On the Exponent of Boolean Primitive Circulant Matrices
653 i
b is a divisor of n; ii
b = j
n j
k + 1, for some nonnegative integer j. In this case, j is unique and is given
by ⌊nb⌋ + 1.
Proof. If b is a divisor of n, then Lemma 5.6 implies that S is 0-periodic. Now suppose
that b is not a divisor of n. Let i =
⌊nb⌋ and n = bi + t, with 0 t b. If t i then b =
⌊
n i
⌋, otherwise b ⌊
n i
⌋. Since n = i + 1b + t − b and t − b 0, it follows that ⌊
n i+1
⌋ b. Therefore, ¹
n i + 1
º b
≤ j
n i
k .
Suppose that b =
j
n i+1
k + 1. Since
b − 1 =
¹ n
i + 1 º
≤ n
i + 1 ,
we have n
≥ b − 1i + 1. Therefore, by Lemma 5.6,
S is 0-periodic. Now suppose that
¥
n i
¦ ≥ b ≥ j
n i+1
k + 2. Then
n i + 1
b − 1
and, by Lemma 5.6, S is not 0-periodic.
6. The conjecture for bases in
S
n,3
. In this section, we prove the following result.
T
HEOREM
6.1. Let S
∈ S
n,3
, n ≥ 3. Conjecture 1 holds if S satisfies one of the
following conditions: i
S is equivalent to {0, a, b}, where a is a divisor of n and a ≥ c
n
; ii
S is equivalent to {0, 1, b} for some b ≤ min
n p
n
, j
n c
n
−1
k − 1
o ;
iii S is equivalent to a 0-periodic basis.
The rest of this section is dedicated to the proof of Theorem 6.1. We first observe that, in general, if
i and j are positive numbers, then n
i + i
n j
+ j can be written as
i − jn − ij 0,
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654
M.I. Bueno and S. Furtado
which is equivalent to j min
n i,
n i
o ∨ j max
n i,
n i
o .
L
EMMA
6.2. Let S =
{0, a, b} ∈ S
n,3
, where a is a divisor of n such that a ≥ c
n
. Then either order
S ≤
j
n c
n
k + c
n
− 2 or n
j − 1 ≤ orderS ≤
n j
+ j − 2,
6.1 for some
j ∈ {1, 2, . . . , c
n
− 1}. Proof. Note that
n 6= 3. If n = 4 then c
n
= a = 2 and orderS = 2, which implies that order
S ≤
j
n c
n
k + c
n
− 2. If n = 8 then c
n
= 3 and a = 4; a direct computation considering all possible values of
b, namely 1, 3, 5, 7, shows that orderS = 4 in fact, in this case,
S ∼ {0, 1, b
′
} for some b
′
∈ Z
n
. Then 6.1 holds with j = 2. Now suppose that n
6= 4 and n 6= 8. Suppose that orderS j
n c
n
k + c
n
− 2. By Theorem 5.2, a
− 1 ≤ orderS ≤ n
a + a
− 2. Then
n c
n
+ c
n
n a
+ a. Taking into account the observation before this lemma and the fact that for
n 6= 3, 4, 8,
c
n
nc
n
, it follows that a nc
n
, as, by hypothesis, a ≥ c
n
. Then, because a divides n, a =
n i
for some i
∈ {2, 3, . . . , c
n
− 1}. Then 6.1 holds with j = i. L
EMMA
6.3. Let S =
{0, 1, b} ∈ S
n,3
, with b ≤ p
n
. If S is 0-periodic and b
≥ ⌊nc
n
⌋ + 2, then there exists i ∈ [2, c
n
− 1] such that i +
j n
i k
− 3 ≤ orderS ≤ i + j
n i
k − 2,
Proof. Note that n 3 and n
6= 8. If b is a divisor of n, then b = ni for some i
∈ [2, c
n
− 1]. By Corollary 5.4, orderS =
n i
+ i − 2.
If b is not a divisor of n, then, taking into account Theorem 5.7 , b =
⌊ni⌋ + 1 for some i
∈ [2, c
n
− 1]. Let m = ⌊ni⌋ and n = mi + t, 0 ≤ t i. By Corollary 5.4, order
S = ¹ mi + t
m + 1 º
+ j
n i
k − 1 =
¹ t
− i m + 1
º + i +
j n
i k
− 1.
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On the Exponent of Boolean Primitive Circulant Matrices
655 If
m + 1 ≥ i − t, then
¹ t
− i m + 1
º =
−1. If
m + 1 i − t, then
¹ t
− i m + 1
º =
−2, as
i − t ≤ i ≤ c
n
− 1 ≤ j
n c
n
k − 1 ≤ b − 3 ≤ 2b = 2m + 2. Note that c
n
≤ j
n c
n
k for
n 6= 3, 8. Thus, the result follows.
L
EMMA
6.4. Let S =
{0, 1, b} ∈ S
n,3
. If b ∈
h c
n
+ 1, j
n c
n
k + 1
i , then orderS
≤ j
n c
n
k + c
n
− 2. Proof. By Theorem 5.3, order
S ≤
¥
n b
¦ + b − 2. Thus, it is enough to show that b +
j n
b k
≤ c
n
+ ¹ n
c
n
º .
6.2 Taking into account the observation before Lemma 6.2, if
c
n
min {b,
n b
} then b +
j n
b k
≤ b + n
b c
n
+ n
c
n
, which implies 6.2, as
c
n
is an integer. Since c
n
min n
b, n
b o
⇔ c
n
b ≤
j
n c
n
k if
c
n
is not a divisor of n
c
n
b ≤
n c
n
− 1 if c
n
is a divisor of n
, we now need to show that 6.2 holds if either
b = j
n c
n
k + 1 or b =
n c
n
and c
n
divides n. The
latter is immediate. For the first case, note that
n
j
n c
n
k + 1
= c
n
− 1, as, if
n = j
n c
n
k c
n
+ t, with 0 ≤ t c
n
, then n = c
n
− 1 µ¹ n
c
n
º + 1
¶ +
µ t
− c
n
+ ¹ n
c
n
º + 1
¶ ,
with ≤ t − c
n
+ j
n c
n
k + 1
j
n c
n
k + 1.
Next we give a result that allows us to show that the conjecture holds if S =
{0, 1, b}, with
b ≤ p
n
and b
∈ [⌊nc
n
⌋ + 2, ⌊nc
n
− 1⌋ − 1]. Note that ⌊nc
n
⌋ + 2 ≤ p
n
if and only
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656
M.I. Bueno and S. Furtado
if c
n
≥ 3. Also, by Lemma 2.5, for c
n
≥ 3 the previous interval is nonempty if and only if n = 14 or n
≥ 16. Finally, observe that ⌊nb⌋ = c
n
− 1. We think the method used to prove the conjecture in this case might be generalizable to
the cases in which b
∈ [⌊
n c
n
−k
⌋, ⌊
n c
n
−k+1
⌋ − 1], with 1 ≤ k ≤ c
n
− 3, when this interval is nonempty. Some results presented in Section 2 will be used.
L
EMMA
6.5. Let n be a positive integer such that c
n
≥ 3, b ∈ h
⌊
n c
n
⌋ + 2, ⌊
n c
n
−1
⌋ − 1 i
and t = n
− c
n
− 1b. If ¹ n
c
n
º + 1 b
− t, 6.3
then either c
n
= ⌊
3
√ n
⌋, or c
n
= ⌊
3
√ n
⌋ + 1 and n = c
3 n
− 1. Moreover, 3c
n
≤ ⌊nc
n
⌋ + 2. Proof. Let
m = c
n
− 1 and r = ⌊nc
n
⌋ + c
n
− 2. First we show that if 6.3 holds, then b =
j
n c
n
−1
k − 1. Suppose that b ≤
j
n c
n
−1
k − 2. Then t ≥ 2c
n
− 1 and, taking into account Lemma 2.6, we get
b − t ≤
¹ n
c
n
− 1 º
− 2 − 2c
n
− 1 ≤ ¹ n
c
n
º + 1,
a contradiction. Now suppose that b =
j
n c
n
−1
k − 1 and 6.3 holds. Then t ≥ c
n
− 1. If c
n
= ⌊
3
√ n
⌋ + 1 and n ≤ c
3 n
− 2, then, taking into account Lemma 2.6, b
− t ≤ ¹
n c
n
− 1 º
− 1 − c
n
− 1 ≤ ¹ n
c
n
º + 1,
a contradiction. Thus, c
n
= ⌊
3
√ n
⌋ or c
n
= ⌊
3
√ n
⌋ + 1 and n = c
3 n
− 1. Taking into account Lemma 2.7, the result follows.
L
EMMA
6.6. Let S =
{0, 1, b} ∈ S
n,3
, with c
n
≥ 3. Suppose that b ∈ [⌊nc
n
⌋ + 2,
⌊nc
n
− 1⌋ − 1]. Then order
{0, 1, b} ≤ ¹ n
c
n
º + c
n
− 2.
Proof. Note that n = 14 or n
≥ 16 for the interval [⌊nc
n
⌋ + 2, ⌊nc
n
− 1⌋ − 1] not to be empty. Let
r = j
n c
n
k + c
n
− 2 and n = c
n
− 1b + t for some 0 t b. Note that t
≥ c
n
− 1. Let m = ⌊nb⌋ = c
n
− 1. Since c
n
≤ ⌊nc
n
⌋, 2m ≤ r. Let
k
1
i = ib and k
2
i = ib − 1 + r, for i ∈ {0, 1, . . . , m},
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On the Exponent of Boolean Primitive Circulant Matrices
657 k
1
i = ib − n and k
2
i = ib − 1 + r − n, for i ∈ {m + 1, m + 2, . . . , 2m},
and k
1
i = ib − 2n and k
2
i = ib − 1 + r − 2n, for i ∈ {2m + 1, 2m + 2, . . . , 3m}.
Consider the following intervals in Z : I
i
= [k
1
i, k
2
i], i ∈ {0, 1, . . . , 3m}. Note that, for
each i = 0, 1, . . . , 3m, k
1
i k
2
i and k
1
i n. Also, 0 ≤ k
1
i, for i = 0, 1, . . . , 3m, i
6= 2m + 1. We have rS ≡ S
r i=0
I
i
mod n. We next show that if b
− t ≤ ⌊nc
n
⌋ + 1, 6.4
then S
2 m
i=0
I
i
≡ Z
n
mod n; if ⌊nc
n
⌋ + 1 b − t, 6.5
then S
3 m
i=0
I
i
≡ Z
n
mod n, which implies rS = Z
n
. Note that 2m r and, by Lemma 6.5, if 6.5 holds,
3m ≤ r.
Consider the intervals I
i
, i = 0, 1, . . . , 3m, ordered in the following way: I
, I
2 m+1
, I
m+1
, I
1
, I
2 m+2
, I
m+2
, . . . , I
3 m
, I
2 m
, I
m
. Clearly, for
j = 1, 2, . . . , m, k
1
j − 1 ≤ k
1
m + j ≤ k
1
j and k
1
2m + j ≤ k
1
m + j. We show that, for each
j = 1, 2, . . . , m, i
k
2
m + j + 1 ≥ k
1
j; ii
k
2
j − 1 + 1 ≥ k
1
m + j if 6.4 holds; iii
k
2
j − 1 + 1 ≥ k
1
2m + j ≥ k
1
j − 1 if 6.5 holds;
iv k
2
2m + j + 1 ≥ k
1
m + j if 6.5 holds; v
k
2
m ≥ n − 1,
which completes the proof. Condition i follows easily taking into account that
c
n
+ t ≤ ⌊
n c
n
⌋ + 1, as t
− c
n
− 1b ≤ n − c
n
− 1 µ¹ n
c
n
º + 2
¶ =
µ n
− c
n
¹ n c
n
º¶ +
¹ n c
n
º − 2c
n
− 1
≤ c
n
− 1 + ¹ n
c
n
º − 2c
n
− 1 = ¹ n
c
n
º − c
n
+ 1. Condition ii follows from a simple calculation.
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658
M.I. Bueno and S. Furtado
Now suppose that 6.5 holds. The first inequality in condition iii holds as b
− 2t ≤ ¹
n c
n
− 1 º
− 1 − 2c
n
− 1 ≤ ¹ n
c
n
º + 1
≤ ¹ n
c
n
º + c
n
− j, where the second inequality follows from Lemma 2.6. Since we have shown in i that
t ≤
j
n c
n
k − c
n
+ 1, then b
⌊nc
n
⌋ + 1 + t 2t, which implies the second inequality.
Condition iv holds if 2c
n
+ t ≤ ⌊nc
n
⌋ + 2. By Lemma 6.5 either c
n
= ⌊
3
√ n
⌋ or c
n
= ⌊
3
√ n
⌋ + 1 and n = c
3 n
− 1. If c
n
= ⌊
3
√ n
⌋, by Lemma 2.6, t
≤ ¹
n c
n
− 1 º
− 1 − ¹ n
c
n
º − 1 ≤ c
n
+ 3 − 2.
Thus, taking into account Lemma 2.7, 2c
n
+ t ≤ 3c
n
+ 1 ≤
¹ n c
n
º + 2.
If c
n
= ⌊
3
√ n
⌋ + 1 and n = c
3 n
− 1, by Lemma 2.6, t
≤ ¹
n c
n
− 1 º
− 1 − ¹ n
c
n
º − 1 ≤ c
n
+ 2 − 2.
By Lemma 2.7, 2c
n
+ t ≤ 3c
n
≤ ¹ n
c
n
º + 2.
Finally, note that condition v is equivalent to t
≤ ⌊nc
n
⌋, which holds as c
n
≥ 3 and we have shown that
t + c
n
≤ ⌊nc
n
⌋ + 1. Proof of Theorem 6.1. It follows from Lemma 6.2 that if condition i holds then Conjec-
ture 1 is satisfied. Now suppose that
S = {0, 1, b}, with b ≤ p
n
. If b ≤ c
n
, Conjecture 1 holds by Theorem 5.3; if
b ∈ [c
n
+ 1, j
n c
n
k + 1], the conjecture holds by Lemma 6.4; if b
∈ hj
n c
n
k + 2,
j
n c
n
−1
k − 1
i then
c
n
≥ 3 and Conjecture 1 holds by Lemma 6.6. Finally, if b
≥ ⌊nc
n
⌋+2 and S is 0 -periodic, Conjecture 1 holds by Lemma 6.3. Since two equivalent bases have the same order, the result follows.
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On the Exponent of Boolean Primitive Circulant Matrices
659
7. The conjecture for bases with cardinality larger than 3. In this section, we include
