Sequences of bent functions and near-bent functions

Abstract

We introduce infinite sequences of Boolean functions whose terms all are bent functions or all are near-bent functions.

1 Introduction

$\mathbb {F}_{2}$ is the finite field of order 2 and an m-Boolean function is a map from ${\mathbb {F}_{2}^{m}}$ to $\mathbb {F}_{2}$. As usual, in order to benefit from the properties of a finite field we identify the $\mathbb {F}_{2}$-vector space ${\mathbb {F}_{2}^{m}}$ with the finite field $\mathbb {F}_{2^{m}}$.

The Walsh transform (Fourier transform) $\hat {F}$ of an m-Boolean function F is the map from $\mathbb {F}_{2^{m}}$ into $\mathbb {Z}$ defined by:

$$\hat{F}(v)=\sum\nolimits_{x\in\mathbb{F}_{2^{m}}}(-1)^{F(x)+ tr(vx)} $$

where tr is the trace of $\mathbb {F}_{2^{m}}$ over $\mathbb {F}_{2}$ defined by $tr(x)={\sum }_{i=0}^{m-1}x^{2^{i}}$.$\hat {F}(v)$ is called the Fourier coefficient of v. Notation: If $e\in \mathbb {F}_{2^{m}}$ then t _e (x)=t r(e x) where tr is the trace of $\mathbb {F}_{2^{m}}$ over $\mathbb {F}_{2}$.

It is easy to prove that:

$$ \hat{F}(v)=2^{m}-2w(F+t_{v}) $$

(1)

where w denotes the weight of a Boolean function.

F is bent if all its Fourier coefficients are in {−2^m/2,2^m/2}. F is near-bent if all its Fourier coefficients are in {−2^(m+1)/2,0,2^(m+1)/2}.Since the Fourier coefficients are in $\mathbb {Z}$, bent functions exist only when m is even and near-bent functions exist only when m is odd.

If m=2t and if F is a bent function then the dual $\tilde {F}$ of F is the (2t)-Boolean function defined by $\hat {F}(v)=(-1)^{\tilde {F}(v)}2^{t}$ where $\hat {F}$ is the Fourier transform of F. It is well-known and easy to prove that the dual $\tilde {F}$ of a bent function F is a bent function and that the dual of $\tilde {F}$ is F.

Bent functions were introduced by Rothaus in [6] .They are interesting for coding theory, cryptology, sequences and were the topic of a lot of works. See for instance [2, 5] Chap. 14, [7] and [1].

2 Special representation of $\mathbb {F}_{2}^{2t}$

In this paper we describe every (2t)-bent function by means of two (2t−1)-near-bent functions. This approach was already used in [4, 8–10] and for the construction of the famous Kerdock code. (see [3] and [5] Chap.15).

We describe every 2t-Boolean function F by means of two (2t−1)-Boolean functions as follows: we identify the finite field $\mathbb {F}_{2^{2t}}$ with:

$$\mathbb{F}_{2^{2t-1}}\times \mathbb{F}_{2}=\lbrace X=(x,\nu)\mid x\in \mathbb{F}_{2^{2t-1}},\nu \in \mathbb{F}_{2}\rbrace. $$

In this way a (2t)-Boolean function F is now defined by:

$$ \quad F(x,\nu)=(\nu+1)f(x)+\nu g(x) $$

(*)

where f and g are the two (2t−1)-Boolean functions such that

$$f(x)=F(x,0) \text{ and } g(x)=F(x,1). $$

It is easy to check that for every (x,ν) the rigth hand side of (*) is equal to F(x,ν).Conversely, if f and g are any two (2t−1)-Boolean functions then (*) defines a (2t)-Boolean function F and f and g are the restriction of F respectively to $\mathbb {F}_{2^{2t-1}}\times \lbrace 0 \rbrace$ and $\mathbb {F}_{2^{2t-1}}\times \lbrace 1 \rbrace$.

We denote such a function by F=[f,g].

We now characterize the (2t−1)-Boolean functions f and g such that F=[f,g] is a bent function. The next proposition is a special version of a well-known result on the hyperplane section of a support of a bent function. A proof is given in ([9],Proposition 14).

Proposition 1

Let f an g be two (2t−1)-Boolean functions and let $\hat {F}$ and $\hat {g}$ be respectively their Walsh transforms. F=[f,g] is a bent function if and only if:

• (a) f and g are near-bent.

• (b) $\forall a\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(a)\mid +\mid \hat {g}(a)\mid =2^{t}$.

where ∣.∣ denotes the absolute value.

Remark: (b) means that one of $\mid \hat {f}(a)\mid$ and $\mid \hat {g}(a)\mid$ is equal to 2^t and the other one is equal to 0.

3 Sequences of bent functions or near-bent functions

In order to present the main result of this paper we first recall a well-known definition and introduce a new one.

Definition 2

If f is an m-Boolean function and if $\omega \in \mathbb {F}_{2^{m}}$, the derivative of f with respect to ω, denoted by D _ω (f), is the m-Boolean function defined by D _ω (f)(x) = f(x) + f(x + ω).

Definition 3

If f is a (2t−1)-Boolean function then:$\hat {I}_{f}^{0}$ is the indicator of the set $\lbrace x\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(x)=0\rbrace$.$\hat {I}_{f}^{-}$ is the indicator of the set $\lbrace x\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(x)=-2^{t}\rbrace$.$\hat {I}_{f}^{+}$ is the indicator of the set $\lbrace x\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(x)=2^{t}\rbrace$.

In other words: $\hat {I}_{f}^{0}(x)=1$ if and only if $\hat {f}(x)=0$, $\hat {I}_{f}^{-}(x)=1$ if and only if $\hat {f}(x)=-2^{t}$ and $\hat {I}_{f}^{+}(x)=1$ if and only if $\hat {f}(x)=2^{t}$.

We now state the main result.

The proof will be given in Subsection 3.3. Starting from a initial special near-bent function it gives rise to a construction of sequences of bent functions and sequences of near-bent functions..

Theorem 4

We define two sequences of (2t)-Boolean functions and two sequences of (2t−1)-Boolean functions.

• 1) f ₀ is a (2t−1)-near-bent function and $r_{0}\in \mathbb {F}_{2^{2t-1}}$ such that $D_{r_{0}}(\hat {I}_{f_{0}}^{0})=1$.

  The sequence $(F_{i})_{i\in \mathbb {N}}$ is defined by:

  $$\begin{array}{@{}rcl@{}} F_{0}&=&\lbrack f_{0},f_{0}+t_{r_{0}}\rbrack \text{ and for } i\geq 1:\\ F_{i}&=&\lbrack f_{i},f_{i}+t_{r_{i}}\rbrack \text{ with } f_{i}(x)=\hat{I}_{f_{i-1}}^{-}(x)+\hat{I}_{f_{i-1}}^{-}(x+r_{i-1}) \end{array} $$

  where tr(r _i−1 r _i )=1.

• 2) $\bar {f}_{0}=f_{0}$ and $\bar {r}_{0}=r_{0}$ and for i≥1:

  The sequence $(\bar {F}_{i})_{i\in \mathbb {N}}$ is defined by:

  $$\begin{array}{@{}rcl@{}} \bar{F}_{0}&=&F_{0}=\lbrack f_{0},f_{0}+t_{r_{0}}\rbrack \text{ and for } i\geq 1:\\ \bar{F}_{i}&=&\lbrack \bar{f}_{i},\bar{f}_{i}+t_{\bar{r}_{i}}\rbrack \text{ with }\bar{f}_{i}(x)=\hat{I}_{\bar{f}_{i-1}}^{-}(x)+\hat{I}_{\bar{f}_{i-1}}^{+}(x+\bar{r}_{i-1}) \end{array} $$

  where $tr(\bar {r}_{i-1}\bar {r}_{i})=1$.

Results: A) F _i is a bent function for every $i\in \mathbb {N}$.

A’) f _i is a near-bent function for every $i\in \mathbb {N}$.

B) $\quad \bar {F}_{i}$ is a bent function for every $i\in \mathbb {N}$.

B’) $\quad \bar {F}_{i}$ is a near-bent function for every $i\in \mathbb {N}$.

Examples of possible initial near-bent functions f ₀:

We present two special cases.

• 1) Kerdock

  • f ₀=Q _u , with $Q(x)={\sum }_{j=1}^{t-1}tr\big (x^{2^{j}+1}\big )\, ,Q_{e}(x)=Q(ex)$.

  • r ₀=u.

  In this case f ₀ is a Kerdock Bent Function (see [3] and [5]).

• 2) Kasami-Welch

  • $f_{_{0}}(x)=tr(x^{4^{s}-2^{s}+1})$ with 2t−1≢0 mod 3 and 3s≡±1 mod (2t−1), s<t,

  • r ₀=1.

It is proved in [4] that f ₀ is a near-bent function.

3.1 The machinery

In order to establish further properties we need some technical results.

Lemma 5

Let F=[f,g] be a (2t)-bent function and let $\hat {F}$ be the Walsh transform of F.

• a) $\hat {F}(u,0)=\hat {f}(u)+\hat {g}(u)$.

• b) $\hat {F}(u,1)=\hat {f}(u)-\hat {g}(u)$.

• c) If f+g=t _u where t _u (x)=tr(ux) then $\hat {g}(a)=\hat {f}(a+u).$

Proof

See [9], Lemma 13. □

We now introduce a connexion between the dual of a bent function [f ₀,f ₁] and the indicators $\hat {I}_{f}^{0}$, $\hat {I}_{f}^{-}$ and $\hat {I}_{f}^{+}$ .

Theorem 6

Let F=[f,g] be a (2t)-bent function and let $\tilde {F}=\lbrack \tilde {f},\tilde {g}\rbrack$ be its dual function. Then:

• a) $\tilde {f}=\hat {I}_{f}^{-}+\hat {I}_{g}^{-}$.

• b) $\tilde {f}+\tilde {g}=\hat {I}_{f}^{0}$.

• c) $\hat {I}_{f}^{0}+\hat {I}_{g}^{0}=1$.

• d) If f+g=t _u where t _u (x)=tr(ux) then:

  $$\begin{array}{@{}rcl@{}} \tilde{f}(x)&=&\hat{I}_{f}^{-}(x)+\hat{I}_{f}^{-}(x+u) (in\,\,other\,\,words\,\, \tilde{f}=D_{u}(\hat{I}_{f}^{-})).\\ \tilde{g}(x)&=&\hat{I}_{f}^{-}(x)+\hat{I}_{f}^{+}(x+u). \end{array} $$

Proof

Proposition 1 says that one of $\mid \hat {f}(a)\mid$ and $\mid \hat {g}(a)\mid$ is equal to 2^t and the other one is equal to 0.

It follows that every a in $\mathbb {F}_{2^{2t-1}}$ belongs to one of the following sets:

$$\begin{array}{@{}rcl@{}} \mathcal{A}_{1}&=&\lbrace a\in \mathbb{F}_{2^{2t-1}}\mid \hat{f}(a)=-2^{t} \text{ and } \hat{g}(a)=0\rbrace.\\ \mathcal{A}_{2}&=&\lbrace a\in \mathbb{F}_{2^{2t-1}}\mid \hat{f}(a)=0 \text{ and } \hat{g}(a)=-2^{t}\rbrace.\\ \mathcal{A}_{3}&=&\lbrace a\in \mathbb{F}_{2^{2t-1}}\mid \hat{f}(a)=2^{t} \text{ and } \hat{g}(a)=0\rbrace.\\ \mathcal{A}_{4}&=&\lbrace a\in \mathbb{F}_{2^{2t-1}}\mid \hat{f}(a)=0 \text{ and } \hat{g}(a)=2^{t}\rbrace. \end{array} $$

Remark that $\mathcal {A}_{1}$ is the set of elements a of $\mathbb {F}_{2^{2t-1}}$ such that $\hat {f}(a)=-2^{t}$. In other words $\mathcal {A}_{1}$ is the support of $\hat {I}_{f}^{-}$.

Similarly: $\mathcal {A}_{2}$ is the support of $\hat {I}_{g}^{-}$, $\mathcal {A}_{3}$ is the support of $\hat {I}_{f}^{+}$, $\mathcal {A}_{4}$ is the support of $\hat {I}_{g}^{+}$.

The distribution of the Fourier coefficients of a near bent function is well known (see for instance Prop. 4 in [1]). This means that $\mathcal {A}_{i}$ is nonempty for i=1,2,3,4. Furthermore, obviously:

If i≠j then $\mathcal {A}_{i}\cap \mathcal {A}_{j}=\emptyset$ and $\mathcal {A}_{1}\cup \mathcal {A}_{2}\cup \mathcal {A}_{3}\cup \mathcal {A}_{4}=\mathbb {F}_{2^{2t-1}}$.Proof of a):

The definition of the dual of F means that (a,η) is in the support of $\tilde {F}$ if and only if $\hat {F}(a,\eta )=-2^{t}$. Since $\hat {F}(a,0)=\hat {f}(a)+\hat {g}(a)$ (Lemma 5) then:

$\hat {F}(a,0)=-2^{t}$ if $a\in \mathcal {A}_{1}$ or $a\in \mathcal {A}_{2}$. In other words: $\tilde {f}(a)=1$ if and only if $a \in \mathcal {A}_{1}\cup \mathcal {A}_{2}$. Therefore $\mathcal {A}_{1}\cup \mathcal {A}_{2}$ is the support of $\tilde {F}$.

Since $\mathcal {A}_{1}$ is the support of $\hat {I}_{f}^{-}$ and $\mathcal {A}_{2}$ is the support of $\hat {I}_{g}^{-}$ and because these two sets are disjoint then $\mathcal {A}_{1}\cup \mathcal {A}_{2}$ is the support of $\hat {I}_{f}^{-}+\hat {I}_{g}^{-}$. This means $\tilde {f}=\hat {I}_{f}^{-}+\hat {I}_{g}^{-}$.Proof of b):

$\hat {I}_{f}^{0}$ is the indicator of the set $\lbrace x\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(x)=0\rbrace$.

We know that (a,0) is in the support of $\tilde {F}$ if and only if $a \in \mathcal {A}_{1}\cup \mathcal {A}_{2}$.and (a,1) is in the support of $\tilde {F}$ if and only if $a \in \mathcal {A}_{1}\cup \mathcal {A}_{4}$.

Hence the support of $\tilde {F}$ is $\mathcal {A}_{1}\cup \mathcal {A}_{2}$ and the support of $\tilde {g}$ is $\mathcal {A}_{1}\cup \mathcal {A}_{4}$.

Consequently the support of $\tilde {f}+\tilde {g}$ is $\mathcal {A}_{2}\cup \mathcal {A}_{4}$. This set is also the support of $\hat {I}_{f}^{0}$ and this means $\tilde {f}+\tilde {g}=\hat {I}_{f}^{0}$.Proof of c):

$\hat {I}_{f}^{0}$ is the indicator of the set $\lbrace a\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(a)=0\rbrace$.

Then $\mathcal {A}_{2}\cup \mathcal {A}_{4}$ is the support of $\hat {I}_{f}^{0}$.

$\hat {I}_{g}^{0}$ is the indicator of the set $\lbrace a\in \mathbb {F}_{2^{2t-1}}\mid \hat {g}(a)=0\rbrace$. We see that $\mathcal {A}_{1}\cup \mathcal {A}_{3}$ is the support of $\hat {I}_{g}^{0}$.

Since $\mathcal {A}_{2}\cup \mathcal {A}_{4}$ and $\mathcal {A}_{1}\cup \mathcal {A}_{3}$ are disjoint then $\mathcal {A}_{2}\cup \mathcal {A}_{4}\cup \mathcal {A}_{1}\cup \mathcal {A}_{3}$ is the support of $\hat {I}_{f}^{0}+\hat {I}_{g}^{0}$. We know that $\mathcal {A}_{2}\cup \mathcal {A}_{4}\cup \mathcal {A}_{1}\cup \mathcal {A}_{3}=\mathbb {F}_{2^{2t-1}}$ and this proves that $\hat {I}_{f}^{0}+\hat {I}_{g}^{0}=1$.Proof of d):

$\hat {I}_{f}^{0}$ is the indicator of the set $\lbrace x\in \mathbb {F}_{2^{2t-1}}\mid \hat {f}(x)=0\rbrace$.

Now assume f+g=t _u . From the descriptions of $\mathcal {A}_{1}$ and $\mathcal {A}_{2}$, if $a\in \mathcal {A}_{1}$ then $a+u\in \mathcal {A}_{2}$ and if $b\in \mathcal {A}_{2}$ then b=a+u with $a=b+u\in \mathcal {A}_{1}$.

Hence $\mathcal {A}_{2}=\lbrace a+u\mid a\in \mathcal {A}_{1}\rbrace$. We know that the support of $\tilde {F}$ is $\mathcal {A}_{1}\cup \mathcal {A}_{2}$ and $\mathcal {A}_{1}$ is the support of $\hat {I}_{f}^{-}$. It follows that:

$$\tilde{f}(x)= \hat{I}_{f}^{-}(x)+\hat{I}_{f}^{-}(x+u)=D_{u}\hat{I}_{f}^{-}(x). $$

□

3.2 Preliminary results

The next results are important tools for the proof of the main theorems.

The first one is given and proved in [4].

Theorem 7

(McGuire and Leander)

tr is the trace function of $\mathbb {F}_{2^{2t-1}}$ and t _e is defined by t _e (x)=tr(ex).

Let f be a (2t−1)-near-bent function.

[f,f+t _e ] is a bent-function if and only if $D_{e}(\hat {I}_{f}^{0})=1$.

Theorem 8

Let f be a (2t−1)-near-bent function.

Let ω be in $\mathbb {F}_{2^{2t-1}}$ and let 𝜖 be in $\mathbb {F}_{2}$.

$$If\, D_{\omega}f=\epsilon \,\,then\,\, \hat{I}_{f}^{0}=t_{\omega}+\epsilon. $$

Remark: According to the definition of $\hat {I}_{f}^{0}$ this lemma means that if D _ω f=𝜖 then $\hat {f}(x)=0$ if and only if t _ω (x)=1+𝜖.

Proof

$\hat {f}(u)={\sum }_{x\in \mathbb {F}_{2^{2t-1}}}(-1)^{f(x)+ tr(ux)}=2^{2t-1}-2w(f+t_{u})$ where w denotes the weight of a Boolean function.

$$\begin{array}{@{}rcl@{}} \hat{f}(u)=0 \text{ if and only if } w(f+t_{u})=2^{2t-2}.\\ D_{\omega}f=\epsilon \text{ means that } f(x+\omega)=f(x)+\epsilon. \end{array} $$

The transform τ: x↦x+ω is a permutation of $\mathbb {F}_{{ }_{2^{2t-1}}}$ and then preserves the weight of every (2t−1)-Boolean function. Thus:

$$\sharp\lbrace x\mid f(x)+tr(ux)=1\rbrace=\sharp\lbrace x\mid f(x+\omega)+tr(u(x+\omega))=1\rbrace. $$

$$\sharp\lbrace x\mid f(x)+tr(ux)=1\rbrace=\sharp\lbrace x\mid f(x)+\epsilon+tr(ux)+tr(u\omega)=1\rbrace. $$

(E)

If t r(u ω)+𝜖=1 the right hand side of (E) is:

$$\sharp\lbrace x\mid f(x)+tr(ux)=0\rbrace=2^{2t-1}-\sharp\lbrace x\mid f(x)+tr(ux)=1\rbrace. $$

Hence (E) becomes:

$$\sharp\lbrace x\mid f(x)+tr(ux)=1\rbrace=2^{2t-1}-\sharp\lbrace x\mid f(x)+tr(ux)=1\rbrace. $$

In other words w(f+t _u )=2^2t−1−w(f+t _u ) .

If t r(u ω)+𝜖=1 then w(f+t _u )=2^2t−2, which is equivalent to $\hat {f}(u)=0$.

For every 𝜖 the number of u such that t r(u ω)+𝜖=1 is 2^2t−2 and this is also the number of u such that $\hat {f}(u)=0$ (see Prop. 4 in [1]). Then, immediately: $\hat {f}(u)=0$ if and only if t r(u ω)+𝜖=1. This means $\hat {I}_{f}^{0}=t_{\omega }+\epsilon$. □

Theorem 9

Let F=[f,f+t _u ] be a bent function with u in $\mathbb {F}_{2^{2t-1}}$ and let

$\tilde {F}=\lbrack \tilde {f}, \tilde {g} \rbrack$ be its dual. Let r be in $\mathbb {F}_{2^{2t-1}}$.

• 1) $\lbrack \tilde {f}, \tilde {f}+ t_{r}\rbrack$ is a bent function if and only if tr(ur)=1.

• 2) $\lbrack \tilde {g}, \tilde {g}+ t_{r}\rbrack$ is a bent function if and only if tr(ur)=1.

Proof

• 1) Since the dual of $\tilde {F}$ is F then in view of Theorem 6, b),$f+f+t_{u}=\hat {I}_{\tilde {f}}^{0}$ that is $\hat {I}_{\tilde {f}}^{0}=t_{u}$.Now we know from McGuire and Leander (Theorem 7) that $\lbrack \tilde {f}, \tilde {f}+t_{r} \rbrack$ is a bent-function if and only if $D_{r}(\hat {I}_{\tilde {f}}^{0})=1$.

  $$\begin{array}{@{}rcl@{}} D_{r}(\hat{I}_{\tilde{f}}^{0})(x)&=&D_{r}(t_{u})(x)\\ &=& t_{u}(x)+ t_{u}(x+r)\\ &=& tr(ux)+tr(u(x+r))\\ &=& tr(ur). \end{array} $$

  It follows that $\lbrack \tilde {f}, \tilde {f}+ t_{r}\rbrack$ is a bent function if and only if t r(u r)=1.

• 2) According to Theorem 6,c), $\hat {I}_{\tilde {f}}^{0}+\hat {I}_{\tilde {g}}^{0}=1$. We deduce that $D_{r}(\hat {I}_{\tilde {f}}^{0})= D_{r}(\hat {I}_{\tilde {g}}^{0})$, whence $D_{r}(\hat {I}_{\tilde {g}}^{0})(x)=tr(ur)$. As previously it follows that $\lbrack \tilde {g}, \tilde {g}+ t_{r}\rbrack$ is a bent function if and only if t r(u r)=1.

□

Remark 10

From a bent function F = [f,f + t _u ] with u in $\mathbb {F}_{2^{2t-1}}$, we obtain 2^2t−2 − 1 distinct bent functions $\lbrack \tilde {f}, \tilde {f}\,+\, t_{r}\rbrack$ and 2^2t−2 − 1 distinct bent functions $\lbrack \tilde {g}, \tilde {g}+ t_{r}\rbrack$ such that t r(u r) = 1.

3.3 Proof of the main theorem

We are now ready to prove Theorem 4.

Proof

We prove the result by induction.Proof of A):Step 0. For i=0 the Boolean function F ₀ is bent as a consequence of Theorem 7.Step j. Assume that for $j\in \mathbb {N}$: F _j−1=[f _j−1,f _j−1+t _{r j−1}] is bent.

Let $\lbrack \tilde {f}_{j-1},\tilde {g}_{j-1} \rbrack$ be the dual of F _j−1. Applying Theorem 9 to F _j−1, it follows that $\lbrack \tilde {f}_{j-1},\tilde {f}_{j-1}+ t_{r_{j}}\rbrack$ is bent if and only if t r(r _j−1 r _j )=1.

By Theorem 6,d),we know that $\tilde {f}_{j-1}=\hat {I}_{f_{j-1}}^{-}(x)+\hat {I}_{f_{j-1}}^{-}(x+r_{j-1})$.Then $\lbrack \tilde {f}_{j-1},\tilde {f}_{j-1}+ r_{j}\rbrack$ is nothing but F _j and this proves that F _j is bent if and only if t r(r _j−1 r _j )=1.Proof of A’): Proposition 1 implies that f _i is a near-bent function for every $i\in \mathbb {N}$.Proof of B): Step 0. For i=0 the Boolean function $\bar {F}_{0}$ is bent because $\bar {f}_{0}=f_{0}$.Step j. Assume that for $j\in \mathbb {N}$: $\bar {F}_{j-1}=\lbrack \bar {f}_{j-1},\bar {f}_{j-1}+t_{\bar {r}_{j-1}}\rbrack$ is bent.

Let $\lbrack \tilde {\bar {f}}_{j-1},\tilde {\bar {g}}_{j-1} \rbrack$ be the dual of $\bar {F}_{j-1}$. Applying Theorem 9 to $\bar {F}_{j-1}$, it follows that $\lbrack \tilde {\bar {g}}_{j-1},\tilde {\bar {g}}_{j-1}+ t_{\bar {r}_{j}}\rbrack$ is bent if and only if $tr(\bar {r}_{j-1}\bar {r}_{j})=1$.

By Theorem 6,d),we know that $\tilde {\bar {g}}_{j-1}=\hat {I}_{\bar {f}_{j-1}}^{-}(x)+\hat {I}_{\bar {f}_{j-1}}^{+}(x+\bar {r}_{j-1})$. Then$\bar {F}_{j}=\lbrack \tilde {\bar {g}}_{j-1},\tilde {\bar {g}}_{j-1}+ t_{\bar {r}_{j}}\rbrack$ hence $\bar {F}_{j}$ is bent if and only if $tr(\bar {r}_{j-1}\bar {r}_{j})=1$.Proof of B’): Proposition 1 also implies that $\bar {f}_{i}$ is a near-bent function for every $i\in \mathbb {N}$. □

Important Remark

Of course we wish to find distinct bent functions as terms of F. If F _i =[f _i ,f _i +t _{r i} ] is equal to F _l with l<i then according to Remark 10, just replace r _i by any other s _i such that t r(r _i−1 s _i )=1. We have 2^2t−2−1 possibilities. In this way we can expect to find a lot of bent functions as elements of F.

An important question is to know if different terms of a sequence introduced above are inequivalent or not under the action of the affine group of $\mathbb {F}_{2^{2t}}$. This give rise to an open problem.

Open problem

What is the maximum number of inequivalent bent functions as terms of a sequence $(F_{i})_{i\in \mathbb {N}}$ (or $(\bar {F}_{i})_{i\in \mathbb {N}}$)?

4 Conclusion

Starting from special near-bent function we have constructed some seqences of bent fuctions with a lot of distinct terms. This leads to an open questions about the maximum number of non-equivalent bent functions as terms of such a sequence.