Asymptotic bounds on the numbers of certain bent functions

Abstract

Using recent results of Keevash et al. [10] and Eberhard et al. [8] together with further new detailed techniques in combinatorics, we present constructions of two concrete families of generalized Maiorana-McFarland bent functions. Our constructions improve the lower bounds on the number of bent functions in n variables over a finite field ${\mathbb F}_p$ if p is odd and n is odd in the limit as n tends to infinity. Moreover we obtain the asymptotically exact number of two dimensional vectorial Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$ as n tends to infinity.

1 Introduction

Let p be a prime. Let ${\mathbb F}_p$ be the finite field with p elements. For a set A, let |A| denote its cardinality. Let $\ln (\cdot )$ be the natural logarithm function.

Bent functions were first introduced by Rothaus in 1976 [17] over ${\mathbb F}_2$. In 1985, Kumar et al. generalized the notion of bent function to arbitrary finite fields [11]. We prefer to introduce bent functions as a special class of functions, namely, plateaued functions.

For a function $f: {\mathbb F}_p^n \rightarrow {\mathbb F}_p$ and $\alpha \in {\mathbb F}_p^n$, let $\hat{f}: {\mathbb F}_{p^n} \rightarrow \mathbb {C}$ be the Walsh Transform of f at $\alpha$ defined as

$$\begin{aligned} \hat{f}(\alpha )=\sum _{x \in {\mathbb F}_p^n} e^{\frac{2\pi \sqrt{-1}}{p}\left( f(x)-\alpha \cdot x\right) }, \end{aligned}$$

where $\alpha \cdot x$ is the inner product $\alpha _1 x_1+ \cdots +\alpha _n x_n$ of $\alpha =(\alpha _1, \ldots , \alpha _n)$ and $x=(x_1, \ldots ,x_n)$.

Let $0 \le m$ be an integer. We say that f is m-plateaued if

$$\begin{aligned} |\hat{f}(\alpha )| \in \{0,p^{\frac{n+m}{2}}\} \end{aligned}$$

for all $\alpha \in {\mathbb F}_{p^n}$. Here $| \cdot |$ denotes the absolute value in complex numbers¹. Let $\textrm{Supp}(\hat{f})$ denote the subset of ${\mathbb F}_{p^n}$ consisting of $\alpha$ such that $\hat{f}(\alpha ) \ne 0$. The following facts (definitions) are well known (see, for example, [5, 15])

• f is bent if and only if f is 0-plateaued.

• If f is m-plateaued, then $|\textrm{Supp}(\hat{f})|=p^{n-m}$.

It seems we have rather limited knowledge in construction of plateaued functions over an arbitrary finite field (see, for example, [4, 9]). A direct, but still very powerful construction of a strict subclass of plateaued functions is for the class of partially bent functions [3]. If $f:{\mathbb F}_{p^s} \rightarrow {\mathbb F}_p$ is a bent function, then for any integer $m \ge 1$, the function

$$\begin{aligned} \begin{array}{rcl} g: {\mathbb F}_{p^s} \times {\mathbb F}_{p^m} & \rightarrow & {\mathbb F}_p \\ (x,y) & \mapsto & f(x) \end{array} \end{aligned}$$

is a partially bent function and m-plateaued function in $m+s$ many variables over ${\mathbb F}_p$. Moreover, given any affine space $U_1$ of dimension s in ${\mathbb F}_q^{m+s}$, it is easy to modify g to $g_1$ such that $\textrm{Supp}(\hat{g}_1)$ is $U_1$.

Bent functions and plateaued functions are central objects for a variety of topics related to cryptography, coding theory and combinatorics. We refer, for example, to [5, 14, 15] and the references therein for further information.

It is an interesting open problem to count bent functions, even for rather moderate values of n (see, [12, 16, 19]). Hence the asymptotic number of bent functions is a natural and actually difficult problem to consider (see [16] and the references therein).

An important class of bent functions is the class of Maiorana-McFarland bent functions [13] (see also [5] and [15] for further details). Let $n \ge 2$ be an even integer and put $n=2m$. In this case, the class $\mathcal {M}(p,n)$ of Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ is the class given by

$$\begin{aligned} \mathcal {M}(p,n)= \left\{ \begin{array}{l} f: {\mathbb F}_{p^m} \times {\mathbb F}_{p^m} \rightarrow {\mathbb F}_p \;\; \text{ such } \text{ that } \\ f(x_1,x_2)=\pi (x_1) \cdot x_2 + g(x_2) \end{array} \right\} , \end{aligned}$$

(1)

where $\pi$ is a permutation map on ${\mathbb F}_p^m$, $\cdot$ is the standard inner product on ${\mathbb F}_p^m$, and $g: {\mathbb F}_{p^m} \rightarrow {\mathbb F}_p$ is an arbitrary function. Let $n \ge 3$ be an odd integer, p is an odd prime, and put $n=2m+1$. In this case, the class $\mathcal {M}(p,n)$ of Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ is the class given by

$$\begin{aligned} \mathcal {M}(p,n)= \left\{ \begin{array}{l} f: {\mathbb F}_p \times {\mathbb F}_p^m \times {\mathbb F}_p^m \rightarrow {\mathbb F}_p \;\; \text{ such } \text{ that } \\ f(x_0,x_1,x_2)=x_0^2+ \pi (x_1) \cdot x_2 + g(x_2) \end{array} \right\} , \end{aligned}$$

(2)

where $\pi$, $\cdot$ and g are defined as in (1). Here we identify ${\mathbb F}_p^n$ to ${\mathbb F}_p^m \times {\mathbb F}_p^m$ in (1), and we identify ${\mathbb F}_p^n$ to ${\mathbb F}_p \times {\mathbb F}_p^m \times {\mathbb F}_p^m$ in (2).

Let $n \ge 2$ be an integer and let p be a prime. We further assume that n is even if $p=2$. Let $f,g: {\mathbb F}_p^n \rightarrow {\mathbb F}_p$ be two functions. We say that f is extended affine equivalent to g if there exist an affine permutation $A: {\mathbb F}_p^n \rightarrow {\mathbb F}_p$, an affine function $\ell : {\mathbb F}_p^n \rightarrow {\mathbb F}_p$, and a nonzero constant $c \in {\mathbb F}_p^*$ such that $g(x)=cf\left( A(x)\right) + \ell (x)$. Assume that $f,g: {\mathbb F}_p^n \rightarrow {\mathbb F}_p$ are two functions extended equivalent to each other. Then it is well known that f is bent if and only if g is bent. The set $\mathcal {M^\sharp }(p,n)$ of completed Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ is the set given by

$$\begin{aligned} \mathcal {M^\sharp }(p,n)= \left\{ \begin{array}{l} f: {\mathbb F}_p^n \rightarrow {\mathbb F}_p \;\; \text{such that there exists} g \in \mathcal {M}(p,n), \\ \text{ which is extended affine equivalent to f } \end{array} \right\} . \end{aligned}$$

The following are well known (see, for example, [5, 15] and [16]):

• Case n is even:

  $$\begin{aligned} \ln \left| \mathcal {M^\sharp }(p,n) \right| = \frac{n}{2} p^{n/2} \ln (p) \left( 1 + o(1)\right) \end{aligned}$$

  (3)

  as $n \rightarrow \infty$ and n is even.

• Case n is odd:

  $$\begin{aligned} \ln \left| \mathcal {M^\sharp }(p,n) \right| = \frac{n-1}{2} p^{(n-1)/2} \ln (p) \left( 1 + o(1)\right) \end{aligned}$$

  (4)

  as $n \rightarrow \infty$ and n is odd.

Here and throughout the paper $o(\cdot )$ stands for the small o notation as $n \rightarrow \infty$.

The definition of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ requires some details and we present them in Section 2. We find it more appropriate to define generalized Maiorana-McFarland bent functions in Section 2 as we also use the details of their definitions in our arguments in Section 2. Hence we refer the reader to Section 2 for an explicit definition of a generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$. Let $\mathcal {GMM}(p,n)$ denote the family of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ (see also [1] and [6]). Note that the notions of completed Maiorana-McFarland bent functions (see also [5]) and generalized Maiorana-McFarland bent functions are different.

We have the obvious bounds that

$$\begin{aligned} |\mathcal {B}(p,n)| \ge | \mathcal {M^\sharp }(p,n)| \;\; \text{ and } \;\; |\mathcal {B}(p,n)| \ge |\mathcal {GMM}(p,n)|. \end{aligned}$$

(5)

In this paper, we use the family of generalized Maiorana-McFarland bent functions in order to improve the lower bounds on bent functions using (5). When we state that we improve (3) or (4) using the family of generalized Maiorana-McFarland bent functions, we actually mean we improve their implications on the lower bounds on bent functions using the obvious bounds in (5).

In [16], the authors obtain that, if $p=2$, then

$$\begin{aligned} \ln \left( |\mathcal {GMM}(p,n)| \right) \ge \frac{3}{4} np^{n/2} \ln (p) \left( 1 + o(1) \right) \end{aligned}$$

(6)

as $n \rightarrow \infty$ and n is even.

In particular they improve the lower bound in (3) so that the coefficient of the main term $np^{n/2}\ln (p)$ is increased from $\frac{1}{2}$ to $\frac{3}{4}$.

Combining (5) and (6) we obtain an asymptotic lower bound on the number of bent functions over ${\mathbb F}_2$, which is the best known asymptotic lower bound on the number of bent functions over ${\mathbb F}_2$, see [16, Table 1].

In [2], among other results, the authors provided an iterative construction of bent functions over ${\mathbb F}_2$. In [19], the author obtained an asymptotic lower bound on the iterative construction based on [2], which is in [19, Corollary 2]. This bound has type $2^{c2^{n/2}}$, where c is a constant and n is the number of variables. The same bound is indicated in [16, Table 1]. This bound is essentially less then the lower bound obtained from the class of Maiorana-McFarland bent functions (see also [16]). In [20, Theorem 2], there is a lower bound for bent functions obtained from the newest iterative construction of bent functions over ${\mathbb F}_2$. This bound is better than the lower bound for the number of iterative bent functions given in [19]. But note that, again, the bound in [20, Theorem 2] has the same type $2^{c2^{n/2}}$, where c is a constant and n is the number or variables. Therefore, this bound is again less than the lower bound obtained from the class of Maiorana-McFarland bent functions.

The methods of [16] do not generalize to odd characteristic. In this paper we improve (4) and we obtain an asymptotic lower bounds on the number of bent functions in odd n variables over ${\mathbb F}_p$ as $n \rightarrow \infty$ and p is odd.

We construct two families of generalized bent functions using two different methods related to the results of [10] and [8], respectively.

Using results of [10] and further detailed techniques we state our first main result in the following.

Theorem 1.1

Let p be an odd prime. There exists a sequence of odd integers n (moreover $n \equiv 1 \mod 4$), $n \rightarrow \infty$ and a corresponding sequence of families $\mathcal {F}_1(n)$ of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ satisfying

$$\begin{aligned} \ln \left( \left| \mathcal {F}_1(n)\right| \right) \ge \frac{n p^{n/2}}{\sqrt{p}} \left( 1 - \frac{1}{2(p^2-1)}\right) \ln (p)(1 + o(1)) \end{aligned}$$

as $n \rightarrow \infty$.

We present a proof of Theorem 1.1 in Section 3 below.

Remark 1.1

In Theorem 1.1, we improve the lower bound in (4) by increasing the coefficient of the main term $n p^{n/2} \ln (p)$ from $\frac{1}{2\sqrt{p}}$ to $\frac{1}{\sqrt{p}}\left( 1 - \frac{1}{2(p^2-1)}\right)$. Note that if $p=3$, then $\frac{1}{\sqrt{p}}\left( 1 - \frac{1}{2(p^2-1)}\right) =\frac{1}{\sqrt{3}}\frac{15}{16}$. This also gives an improved lower bound in the number of bent functions over ${\mathbb F}_p$ for an odd number of variables n using (5) in the limit as $n \rightarrow \infty$ if $p>3$.

Using results of [8] and further different detailed techniques we state our second main result in the following.

Theorem 1.2

Recall that ${\mathbb F}_3$ is the finite field with 3 elements. There exists a sequence of odd integers $n \rightarrow \infty$ and a corresponding sequence of families $\mathcal {F}_2(n)$ of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_3$ satisfying

$$\begin{aligned} \ln \left( \left| \mathcal {F}_2(n)\right| \right) \ge \frac{n 3^{n/2}}{\sqrt{3}}\ln (3)(1 + o(1)) \end{aligned}$$

as $n \rightarrow \infty$.

We present a proof of Theorem 1.2 in Section 4 below.

Remark 1.2

In Theorem 1.2, we improve the lower bound in Theorem 1.1 (and hence the lower bound in (4)) by increasing the coefficient of the main term $n 3^{n/2} \ln (3)$ from $\frac{1}{\sqrt{3}}\frac{15}{16}$ to $\frac{1}{\sqrt{3}}$. This also gives an improved lower bound in the number of bent functions over ${\mathbb F}_3$ for an odd number of variables n using (5) in the limit as $n \rightarrow \infty$.

Remark 1.3

In Theorem 1.2, our results hold for all sufficiently large odd integers n. However, in Theorem 1.1, our results hold only for sufficiently large odd integers n with $n \equiv 1 \mod 4$.

In order to state our third main result we need to introduce further notation. Let $n \ge 4$ be an even integer. Put $m=n/2$. Let $\mathcal{M}\mathcal{M}(n;1)$ denote the set of Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$ defined as

$$\begin{aligned} \begin{array}{rcl} \displaystyle \mathcal{M}\mathcal{M}(n;1) & = & \displaystyle \left\{ \begin{array}{l} f: {\mathbb F}_2^m \times {\mathbb F}_2^m \rightarrow {\mathbb F}_2: f(x,y)=\pi (x) \cdot y + g(y), \\ \text{ where } \pi \,\text{is a permutation map on}\,{\mathbb F}_2^m, \\ \cdot \text{ is the standard inner product on}\, {\mathbb F}_2^m, \\ \text{ and } g:{\mathbb F}_2^m \rightarrow {\mathbb F}_2 \,\text{is an arbitrary map} \end{array} \right\} . \end{array} \end{aligned}$$

Note that if $f \in \mathcal{M}\mathcal{M}(n;1)$, then it is well known that f is a bent function.

Let $\mathcal{M}\mathcal{M}(n;2)$ denote the set of two dimensional vectorial Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$ defined as

$$\begin{aligned} \begin{array}{rcl} \displaystyle \mathcal{M}\mathcal{M}(n;2) & = & \displaystyle \left\{ \begin{array}{l} (f_1,f_2): \;\; f_1,f_2 \in \mathcal{M}\mathcal{M}(n;1), \\ \text{ the map } af_1+bf_2: {\mathbb F}_2^m \times {\mathbb F}_2^m \rightarrow {\mathbb F}_2, \\ \text{ defined as } x \mapsto af_1(x) + bf_2(x), \text{ is a bent function } \\ \text{ for each } (a,b) \in {\mathbb F}_2 \times {\mathbb F}_2 \setminus \{(0,0)\} \\ \end{array} \right\} . \end{array} \end{aligned}$$

(7)

In other words; $(f_1,f_2) \in \mathcal{M}\mathcal{M}(n,2)$ if and only if $f_1 \in \mathcal{M}\mathcal{M}(n,1)$, $f_2 \in \mathcal{M}\mathcal{M}(n,1)$ and the map

$$\begin{aligned} \begin{array}{rcl} (f_1,f_2): {\mathbb F}_2^m \times {\mathbb F}_2^m & \rightarrow & {\mathbb F}_2 \times {\mathbb F}_2 \\ z & \mapsto & (f_1(z),f_2(z)), \end{array} \end{aligned}$$

where $z=(x,y) \in {\mathbb F}_2^m \times {\mathbb F}_2^m$ is a two dimensional vectorial bent function. Note that there is no two dimensional vectorial Maiorana-McFarland bent function in 2 variables and hence we assume that $n \ge 4$.

Using again results of [8] and further different detailed techniques we state our third main result in the following.

Theorem 1.3

Let $n \ge 4$ be an even integer. Recall that $\mathcal{M}\mathcal{M}(n;2)$ denotes the set of two dimensional vectorial Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$, which is defined in (7). We have

$$\begin{aligned} |\mathcal{M}\mathcal{M}(n;2)|= \frac{\left( 2^{n/2} ! \right) ^3}{2^{\left( 2^{n/2}\left( n/2-2\right) -n/2\right) }} \left( e^{-1/2} + o(1)\right) \end{aligned}$$

(8)

as $n \rightarrow \infty$ and $2 \mid n$. Moreover we have

$$\begin{aligned} \ln |\mathcal{M}\mathcal{M}(n;2)|= n2^{n/2} \ln (2) \left( 1 + o(1)\right) \end{aligned}$$

as $n \rightarrow \infty$ and $2 \mid n$.

We present a proof of Theorem 1.3 in Section 5 below.

The rest of the paper is organized as follows. We explain why we use only partially bent functions in Section 2 below. We complete the proofs of Theorems 1.1, 1.2 and 1.3 in Sections 3, 4 and 5 consequently. We also present related results in these sections.

2 Why do we use only partially bent functions?

In this section we briefly explain why we only use partially bent functions and not arbitrary plateaued functions. Let $s \ge 1$ be an integer. Let $n_1 \ge 1$ be a variable integer which runs and tends to infinity over a sequence. We construct bent functions with $2n_1 + s$ many variables over ${\mathbb F}_p$. Hence our number of variables tends to infinity as $n_1$ tends to infinity.

Let $\mathcal {P}=(A_1, \ldots , A_{p^{n_1}})$ be an ordered partition of ${\mathbb F}_{p^{n_1+s}}$ into subsets of size exactly $p^s$. We will need a huge number of such partitions that we can control.

By control we mean the following. Given such $\mathcal {P}$, we need to design a corresponding ordered set of $n_1$-plateaued functions $(g_1, \ldots , g_{p^{n_1}})$ such that $g_i: {\mathbb F}_{p^{s+n_1}} \rightarrow {\mathbb F}_p$ and

$$\begin{aligned} \textrm{Supp}(\hat{g}_i)=A_i \end{aligned}$$

(9)

for each $1 \le i \le p^{n_1}$.

Let $\phi : {\mathbb F}_{p^{n_1}} \rightarrow \{1,2, \ldots , p^{n_1}\}$ be a fixed bijection. A generalized Maiorana-McFarland bent function in $(2n_1+s)$ variables over ${\mathbb F}_p$ is defined as (see [1, 6])

$$\begin{aligned} \begin{array}{rcl} f: {\mathbb F}_p^{s+n_1} \times {\mathbb F}_p^{n_1} & \rightarrow & {\mathbb F}_p \\ (y,z) & \mapsto & g_{\phi (z)}(y). \end{array} \end{aligned}$$

If $(A_1, \ldots , A_{p^{n_1}})$ and $(B_1, \ldots , B_{p^{n_1}})$ are two distinct ordered partitions of ${\mathbb F}_{p^{n_1+s}}$ into subsets of size exactly $p^s$, i.e. $A_i \ne B_i$ for at least one i, then independent from the corresponding ordered set of $n_1$-plateaued functions (provided they exist), the constructed bent functions $f_A$ and $f_B$ in $(2n_1+s)$ variables are distinct. Moreover assume that we fix an ordered partition $(A_1, \ldots , A_{p^{n_1}})$ of ${\mathbb F}_{p^{n_1+s}}$ into subsets of size exactly $p^s$. Assume also that there are two corresponding ordered set of $n_1$-plateaued functions $(g_1, \ldots , g_{p^{n_1}})$ and $(h_1, \ldots , h_{p^{n_1}})$ such that $g_i, h_i: {\mathbb F}_{p^{s+n_1}} \rightarrow {\mathbb F}_p$ and

$$\begin{aligned} \textrm{Supp}(\hat{g}_i)=\textrm{Supp}(\hat{h}_i)=A_i \end{aligned}$$

(10)

for each $1 \le i \le p^{n_1}$. Then if $g_i \ne h_i$ for some i, then the constructed bent functions $f_g$ and $f_h$ in $(2n_1+s)$ variables are distinct.

An important problem is to have a large number of such partitions $\mathcal {P}$ that we make sure existence of a large number of corresponding ordered sequences of $n_1$-plateaued functions.

We know a sufficiently large number of such partitions using affine subspaces of ${\mathbb F}_{p^{n_1+s}}$ of dimension s. This implies that we use only partially bent functions [3]. It is still not an easy problem to count even this particular subject as $n_1$ tends to infinity. We use methods from [8, 10] together with many new and further techniques to have good asymptotic lower bounds. It seems difficult to improve these asymptotic lower bounds making also use of non partially bent but plateaued functions.

3 Proof of Theorem 1.1 and related results

In this section we prove Theorem 1.1. First we give related results that we use in the proof.

Let $s \ge 1$ be an integer. We consider positive integers $n_1 \rightarrow \infty$ such that $(s_1+1) \mid n_1$. For each such $n_1$, let $\mathfrak {S}(n_1,s_1)$ be the set of spreads in ${\mathbb F}_{p^{n_1+s_1+1}}$ of dimension $(s_1+1)$ over ${\mathbb F}_p$. Recall that

$$\begin{aligned} \mathcal {A}=\{A_1, \ldots , A_N\} \in \mathfrak {S}(n_1,s_1) \end{aligned}$$

if and only if

• $N=\frac{p^{n_1+s_1+1}-1}{p^{s_1+1}-1}$,

• $\dim A_i= s_1+1$ for $1 \le N$,

• $A_i \cap A_j= \{0\}$ for $1 \le i < j \le N$, and

• $\bigcup _{i=1}^N A_i = {\mathbb F}_{p^{n_1+s_1+1}}$.

We fix $N=\frac{p^{n_1+s_1+1}-1}{p^{s_1+1}-1}$ throughout this section.

Let $M_1(n_1,s_1)=|\mathfrak {S}(n_1,s_1)|$ be the number of spreads in ${\mathbb F}_{p^{n_1+s_1+1}}$ of dimension $(s_1+1)$ over ${\mathbb F}_p$. By a result of Keevash et al. [10], we have

$$\begin{aligned} \ln \left( M_1(n_1,s_1)\right) =p^{n_1} (n_1+s_1) s_1 \ln (p) (1+o(1)) \end{aligned}$$

(11)

as $n_1 \rightarrow \infty$ and $(s_1+1) \mid n_1$.

Let $\Gamma \subseteq {\mathbb F}_{p^{n_1+s_1+1}}$ be an affine hyperplane with $0 \not \in \Gamma$. Equivalently, there exists an ${\mathbb F}_p$-linear subspace $\Gamma _0 \subseteq {\mathbb F}_{p^{n_1+s_1+1}}$ of dimension $(n_1+s_1)$ and $t_0 \in {\mathbb F}_{p^{n_1+s_1+1}} \setminus \{0\}$ such that $\Gamma =t_0 + \Gamma _0=\{t_0 + x: x \in \Gamma _0\}$.

The following lemma is simple but useful. We also fix $M=p^{n_1}$ throughout this section.

Lemma 3.1

We keep the notation and assumptions as above. Let $\mathcal {A} \in \mathfrak {S}(n_1,s_1)$ and put

$$\begin{aligned} \mathcal {C}=\{A \cap \Gamma : A \in \mathcal {A}, \;\; \text{ and } \;\; A \cap \Gamma \ne \emptyset \}. \end{aligned}$$

Then we have the followings:

• 1.) $|\mathcal {C}|=M$. Let $\mathcal {C}=\{C_1, \ldots , C_M\}$ be an enumeration of its elements.

• 2.) For $1 \le i \le M$, we have that $C_i \subseteq {\mathbb F}_{p^{n_1+s_1+1}}$ is an affine space of dimension $s_1$.

• 3.) $C_i \cap C_j = \emptyset$ for $1 \le i < j \le M$.

• 4.) $\bigcup _{i=1}^M C_i = \Gamma$.

Proof

We identify ${\mathbb F}_{p^{n_1+s_1+1}}$ with ${\mathbb F}_p^{n_1+s_1+1}$ for simplicity in this proof. Hence $\Gamma$, elements of $\mathcal {A}$ and elements of $\mathcal {C}$ are all subsets of ${\mathbb F}_p^{n_1+s_1+1}$ throughout this proof.

Let $(a_0, a_1, \ldots , a_{n_1+s_1}) \in {\mathbb F}_p^{n_1+s_1+1}$ and $b \in {\mathbb F}_p$ such that

$$\begin{aligned} \Gamma =\left\{ (x_0, x_1, \ldots , x_{n_1+s_1}) \in {\mathbb F}_p^{n_1+s_1+1}: a_0x_0 + a_1x_1 + \cdots a_{n_1+s_1}x_{n_1+1}=b\right\} . \end{aligned}$$

As $\Gamma$ is an hyperplane, we have $(a_0, a_1, \ldots , a_{n_1+s_1}) \ne (0, \ldots , 0)$. As $(0, \ldots , 0) \not \in \Gamma$, we have $b \ne 0$.

Let $A \subseteq {\mathbb F}_p^{n_1+s_1+1}$ be an $(s_1+1)$-dimensional subspace. There exists an $n_1 \times (n_1+s_1+1)$ matrix D over ${\mathbb F}_p$ such that $\textrm{rank}(D)=n_1$ and

$$\begin{aligned} A=\left\{ (x_0, x_1, \ldots , x_{n_1+s_1}) \in {\mathbb F}_p^{n_1+s_1+1}: D \left[ \begin{array}{c} x_0 \\ x_1 \\ \vdots \\ x_{n_1+s_1+1} \end{array} \right] = \left[ \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} \right] \right\} . \end{aligned}$$

Let $D'$ be the $(n_1+1) \times (n_1+s_1+1)$ matrix over ${\mathbb F}_p$ obtained by adding row $(a_0,a_1, \ldots ,a_{n_1+s_1})$ at the end of D so that

$$\begin{aligned} D'=\left[ \begin{array}{c} D \\ \begin{array}{cccc} a_0 & a_1 & \cdots & a_{n_1+s_1} \end{array} \end{array} \right] . \end{aligned}$$

If $\textrm{rank}D'=\textrm{rank}D$, then $A \cap \Gamma =\emptyset$ as $b \ne 0$.

If $\textrm{rank}D'=\textrm{rank}D+1=n_1+1$, then $(n_1+s_1+1)-\textrm{rank}D'=s_1$ and hence $|A \cap \Gamma | =p^{s_1}$.

Let $L=|\mathcal {C}|$ and $\{C_1, \ldots , C_L\}$ be an enumeration of $\mathcal {C}$. By definition of $\mathcal {C}$, as $\mathcal {A}$ is a spread of ${\mathbb F}_p^{n_1+s_1+1}$ of dimension $s_1+1$ we obtain

$$\begin{aligned} \bigcup _{i=1}^L C_i = \Gamma . \end{aligned}$$

Moreover as $(0, \ldots , 0) \not \in \Gamma$ we have that

$$\begin{aligned} C_i \cap C_j = \emptyset \;\; for 1 \le i < j \le L. \end{aligned}$$

These imply that

$$\begin{aligned} |\bigcup _{i=1}^L C_i|=L|C_1|=Lp^{s_1}=|\Gamma |=p^{n_1+s_1}, \end{aligned}$$

and hence $L=M=p^{n_1}$. This completes the proof.

$\square$

We choose and fix $\mathcal {A}_0 \in \mathfrak {S}(n_1,s_1)$. Let $\mathcal {C}_0=\{A \cap \Gamma : A \in \mathcal {A}_0, \;\; \text{ and } \;\; A \cap \Gamma \ne \emptyset \}$. Let $\mathfrak {S}_0(n_1,s_1) \subseteq \mathfrak {S}(n_1,s_1)$ be the subset of all spreads consisting of $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$ if and only if

$$\begin{aligned} \mathcal {C}_0=\{B \cap \Gamma : B \in \mathcal {B}, \;\; \text{ and } \;\; B \cap \Gamma \ne \emptyset \} \end{aligned}$$

(see Lemma 3.1). Namely $\mathfrak {S}_0(n_1,s_1)$ consists of all spreads $\mathcal {B}$ which give the set $\mathcal {C}_0$, when the elements of $\mathcal {B}$ intersect with $\Gamma$. Note that $\mathcal {A}_0 \in \mathfrak {S}_0(n_1,s_1)$.

Using Lemma 3.1 we enumerate its elements as $\mathcal {C}_0=\{C_1, \ldots , C_M\}$. In particular we fix the sets $C_1, C_2, \ldots , C_M$. For each $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$, we enumerate its elements as $\mathcal {B}=\{B_1,B_2, \ldots , B_M, B_{M+1}, \ldots , B_N\}$ such that

$$\begin{aligned} B_i \cap \Gamma = C_i \;\; \text{ for } \;\; 1 \le i \le M, \end{aligned}$$

and

$$\begin{aligned} B_i \cap \Gamma = \emptyset \;\; \text{ for } \;\; M+1 \le i \le N. \end{aligned}$$

Here we again use Lemma 3.1.

Let $1 \le i \le M$. Note that $B_i$ is an ${\mathbb F}_p$-linear space of dimension $s_1+1$, $B_i \cap \Gamma =C_i$ is an $s_1$-dimensional affine space and $0 \not \in C_i$ as $0 \not \in \Gamma$. These arguments imply that

$$\begin{aligned} B_i=\textrm{Span}C_i=\textrm{Span}\{x: x \in C_i\}. \end{aligned}$$

Consequently if $\mathcal {A}, \mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$, then we have

$$\begin{aligned} A_1=B_1, \; A_2=B_2, \; \ldots \;, A_M=B_M. \end{aligned}$$

Moreover $\mathcal {A} \ne \mathcal {B}$ if and only if

$$\begin{aligned} \{A_j: M+1 \le j \le N\} \ne \{B_j: M+1 \le j \le N\}. \end{aligned}$$

Let $M'(n_1,s_1)$ be the cardinality of $\mathfrak {S}_0(n_1,s_1)$. We obtain an upper bound on $M'(n_1,s_1)$ using some methods on hypergraphs in Proposition 3.2 below. We need some preparations before its statement.

Let $V \subseteq {\mathbb F}_{p^{n_1+s_1+1}}$ be the subset given by

$$\begin{aligned} V={\mathbb F}_{p^{n_1+s_1+1}} \setminus \left\{ \bigcup _{i=1}^M \textrm{Span}C_i\right\} . \end{aligned}$$

(12)

Let $\mathcal {E}$ be the collection of certain subsets of ${\mathbb F}_{p^{n_1+s_1+1}}$ given by

$$\begin{aligned} \mathcal {E}= \bigcup _{\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)} \bigcup _{j=M+1}^N \left\{ B_j \setminus \{0\}\right\} . \end{aligned}$$

(13)

Here, for each spread $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$, we assume an enumeration $\mathcal {B}=\{B_1, \ldots , B_N\}$ such that $B_i \cap \Gamma = C_i$ for $1 \le i \le M$ and $B_j \cap \Gamma =\emptyset$ for $m+1 \le j \le N$.

The following lemma is also useful.

Lemma 3.2

We keep the notation and assumptions as above. We have the followings:

• 1.) $|V|=p^{n_1}-1$.

• 2.) If $E \in \mathcal {E}$, then $E \subseteq V$.

• 3.) If $E \in \mathcal {E}$, then $|E|=p^{s_1+1}-1$.

• 4.) If $v \in V$, then

  $$\begin{aligned} |\left\{ E \in \mathcal {E}: v \in E\right\} | \le \frac{(p^{n_1}-p)(p^{n_1} -p^2) \cdots (p^{n_1}-p^{s_1})}{(p^{s_1+1}-p)(p^{s_1+1} -p^2) \cdots (p^{s_1+1}-p^{s_1})}. \end{aligned}$$

Proof

Note that

$$\begin{aligned} |\bigcup _{i=1}^M \textrm{Span}C_i|= |\bigcup _{i=1}^M A_i|= 1 + M(p^{s_1+1}-1)= 1 + p^{n_1}(p^{s_1+1}-1)=p^{n_1+s_1+1}-p^{n_1}+1, \end{aligned}$$

where we use the fact that $\{A_1, \ldots , A_M\}$ is a subset of the spread $\mathcal {A}_0$ fixed above. Hence we get that

$$\begin{aligned} |V|=p^{n_1+s_1+1} - \left( p^{n_1+s_1+1}-p^{n_1}+1 \right) =p^{n_1}-1. \end{aligned}$$

If $E \in \mathcal {E}$, then there exists $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$ such that $\mathcal {B}=\{B_1, \ldots , B_N\}$ with an enumeration that $B_i=A_i$ for $1 \le i \le M$, and $E=B_j \setminus \{0\}$ for $M_1+1 \le j \le N$. These imply that $E \cap A_i= E \cap \textrm{Span}C_i=\emptyset$ for each $1 \le i \le M$ and hence $E \subseteq V$. Moreover we have $|E|=|B_j|-1=p^{s_1+1}-1$.

Let $v \in V$ and $k=|\{E \in \mathcal {E}: v \in E\}|$. In order to complete the proof, it is enough to prove that the number of $(s_1+1)$-dimensional subspaces U in ${\mathbb F}_{p^{n_1+s_1+1}}$ with the extra properties that $v \in U$ and $(U \setminus \{0\}) \subseteq V$ is at most

$$\begin{aligned} \frac{(p^{n_1}-p)(p^{n_1} -p^2) \cdots (p^{n_1}-p^{s_1})}{(p^{s_1+1}-p)(p^{s_1+1} -p^2) \cdots (p^{s_1+1}-p^{s_1})}. \end{aligned}$$

Put $P_1=v$. Note that $P_1 \ne 0$. Let $P_2 \in V \setminus \textrm{Span}\{P_1\}$. As $|V|=p^{n_1}-1$ we have $|V \setminus \textrm{Span}\{P_1\}|=p^{n_1}-p$. Let $P_3 \in V \setminus \textrm{Span}\{P_1,P_2\}$. Similarly we have $|V \setminus \textrm{Span}\{P_1,P_2\}|=p^{n_1}-p^2$. If U is an $(s_1+1)$ dimensional subspace in ${\mathbb F}_{p^{n_1+s_1+1}}$ with the extra properties that $v \in U$ and $(U \setminus \{0\}) \subseteq V$, then there exist an ordered set of points $(P_2, \ldots , P_{s_1+1})$ such that $v=P_1$, $P_2 \in V \setminus \textrm{Span}\{P_1\}$, $P_3 \in V \setminus \textrm{Span}\{P_1,P_2\}$,..., $P_{s_1+1} \in V \setminus \textrm{Span}\{P_1, \ldots , P_{s_1}\}$ such that $U=\textrm{Span}\{P_1,P_2, \ldots , P_{s_1+1}\}$. Moreover let $(P_2, \ldots , P_{s_1+1})$ be an ordered set of points chosen in this way leading to the $(s_1+1)$-dimensional subspace $U=\textrm{Span}\{P_1,P_2, \ldots , P_{s_1+1}\}$. Then another the ordered set of points $(P'_2, \ldots , P'_{s_1+1})$ chosen in the same way lead to the same U if and only if $P'_2 \in U \setminus \textrm{Span}{P_1}$, $P'_3 \in U \setminus \textrm{Span}\{P_1,P_2\}$, ..., $P'_{s_1+1} \in U \setminus \textrm{Span}\{P_1, \ldots , P_{s_1}\}$. These arguments complete the proof.$\square$

Next we use some methods from hypergraphs. We recall some definitions and results on finite hypergraphs that we use in this paper. Let V be a finite set with $|V|=n \ge 1$. Let $\mathcal {E}$ be a nonempty subset of the power set of V such that $|E| \ge 2$ for each $E \in \mathcal {E}$. We consider $G=(V , \mathcal {E})$ as a hypergraph on n vertices.

• An element of V is called a vertex of G.

• An element of $\mathcal {E}$ is called a hyperedge of G.

• For $v \in V$ and $E \in \mathcal {E}$, we say that v is in E if $v \in E$.

Let $d \ge 2$ be an integer. We say that G is d-uniform if $|E|=d$ for each $E \in \mathcal {E}$.

For each vertex $v \in V$, let the degree $\deg (v)$ of v be the number of hyperedges $E \in \mathcal {E}$ such that $v \in E$. We say that g is with maximum degree k if $k=\max \{\deg (v): v \in V\}$.

A subset $\mathcal {U} \subseteq \mathcal {E}$ of the hyperedges of G is called a perfect matching of G if for each vertex $v \in V$ there exists exactly one edge $E \in \mathcal {U}$.

We are ready to state the following useful result of Taranenko. It follows from Proposition 1 and Theorem 4 of [18].

Proposition 3.1

Assume that G is a hypergraph with n vertices. Assume also that G is d-regular and the maximum degree of G is k. Put

$$\begin{aligned} \mu =\mu (d)=\frac{d^d (d!)^{\frac{1}{d}}}{(d!)^2}. \end{aligned}$$

If $d \mid n$, then the number of perfect matchings of G is at most

$$\begin{aligned} \left( \mu k\right) ^{\frac{n}{d}}. \end{aligned}$$

Moreover $\mu =\mu (d) < 1$ if $d \ge 4$.

Recall that $V \subseteq {\mathbb F}_{p^{n_1+s_1+1}}$ is the set defined in (12). Using Lemma 3.2 we observe that $\mathcal {E}$ defined in (13) is a subset of the power set of V. Let G be the hypergraph

$$\begin{aligned} G=(V , \mathcal {E}) \end{aligned}$$

with the set of vertices V and the set of hyperedges $\mathcal {E}$. Using Lemma 3.2 we conclude that G is a d-regular graph with n vertices, where $n=p^{n_1}-1$ and $d=p^{s_1+1}-1$. Moreover the maximum degree k of G satisfies

$$\begin{aligned} k \le \frac{(p^{n_1}-p)(p^{n_1} -p^2) \cdots (p^{n_1}-p^{s_1})}{(p^{s_1+1}-p)(p^{s_1+1} -p^2) \cdots (p^{s_1+1}-p^{s_1})}. \end{aligned}$$

(14)

Now we are ready to present the next proposition. Recall that $M'(n_1,s_1)$ is the cardinality of $\mathfrak {S}_0(n_1,s_1)$.

Proposition 3.2

We keep the notation and assumptions as above. Let $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$ be an arbitrary spread in $\mathfrak {S}_0(n_1,s_1)$. Depending on $\mathcal {B}$, we consider the subset of hyperedges of G given by

$$\begin{aligned} \left\{ B_j \setminus \{0\}: M+1 \le j \le N \right\} \subseteq \mathcal {E}. \end{aligned}$$

(15)

Here we assume an enumeration $\mathcal {B}=\{B_1, \ldots , B_N\}$ such that $B_i \cap \Gamma = C_i$ for $1 \le i \le M$ and $B_j \cap \Gamma =\emptyset$ for $M+1 \le j \le N$.

Then the subset of the hyperedges given in (15) is a perfect matching of G. Moreover we have

$$\begin{aligned} M'(n_1,s_1) \le p^{n_1s_1 \delta (s_1)}, \end{aligned}$$

where $\displaystyle \delta (s_1)=\frac{p^{n_1}-1}{p^{s_1+1}-1}$.

Proof

We first show that the set of hyperedges in (15) is a perfect matching of G. As $\mathcal {B} \in \mathfrak {S}_0(n_1,s_1)$, by definition we have

$$\begin{aligned} \bigcup _{i=1}^M B_i = \bigcup _{i=1}^M \textrm{Span}C_i. \end{aligned}$$

As $\mathcal {B}$ is a spread of ${\mathbb F}_{p^{n_1+s_1+1}}$, it follows from the definition of V that

$$\begin{aligned} \bigcup _{j=M+1}^N \left( B_j \setminus \{0\}\right) = V. \end{aligned}$$

Moreover $(B_{j_1}\setminus \{0\}) \cap (B_{j_2} \setminus \{0\})=\emptyset$ for $M+1 \le j_1 < j_2 \le N$ as $\mathcal {B}$ is a spread. Hence we complete the proof of the statement that the subset of the hyperedges given in (15) is a perfect matching of G.

Let PM denote the number of perfect matching of G. These arguments imply that

$$\begin{aligned} M'(n_1,s_1)=|\mathfrak {S}_0(n_1,s_1)| \le PM. \end{aligned}$$

Put $n=p^{n_1}-1$, $d=p^{s_1+1}$, and let k be the maximum degree of G. It follows from Lemma 3.2 that

$$\begin{aligned} k \le \frac{(p^{n_1}-p)(p^{n_1} -p^2) \cdots (p^{n_1}-p^{s_1})}{(p^{s_1+1}-p)(p^{s_1+1} -p^2) \cdots (p^{s_1+1}-p^{s_1})}. \end{aligned}$$

(16)

It is clear that we have

$$\begin{aligned} \frac{(p^{n_1}-p)(p^{n_1} -p^2) \cdots (p^{n_1}-p^{s_1})}{(p^{s_1+1}-p)(p^{s_1+1} -p^2) \cdots (p^{s_1+1}-p^{s_1})} \le p^{n_1s_1}. \end{aligned}$$

(17)

As $(s_1+1) \mid n_1$, we have $d \mid n$. Recall that G is a d-regular graph with n vertices, where $n=p^{n_1}-1$ and $d=p^{s_1+1}-1$. Using Proposition 3.1 we conclude that

$$\begin{aligned} PM \le (\mu k)^{\frac{n}{d}} = (\mu k)^{\frac{p^{n_1}-1}{p^{s_1+1}-1}}, \end{aligned}$$

(18)

where $\mu =\mu (d)=\frac{d^d (d!)^{1/d}}{(d!)^2}$. Note that $d=p^{s_1+1}-1 \ge 3^2-1=8$ as p is odd and $s_1 \ge 1$ is an integer. This implies that $\mu < 1$ (see also Proposition 3.1). Moreover combining (16) and (17) we obtain that

$$\begin{aligned} k \le p^{n_1 s_1}. \end{aligned}$$

(19)

Using (18), (19) and the fact that $\mu < 1$ we conclude that

$$\begin{aligned} PM \le p^{n_1s_1 \delta (s_1)}, \end{aligned}$$

where $\delta (s_1)=\frac{n}{d}=\frac{p^{n_1}-1}{p^{s_1+1}-1}$. This completes the proof.

$\square$ Let $M_2(n_1,s_1)$ be the number of ordered affine partitions of $\Gamma$ of dimension $s_1$. Recall that $\mathcal {U}=(U_1, \ldots , U_L)$ is an ordered affine partition of $\Gamma$ of dimension $s_1$ if and only if the followings holds:

• $L=M$,

• $U_i$ is an affine space of dimension $s_1$ for $1 \le i \le M$,

• $U_i \cap U_j = \emptyset$ for $1 \le i < j \le M$,

• $\bigcup _{i=1}^M U_i = \Gamma ,$ and

• the order in $\mathcal {U}$ is important.

Now we are ready to give a lower bound on $M_2(n_1,s_1)$ as $n_1 \rightarrow \infty$ and $(s_1+1) \mid n_1$.

Proposition 3.3

We keep the notation and assumptions as above. We have

$$\begin{aligned} \ln \left( M_2(n_1,s_1) \right) \ge \left( p^{n_1} n_1 (s_1 +1) - \frac{n_1s_1p^{n_1}}{p^{s_1+1}-1} \right) \ln (p)(1+o(1)) \end{aligned}$$

as $n_1 \rightarrow \infty$ and $(s_1+1) \mid n_1$.

Proof

The number of different orderings of a given $p^{n_1}$-tuple consisting of different entries is $p^{n_1}!$. It follows from Stirling’s formula that we have

$$\begin{aligned} \ln (p^{n1}!)= p^{n_1} n_1 \ln (p) (1 + o(1)) \end{aligned}$$

(20)

as $n_1 \rightarrow \infty$.

Using the results Keevash et al. in (11), (20), Proposition 3.2, and the arguments above in this section, we conclude that

$$\begin{aligned} \begin{array}{rcl} \ln \left( M_2(n_1,s_1) \right) & \ge & p^{n_1}(n_1+s_1) s_1 \ln (p) (1+o(1)) + p^{n_1}n_1\ln (p)(1+o(1))-n_1s_1\frac{p^{n_1}-1}{p^{s_1+1}-1}\ln (p) \\ \\ & = & \left( p^{n_1} n_1 s_1 + p^{n_1}n_1 - \frac{n_1s_1 p^{n_1}}{p^{s_1+1}-1}\right) \ln (p)(1+o(1)) \\ \\ & = & \left( p^{n_1} n_1(s_1+1) - \frac{n_1s_1 p^{n_1}}{p^{s_1+1}-1}\right) \ln (p)(1+o(1)). \end{array} \end{aligned}$$

This completes the proof.

$\square$

We recall and fix some further notation and definitions that we use in the rest of this paper. Let $\textrm{Tr}:{\mathbb F}_{p^{n_1+1}} \rightarrow {\mathbb F}_p$ be the trace map given by $\textrm{Tr}(x)=x+x^p + \cdots + x^{p^{n_1}}$. Let $\{\theta _0, \theta _1, \ldots , \theta _{n_1}\}$ be an ordered basis of ${\mathbb F}_{p^{n_1+1}}$ over ${\mathbb F}_p$. We call that $\{w_0,w_1, \ldots , w_{n_1} \}$ is the (ordered) trace dual basis of ${\mathbb F}_{p^{n_1+1}}$ over ${\mathbb F}_p$ corresponding to $\{\theta _0, \theta _1, \ldots , \theta _{n_1}\}$ if

$$\begin{aligned} \textrm{Tr}(w_i \theta _j) = \left\{ \begin{array}{cl} 1 & \text{ if } i=j\text{, } \\ 0 & \text{ if } i \ne j, \end{array} \right. \end{aligned}$$

(21)

for $0 \le i,j \le n_1$

Let $h : {\mathbb F}_{p^{n_1+1}} \rightarrow {\mathbb F}_p$ be an arbitrary map. We define the Walsh transform of h at $\alpha \in {\mathbb F}_{p^{n_1+1}}$ as

$$\begin{aligned} \hat{h}(\alpha )=\sum _{x \in {\mathbb F}_{p^{n_1+1}}} e^{\frac{2 \pi \sqrt{-1}}{p} \left( h(x) - \textrm{Tr}(\alpha x)\right) }. \end{aligned}$$

Note that this definition is equivalent to the definition of Walsh transform in Section 1 as the inner product $<\alpha , x>=\textrm{Tr}(\alpha x)$ on ${\mathbb F}_{p^{n_1+1}}$ is a nondegenerate inner product equivalent to the standard inner product on ${\mathbb F}_p^{n_1+1}$ up to a choice of a pair of trace dual bases.

Now we are ready to complete the proof of Theorem 1.1. First we recall the formulation of Theorem 1.1 for the sake of reader:

“Let p be an odd prime. There exists a sequence of odd integers n (moreover $n \equiv 1 \mod 4$), $n \rightarrow \infty$ and a corresponding sequence of families $\mathcal {F}_1(n)$ of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_p$ satisfying

$$\begin{aligned} \ln \left( \left| \mathcal {F}_1(n)\right| \right) \ge \frac{n p^{n/2}}{\sqrt{p}} \left( 1 - \frac{1}{2(p^2-1)}\right) \ln (p)(1 + o(1)) \end{aligned}$$

as $n \rightarrow \infty$".

Proof of Theorem

1.1 We keep the notation and assumptions as above. Recall that $\Gamma$ is an hyperplane in ${\mathbb F}_{p^{n_1+s_1+1}}$. This implies that the number of ordered affine partitions of ${\mathbb F}_{p^{n_1+s_1}}$ of dimension $s_1$ is the same as the number of ordered affine partitions of $\Gamma$ of dimension $s_1$, which is $M_2(n_1,s_1)$.

Let $s_1=1$. Using Proposition 3.3 we obtain that

$$\begin{aligned} \ln \left( M_2(n_1,1)\right) \ge \left( p^{n_1}2n_1 - \frac{n_1 p^{n_1}}{p^2-1}\right) \ln (p)(1+o(1)) \end{aligned}$$

(22)

as $n_1 \rightarrow \infty$ and $2 \mid n_1$.

Assume that $n_1 \ge 2$ is an integer with $2 \mid n_1$, and for each affine subspace $U \subseteq {\mathbb F}_{p^{n_1+1}}$ of dimension 1, there exists an $n_1$-plateaued function $g : {\mathbb F}_{p^{n_1+1}} \rightarrow {\mathbb F}_p$ such that

$$\begin{aligned} \textrm{Supp}(\hat{g})=U, \end{aligned}$$

(23)

where $\hat{g}: {\mathbb F}_{p^{n_1+1}} \rightarrow \mathbb {C}$ is the Walsh transform of g. We will prove the assumption in (23) at the end.

Let $\mathcal {U}=(U_1,U_2, \ldots , U_{p^{n_1}})$ be an ordered affine partitions of ${\mathbb F}_{p^{n_1+s_1}}$ of dimension 1. Using the assumption in (23) we obtain $n_1$-plateaued function $g_i : {\mathbb F}_{p^{n_1+1}} \rightarrow {\mathbb F}_p$ such that $\textrm{Supp}(\hat{g}_i)=U_i$ for each $1 \le i \le p^{n_1}$. Hence using Maiorana-McFarland bent functions we obtain at least $M_2(n_1,1)$ distinct bent functions in $\mathcal {F}_1(n)$, where $n=2n_1+1$. This implies that

$$\begin{aligned} \ln (|\mathcal {F}_1(n)|) \ge \ln \left( M_2\left( \frac{n-1}{2},1\right) \right) , \end{aligned}$$

(24)

where $n \ge 5$ is an integer with $n \equiv 1 \mod 4$.

Combining (22) and (24) we obtain that

$$\begin{aligned} \begin{array}{rcl} \displaystyle \ln (|\mathcal {F}_1(n)|) & \ge & \displaystyle \left( p^{\frac{n-1}{2}}2\left( \frac{n-1}{2}\right) - \frac{\left( \frac{n-1}{2}\right) p^{\frac{n-1}{2}}}{p^2-1}\right) \ln (p)(1+o(1)) \\ \\ & = & \displaystyle \frac{np^{n/2}}{\sqrt{p}} \left( 1-\frac{1}{2(p^2-1)} \right) \ln (p)(1+o(1)) \end{array} \end{aligned}$$

as $n \rightarrow \infty$ with $n \equiv 1 \mod 4$.

It remains to prove the assumption in (23). Let $U \subseteq {\mathbb F}_{p^{n_1+1}}$ be an affine subspace of dimension 1. There exist $\theta _0,u \in {\mathbb F}_{p^{n_1+1}}$ with $\theta _0 \ne 0$ such that

$$\begin{aligned} U=\theta _0 {\mathbb F}_p + u=\{\theta _0 x + u: x \in {\mathbb F}_p\}. \end{aligned}$$

We extend $\{\theta _0\}$ to a basis $\{\theta _0,\theta _1, \ldots , \theta _{n_1}\}$ of ${\mathbb F}_{p^{n_1+1}}$ over ${\mathbb F}_p$. Let $u_0,u_1, \ldots , u_{n_1} \in {\mathbb F}_p$ such that $u=u_0 \theta _0 + u_1 \theta _1 + \cdots + u_{n_1}\theta _{n_1}$. Note that we also have

$$\begin{aligned} A=\theta _0 {\mathbb F}_p + (u_1 \theta _1 + \cdots + u_{n_1}\theta _{n_1}). \end{aligned}$$

(25)

Let $f:{\mathbb F}_p \rightarrow {\mathbb F}_p$ be the map $f(x)=x^2$. It is well known that f is a bent function as p is odd.

Let $g: {\mathbb F}_{p^{n_1+1}} \rightarrow {\mathbb F}_p$ be the map defined as follows: Let $\{w_0,w_1, \ldots , w_{n_1} \}$ be the trace dual basis of ${\mathbb F}_{p^{n_1+1}}$ over ${\mathbb F}_p$ corresponding to $\{\theta _0, \theta _1, \ldots , \theta _{n_1}\}$ (see (21) for a definition). For $x_0,x_1, \ldots , x_{n_1}$ and $x=x_0 w_0 + x_1 w_1 +\cdots + x_{n_1} w_{n_1}$, let g(x) be defined as

$$\begin{aligned} g(x)=f(x_0) + u_1x_1 + \cdots + u_{n_1}x_{n_1} \end{aligned}$$

using basis $\{w_0, w_1, \ldots , w_{n_1}\}$.

For $\alpha _0,\alpha _1, \ldots ,\alpha _{n_1}$, let $\alpha =\alpha _0 \theta _0 + \alpha _1 \theta _1 + \cdots + \alpha _{n_1} \theta _{n_1}$ and consider

$$\begin{aligned} \begin{array}{rcl} \displaystyle \hat{g}(\alpha ) & = & \displaystyle \sum _{x \in {\mathbb F}_{p^{n_1+1}}} e^{\frac{2 \pi \sqrt{-1}}{p} \left( g(x) - \textrm{Tr}(\alpha x)\right) } \\ \\ & = & \displaystyle \sum _{x_0,x_1, \ldots , x_{n_1} \in {\mathbb F}_p} e^{\frac{2 \pi \sqrt{-1}}{p} \left( g(x) - (\alpha _0 x_0 + \alpha _1 x_1 + \cdots + \alpha _{n_1} x_{n_1} \right) } \\ \\ & = & \displaystyle \sum _{x_0,x_1, \ldots , x_{n_1} \in {\mathbb F}_p} e^{\frac{2 \pi \sqrt{-1}}{p} \left( f(x_0) + (u_1x_1 + \cdots + u_{n_1}x_{n_1}) - (\alpha _0 x_0 + \alpha _1 x_1 + \cdots + \alpha _{n_1} x_{n_1} \right) } \\ \\ & = & \displaystyle \sum _{x_0 \in {\mathbb F}_p} e^{\frac{2 \pi \sqrt{-1}}{p} \left( f(x_0) - \alpha _0x_0\right) } \sum _{x_1, \ldots , x_{n_1} \in {\mathbb F}_p} e^{\frac{2 \pi \sqrt{-1}}{p} \left( (u_1-\alpha _1)x_1 + \cdots + (u_{n_1}-\alpha _{n_1})x_{n_1} \right) } \\ \\ & = & \displaystyle \hat{f}(\alpha _0) S(u_1-\alpha _1,\ldots , u_{n_1}-\alpha _{n-1}), \end{array} \end{aligned}$$

where $S(u_1-\alpha _1,\ldots , u_{n_1}-\alpha _{n-1})=\sum _{x_1, \ldots , x_{n_1} \in {\mathbb F}_p} e^{\frac{2 \pi \sqrt{-1}}{p} \left( (u_1-\alpha _1)x_1 + \cdots + (u_{n_1}-\alpha _{n_1})x_{n_1} \right) }$.

Note that $|\hat{f}(\alpha _0)|=p^{1/2}$ for all $\alpha _0 \in {\mathbb F}_p$ as f is bent. Also we have $S(u_1-\alpha _1,\ldots , u_{n_1}-\alpha _{n_1})=0$ if $(u_1, \ldots , u_{n_1}) \ne (\alpha _1, \ldots , \alpha _{n_1})$. Moreover $S(u_1-\alpha _1,\ldots , u_{n_1}-\alpha _{n_1})=p^{n_1}$ if $(u_1, \ldots , u_{n_1}) = (\alpha _1, \ldots , \alpha _{n_1})$. These arguments prove the assumption in (23). This completes the proof.

$\square$

4 Proof of Theorem 1.2 and related results

In this section we prove Theorem 1.2. First we give related results that we use in the proof.

Let $m \ge 1$ be an integer. Recall that a map

$$\begin{aligned} f: {\mathbb F}_3^m \rightarrow {\mathbb F}_3^m \end{aligned}$$

is called a complete map if and only if

• 1) f is a permutation map, and

• 2) the map $x \mapsto f(x)+x$ is a permutation map.

Note that the map $x \mapsto 2x=x+x$ is also a permutation map on ${\mathbb F}_3^m$. Therefore equivalently, f is a complete map on ${\mathbb F}_3^m$ if and only if item 1) above and

• 2*) the map $x \mapsto 2(f(x)+x)$ is a permutation map

hold, instead of item 2) above.

It is well known that there is a one-to-one correspondence between complete maps on ${\mathbb F}_3^m$ and the transversals of the Latin square obtained by the Cayley table of the elementary abelian group $({\mathbb F}_3^m,+)$. We refer to [7] for further details.

Let $f: {\mathbb F}_3^m \rightarrow {\mathbb F}_3^m$ be a complete map. Let $a \in {\mathbb F}_3^m$. Let $A(f;a) \subseteq {\mathbb F}_3^{m+1}$ be the set consisting of 3 elements defined as

$$\begin{aligned} A(f;a)=\left\{ (a,0),(f(a),1),(2f(a)+2a,2) \right\} . \end{aligned}$$

(26)

We start with a simple and useful lemma.

Lemma 4.1

We keep the notation and assumptions as above. Let $f,g: {\mathbb F}_3^m \rightarrow {\mathbb F}_3^m$ be complete maps. Let $a,b \in {\mathbb F}_3^m$. We have the followings:

• 1) A(f; a) is an affine set of dimension 1 over ${\mathbb F}_3$. Namely there exist $a_0,a_1 \in {\mathbb F}_3^m$ with $a_1 \ne 0$ such that $A(f;a)=a_0 + a_1 {\mathbb F}_3=\{a_0, a_0+a_1, a_0+2a_1\}$.

• 2) If $a \ne b$, then $A(f;a) \ne A(g;b)$ independent from the choice of f and g.

• 3) Let f and g be distinct. Let $a \in {\mathbb F}_3^m$ such that $f(a) \ne g(a)$. Then we have that $A(f;a) \ne A(g;a)$.

• 4) We have $\bigcup _{a \in {\mathbb F}_3^m} A(f;a)={\mathbb F}_3^{m+1}$.

Proof

Let $P_1=(a,0)$, $P_2=(f(a),1)$ and $P_3=(2f(a)+2a,2)$ be three elements in ${\mathbb F}_3^{m+1}$. In order to prove item 1), it is enough to show that

$$\begin{aligned} P_2-P_1=P_3-P_2. \end{aligned}$$

Indeed we have

$$\begin{aligned} P_2-P_1=(f(a)-a,1), \;\; P_3-P_2=(f(a)+2a,1). \end{aligned}$$

As $-a=2a$ in characteristic 3 we complete the proof of item 1).

Next we prove item 2). Assume that $a,b \in {\mathbb F}_3^m$ are distinct elements and $A(f;a)=A(g;b)$. The only elements in A(f; a) and A(g; b) with the last entry 0 are (a, 0) and (b, 0), respectively. As $a \ne b$, we get a contradiction to the assumption that $A(f;a)=A(g;b)$. This completes the proof of item 2).

Now we prove item 3). Assume that f and g are distinct complete maps on ${\mathbb F}_3^m$. Let $a \in {\mathbb F}_3^m$ such that $f(a) \ne g(a)$. Assume that $A(f;a)=A(g;a)$. The only elements in A(f; a) and A(g; b) with the last entry 1 are (f(a), 1) and (g(a), 1), respectively. As $f(a) \ne g(a)$, we get a contradiction to the assumption that $A(f;a)=A(g;a)$. This completes the proof of item 3).

Finally we prove item 4). Note that $|A(f;a)|=3$ and $A(f;a) \subseteq {\mathbb F}_3^{m+1}$ for each $a \in {\mathbb F}_3^m$. Let a and b be distinct elements of ${\mathbb F}_3^m$. We show that $A(f;a) \cap A(f;b) =\emptyset$. Indeed, otherwise one element of A(f; a) is also in A(f; b). As $a\ne b$, we have that $(a,0) \not \in A(f;b)$. As f is a permutation map and $a \ne b$, we have that $f(a) \ne f(b)$ and hence $(f(a),1) \not \in A(f;b)$. As f is a complete map and $a \ne b$, using item 2*) above in the definition of complete maps, we have that $2f(a)+2a \ne 2f(b)+2b$ and hence $(2f(a)+2a,2) \not \in A(f;b)$. These arguments show that $A(f;a) \cap A(f;b) =\emptyset$. Counting the cardinalities we complete the proof of item 4).

$\square$ For a complete map $f: {\mathbb F}_3^m \rightarrow {\mathbb F}_3^m$, using Lemma 4.1, let $\mathcal {A}_f$ be the affine partition of ${\mathbb F}_3^{m+1}$ of dimension 1 given by

$$\begin{aligned} \mathcal {A}_f=\left\{ A(f;a): a \in {\mathbb F}_3^{m+1} \right\} , \end{aligned}$$

(27)

where the affine set A(f; a) is defined in (26). It follows from Lemma 4.1 that we have

$$\begin{aligned} \mathcal {A}_f \ne \mathcal {A}_g \end{aligned}$$

if f and g are distinct complete maps on ${\mathbb F}_3^m$.

The following proposition is due to Eberhard et al. [8], and it is a direct consequence of [8, Theorem 1.2].

Proposition 4.1

Let G be a finite abelian group of odd order N. Then the number of complete mappings of G is $(e^{-1/2}+o(1))N\cdot N!^2/N^N$.

Let $L_1(m+1,1)$ be the number of unordered affine partitions of ${\mathbb F}_3^m$ of dimension 1. Let $L_0(m)$ be the number of complete maps on ${\mathbb F}_3^m$. The arguments above imply that

$$\begin{aligned} L_0(m) \le L_1(m+1,1). \end{aligned}$$

(28)

It follows from Proposition 4.1 that we have

$$\begin{aligned} L_0(m)=\left( e^{-1/2} + o(1)\right) \frac{3^m \left( 3^m!\right) ^2}{\left( 3^m\right) ^{\left( 3^m\right) }} \end{aligned}$$

(29)

as $m \rightarrow \infty$.

Let $L_2(m+1,1)$ be the number of ordered affine partitions of ${\mathbb F}_3^m$ of dimension 1. We obtain a lower bound on $L_2(m+1,1)$ in the following.

Proposition 4.2

We keep the notation and assumptions as above. We have

$$\begin{aligned} \ln \left( L_2(m+1,1) \right) \ge 2m3^m \ln (3)\left( 1 + o(1)\right) \end{aligned}$$

as $m \rightarrow \infty$.

Proof

An ordered affine partition of ${\mathbb F}_3^{m+1}$ of dimension 1 is a tuple consisting of distinct entries of length $3^m$. The number of different orderings of such a tuple is $3^m!$. It follows from Stirling’s formula that we have

$$\begin{aligned} \ln \left( 3^m!\right) = 3^m m \ln (3) \left( 1 + o(1)\right) \end{aligned}$$

(30)

as $m \rightarrow \infty$.

Using (29) we get

$$\begin{aligned} \begin{array}{rcl} \displaystyle \ln \left( L_0(m) \right) & = & \displaystyle m\ln (3) + 2 \cdot 3^m m \ln (3) \left( 1 + o(1)\right) \\ & & \displaystyle -3^m m \ln (3) \\ \\ & = & \displaystyle 3^m m \ln (3) \left( 1 + o(1)\right) \end{array} \end{aligned}$$

(31)

as $m \rightarrow \infty$.

Combining (28) (30) and (31) we complete the proof.$\square$

Remark 4.1

Our arguments in Proposition 4.2 are specific to ${\mathbb F}_3$. In particular the same arguments using complete maps on ${\mathbb F}_p^m$ do not work and we do not have a similar lower bound if p is a prime with $p \ge 5$. This is the reason that we state Theorem 1.2 only for $p=3$.

Now we are ready to complete the proof of Theorem 1.2. First we recall the formulation of Theorem 1.2 for the sake of reader:

“Recall that ${\mathbb F}_3$ is the finite field with 3 elements. There exists a sequence of odd integers $n \rightarrow \infty$ and a corresponding sequence of families $\mathcal {F}_2(n)$ of generalized Maiorana-McFarland bent functions in n variables over ${\mathbb F}_3$ satisfying

$$\begin{aligned} \ln \left( \left| \mathcal {F}_2(n)\right| \right) \ge \frac{n 3^{n/2}}{\sqrt{3}}\ln (3)(1 + o(1)) \end{aligned}$$

as $n \rightarrow \infty$."

Proof of Theorem

1.2 We keep the notation and assumptions as above. Using Proposition 4.2 we obtain that

$$\begin{aligned} \ln L_2(m+1,1) \ge 2m3^m \ln (3)\left( 1 + o(1)\right) \end{aligned}$$

as $m \rightarrow \infty$. Put $n=2m+1$ and let $m \rightarrow \infty$. Using the methods in the proof of Theorem 1.1 above, we obtain at least $L_2(m+1,1)$ distinct bent functions in $\mathcal {F}_2(n)$. This implies that

$$\begin{aligned} \begin{array}{rcl} \displaystyle \ln |\mathcal {F}_2(n)| & \ge & \displaystyle 2 \left( \frac{n-1}{2}\right) 3^{\frac{n-1}{2}} \ln (3) \left( 1 + o(1) \right) \\ \\ & = & \displaystyle \frac{n 3^{n/2}}{\sqrt{3}} \ln (3) \left( 1 + o(1) \right) \end{array} \end{aligned}$$

as $n \rightarrow \infty$ and n is odd. This completes the proof.

5 Proof of Theorem 1.3 and related results

In this section we prove Theorem 1.3. First we give a related result that we use in the proof.

Let $n \ge 4$ be an even integer. Put $m=n/2$. Recall that a map f on ${\mathbb F}_2^m$ is a complete map if both f and the map $x \mapsto f(x)+x$ are permutation maps on ${\mathbb F}_2^m$. Let L(m) be the number of complete maps on ${\mathbb F}_2^m$. Note that there is no complete map on ${\mathbb F}_2$.

The following proposition follows from definitions.

Proposition 5.1

Let $n \ge 4$ be an even integer. Recall that $\mathcal{M}\mathcal{M}(n;2)$ denotes the set of two dimensional vectorial Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$, which is defined in (7). We have

$$\begin{aligned} |\mathcal{M}\mathcal{M}(n;2)|=\left( 2^{n/2}!\right) L(n/2) 2^{\left( 2^{n/2+1}\right) }. \end{aligned}$$

Proof

Let $f_1,f_2 \in \mathcal{M}\mathcal{M}(n,1)$. Put $f_1(x,y)=\pi _1(x) \cdot y + g_1(y)$ and $f_2(x,y)=\pi _2(x) \cdot y + g_2(y)$. Here $g_1,g_2:{\mathbb F}_2^m \rightarrow {\mathbb F}_2$ are arbitrary maps and $\pi _1, \pi _2$ are permutation maps on ${\mathbb F}_2^m$. It follows from the definitions that $(f_1,f_2) \in \mathcal{M}\mathcal{M}(n,2)$ if and only if the map

$$\begin{aligned} \pi _1 + \pi _2 \end{aligned}$$

(32)

is a permutation map on ${\mathbb F}_2^m$. Let $\mu =\pi _2 \circ \pi ^{-1}$. Note that $\mu$ is a permutation map on ${\mathbb F}_2^m$. Composing with $\pi _1^{-1}$ we obtain that the map in (32) is a permutation map if and only if

$$\begin{aligned} x \mapsto x+ \mu (x) \end{aligned}$$

is a permutation map on ${\mathbb F}_2^m$. Consequently $(f_1,f_2) \in \mathcal{M}\mathcal{M}(n,2)$ if and only if $\mu$ is a complete map. There is a one-to-one correspondence between $\mathcal{M}\mathcal{M}(n,2)$ between the set

$$\begin{aligned} \begin{array}{rl} \displaystyle \left\{ (\pi _1, \mu ,g_1,g_2): \right. & \displaystyle \pi _1 \text{ is } \text{ a } \text{ permutation } \text{ map } \text{ on } {\mathbb F}_2^m, \\ & \displaystyle \mu \text{ is } \text{ a } \text{ complete } \text{ map } \text{ on } {\mathbb F}_2^m, \\ & \left. \displaystyle \text{ and } g_1,g_2:{\mathbb F}_2^m \rightarrow {\mathbb F}_2 \text{ are } \text{ arbitrary } \text{ maps } \right\} . \end{array} \end{aligned}$$

(33)

Here $\pi _2=\mu \circ \pi _1$. The number of arbitrary functions $g: {\mathbb F}_2^m \rightarrow {\mathbb F}_2$ is $2^{\left( 2^m\right) }$. Counting the cardinality of the set in (33) we complete the proof.

$\square$

Now we are ready to complete the proof of Theorem 1.3. First we recall the formulation of Theorem 1.3 for the sake of reader:

“Let $n \ge 4$ be an even integer. Recall that $\mathcal{M}\mathcal{M}(n;2)$ denotes the set of two dimensional vectorial Maiorana-McFarland bent functions in n variables over ${\mathbb F}_2$, which is defined in (7). We have

$$\begin{aligned} |\mathcal{M}\mathcal{M}(n;2)|= \frac{\left( 2^{n/2} ! \right) ^3}{2^{\left( 2^{n/2}\left( n/2-2\right) -n/2\right) }} \left( e^{-1/2} + o(1)\right) \end{aligned}$$

(34)

as $n \rightarrow \infty$ and $2 \mid n$. Moreover we have

$$\begin{aligned} \ln |\mathcal{M}\mathcal{M}(n;2)|= n2^{n/2} \ln (2) \left( 1 + o(1)\right) \end{aligned}$$

as $n \rightarrow \infty$ and $2 \mid n$."

Proof of Theorem

1.3 We keep the notation and assumptions as above. Using Proposition 4.1, we obtain that for the number of complete maps on ${\mathbb F}_2^{n/2}$ we have

$$\begin{aligned} L(n/2)=\frac{\left( 2^{n/2}!\right) ^2 2^{n/2}}{2^{\left( \frac{n}{2}2^{n/2}\right) }}\left( e^{-1/2}+o(1)\right) . \end{aligned}$$

(35)

as $n \rightarrow \infty$ and $2 \mid n$. Combining Proposition 5.1 and (35) we complete the proof of (8), which is the same as (34).

Using Stirling’s formula we have

$$\begin{aligned} \ln \left( 2^{n/2}!\right) ^3 = 3 \cdot 2^{n/2} \frac{n}{2} \ln (2) \left( 1 + o(1)\right) \end{aligned}$$

(36)

as $n \rightarrow \infty$ and $2 \mid n$. Using (8) (which is the same as (34)) and (36) we get

$$\begin{aligned} \begin{array}{rcl} \displaystyle \ln \left| \mathcal{M}\mathcal{M}(n;2)\right| & = & \displaystyle \frac{3n}{2} 2^{n/2} \ln (2) \left( 1 + o(1) \right) \\ & & \displaystyle - \left( 2^{n/2}\left( n/2-2\right) -n/2\right) \ln (2) \\ & = & n2^{n/2}\ln (2)\left( 1 + o(1)\right) \end{array} \end{aligned}$$

as $n \rightarrow \infty$ and $2 \mid n$. This completes the proof.

$\square$