On the Exponent of Boolean Primitive Circulant Matrices
647
3. Maximal generalized gaps. Let n be a positive integer. Let E
n
= {orderS : S ∈
S
n
}. It is well known [2] that E
n
⊂ [1, n − 1]. We call a gap in E
n
a nonempty interval A
⊂ [1, n − 1] such that A ∩ E
n
= ∅. We say that a gap A in E
n
is maximal if A
′
∩ E
n
6= ∅ for any interval
A
′
⊂ [1, n − 1], with A strictly contained in A
′
. For each positive integer
n and each j ∈ {1, 2, . . . , c
n
− 1}, if ¹
n j + 1
º + j
≤ ¹ n
j º
− 2, let
B
j,n
= ·¹
n j + 1
º + j,
¹ n j
º − 2
¸ ,
3.1 otherwise let
B
j,n
= ∅.
Clearly, if Conjecture 1 is true and B
j,n
is nonempty, B
j,n
is a gap in E
n
. Though the intervals
B
j,n
are not necessarily maximal gaps in E
n
, the next theorem shows that, for each positive integer
j, there is an integer n, with j ≤ c
n
−1, such that B
j,n
is a maximal gap in E
n
. Here, we use the result that, if
b 1 is a divisor of n, then order {0, 1, b} =
¥
n b
¦ + b − 2, which is a particular case of Corollary 5.4. If
a ∈ Z
n
, we denote by hai the cyclic group
generated by a in Z
n
. T
HEOREM
3.1. For each positive integer j, there is an integer n, with j
≤ c
n
− 1, such that
B
j,n
is a maximal gap in E
n
. Proof. We show that for each
j there is an integer n, with j ≤ c
n
− 1, and two bases for Z
n
, say S
1
and S
2
, such that orderS
1
= j
n j+1
k + j
− 1 and orderS
2
= j
n j
k − 1.
Let n = jj + 1j + 3. First, we show that c
n
− 1 ≥ j. If j = 1, then n = 8 and c
n
= 3. If j 1, then n − j + 1
3
= j
2
− 1 0, which implies that
3
√ n j + 1. Then
c
n
− 1 ≥ ⌊
3
√ n
⌋ − 1 ≥
3
√ n
− 2 j − 1, where the first inequality follows from Theorem 2.3. Since
j+1 divides n, by Corollary 5.4, for S
1
= {0, 1, j+1}, orderS
1
= j
n j+1
k +j
−1. If
j 1, let S
2
= hj + 1j + 3i ∪ 1 + hj + 1j + 3i. Then, for k 0,
kS
2
=
k
[
i=1
i + hj + 1j + 3i .
If j = 1, let S
2
= {0, 1}. In any case, it is easy to see that orderS
2
= j + 1j + 3 − 1 =
j
n j
k − 1.
4. Order of bases for Z
n
. Let
T be a subset of the additive group Z
n
, and let q
∈ Z
n
. We define
q + T = {q + t : t ∈ T } and q ∗ T = {qt : t ∈ T }.
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648
M.I. Bueno and S. Furtado
Clearly, if S
⊂ Z
n
and q
∈ Z
n
, S is a basis for Z
n
if and only if q + S is a basis for Z
n
. Moreover, if
S is a basis, orderS = orderq + S. L
EMMA
4.1. Let n and q be positive integers. If S
∈ S
n
and gcdq, n = 1, then
q ∗ S ∈ S
n
and order S = orderq
∗ S. Proof. It is enough to show that, for all
k ≥ 1, |q ∗ kS| = |kS|, as kq ∗ S = q ∗ kS.
Let T = kS. Clearly,
|q ∗ T | ≤ |T |. Now suppose that t
1
, t
2
∈ T with t
1
6= t
2
. Suppose that
qt
1
= qt
2
mod n. Then t
1
− t
2
q = 0 mod n, or equivalently, t
1
− t
2
q = kn for some positive integer
k. Since gcdq, n = 1, t
1
− t
2
= 0 mod n. As 0 ≤ t
1
, t
2
n, then t
1
− t
2
= 0, which is a contradiction. Thus, |q ∗ T | ≥ |T |, which completes the proof.
We note that, if gcdn, q
6= 1, then q ∗ S is not a basis for Z
n
. Let
S
1
, S
2
⊂ Z
n
. We say that S
1
and S
2
are equivalent, and we write S
1
∼ S
2
, if there exist integers
q
1
and q
2
, where gcdq
1
, n = 1, such that S
2
= q
2
+ q
1
∗ S
1
. Note that ∼ is
an equivalence relation. Clearly, from the observations above, if S
1
∈ S
n
and S
1
∼ S
2
, then S
2
∈ S
n
and order S
1
= orderS
2
. Note that if
S = {s
1
, s
2
, . . . , s
t
} ∈ S
n
, then S ∼ {0, s
2
− s
1
, . . . , s
t
− s
1
}. Therefore, in what follows we assume that
∈ S. R
EMARK
1. Let S =
{0, a} ∈ S
n
. Then orderS = n − 1 since S ∼ {0, 1}, as a is a
unit for Z
n
. We now introduce some definitions that will be used in the next sections.
Let S =
{0, s
1
, . . . , s
t
} ∈ S
n
. Then any element q
∈ Z
n
can be expressed as x
1
s
1
+ · · ·+x
t
s
t
mod n, for some nonnegative integers x
1
, . . . , x
t
. Moreover, if q 6= 0, the smallest
k such that q ∈ kS is the minimum x
1
+ · · ·+x
t
among all the solutions x
1
, . . . , x
t
, x
i
≥ 0, to
x
1
s
1
+ · · · + x
t
s
t
= q mod n. D
EFINITION
4.2. Let S =
{0, s
1
, . . . , s
t
} ∈ S
n
and q
∈ Z
n
. If q = 0, then we define expq; S, n = 1; otherwise we define
expq; S, n := min {x
1
+ · · · + x
t
: x
1
s
1
+ · · · + x
t
s
t
= q mod n, x
i
≥ 0}.
Clearly, if S is a basis for Z
n
and ∈ S, orderS = max{expq; S, n : q ∈ Z
n
}. D
EFINITION
4.3. Let S =
{0, s
1
, . . . , s
t
} ∈ S
n
, q ∈ Z
n
and k be a positive integer. If
q 6= 0, we say that q is k; S, n-periodic if k is the smallest nonnegative integer for which
there are nonnegative integers x
1
, . . . , x
t
satisfying x
1
+ · · · + x
t
= expq; S, n and x
1
s
1
+ · · · + x
t
s
t
= q + kn. If
q = 0, we say that q is 0; S, n-periodic. We say that S is K-periodic if there exist
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On the Exponent of Boolean Primitive Circulant Matrices
649 K; S, n-periodic elements in Z
n
and there are no k; S, n-periodic elements in Z
n
for k K.
R
EMARK
2. If the minimum nonzero element of a basis S of Z
n
, say b, is not 1, then S
is not 0-periodic as any b
′
, with 0 b
′
b, is not 0; S, n-periodic. We finish this section with the following lemma.
L
EMMA
4.4. [2] Let S
∈ S
n
and m be a divisor of n. Suppose that S contains an
element of order m. Then
order S
≤ n
m + m
− 2.
By Remark 1, all bases for Z
n
, n
≥ 3, with cardinality 2 have order n − 1, and therefore, they satisfy Conjecture 1. In the next section we focus on bases with cardinality
3.
5. Order of bases for Z
Kategori