A Class of Nonbinary Codes and Sequence Families - Sequences and Their Applications-SETA 2008 - page 84

Theorem 3. For n =2 m 4 and any positive integer k satisfying Equation

(2), the weight distribution of the nonbinary Kasami codes

≥

k is given as follows:

C

Table 2. The weight distribution of code C k

Weight Frequency

0 1

( p − 1) p n − 1 ( p n − 1)(1 + p n − 2 d ( p m + d − p m + p m − 1 + p m − d − 1))

n − 2

2 )

n − 2

2 ) / (2 p d +2)

( p − 1)( p n − 1 − p p d ( p m +1)( p n − 1)( p n − 1 +( p − 1) p

n − 2
2 )

n + d n d n − 2

2 )

p

− 2 p

+ p

( p − 1)( p n − 1 + p ( p m − 1)( p n − 1 − ( p − 1) p

d

2( p − 1)

n − 2 n − 2

2 ) / (2 p d +2)

( p − 1) p n − 1 + p p d ( p m +1)( p n − 1)( p − 1)( p n − 1 − p

2

n − 2
2

n + d n d n − 2

2 )

( p − 1) p n − 1 − p

p

− 2 p

+ p

( p m − 1)( p − 1)( p n − 1 + p

d

2( p − 1)

n + d − 1 n − d − 1

( p − 1) p n − 1 − p p m − d ( p n − 1)( p − 1)( p n − d − 1 + p ) / 2

2 2

n + d − 1
2

n − d − 1
2

( p − 1) p n − 1 + p p m − d ( p n − 1)( p − 1)( p n − d − 1 − p

) / 2

m − d

n +2 d − 2
2

n n − 2 d − 2

2

( p

− 1)( p − 1)

( p − 1)( p n − 1 + p ( p n − 2 d − 1 − ( p − 1) p ) )

2 d

p − 1

n +2 d − 2 m − d n n − 2 d − 2

( p

− 1)( p − 1)

( p − 1) p n − 1 − p ( p − 1)( p n − 2 d − 1 + p )

2 2

2 d

p

− 1

It will be proven by the techniques developed in the next two sections.

3 Rank Distribution of Quadratic Form Π γ,δ ( x )

This section investigates the rank distribution of the quadratic form Π γ,δ ( x )
defined by (1) for either γ =0or δ =0.

Lemma 4 (Theorems 5.4 and 5.6 of [1]). Let h c ( x )= x p s +1

∈ F p l .

− cx + c , c

Then h c ( x )=0 has either 0 , 1 , 2 ,or p gcd( s,l ) +1 roots in F p l . Further, let N 1

∈ F p l such that h c ( x )=0 has exactly one solution in
F p l ,then N 1 = p l− gcd( s,l ) and if x 0 ∈ F p l is the unique solution of the equation,

denote the number of c

p l − 1

p gcd( s,l )

then ( x 0 − 1) =1 .

− 1

Proposition 5. Let g δ,γ ( y )= δ p n − k y p m − k +1 + γy + δ with γδ =0 ,and d be

defined as in (2). Then

(1) The equation g δ,γ ( y )=0 has either 0 , 1 , 2 ,or p d +1 roots in F p n ;

(2) If y 1 , y 2 ∈ F p n are different solutions of g δ,γ ( y )=0 ,then ( y 1 y 2 ) p n − 1

=1 ;

p d

− 1

p n − 1

p d

(3) If g δ,γ ( y )=0 has exactly one solution y 0 ∈ F p n ,then y

− 1

=1 .

0

γ p m − k +1
δ p m − k ( p m +1) .Then g δ,γ ( y ) = 0 becomes

δ
γ x and c =

Proof. (1) Let y = −

x p m − k +1 − cx + c =0 . (4)