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i<
2
n
. Obviously,
ζ
i
+2
n
−
1
=1+
ζ
i
,0
i<
2
n−
1
. Applying
Let
ζ
i
=
μ
(
η
i
), 0
≤
≤
the modified Gray map to the sequences in
, and doubling the
size by the technique proposed in [4] and [8], we obtain the binary family of
Kerdock sequences
{
a
0
,
···
,a
2
n
−
1
}
−
1
V
.
,
2
n
Definition 2.
The binary Family
V
of sequences
{
v
i
,i
=0
,
1
,
···
−
1
}
of
length
2(2
n
−
1)
is defined as
i<
2
n−
1
:
v
i
(
t
)=
tr
1
(
ζ
i
α
t
1
)+
p
(
α
t
1
)
, t
=2
t
1
-
0
≤
tr
1
((1 +
ζ
i
)
α
t
1
+2
n
−
1
)+
p
(
α
t
1
+2
n
−
1
)
,t
=2
t
1
+1
,
(7)
-
2
n−
1
i<
2
n
:
v
i
(
t
)=
tr
1
(
ζ
i−
2
n
−
1
α
t
1
)+
p
(
α
t
1
)+1
, t
=2
t
1
≤
(8)
tr
1
((1 +
ζ
i−
2
n
−
1
)
α
t
1
+2
n
−
1
)+
p
(
α
t
1
+2
n
−
1
)
,t
=2
t
1
+1
.
Remark 1.
When
2
n−
1
i<
2
n
, the sequence
v
i
could be viewed as the modi-
fied Gray map sequence of
a
i
+3 = (
a
i
(
t
)+3 =
Tr
1
((1 + 2
η
i
)
β
t
)+3
,
0
≤
t<
2
n
)
.
From
(4)
, the Gray map of
a
i
(
t
)+3=(
π
(
a
i
(
t
)+3)
,ν
(
a
i
(
t
)+3))
where
≤
π
(
a
i
(
t
)+3)=
tr
1
(
ζ
i
α
t
1
)+
p
(
α
t
1
)+
tr
1
(
α
t
1
)+1
,
ν
(
a
i
(
t
)+3)=
tr
1
(
ζ
i
α
t
1
)+
p
(
α
t
1
)
.
Hence, the expression of the modified Gray map sequence in
(8)
follows from the
fact that
ζ
i
=1+
ζ
i−
2
n
−
1
,
2
n−
1
i<
2
n
.
This connection was firstly observed in [4].
≤
We have the following main result on the correlation distribution of Kerdock
sequences family.
Theorem 6.
Fami ly
V
has the following correlation distribution:
⎧
⎨
2+2
n
+1
,
2
n
times
0
,
−
2
2
n
times
2
2
n−
2
2
n
−
2
,
3
·
−
times
2
2
n−
2
times
2
,
2+2
n
+
2
,
3
2)(2
n−
2
+2
n
−
2
)
times
2
n−
1
(2
n
−
·
−
R
i,j
(
τ
)=
2
n
−
2
,
3
2
n
−
2
)
times
2
n−
1
(2
n
2)(2
n−
2
−
2
−
·
−
−
⎩
2
n
+
2
,
2)(2
n−
2
+2
n
−
2
)
times
−
2
n−
1
(2
n
−
2
2+2
n
+
2
,
2
n
−
2
)
times
2
n−
1
(2
n
2)(2
n−
2
−
−
2
n
+
2
,
2
2
n−
1
(2
n
−
2)
times
2
n
+
2
,
2
2
n−
1
(2
n
−
−
2)
times.
To prove Theorem 6, we need the following lemma.
τ
=2
τ
1
+
τ
2
<
2(2
n
τ
1
<
2
n
Lemma 7.
Let
0
≤
−
1)
where
0
≤
−
1
and
τ
2
=0
,
1
.
Then
