On the 4-Adic Complexity of Quaternary Sequences with Ideal Autocorrelation

I Introduction

With the development of correlation attack and algebraic attack, it is becoming the main trend to use the nonlinear feedback shift register sequences with pseudorandom property as the driving sequences in stream cipher design. The feedback with carry shift register (FCSR) proposed by [6] and [7] is a kind of generator which can produce nonlinear sequences quickly.

Balanced binary and quaternary sequences with good autocorrelation play important roles in communication and cryptography systems. The $d$ -adic complexity $Φ_{d} (s)$ measures the smallest length of FCSR which generates the sequence $s$ over $Z / (d)$ . Sequences over $Z / (d)$ with low $d$ -adic complexity are susceptibly decoded by the rational approximation algorithm, see [6], [8-9]. Particularly, a quaternary sequence $s$ can be decoded by the rational approximation algorithm with $6 Φ_{4} (s) + 16$ consecutive bits. Hence, the 4-adic complexity $Φ_{4} (s)$ of a safe sequence $s$ with period $N$ should exceed $\frac{N - 16}{6}$ . There are numerous results about the 2-adic complexity of binary sequences with good autocorrelation, see [2-3], [11-14], for example. However, the 4-adic complexity of quaternary sequences with good autocorrelation has not been studied so fully and there are few quaternary sequences with good autocorrelation whose 4-adic complexity is known, see [10]. This may pose risk to communication and cryptography system.

In this paper, we determine the 4-adic complexity of the balanced quaternary sequences of even period $2 p$ and $2 (2^{n} - 1)$ with ideal autocorrelation defined in [5] and [4], respectively. Our results show that the 4-adic complexity of the quaternary sequences with period $2 p$ and $2 (2^{n} - 1)$ defined in these two papers is larger than $\frac{2 p - 16}{6}$ and $\frac{2 (2^{n} - 1) - 16}{6}$ respectively. Hence they are safe enough to resist the attack of the rational approximation algorithm.

Ii Preliminaries

In the application of communication and cryptography, balanced sequences with good autocorrelation property are preferred.

For a sequence $g = (g_{0}, g_{1}, \dots, g_{N - 1})$ over $Z / (d)$ with period $N$ , it is said to be balanced if $| A_{i} - A_{j} | \leq 1$ for any pair of $i, j$ with $0 \leq i \neq j \leq N - 1$ ,where

A_{k} = {t | g_{t} = k, 0 \leq t < N}, k = 0, 1, \dots, d - 1.

The autocorrelation function of a sequence $s = (s_{0}, s_{1}, \dots, s_{N - 1})$ over $Z / (d)$ with period $N$ is defined by

C_{s} (τ) = N - 1 \sum i = 0 ζ_{d}^{s_{i} - s_{i + τ}}, 0 \leq τ < N,

where $ζ_{d}$ is a complex $d$ -th primitive root of unity.

The maximal out-of-phase autocorrelation magnitude should be as small as possible and the number of the occurrences of the maximal out-of-phase autocorrelation magnitude should be minimized. A sequence with the possible minimum value of the maximal out-of-phase autocorrelation magnitude and the minimum number of occurrences of the maximal out-of-phase autocorrelation magnitude is said to have the ideal autocorrelation property.

For a binary sequence $s$ with period $N$ , it is well known that if

C_{s} (τ) = - 1 for all 0 < τ < N,

(1)

then $s$ is an ideal autocorrelation sequence.

The autocorrelation distribution of a quaternary sequence $s$ of even period $N$ with ideal autocorrelation and balance property is given by

C_{s} (τ) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} N, & 1 times, 0, & \frac{N}{2} - 1 times, - 2, & \frac{N}{2} times . \end{matrix}

in [5].

By using the Legendre sequences and the Gray mapping, two classes of balanced quaternary sequences of even period $2 p$ with ideal autocorrelation were constructed in [5]. Balanced quaternary sequences of period $2 (2^{n} - 1)$ with ideal autocorrelation were constructed in [4] by using the binary sequences of period $2^{n} - 1$ with ideal autocorrelation and the Gray mapping.

For an odd prime

p

, let

Q R

and

Q N R

be the set of quadratic residues and quadratic non-residues in the set

Z_{p}^{*} = Z / (p) ∖ {0} = {1, 2, \dots, p - 1}

, respectively. Two classes of Legendre sequences

b

and

c

of period

p

are defined by

b_{t} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 0, & for t = 0 0, & for t \in Q R 1, & for t \in Q N R \end{matrix}

c_{t} = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 1, & for t = 0 0, & for t \in Q R 1, & for t \in Q N R \end{matrix}

respectively.

The Gray mapping $ϕ$ is defined by

ϕ (0, 0) = 0, ϕ (0, 1) = 1, ϕ (1, 1) = 2, ϕ (1, 0) = 3.

According to the definition of the Gray mapping, we can get

ϕ (a, e) = 2 a - a (e - 1) - (a - 1) e

(2)

where $0 \leq a \leq 1$ and $0 \leq e \leq 1$ .

The following two classes of quaternary sequences $g^{1}$ and $g^{2}$ of even period $2 p$ defined by using the Gray mapping and the Legendre sequences were shown to have ideal autocorrelation and balance property in [5].

Definition 1.

([5]) For an odd prime $p$ with $p \equiv 1 (mod 4)$ , let $s^{0}$ and $s^{1}$ be two binary sequences of the same period $2 p$ defined by

s_{t}^{0} = {\begin{matrix} b_{t}, & for t \equiv 0 mod 2 c_{t}, & for t \equiv 1 mod 2 \end{matrix}

s_{t}^{1} = {\begin{matrix} b_{t}, & for t \equiv 0 mod 2 1 - c_{t}, & for t \equiv 1 mod 2 . \end{matrix}

The quaternary sequence $g^{1}$ of period $2 p$ is defined by $g_{t}^{1} = ϕ (s_{t}^{0}, s_{t}^{1})$ .

Definition 2.

([5]) For an odd prime $p$ with $p \equiv 3 (mod 4)$ , let $s^{2}$ and $s^{3}$ be two binary sequences of the same period $2 p$ defined by

s_{t}^{2} = {\begin{matrix} b_{t}, & for 0 \leq t < p b_{t}, & for p \leq t < 2 p \end{matrix}

s_{t}^{3} = {\begin{matrix} c_{t}, & for t \equiv 0 mod 2 1 - c_{t}, & for t \equiv 1 mod 2 . \end{matrix}

The quaternary sequence $g^{2}$ of period $2 p$ is defined by $g_{t}^{2} = ϕ (s_{t}^{2}, s_{t}^{3})$ .

Let $Z_{2^{n} - 1} = Z / (2^{n} - 1) = {0, 1, 2, \dots, 2^{n} - 2}$ . Assume that $s$ is a binary sequence of period $2^{n} - 1$ with ideal autocorrelation. Let $D_{0}$ be the characteristic set of $s$ defined by

D_{0} = {t | s_{t} = 1, 0 \leq t \leq 2^{n} - 2}

and ${¯ ¯¯¯ ¯ D}_{0} = Z_{2^{n} - 1} ∖ D_{0}$ . By the Chinese remainder theorem, we have the isomorphism

ϕ : Z_{2 \times (2^{n} - 1)} ≃ Z_{2} \times Z_{2^{n} - 1}, h \mapsto (h mod 2, h mod 2^{n} - 1) .

The following class of quaternary sequences $g^{3}$ of even period $2 (2^{n} - 1)$ defined by using the Gray mapping and the ideal autocorrelation sequences with period $2^{n} - 1$ were shown to have ideal autocorrelation and balance property in [4].

Definition 3.

([4]) Let $s$ be binary sequence of period $2^{n} - 1$ with ideal autocorrelation and $D_{0}$ a characteristic set of $s$ . Let $g^{3}$ be the quaternary sequence defined by

g_{t}^{3} = ϕ (u_{t}, v_{t}),

where $u$ and $v$ are the binary sequences of period $2^{n + 1} - 2$ defined by

u_{t} = {\begin{matrix} 1, & if t \in {0, 1} \times D_{0} 0, & if t \in {0, 1} \times {¯ ¯¯¯ ¯ D}_{0} \end{matrix}

v_{t} = {\begin{matrix} 1, & if t \in {0} \times D_{0} ⋃ {1} \times {¯ ¯¯¯ ¯ D}_{0} 0, & if t \in {0} \times {¯ ¯¯¯ ¯ D}_{0} ⋃ {1} \times D_{0} . \end{matrix}

The definition about the $4$ -adic complexity of quaternary sequences with period $N$ is defined as follows.

Definition 4.

([6, 9]) For a quaternary sequence $s = (s_{0}, s_{1}, \dots, s_{N - 1})$ with period $N$ , let $S (4) = \sum_{i = 0}^{N - 1} s_{i} 4^{i}$ . The 4-adic complexity $Φ_{4} (s)$ is defined by ${log}_{4} \frac{4^{N} - 1}{gcd (4^{N} - 1, S (4))},$ where $gcd (a, b)$ denotes the greatest common divisor of $a$ and $b$ . (The exact value of the smallest length of FCSR which generates the quaternary sequence is $⌊ {log}_{4} ((4^{N} - 1) / gcd (4^{N} - 1, S (4)) + 1) ⌋$ .

According to Definition 4, determining the $4$ -adic complexity of quaternary sequences is equivalent to determining $gcd (4^{N} - 1, S (4))$ .

Iii Main result

In this section, we study the 4-adic complexity of the quaternary sequences of period $2 p$ and $2 (2^{n} - 1)$ with ideal autocorrelation in Section II.

For $i \in Z_{p}^{*}$ , the Legendre symbol $(\frac{i}{p})$ is defined by

(\frac{i}{p}) = {\begin{matrix} 1, & if % i \in Q R - 1, & otherwise . \end{matrix}

The following four lemmas are useful in the sequel.

Lemma 1.

([1], Theorem 7.3) If $s$ is a periodic binary sequence of odd period $2^{n} - 1$ with ideal autocorrelation, then the number of nonzero bits in one period of $s$ is $2^{n - 1}$ .

The proof of the lemma is similar to that of Lemma 2(1) in [14]. For the completeness of the paper, we give a simple proof.

Lemma 2.

Let $p$ be an odd prime. Then

{(p - 1 \sum i = 1 (\frac{i}{p}) 4^{i})}^{2} \equiv - (\frac{- 1}{p}) \frac{4^{p} - 1}{3} + (\frac{- 1}{p}) p (mod 4^{p} - 1) .

Proof.

Since

	${(p - 1 \sum i = 1 (\frac{i}{p}) 4^{i})}^{2}$	$= p - 1 \sum i = 1 (\frac{i}{p}) 4^{i} p - 1 \sum j = 1 (\frac{j}{p}) 4^{j}$
		$= p - 1 \sum i, j = 1 (\frac{i j}{p}) 4^{i + j} (% let j = i k)$
		$= p - 1 \sum i, k = 1 (\frac{k}{p}) 4^{i (k + 1)}$
		$= p - 2 \sum k = 1 (\frac{k}{p}) p - 1 \sum i = 1 4^{i (k + 1)} + p - 1 \sum i = 1 4^{i p} (\frac{p - 1}{p}) .$

Then from $\sum_{k = 1}^{p - 2} (\frac{k}{p}) = - (\frac{- 1}{p})$ and

p - 1 \sum i = 1 4^{i (k + 1)} \equiv p - 1 \sum i = 1 4^{i} (mod 4^{p} - 1) (1 \leq k \leq p - 2),

we get

	${(p - 1 \sum i = 1 (\frac{i}{p}) 4^{i})}^{2}$
		$\equiv - (\frac{- 1}{p}) \frac{4^{p} - 1}{3} + (\frac{- 1}{} p) p (mod 4^{p} - 1) .$

∎

Lemma 3.

For a prime $p$ , if $25 | (4^{p} + 1)$ , then we have $p = 5$ .

Proof.

Since $4^{10} \equiv 1 (mod 25)$ and $4^{i} \equiv - 1 (mod 25)$ has only one solution $i = 5$ in the set ${i | 1 \leq i \leq 9}$ , then from $4^{p} \equiv - 1 (mod 25)$ , we get $p = 5 + 10 k (k \in Z)$ which implies $5 | p$ . Hence we get $p = 5$ . ∎

Lemma 4.

For an odd prime $p$ , we have

p - 1 \sum t = 1 (\frac{2 t}{p}) 4^{2 t} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (\frac{2 t + 1}{p}) 4^{2 t + 1} \equiv ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} (mod 4^{p} - 1) 0 (mod 4^{p} + 1) \end{matrix}

Proof.

From

p - 1 \sum t = 1 (\frac{2 t}{p}) 4^{2 t} \equiv p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} \equiv p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (\frac{2 t + 1}{p}) 4^{2 t + 1} (mod 4^{p} - 1)

we get

p - 1 \sum t = 1 (\frac{2 t}{p}) 4^{2 t} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (\frac{2 t + 1}{p}) 4^{2 t + 1} \equiv 2 p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} (mod 4^{p} - 1) .

Since

\frac{p - 1}{2} \sum t = 1 (\frac{2 t}{p}) 4^{2 t} \equiv - p - 1 \sum t = \frac{p - 3}{2} + 2 (\frac{2 t + 1}{p}) 4^{2 t + 1} (mod 4^{p} + 1)

p - 1 \sum t = \frac{p - 1}{2} + 1 (\frac{2 t}{p}) 4^{2 t} \equiv - \frac{p - 3}{2} \sum t = 0 (\frac{2 t + 1}{p}) 4^{2 t + 1} (mod 4^{p} + 1)

then the rest result follows from

p - 1 \sum t = 1 (\frac{2 t}{p}) 4^{2 t}

= \frac{p - 1}{2} \sum t = 1 (\frac{2 t}{p}) 4^{2 t} + p - 1 \sum t = \frac{p - 1}{2} + 1 (\frac{2 t}{p}) 4^{2 t}

and

p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (\frac{2 t + 1}{p}) 4^{2 t + 1} = \frac{p - 3}{2} \sum t = 0 (\frac{2 t + 1}{p}) 4^{2 t + 1} + p - 1 \sum t = \frac{p - 3}{2} + 2 (\frac{2 t + 1}{p}) 4^{2 t + 1} .

∎

Now we study the 4-adic complexity of the quaternary sequence $g^{1}$ in Definition 1.

Theorem 5.

For the quaternary sequence $g^{1}$ in Definition 1, we have

Φ_{4} (g^{1}) = ⎧ ⎨ ⎩ \begin{matrix} {log}_{4} \frac{4^{2 p} - 1}{} 15, & if 5 | (p + 2) {log}_{4} \frac{4^{2 p} - 1}{3}, & else . \end{matrix}

Proof.

(i) Firstly, we prove

gcd (g^{1} (4), 4^{p} - 1) = 3.

Let the symbols be the same as before. Then we get

$g^{1} (4)$	$= 2 p - 1 \sum t = 0 ϕ (s_{t}^{0}, s_{t}^{1}) 4^{t}$
	$= p - 1 \sum t = 0 ϕ (b_{2 t}, b_{2 t}) 4^{2 t} + p - 1 \sum t = 0 ϕ (c_{2 t + 1}, 1 - c_{2 t + 1}) 4^{2 t + 1}$
	$= p - 1 \sum t = 0 [2 b_{2 t} - b_{2 t} (b_{2 t} - 1) - (b_{2 t} - 1) b_{2 t}] 4^{2 t} + p - 1 \sum t = 0 [2 c_{2 t + 1} - c_{2 t + 1} (- c_{2 t + 1})$
	$- (c_{2 t + 1} - 1) (1 - c_{2 t + 1})] 4^{2 t + 1} (by (???))$
	$= p - 1 \sum t = 0 [2 b_{2 t} - {(b_{2 t})}^{2} + b_{2 t} - {(b_{2 t})}^{2} + b_{2 t}] 4^{2 t} + p - 1 \sum t = 0 [2 c_{2 t + 1} + (c_{2 t + 1})^{2} + 1 - 2 c_{2 t + 1} + (c_{2 t + 1})^{2}] 4^{2 t + 1}$
	$= p - 1 \sum t = 0 2 b_{2 t} 4^{2 t} + p - 1 \sum t = 0 (2 c_{2 t + 1} + 1) 4^{2 t + 1} (since a^{2} = a (0 \leq a \leq 1))$
	$= 2 b_{0} 4^{0} + p - 1 \sum t = 1 2 b_{2 t} 4^{2 t} + (2 c_{p} + 1) 4^{p} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (2 c_{2 t + 1} + 1) 4^{2 t + 1}$
	$= p - 1 \sum t = 1 2 b_{2 t} 4^{2 t} + 3 \cdot 4^{p} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (2 c_{2 t + 1} + 1) 4^{2 t + 1} (since% b_{0} = 0, c_{0} = c_{p} = 1)$
	$= p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} 4^{2 t + 1} + 3 \cdot 4^{p} + p - 1 \sum t = 1 2 b_{2 t} 4^{2 t} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} 2 c_{2 t + 1} 4^{2 t + 1}$
	$= 4 p - 1 \sum t = 0 4^{2 t} - 4^{p} + 3 \cdot 4^{p} + 2 p - 1 \sum t = 1 \frac{1 - (\frac{2 t}{p})}{2} 4^{2 t} + 2 p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} \frac{1 - (\frac{2 t + 1}{p})}{2} 4^{2 t + 1}$
	$\equiv ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 9 p - 1 \sum t = 0 4^{2 t} - 2 p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} (mod 4^{p} - 1) 9 p - 1 \sum t = 0 4^{2 t} - 2 (mod 4^{p} + 1) . \end{matrix} (by Lemma ???)$	(3)

Since $3 | (4^{p} - 1$ ), then from (3) we know

g1(4)≡p−1∑t=1(tp)4t≡0(mod3) \bigg{(}since p−1∑t=1(tp)=0\bigg{)}.

It then follows that $3 | gcd (g^{1} (4), 4^{p} - 1)$ . If $9 | (4^{p} - 1)$ , then from $4^{3} \equiv 1 (mod 9)$ and $4^{p} \equiv 1 (mod 9)$ , we get $p = 3$ which contradicts with $p \equiv 1 (mod 4)$ . Therefore $9 ∤ gcd (g^{1} (4), 4^{p} - 1)$ .

Assume that $d_{1}$ is a prime divisor of $gcd (g^{1} (4), 4^{p} - 1)$ such that $d_{1} \neq 3$ . By (3) we get

g^{1} (4) \equiv - 2 p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} (mod d_{1}) .

Then we have $d_{1} ∣ ∣ {(p - 1 \sum t = 1 (\frac{t}{p}) 4^{t})}^{2}$ . Combining with Lemma 2, we have $d_{1} ∣ p$ which implies $d_{1} = p$ . Hence, we have $4^{p} \equiv 1 (mod p)$ . By Fermat’s little Theorem, we get $4^{p - 1} \equiv 1 (mod p)$ . Then we have $p | (p - 1)$ which is a contradiction. Hence, we know $d_{1} = 1$ .

Therefore we get

gcd (g^{1} (4), 4^{p} - 1) = 3.

(4)

(ii) Next, we prove

gcd (g^{1} (4), 4^{p} + 1) = {\begin{matrix} 5, & if 5 | (p + 2) 1, & else . \end{matrix}

By (3) we have

g^{1} (4) \equiv - p - 2 (mod 5) .

Then we get $5 | gcd (g^{1} (4), 4^{p} + 1)$ only when $5 | (p + 2)$ .

Assume that $5 | (p + 2)$ and $25 | (4^{p} + 1)$ , then by Lemma 3 we get $p = 5$ which contradicts with $5 | (p + 2)$ . It then follows that $25 ∤ gcd (g^{1} (4), 4^{p} + 1)$ .

Let $d_{2}$ be a divisor of $gcd (g^{1} (4), 4^{p} + 1)$ such that $5 ∤ d_{2}$ . Then from (3) we have $g^{1} (4) \equiv - 2 (mod d_{2}) .$ Thus $d_{2} ∣ 2$ which implies $d_{2} = 1.$ Therefore

gcd (g^{1} (4), 4^{p} + 1) = {\begin{matrix} 5, & if 5 | (p + 2) 1, & else . \end{matrix}

(5)

Combining with (4-5) and the definition of the 4-adic complexity, the result is proven. ∎

The 4-adic complexity of the sequence $g^{2}$ in Definition 2 is given by the following theorem.

Theorem 6.

For the quaternary sequence $g^{2}$ in Definition 2, we have

Φ_{4} (g^{2}) = {\begin{matrix} {log}_{4} \frac{4^{2 p} - 1}{} 5, & if 5 | (p - 2) {log}_{4} (4^{2 p} - 1), & else . \end{matrix}

Proof.

(i) Firstly, we determine the exact value of $gcd (g^{2} (4), 4^{p} - 1)$ .

Assume that the symbols are the same as before. Then we have

$g^{2} (4)$	$= 2 p - 1 \sum t = 0 ϕ (s_{t}^{2}, s_{t}^{3}) 4^{t}$
	$= p - 1 \sum t = 0 ϕ (b_{2 t}, c_{2 t}) 4^{2 t} + p - 1 \sum t = 0 ϕ (b_{2 t + 1}, 1 - c_{2 t + 1}) 4^{2 t + 1}$
	$= p - 1 \sum t = 0 [2 b_{2 t} - b_{2 t} (c_{2 t} - 1) - (b_{2 t} - 1) c_{2 t}] 4^{2 t} + p - 1 \sum t = 0 [2 b_{2 t + 1} + b_{2 t + 1} c_{2 t + 1}$
	$- (b_{2 t + 1} - 1) (1 - c_{2 t + 1})] 4^{2 t + 1} (by (???))$
	$= (2 \times 0 - 0 \times (1 - 1) - (0 - 1) \times 1) \times 4^{0} + p - 1 \sum t = 1 [2 b_{2 t} - b_{2 t} c_{2 t} + b_{2 t} - b_{2 t} c_{2 t} + c_{2 t}] 4^{2 t}$
	$+ [2 b_{p} + b_{p} c_{p} - (b_{p} - 1) (1 - c_{p})] 4^{p} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} [2 b_{2 t + 1} + b_{2 t + 1} c_{2 t + 1} + {(b_{2 t + 1})}^{2} + 1 - 2 b_{2 t + 1}] 4^{2 t + 1}$
	$= 1 + p - 1 \sum t = 1 2 b_{2 t} 4^{2 t} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} (2 b_{2 t + 1} + 1) 4^{2 t + 1} (since% a^{2} = a (0 \leq a \leq 1))$
	$= 1 + p - 1 \sum t = 1 2 b_{2 t} 4^{2 t} + 4 p - 1 \sum t = 0 4^{2 t} - 4^{2 \cdot \frac{p - 1}{2} + 1} + p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} 2 b_{2 t + 1} 4^{2 t + 1}$
	$= 2 p - 1 \sum t = 1 \frac{1 - (\frac{2 t}{p})}{2} 4^{2 t} + 2 p - 1 \sum \begin{matrix} t = 0 t \neq \frac{p - 1}{2} \end{matrix} \frac{1 - (\frac{2 t + 1}{p})}{2} 4^{2 t + 1} + 1 - 4^{p} + 4 p - 1 \sum t = 0 4^{2 t}$
	$\equiv ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 9 p - 1 \sum t = 0 4^{2 t} - 2 p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} - 2 (mod 4^{p} - 1) 9 p - 1 \sum t = 0 4^{2 t} + 2 (mod 4^{p} + 1) . \end{matrix} (by % Lemma ???)$	(6)

Since $3 | (4^{p} - 1)$ , then by (6) we know

g^{2} (4) \equiv 1 (mod 3) .

Then we get $3 ∤ gcd (g^{2} (4), 4^{p} - 1)$ .

Let $d_{3}$ be a prime divisor of $gcd (g^{2} (4), 4^{p} - 1)$ . From (6) and $d_{3} \neq 3$ we have

g^{2} (4) \equiv - 2 (p - 1 \sum t = 1 (\frac{t}{p}) 4^{t} + 1) (mod d_{3}) .

Then by Lemma 2 and $p \equiv 3 (mod 4)$ , we have

1 \equiv (p - 1 \sum t = 1 (\frac{t}{p}) 4^{t})^{2} \equiv - p + \frac{4^{p} - 1}{3} \equiv - p (mod d_{3}) .

Hence, we have $d_{3} | (p + 1)$ . From $4^{d_{3} - 1} \equiv 1 (mod d_{3})$ and $4^{p} \equiv 1 (mod d_{3})$ we get $p | (d_{3} - 1)$ which is a contradiction. Therefore

gcd (g^{2} (4), 4^{p} - 1) = 1.

(7)

(ii) Now we determine $gcd (g^{2} (4), 4^{p} + 1)$ .

Since $5 ∣ (4^{p} + 1)$ , then by (6) we get

g^{2} (4) \equiv - p - 1 \sum t = 0 4^{2 t} + 2 \equiv - p + 2 (mod 5) .

Hence we get $5 | gcd (g^{2} (4), 4^{p} + 1)$ only when $5 | (p - 2)$ .

Assume that $5 | (p - 2)$ and $25 | (4^{p} + 1)$ , then from Lemma 3 we have $p = 5$ which contradicts with $5 | (p - 2)$ . It then follows that $25 ∤ gcd (g^{2} (4), 4^{p} + 1)$ . Assume that $d_{4}$ is a prime divisor of $g^{2} (4)$ and $4^{p} + 1$ such that $d_{4} \neq 5$ . Then by (6) we get

g^{2} (4) \equiv 2 (mod d_{4})

which implies $d_{4} = 1$ . Hence we have

gcd (g^{2} (4), 4^{p} + 1) = {\begin{matrix} 5, & if 5 | (p - 2) 1, & otherwise . \end{matrix}

(8)

Combining with (7-8) and the definition of the 4-adic complexity, the result is proven. ∎

The 4-adic complexity of the sequence $g^{3}$ with period $2^{n + 1} - 2$ in Definition 3 is given as follows.

Theorem 7.

For the quaternary sequence $g^{3}$ with period $2^{n + 1} - 2$ in Definition 3, we have

Φ (g^{3}) = {log}_{4} 5 \cdot (4^{2^{n} - 1} - 1) .

Proof.

With the symbols the same as before, we have

	$g^{3} (4)$	$= 2^{n + 1} - 3 \sum t = 0 ϕ (u_{t}, v_{t}) 4^{t}$
		$= 2^{n + 1} - 3 \sum \begin{matrix} t = 0 t \in {1} \times {¯ ¯¯ ¯ D}_{0} \end{matrix} 4^{t} + 2^{n + 1} - 3 \sum \begin{matrix} t = 0 t \in {0} \times D_{0} \end{matrix} 2 \cdot 4^{t} + 2^{n + 1} - 3 \sum \begin{matrix} t = 0 t \in {1} \times D_{0} \end{matrix} 3 \cdot 4^{t}$
		$= 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 ∤ t, s_{t} = 0 \end{matrix} 4^{t} + 2 \cdot 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 \| t, s_{t} = 1 \end{matrix} 4^{t} + 3 \cdot 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 ∤ t, s_{t} = 1 \end{matrix} 4^{t}$
		$= 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 ∤ t \end{matrix} (1 - s_{t}) 4^{t} + 2 \cdot 2^{n + 1} - 3 \sum \begin{matrix} c t = 0 2 \| t \end{matrix} s_{t} \cdot 4^{t} + 3 \cdot 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 ∤ t \end{matrix} s_{t} \cdot 4^{t}$
		$= 2^{n + 1} - 3 \sum \begin{matrix} t = 0 2 ∤ t \end{matrix} 4^{t} + 2 2^{n + 1} - 3 \sum \begin{matrix} t = 0 \end{matrix} s_{t} \cdot 4^{t}$
		$<$

On the 4-Adic Complexity of Quaternary Sequences with Ideal Autocorrelation

Authors

The linear complexity of new binary cyclotomic sequences of period p^n

The 2-Adic Complexity of Two Classes of Binary Sequences with Interleaved Structure

Computing the 2-adic complexity of two classes of Ding-Helleseth generalized cyclotomic sequences of period of twin prime products

Binary sequences derived from differences of consecutive quadratic residues

Sumsets of Wythoff Sequences, Fibonacci Representation, and Beyond

Further results on the 2-adic complexity of a class of balanced generalized cyclotomic sequences

Sorting Permutations with Fixed Pinnacle Set

I Introduction

Ii Preliminaries

Definition 1.

Definition 2.

Definition 3.

Definition 4.

Iii Main result

Lemma 1.

Lemma 2.

Proof.

Lemma 3.

Proof.

Lemma 4.

Proof.

Theorem 5.

Proof.

Theorem 6.

Proof.

Theorem 7.

Proof.

On the 4-Adic Complexity of Quaternary Sequences with Ideal Autocorrelation

Authors

Related Research

The linear complexity of new binary cyclotomic sequences of period p^n

The 2-Adic Complexity of Two Classes of Binary Sequences with Interleaved Structure

Computing the 2-adic complexity of two classes of Ding-Helleseth generalized cyclotomic sequences of period of twin prime products

Binary sequences derived from differences of consecutive quadratic residues

Sumsets of Wythoff Sequences, Fibonacci Representation, and Beyond

Further results on the 2-adic complexity of a class of balanced generalized cyclotomic sequences

Sorting Permutations with Fixed Pinnacle Set

This week in AI

I Introduction

Ii Preliminaries

Definition 1.

Definition 2.

Definition 3.

Definition 4.

Iii Main result

Lemma 1.

Lemma 2.

Proof.

Lemma 3.

Proof.

Lemma 4.

Proof.

Theorem 5.

Proof.

Theorem 6.

Proof.

Theorem 7.

Proof.