of F(x) can also be determined when $\overline {\delta }$ runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$ according to the balance property of trace function. Notice that $\frac {1}{1+v}$ runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$ when v ranges over $\mathbb {F}_{2^{n}}\backslash \{0,1\}$. Then, by the balance property of trace function, one can conclude that $X={\text {tr}_{e}^{n}}(\frac {1}{1+v})\in \{0,1\}$ occurs 2^{n − e}×2−2 times and $X={\text {tr}_{e}^{n}}(\frac {1}{1+v})\not \in \{0,1\}$ occurs 2^{n − e}×(2^e − 2) times when v runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$.

Based on the above discussions, one then can obtain the desired result. This completes the proof.

Appendix B: Proof of Proposition 3

For the case of m is even, one has ${\text {tr}_{e}^{n}}(1) = 0$ and then ${\text {tr}_{e}^{n}}(\frac {v^{2}}{1+v^{2}}) = {\text {tr}_{e}^{n}}(\frac {1}{1+v^{2}}+1) = X^{2}$. Plugging (10) into ${\text {tr}_{e}^{n}}(z) = a$ and ${\text {tr}_{e}^{n}}(vz) = b$, one then gets

$$\begin{array}{@{}rcl@{}} [a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b)]X^{2}+[b+{\text{tr}_{1}^{e}}(b)]X=a \end{array} $$

(14)

and

$$\begin{array}{@{}rcl@{}} [a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b)]X^{2}+[a+{\text{tr}_{1}^{e}}(a)]X=b. \end{array} $$

(15)

Similar as the case of m is odd, we discuss the solutions to (9) via (10), (14) and (15) according to the value of $X={\text {tr}_{e}^{n}}(\frac {1}{1+v})$ as follows:

• Case 1: X = 0. If X = 0, then by (14) and (15) one has a = b = 0, which means in this case (9) has only the zero solution, i.e., the rank of F(x) is n.

• Case 2: X = 1. For this case, from (14) and (15) one has ${\text {tr}_{1}^{e}}(a) = 0$ and ${\text {tr}_{1}^{e}}(b) = 0$. Then, according to the balance property of the trace function, the number of pairs $(a,b)\in \mathbb {F}_{2^{e}}\times \mathbb {F}_{2^{e}}$ satisfying ${\text {tr}_{1}^{e}}(a) = 0$ and ${\text {tr}_{1}^{e}}(b) = 0$ is 2^{e − 1}×2^{e − 1} = 2^{2e − 2}, i.e., the number of solutions to (9) is 2^{2e − 2}. Thus, for this case the rank of F(x) is n − 2e + 2.

• Case 3: $X\in \mathbb {F}_{2^{e}}\backslash \{0,1\}$. In this case (14) and (15) imply

  $$\begin{array}{@{}rcl@{}} a+{\text{tr}_{1}^{e}}(a)+b+{\text{tr}_{1}^{e}}(b) = \frac{a+b}{X}. \end{array} $$

Then, replacing $a+{\text {tr}_{1}^{e}}(a)+b+{\text {tr}_{1}^{e}}(b)$ in (14) and (15) by $\frac {a+b}{X}$, one has

$$\begin{array}{@{}rcl@{}}\left\{\begin{array}{ll} (a+{\text{tr}_{1}^{e}}(b))X=a, \\(b+{\text{tr}_{1}^{e}}(a))X=b. \end{array}\right. \end{array} $$

(16)

By a simple calculation, one then has

$$a=\frac{X}{X+1}{\text{tr}_{1}^{e}}(b),\,b=\frac{X}{X+1}{\text{tr}_{1}^{e}}(a),$$

which implies $a,b\in \{0,\frac {X}{X+1}\}$ since ${\text {tr}_{1}^{e}}(a),{\text {tr}_{1}^{e}}(b)\in \{0,1\}$. Then, we can discuss it as below:

• Case 3.1: If a = 0, then by the second equation in (16) one gets b = 0 since X ≠ 1. Note that (a, b) = (0,0) always is a solution to (16).

• Case 3.2: If $a=\frac {X}{X+1}$, then by (16) one gets ${\text {tr}_{1}^{e}}(b) = 1$ and $b=\frac {X}{X+1}{\text {tr}_{1}^{e}}(\frac {X}{X+1})$. This implies $(a,b) = (\frac {X}{X+1},\frac {X}{X+1})$ is a solution to (16) if and only if ${\text {tr}_{1}^{e}}(\frac {X}{X+1}) = 1$.

Therefore, by the above discussions, one can conclude that

• 1) the rank of F(x) is n if either X = 0 or ${\text {tr}_{1}^{e}}(\frac {X}{X+1}) = 0$ with X ≠ 0,1;

• 2) the rank of F(x) is n − 2e + 2 if $X={\text {tr}_{e}^{n}}(\frac {1}{1+v}) = 1$;

• 3) the rank of F(x) is n − 1 if ${\text {tr}_{1}^{e}}(\frac {X}{X+1}) = 1$ with X ≠ 0,1.

Note that $X={\text {tr}_{e}^{n}}(\frac {1}{1+v})$ and $v=\overline {\delta }$. To determine the rank distribution of F(x), for i = 0, 1, define

$$A_{i}=\{\overline{\delta}\in\mathbb{F}_{2^{n}}\backslash\{0,1\}:{\text{tr}_{1}^{e}}(\frac{X}{X+1}) = i, X\ne 0,1\}.$$

When $\overline {\delta }$ runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$, so does $\frac {1}{1+v}$. This together with the balance property of trace function one can claim that X = 0 occurs 2^{n − e} − 2 times and every nonzero element $X\in \mathbb {F}_{2^{e}}$ occurs 2^{n − e} times when $\overline {\delta }$ runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$ since ${\text {tr}_{e}^{n}}(0) = {\text {tr}_{e}^{n}}(1) = 0$ for even n/e = m. Immediately, one then obtains that rank(F(x)) = n − 2e + 2 occurs 2^{n − e} times when $\overline {\delta }$ runs through $\mathbb {F}_{2^{n}}\backslash \{0,1\}$. On the other hand, since $\frac {X}{X+1}$ runs through $\mathbb {F}_{2^{e}}\backslash \{0,1\}$ when X ranges over $\mathbb {F}_{2^{e}}\backslash \{0,1\}$, ${\text {tr}_{1}^{e}}(0) = 0$ and ${\text {tr}_{1}^{e}}(1) = e\pmod {2}$, thus according to the balance property of trace function, if e is odd, one gets

$$|A_{0}|=|A_{1}|=2^{n-e}(2^{e-1}-1) = 2^{n-1}-2^{n-e}, $$

and for even e, one has

$$|A_{0}|=2^{n-e}(2^{e-1}-2) = 2^{n-1}-2^{n-e+1},\,|A_{1}|=2^{n-e}(2^{e-1}-0) = 2^{n-1}. $$

Hence, the rank distribution of F(x) is determined. This completes the proof.

Appendix C: Proof of Theorem 2

For the generalized maximal length ℤ₄-sequences b _i and b _j defined by (6), their correlation function $R_{b_{i},b_{j}}(\tau )$ can be considered as below, where 0 ≤ τ ≤ 2ⁿ − 2.

• Case 1. i = j = 2ⁿ+1.

  This is a trivial case, by (6), one can get

  $$\begin{array}{@{}rcl@{}} R_{b_{i},b_{j}}(\tau) = \sum\limits_{t=0}^{2^{n}-2}w^{2{\text{Tr}_{1}^{n}}(\beta^{t}-\beta^{t+\tau})}=\sum\limits_{x\in \mathbb{L}}w^{2{\text{Tr}_{1}^{n}}((1-\beta^{\tau})x)}-1. \end{array} $$

Thus, $R_{b_{i},b_{j}}(\tau ) = 2^{n}-1$ if τ = 0, and R _{i, j} (τ) = −1 if τ ≠ 0.

• Case 2. 1 ≤ i ≤ 2ⁿ and j = 2ⁿ + 1.

  For this case, from (6) one can derive

  $$\begin{array}{@{}rcl@{}} R_{b_{i},b_{j}}(\tau) &=&\sum\limits_{t=0}^{2^{n}-2}w^{{\text{Tr}_{1}^{n}}((1+2\eta_{i})\beta^{t})+2P(\beta^{t})+2Q(\beta^{t})-2{\text{Tr}_{1}^{n}}(\beta^{t+\tau})} \\&=& \sum\limits_{t=0}^{2^{n}-2}w^{{\text{Tr}_{1}^{n}}((1+2(\eta_{i}+\beta^{\tau}))\beta^{t})+2P(\beta^{t})+2Q(\beta^{t})} \\&=& \sum\limits_{x\in \mathbb{L}}w^{{\text{Tr}_{1}^{n}}((1+2(\eta_{i}+\beta^{\tau}))x)+2P(x)+2Q(x)}-1 \\&=&\sigma(\eta_{i}+\beta^{\tau},0,0)-1 \end{array} $$

due to (7). Note that 2(η _i + β ^τ) runs through $2\mathbb {L}$ for any fixed τ ∈ {0, 1, ⋯ , 2ⁿ − 2} when η _i ranges over 𝕃, i.e., i ranges from 1 to 2ⁿ. Moreover, by Proposition 2, one has the rank of F(x) is n − e + 1 if δ = 0. Then, for odd n − e + 1=(m − 1)e + 1 (since m is odd), by (7), (8) and Lemma 2, one can claim that $R_{b_{i},b_{j}}(\tau ) = \sigma (\eta _{i}+\beta ^{\tau },0,0)-1$ has the following distribution

Table 4

when i ranges from 1 to 2ⁿ and τ varies from 0 to 2ⁿ − 2.

• Case 3. i = 2ⁿ+1 and 1 ≤ j ≤ 2ⁿ.

  Due to the fact that $R_{b_{i},b_{j}}(\tau ) = R_{b_{j},b_{i}}(-\tau )^{*}$, thus for this case one can easily get the value distribution of $R_{b_{i},b_{j}}(\tau )$ according to Case 2 when i = 2ⁿ+1, j ranges from 1 to 2ⁿ and τ runs through {0, 1, ⋯ , 2ⁿ − 2}.

• Case 4. 1 ≤ i, j ≤ 2ⁿ with τ = 0.

  In this case, δ = β ^τ = 1, then by (6) and (7) one gets

  $$\begin{array}{@{}rcl@{}} R_{b_{i},b_{j}}(0) = \sigma(\eta_{i},\eta_{j},1)-1=\sum\limits_{x\in \mathbb{L}}w^{2{\text{Tr}_{1}^{n}}((\eta_{i}+\eta_{j})x)}-1 \end{array} $$

  equals −1 if i≠j, and 2ⁿ − 1 if i = j. Further, $R_{b_{i},b_{j}}(0) = -1$ occurs 2ⁿ(2ⁿ − 1) times and $R_{b_{i},b_{j}}(0) = 2^{n}-1$ occurs 2ⁿ times, respectively.

• Case 5. 1 ≤ i, j ≤ 2ⁿ with τ ≠ 0.

  For this case, δ = β ^τ and $\overline {\delta }\not =0,1$. Then, from (6) and (7), one can similarly derive

  $$\begin{array}{@{}rcl@{}} R_{b_{i},b_{j}}(\tau) = \sigma(\eta_{i},\eta_{j},\delta)-1. \end{array} $$

Let 1 − δ = δ ₁ + 2δ ₂, then 1 + 2η _i − (1+2η _j )δ = δ ₁ + 2(η _i + η _j (1 − δ ₁) + δ ₂). This implies 2(η _i + η _j (1−δ ₁) + δ ₂) ranges over 𝕃 when η _i runs through 𝕃 for any fixed δ and j. On the other hand, for odd m, one has n − e + 1=(m − 1)e + 1 is odd and n − e + 2 is even. Then, by (7), (8), Lemma 2 and Proposition 2, one can have that $R_{b_{i},b_{j}}(\tau ) = \sigma (\eta _{i},\eta _{j},\delta )-1$ has the following distribution

Table 5

when i and j range from 1 to 2ⁿ and τ varies from 0 to 2ⁿ − 2.

Combining the above cases, the correlation distribution of the generalized maximal length $\mathbb {Z}_{4}$-sequences for odd m is obtained. This completes the proof.