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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