12 PASCALE CHARPIN
Proof: In accordance with Theorem 2, ¢n(x
Oj(Y))
= 0 if and only if R(ai) = 0.
Moreover, by hypothesis, ¢n-i(Y) = 0 for every i such that wp(i) = r unless
n-iEU. Since U={s
1, ...
,s1}wehave:
l
R(ai) = L
¢i(x)¢n-i(y)a-ij
= L ¢n-sr(x)¢sr(y)asrj = 0 ·
n-iEU
r=l
Let
z
E
A.
Recall that the Mattson-Solomon polynomial of
z
is
n-1
MSz(Z)
=
L ¢i(z)zn-i .
i=O
It is well-known that MSz(ai) =
Za;
[16, p.239]. Then
w(z)
=
n- (card {j
E [O,n -1]1
MSz(ai)
=
0})
+E.
where

equals 0 if zo = 0 and
1
otherwise.
Let y' be the codeword of E whose coefficients ¢s, are given by Formula
(15)
(the other coefficients
¢i(Y')
are zero, by definition of the cyclic code E). For
every
j
in [0,
n -
1], we have obviously
l l
MSy'(ai) = L¢s,(Y')(ai)n-s, =
L¢n-s,(x)¢s,(y)(a)-js,
= R(ai).
t=l t=l
Hence R(ai) = 0 if and only if
y~ 3
= 0. Let n(y) be the number of distinct shifts
of y and let n = n(y)n1. Then
card { ai(Y)
I
x,ai(Y) =
0}
=~card
{j
I
x,ai(Y) =
0}
=
~(n-w(y').
n1 n1
Note that
y~
= 0, by definition.
0
Now we want to show how it is possible to achieve the computation of the
weight distribution of the code D; - then of the coset
x
+
C of depth
r.
We will
examine this problem when U contains only one cyclotomic class of p modulo n.
From Proposition 5, we have immediate corollaries.
COROLLARY
2. Let U
= {
s, ps, ... }, x
E
pr \
C ; let y be any element in
E. Set
{3
=
¢n-s(x)¢s(y).
Let t = (n, s) and n = tn1 . Then , for every
j
E
[0,
n-
1]:
¢n(x ai(Y))
=
0 if and only if Tr((3ai
8
)
=
0.
Let y' be the element of E such that ¢
8
(y')
= {3.
Then the number L(y) of
cosets ai(Y)
+
pm(p-l)-r+l
contained in D; equals (n- w(y'))jt.
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