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From there, we can write
r
j
i
ξ
ks
i
.
g
(
X
)=
β
0
+
β
1
X
+
...
+
β
d−
1
X
d
,
where
β
k
=
α
k
i
=1
The condition
g
(
X
) = 0 implies that in particular
β
1
=0.So
r
j
i
ξ
s
i
=0
.
i
=1
Z
2
l
, the above equality im-
Since
ξ
s
1
,ξ
s
2
,...,ξ
s
r
are linearly independent over
plies
j
1
=
j
2
=
...
=
j
r
=0.
5 Partial Period Distributions
In this section we will consider periodic binary sequence of period 2
m
−
1. For
any integer
D
S
D
be a polynomial with non-zero linear term, and
set
c
t
=MSB(Tr(
f
(
ξ
t
)))
,
where
t
=0
,...,n
≥
4, let
f
∈
2,
n
=2
m
.
−
Theorem 5.1.
With notation as above, let
H
be an integer in the range
0
<
k<n−
1
and for any
k
in the range
0
<k<n−
1
, consider the sequence
c
k
,...,c
k
+
H−
1
of length
H
. For any
δ ∈{
0
,
1
}
,let
N
δ
be the number of
δ
in
c
k
,...,c
k
+
H−
1
. Then we have
≤
2
l
π
ln(2) + 1
2
+1
D
√
2
m
.
π
ln
4(2
m
H
2
1
2
−
1)
N
δ
−
π
Proof.
Following [10], we have
k
+
H
−
1
1
2
(1 + (
1)
c
j
+
δ
)
.
N
δ
=
−
j
=
k
Thus
k
+
H
−
1
1)
δ
2
H
2
(
−
1)
c
j
.
N
δ
−
=
(
−
j
=
k
As we have
c
t
=MSB(Tr(
f
(
ξ
t
))) where
t
ranges between 0 and
n
−
2andby
1)
c
t
is equal to:
(1), we obtain that (
−
2
l
2
l
−
1
−
1
μ
(Tr(
f
(
ξ
t
))) =
μ
j
ψ
j
(Tr(
f
(
ξ
t
))) =
μ
j
Ψ
j
(
f
(
ξ
t
))
.
j
=0
j
=0
Changing the order of summation, we obtain that:
2
l
k
+
H−
1
−
1
k
+
H−
1
1)
c
t
Ψ
j
(
f
(
ξ
t
))
.
(
−
=
μ
j
t
=
k
j
=0
t
=
k
