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where S denote complementation of S in
. We can now characterize the AB
exponents simply by these cardinalities: An exponent d is AB if and only if
H d
F
αH ∈{
2 m− 2 , 2 m− 2
2 ( m− 3) / 2
F .
±
}
,
α
(4)
2 Computing
Θ d (1)
Our main result in this section is the following theorem.
Theorem 1. Let
F
=
F 2 m , m odd, and let d be an AB exponent in
F
such that
x d
d is also AB in
F 2 k for all k
|
m , that is the function f :
F 2 k F 2 k ,x
is
AB. Then:
Θ d (1) = +2 ( m +1) / 2
if m
≡±
1(mod8) ,
(5)
2 ( m +1) / 2
if m
≡±
3(mod8) .
Recently Lahtonen et. al. [1] proved the following theorem:
Theorem 2. [1] Let
F
=
F 2 m ,and d
Gold Kasami be an AB exponent in
F
.
Then Θ d (1) satisfies (5).
Actually the Gold case is due to Dillon and Dobbertin [3]. Also we should men-
tion that when d is Gold, Θ d (1) determines all the values in the crosscorrelation
spectrum of d , as shown in [1].
The following theorem proves that (5) is true for all AB exponents d in
F 2 p ,
where p is an odd prime. This will also act as a basis of a further inductive
generalization.
Theorem 3. Let
F
=
F 2 m , m = p be an odd prime, and let d be an AB exponent
.Then Θ d (1) satisfies (5).
Proof. Since p is a prime, c 1 =2and c p =2 p
F
in
2. Now, H and H d
are combi-
nations of cosets (since Tr ( α )= Tr α 2 ,and x d is a permutation which maps
cyclotomic cosets to cyclotomic cosets), and consist of c p / 2 p cosets of cardinality
p and C 0 =
.Then H
H d = lp +1 , where 0
{
0
}
l
c p / 2 p .Nowwehave
c p
2
+1=2 p− 1 ,and lp +1
2 p− 2 , 2 p− 2
2 ( p− 3) / 2
.But lp +1=2 p− 2 leads to
∈{
±
}
a quick contradiction since
1= c p
2
2 lp +2=2 p
+1
2 l )= p ( 2 p
1= p ( c p
2
2 p
)
2 p
2 p
2
2 p
is impossible (note that
is an integer). When we analyse the other possi-
bilities we see that
lp +1=2 p− 2
2 ( p− 3) / 2
±
2 lp +2= c p
2
2 ( p− 1) / 2
+1
±
2 ( p− 1) / 2 = p ( 2 p
1= p ( c p
2
2 ( p− 1) / 2 .
2 p
2 l )
±
)
±
2 p
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