DOI: 10.2478/s12175-014-0242-6
Math. Slovaca 64 (2014), No. 4, 819–842
HOOKED EXTENDED LANGFORD SEQUENCES
OF SMALL AND LARGE DEFECTS
Shai Mor* — Václav Linek**
(Communicated by Peter Horak )
ABSTRACT. It is shown that for m = 2d + 5, 2d + 6, 2d + 7 and d ≥ 1, the set
{1, . . . , 2m + 1} − {k} can be partitioned into differences d, d + 1, . . . , d + m − 1
whenever (m, k) ≡ (0, 1), (1, d), (2, 0), (3, d + 1) (mod (4, 2)) and 1 ≤ k ≤ 2m + 1.
It is also shown that for m = 2d + 5, 2d + 6, 2d + 7, and d ≥ 1, the set
{1, . . . , 2m + 2} − {k, 2m + 1} can be partitioned into differences d, d + 1, . . .
. . . , d + m − 1 whenever (m, k) ≡ (0, 0), (1, d + 1), (2, 1), (3, d) (mod (4, 2)) and
k ≥ m + 2.
These partitions are used to show that if m ≥ 8d + 3, then the set {1, . . .
. . . , 2m+2}−{k, 2m+1} can be partitioned into differences d, d+1, . . . , d+m−1
whenever (m, k) ≡ (0, 0), (1, d + 1), (2, 1), (3, d) (mod (4, 2)).
A list of values m, d that are open for the existence of these partitions (which
are equivalent to the existence of Langford and hooked Langford sequences) is
given in the conclusion.
c
2014
Mathematical Institute
Slovak Academy of Sciences
1. Notation
All of the intervals that occur in this article have integer endpoints and are
understood to be intervals of integers, that is, for integers a and b the convention
is that [a, b] = {a, a + 1, . . . , b}. We write [a, b] = [a, b] ∩ (2Z + 1) for the set of
odd integers in [a, b] and [a, b] = [a, b] ∩ 2Z for the set of even integers in [a, b].
The juxtaposition, S1 S2 , of sequences S1 and S2 is their concatenation.
2. Introduction
For general information about Langford sequences we refer to Shalaby [12].
The problem of partitioning the set S = {1, 2, . . . , 2m} into differences d, d+1,
. . . , d + m − 1 is today known as the problem of constructing a Langford sequence
of defect d and m differences. Langford [3] posed this problem for d = 2 and
2010 M a t h e m a t i c s S u b j e c t C l a s s i f i c a t i o n: Primary 11B83; Secondary 05B05.
K e y w o r d s: Langford sequence, Skolem sequence, Steiner triple system.
Research supported by an NSERC Discovery Grant.
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SHAI MOR — VÁCLAV LINEK
subsequently Priday [11] solved it, although slightly earlier Skolem [14] posed
and solved this problem for d = 1; indeed solutions for d = 1 are called Skolem
sequences. Skolem sequences of n differences were developed in order to generate
cyclic Steiner Triple Systems of order 6n + 1, as shown by Skolem himself [15].
The partition P = {(3, 5), (1, 4), (2, 6)} corresponds to a Langford sequence of
defect d = 2 and m = 3 differences, where the corresponding sequence is 342324,
with j appearing in positions aj , bj with |bj − aj | = j and (aj , bj ) ∈ P .
The problem of partitioning S = {1, 2, . . . , 2m + 1} − {2m} into differences
d, d+1, . . . , d+m−1 is equivalent to constructing a hooked Langford sequence of
defect d and m differences, where the unused position, 2m, is referred to as the
hook. Thus the sequence 456242536 3 is a hooked Langford sequence of defect
d = 2 and m = 5 differences, with a hook at position 2m = 10. When d = 1 a
hooked Langford sequence is also called a hooked Skolem sequence, and O’Keefe
[7] was the first to establish the existence of these sequences.
We denote a Langford sequence of defect d and m differences by Lm
d , so we
3
can write L2 = 342324. Similarly, we denote a hooked Langford sequence of
5
defect d and m differences by hLm
d , so the second given sequence is a hL2 .
For arbitrary d the following results have been obtained
2.1 ([2], [13]) If m ≥ 2d − 1, then [1, 2m] can be partitioned into
differences [d, d+m−1] whenever (m, d) ≡ (0, 1), (1, 1), (0, 0), (3, 0) (mod (4, 2)),
to form a Langford sequence, denoted Lm
d .
2.2 ([13]) If m(m+1−2d)+2 ≥ 0, then [1, 2m+1]−{2}, equivalently
[1, 2m + 1] − {2m}, can be partitioned into differences [d, d + m − 1] whenever
(m, d) ≡ (2, 0), (1, 0), (2, 1), (3, 1) (mod (4, 2)), to form a hooked Langford sequence, denoted hLm
d .
In this paper we consider the problems of constructing extended and hookedextended Langford sequences. A k-extended Langford sequence of defect d and
m differences is a partition of [1, 2m + 1] − {k} into differences [d, d + m − 1]. A
hooked, k-extended Langford sequence of defect d and m differences is a partition
of [1, 2m + 2] − {k, 2m + 1} into differences [d, d + m − 1]. (Note that partitions of
[1, 2m + 2] − {k, 2m + 1} into differences [d, d + m − 1] correspond to partitions of
[1, 2m + 2] − {2, k} into differences [d, d+ m − 1], so sometimes one must take this
into account when referring to the literature.) In the case of a hooked extended
Langford sequence the next to last unused position, 2m + 1, is referred to as the
hook.
Thus 34 3242 is a Langford sequence of defect d = 2 and m = 3 differences, with an extension at position k = 3 indicated by the underscore, “ ”, and
242 34 3 is a hooked extended Langford sequence of defect d = 2 and m = 3
differences, with a hook at position 7 and an extension at position k = 4. We
append the extension k in round brackets to the preceding notation for sequences in order to indicate a k-extension, so we write L32 (3) = 34 3242 and
hL32 (4) = 242 34 3.
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
The problem of constructing k-extended Langford sequences for all possible
k given d and m was solved by Baker [1] for d = 1, by Linek and Jiang [5]
for d = 2, 3; and by Linek and Mor [6] for d = 4. In [6], extended Langford
sequences are constructed for all but a finite number of m for each d. In this
paper we construct hooked extended Langford sequences in Theorem 8.1, where
for each defect d we obtain all possible k-extensions for all but a finite number
of values of m.
3. Necessary conditions
Some necessary conditions for the existence of a partition of a set of 2m
integers into m given differences are well known:
3.1 ([8]) Let P = {p1, p2 , . . . , p2m} be a set of 2m integers with p1 ≤
p2 ≤ · · · ≤ p2m . If the set P can be partitioned into m differences d1 , d2 , . . . , dm ,
then
2m
m
pi ≡
di (mod 2),
(1)
i=1
m
i=1
i=1
di ≤
2m
i=m+1
pi
−
m
pi .
(2)
i=1
We call (1) and (2) the parity and density conditions, respectively. In the
case of an extended Langford sequence these two conditions imply:
3.2 ([5], [6]) The following are necessary for the existence of a
k-extended Langford sequence of defect d and m differences, Lm
d (k):
(a) m ≥ 2d − 3,
(b) k ∈ [1, 2m + 1],
m
(c) m
2 (2d − 1 − m) + 1 ≤ k ≤ 2 (m − 2d + 5) + 1,
(d) (m, k) ≡ (0, 1), (1, d), (2, 0), (3, d + 1) (mod (4, 2)).
Conditions for the existence of a hooked, extended Langford sequence can be
deduced similarly.
3.3 The following are necessary for the existence of a hooked k-extended Langford sequence of defect d and m differences, hLm
d (k):
(a) m + 2/m ≥ 2d − 3,
(b) k ∈ [1, 2m + 2] − {2m + 1},
m
(c) m
2 (2d − 1 − m) ≤ k ≤ 2 (m − 2d + 5) + 2,
(d) (m, k) ≡ (0, 0), (1, d + 1), (2, 1), (3, d) (mod (4, 2)).
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P r o o f. The existence of a hLm
d (k) is equivalent to the existence of a partition
of P = {p1 , p2 , . . . , p2m } = [1, 2m + 2] − {k, 2m + 1} into differences d1 = d,
d2 = d + 1, . . . , dm = d + m − 1. The parity condition (1) yields
(m + 1)(2m + 1) − k + 1 ≡ dm +
(m − 1)m
2
(mod 2),
or equivalently
2(m + 1)(2m + 1) − 2k + 2 ≡ 2dm + (m − 1)m (mod 4).
(3)
Part (d) now follows from (3). For example, if m ≡ 1 (mod 4), then (3) reduces
to 2 − 2k ≡ 2d (mod 4), and hence k ≡ d + 1 (mod 2) in this case.
Now consider the case where k ∈ [1, m + 1].
The density condition (2) yields
(m + 1)(3m + 4)
(m + 1)(m + 2)
m(m − 1)
≤
− (2m + 1) −
−k .
dm +
2
2
2
(4)
The first inequality in part (c) now follows from (4) by isolating k on one side.
Since k ≤ m + 1, it follows that
m
(2d − 1 − m) ≤ m + 1,
(5)
2
which is equivalent to the inequality in part (a). Finally, the inequality in part
(a) is also equivalent to the inequality m + 1 ≤ m
2 (m − 2d + 5) + 2, hence
k≤m
(m
−
2d
+
5)
+
2
and
the
second
inequality
in
part (c) is established.
2
Thus, parts (a) and (c) hold if 1 ≤ k ≤ m + 1, and by similar calculations
parts (a) and (c) also hold if k ∈ [m + 2, 2m + 2] − {2m + 1}.
This concludes the proof of necessity.
2
By Theorem 3.3 we have m ≥ (2d − 3) − m
≥ 2d − 5. If m is “close” to 2d − 5,
m
then the following statements about a hLd (k) follow easily from Theorem 3.3:
• If m = 2d − 5, then (d, m, k) = (3, 1, 2).
• If m = 2d − 4, then (d, m, k) = (3, 2, 3).
• If m = 2d − 3, then 2d − 3 ≤ k ≤ 2d − 1 for all d.
• If m = 2d − 2, then d − 1 ≤ k ≤ 3d − 1 for all d.
m
• If m ≥ 2d − 1, then m
2 (2d − 1 − m) < 1 < 2m + 2 ≤ 2 (m − 2d + 5) + 2.
The position of the k-extension is therefore affected by both parity and density
2
2
when m + m
≤ 2d − 2, while only parity affects the position of k when m + m
≥
2d − 1, as then the density condition is trivially satisfied for any k.
It is convenient, therefore, to refer to the difference set Di = [d, 3d − 2 + i].
Then no hooked extended Langford sequences exists with differences Di , i < −4.
The case of D−4 is exactly the sequence 3 3. The case of D−3 is exactly
the sequence 34 3 4. In the case of D−2 , we can exhibit one infinite family of sequences of the form hL2d−3
(2d − 1) as follows: hL33 (5) = 3453 4 5,
d
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
hL54 (7) = 567845 647 8, hL75 (9) = 789ab567 8596a b, . . . (where a = 10, b = 11).
The case of D−1 is not amenable to the techniques we employ, so we present no
solution to it. Generally speaking, the case m ≥ 2d − 1 is of the most interest,
especially from the point of view of applications in design theory (see Shalaby
[12]).
To give some insight into the structure of the subsequent tables, consider the
partition of Table 1a, corresponding to a Langford sequence L2d−1
.
d
Table 1a. A L2d−1
,d≥1
d
1a
ar
br
|br − ar |
(1)
d+j
2d + 2j
d+j
0≤j ≤d−1
(2)
1+j
2d + 1 + 2j
2d + j
0≤j ≤d−2
Taking d = 5 and writing out the sequence yields
L95 = 10, 11, 12, 13, 5, 6, 7, 8, 9, 5, 10, 6, 11, 7, 12, 8, 13, 9.
(6)
Now, the second occurrence of the symbol 8 can be moved to the front of the
sequence, thereby creating the following extended Langford sequence:
L95 (17) = 8, 10, 11, 12, 13, 5, 6, 7, 8, 9, 5, 10, 6, 11, 7, 12, , 13, 9.
(7)
Furthermore, the second occurrence of any one of 5, 6, 7, or 9 could also have
been moved to the front, allowing us to produce several extended Langford
sequences from just one sequence (this method is sometimes called pivoting).
The sequences of Table 1a have similar properties (i.e., allow pivoting) to
the sequence we examined for d = 5; the extended sequences one obtains from
Table 1a are stored in Table 1b (note the shift in positions by 1).
Table 1b. Sequences L2d−1
(k), d ≥ 1, k ∈ [2d + 1, 4d − 1]
d
1b
ar
br
|br − ar |
(0)
(1)
(2)
1
d+1+j
2+j
1
2d + 1 + 2j or 1
2d + 2 + 2j
0
d+j
2d + j
0 ≤j ≤d−1
0 ≤j ≤d−2
By reversing the sequences of Table 1b, we obtain for each d ≥ 1 all possible
extended Langford sequences of the form L2d−1
(k), that is, all possible k are
d
achieved for each d ≥ 1.
We remark that the L95 of (6) is composed of the two structures
5, 6, 7, 8, 9, 5, , 6, , 7, , 8, , 9
10, 11, 12, 13, , , , , , , 10, , 11, , 12, , 13
(8)
(9)
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which fit together like the jaws of a crocodile. The sequences in the subsequent
tables will also have such jaw-like structures, but with four jaws instead of two,
and with some isolated differences occurring which do not belong to any of the
jaws.
We call [a, b] perfect, and denote this by [a, b]P , when there are an even number
of evens in the interval [a, b] (see Nordh [9]). Otherwise, we call [a, b] nonperfect,
and denote this by [a, b]N P . The significance of this concept is that if D =
2m
[d, d + m − 1] is a perfect interval of differences, then
d≡
i (mod 2), that
d∈D
i=1
is, the parity condition (1) is satisfied by D and P = [1, 2m].
Let m = |b−a+1|. If m ≡ 0 (mod 4), then [a, b] is perfect. If m ≡ 2 (mod 4),
then [a, b] is nonperfect. If m ≡ 1 (mod 4), then [a, b] is perfect precisely when
a is odd, and if m ≡ 3 (mod 4), then [a, b] is perfect precisely when a is even.
Furthermore, a perfect interval can only split into either two perfect subintervals
or into two nonperfect subintervals, while a nonperfect interval can be split only
into a perfect subinterval and a nonperfect subinterval. Note also that if two
consecutive differences are included (or removed) at either end of a difference
interval, then the perfect/nonperfect status of the interval is flipped.
Lemma 3.4 follows from Theorem 3.2, Theorem 3.3, and the preceding comments, and shows the connection between the parity of an extension and the
perfect/nonperfect status of a difference interval.
3.4
Let D = [d, d + m − 1] be an interval of differences.
(a) If [1, 2m + 1] − {k} can be partitioned into differences D, then k is odd
when D is perfect and k is even when D is nonperfect.
(b) If [1, 2m + 2] − {k, 2m + 1} can be partitioned into differences D, then k is
even when D is perfect and k is odd when D is nonperfect.
The last two theorems quoted in this section will be used to prove our main
theorem.
3.5 ([6]) If m = 2d−1, 2d or 2d+1, and d ≥ 1, then [1, 2m + 1]−{k}
can be partitioned into differences [d, d + m − 1] whenever (m, k) ≡ (0, 1), (1, d),
(2, 0), (3, d + 1) (mod (4, 2)).
3.6 ([6]) If m = 2d − 1, 2d or 2d + 1, and d ≥ 1, then [1, 2m + 2] −
{k, 2m + 1} can be partitioned into differences [d, d + m − 1] whenever (m, k) ≡
(0, 1), (1, d), (2, 0), (3, d + 1) (mod (4, 2)), with the exception of (d, m, k) =
(1, 1, 2), (2, 3, 2).
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
4. The result for differences Di , i = 6, 7, 8
Let Di = [d, 3d − 2 + i]. Of particular relevance to the general case will
be constructions of extended and hooked extended Langford sequences with
differences in Di , 6 ≤ i ≤ 8. The main result here is
4.1 If m = 2d+5, 2d+6 or 2d+7, and d ≥ 1, then [1, 2m + 1]−{k}
can be partitioned into differences [d, d + m − 1] whenever (m, k) ≡ (0, 1), (1, d),
(2, 0), (3, d + 1) (mod (4, 2)) and 1 ≤ k ≤ 2m + 1.
P r o o f. Lemma 5.1, Lemma 6.1, Lemma 6.2, and Lemma 7.1 complete the
proof for cases of difference sets D6 , D7 even defect, D7 odd defect, and D8
(respectively).
4.2 If m = 2d + 5, 2d + 6 or 2d + 7, and d ≥ 1, then [1, 2m + 2] −
{k, 2m + 1} can be partitioned into differences [d, d + m − 1] whenever (m, k) ≡
(0, 0), (1, d + 1), (2, 1), (3, d) (mod (4, 2)) and k ≥ m + 2.
P r o o f. Lemma 5.2, Lemma 6.3, Lemma 6.4, and Lemma 7.2 complete the
proof for cases of difference sets D6 , D7 even defect, D7 odd defect, and D8
(respectively).
All k-extensions can be obtained for Langford sequences with 1 ≤ d ≤ 3 and
differences Dj , 6 ≤ j ≤ 8 (see Baker [1], Linek and Jiang [5]).
Also, all k-extensions can be obtained for hooked Langford sequences with
1 ≤ d ≤ 2 and differences in Dj , 6 ≤ j ≤ 8, and this is easily extended by hand
calculation to d = 3 (see Linek and Jiang [4], [5])
Henceforth, we take d ≥ 4.
5. Difference set D6 , m = 2d + 5
By the remarks at the end of the preceding section, in this section we take
d ≥ 4 (although many tables are valid for even smaller defects).
The set D6 = [d, 3d + 4] is a nonperfect difference set, so by Lemma 3.4 any
(k) has k ∈ [1, 4d + 11] .
corresponding L2d+5
d
5.1 For m = 2d + 5 and d ≥ 4, the set [1, 2m + 1] − {k} can be
partitioned into differences [d, d + m − 1] if k is even.
P r o o f. For even defects d ≥ 6, the union of the ranges of extensions in Table 2a
and Table 2b is [2d + 6, 4d + 10] . For d = 4 this range is obtained by appending
(20) = g6ab9df678eca9b7g8d 5f4ce54
the sequence L2d+5
d
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For odd defects d ≥ 7 the union of the ranges of extensions in Table 3a,
Table 3b, and Table 3c, is [2d + 6, 4d + 10] . For d = 5 this range is obtained by
(26) = j5ibh65a7eg6cfb7da9jih8ec g9fd8.
appending the sequence L2d+5
d
By reversing all these sequences, we obtain all extensions k ∈ [1, 4d + 11] ,
for each parity of d.
Table 2a. Sequences L2d+5
(k), d even, k ∈ [2d + 6, 4d + 8] − {3d + 8, 3d + 10}
d
2a
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
1
2
3
4+j
d+4+j
3d/2 + 7 + j
2d + 7
3d + 8
3d + 10
3d/2 + 5
2
3d/2 + 6
2d + 9 + 2j
2d + 6 + 2j or 2
3d + 12 + 2j or 2
4d + 11
4d + 9
4d + 10
3d/2 + 4
0
3d/2 + 3
2d + 5 + j
d+2+j
3d/2 + 5 + j
2d + 4
d+1
d
–
–
–
0≤j ≤ d−1
0 ≤ j ≤ d/2
0 ≤ j ≤ d/2 − 2
–
–
–
Table 2b. Sequences L2d+5
(k), d even, k ∈ [2d + 14, 4d + 10] − {3d + 12}
d
2b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
ai
bi
|bi − ai |
1
2
3
4+j
d/2 + 3
d/2 + 4 + j
d+5
d+6
d+8+j
3d/2 + 7
3d/2 + 8 + j
2d + 12
3d + 7
d+4
2
d+7
2d + 9 + 2j
2d + 8
3d + 9 + 2j
2d + 10
2d + 7
2d + 14 + 2j or 2
4d + 11
3d + 14 + 2j or 2
3d + 12
4d + 9
d+3
0
d+4
2d + 5 + j
3d/2 + 5
5d/2 + 5 + j
d+5
d+1
d+6+j
5d/2 + 4
3d/2 + 6 + j
d
d+2
–
–
–
0≤j
–
0≤j
–
–
0≤j
–
0≤j
–
–
≤ d/2 − 2
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 2
We now construct some hooked extended Langford sequences for D6 .
5.2 For m = 2d + 5 and d ≥ 4 the set [1, 2m + 2] − {k, 2m + 1} can be
partitioned into differences [d, d + m − 1] if (m, k) ≡ (1, d + 1), (3, d) (mod (4, 2))
and k ≥ 2d + 7.
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(2d + 6), d odd
Table 3a. An L2d+5
d
3a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
ai
bi
|bi − ai |
1
2
3
4+j
d+4
d+5+j
(3d + 13)/2 + j
2d + 5
2d + 6
2d + 7
3d + 7
3d + 9
(3d + 9)/2
d+3
(3d + 11)/2
2d + 9 + 2j
4d + 8
2d + 8 + 2j
3d + 11 + 2j
4d + 9
2d + 6
4d + 10
4d + 7
4d + 11
(3d + 7)/2
d+1
(3d + 5)/2
2d + 5 + j
3d + 4
d+3+j
(3d + 9)/2 + j
2d + 4
0
2d + 3
d
d+2
–
–
–
0 ≤j ≤ d−2
–
0 ≤ j ≤ (d − 3)/2
0 ≤ j ≤ (d − 5)/2
–
–
–
–
–
Table 3b. Sequences L2d+5
(k), d odd, k ∈ [2d + 8, 4d + 4] − {3d + 7}
d
3b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
ai
bi
|bi − ai |
1
2
3
4+j
(d + 5)/2
(d + 7)/2
(d + 9)/2 + j
d+5+j
(3d + 9)/2
(3d + 11)/2 + j
2d + 6
2d + 7
3d + 6
3d + 8
2d + 4
2
d+4
2d + 9 + 2j
2d + 5
3d + 7
3d + 10 + 2j
2d + 8 + 2j or 2
4d + 9
3d + 9 + 2j or 2
4d + 8
4d + 11
4d + 6
4d + 10
2d + 3
0
d+1
2d + 5 + j
(3d + 5)/2
(5d + 7)/2
(5d + 11)/2 + j
d+3+j
(5d + 9)/2
(3d + 7)/2 + j
2d + 2
2d + 4
d
d+2
–
–
–
0≤j
–
–
0≤j
0≤j
–
0≤j
–
–
–
–
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 5)/2
P r o o f. For even defects d ≥ 6 this follows from Table 4a and Table 4b, while
for d = 4 we must append the two sequences
hL13
4 (17) = j6iabh86f9egda8b c9ji7hfed5g7c 5,
hL13
4 (21) = j7biah6g78fd6baec89j ihgdf59ce 5.
For odd defects d the lemma follows from Table 5a and Table 5b.
827
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 12, 4d + 10] − {3d + 9, 3d + 11}
Table 3c. Sequences L2d+5
d
3c
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
ai
bi
|bi − ai |
1
2
3
4+j
(d + 9)/2
(d + 11)/2
(d + 13)/2 + j
d+7+j
(3d + 15)/2 + j
2d + 7
2d + 10
3d + 9
(3d + 11)/2
2
d+6
2d + 9 + 2j
2d + 8
(3d + 13)/2
3d + 12 + 2j
2d + 12 + 2j or 2
3d + 13 + 2j or 2
3d + 11
3d + 10
4d + 11
(3d + 9)/2
0
d+3
2d + 5 + j
(3d + 7)/2
d+1
(5d + 11)/2 + j
d+5+j
(3d + 11)/2 + j
d+4
d
d+2
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
≤ (d − 1)/2
≤ (d − 3)/2
≤ (d − 5)/2
≤ (d − 3)/2
Table 4a. Sequences hL2d+5
(k), d even, k ∈ [2d + 7, 4d + 5] − {3d + 5, 3d + 7}
d
4a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
ai
bi
|bi − ai |
1
2
3+j
d/2 + 2
d/2 + 3
d/2 + 4 + j
d+3
d+4+j
3d/2 + 5 + j
2d + 4
2d + 6
2d + 8
3d + 5
3d + 7
4d + 11
1
3d/2 + 4
2d + 10 + 2j
2d + 5
3d/2 + 3
3d + 10 + 2j
3d + 8
2d + 7 + 2j or 1
3d + 9 + 2j or 1
4d + 10
4d + 9
4d + 12
4d + 7
4d + 8
4d + 11
0
3d/2 + 2
2d + 7 + j
3d/2 + 3
d
5d/2 + 6 + j
2d + 5
d+3+j
3d/2 + 4 + j
2d + 6
2d + 3
2d + 4
d+2
d+1
0
–
–
0≤j
–
–
0≤j
–
0≤j
0≤j
–
–
–
–
–
–
≤ d/2 − 2
≤ d/2 − 2
≤ d/2 − 2
≤ d/2 − 2
6. Extended sequences for D7 , m = 2d + 6.
As in the preceding section, we take d ≥ 4 (though many tables are valid for
even smaller defects).
The difference set D7 = [d, 3d + 5] is perfect or nonperfect as the defect, d,
is odd or even (respectively). In the next two lemmas we construct extended
(k), for even and odd defects.
Langford sequences for D7 , namely L2d+6
d
828
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(k), d even, k ∈ [2d + 11, 4d + 9] − {3d + 9, 3d + 11}
Table 4b. Sequences hL2d+5
d
4b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
ai
bi
|bi − ai |
1
2
3
4+j
d/2 + 4
d/2 + 5
d/2 + 6 + j
d+6+j
3d/2 + 7 + j
2d + 6
2d + 7
2d + 8
3d + 9
4d + 11
1
3d/2 + 6
d+5
2d + 10 + 2j
2d + 9
3d/2 + 5
3d + 12 + 2j
2d + 11 + 2j or 1
3d + 13 + 2j or 1
3d + 10
4d + 12
3d + 11
4d + 10
4d + 11
0
3d/2 + 4
d+2
2d + 6 + j
3d/2 + 5
d
5d/2 + 6 + j
d+5+j
3d/2 + 6 + j
d+4
2d + 5
d+3
d+1
0
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
–
–
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 2
≤ d/2 − 2
Table 5a. Sequences hL2d+5
(k), d odd, k ∈ [2d + 7, 4d + 7] − {3d + 6, 3d + 8}
d
5a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
ai
bi
|bi − ai |
1
2
3+j
(d + 5)/2
(d + 7)/2
(d + 9)/2 + j
d+3
d+4+j
(3d + 11)/2 + j
2d + 5
2d + 8
3d + 6
3d + 8
4d + 11
1
(3d + 9)/2
2d + 10 + 2j
2d + 6
(3d + 7)/2
3d + 11 + 2j
3d + 9
2d + 7 + 2j or 1
3d + 10 + 2j or 1
4d + 10
4d + 12
4d + 8
4d + 9
4d + 11
0
(3d + 5)/2
2d + 7 + j
(3d + 7)/2
d
(5d + 13)/2 + j
2d + 6
d+3+j
(3d + 9)/2 + j
2d + 5
2d + 4
d+2
d+1
0
–
–
0≤j
–
–
0≤j
–
0≤j
0≤j
–
–
–
–
–
≤ (d − 3)/2
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d − 3)/2
6.1
For m = 2d + 6 and even d ≥ 4, the set [1, 2m + 1] − {k} can be
partitioned into differences [d, d + m − 1] if k is even.
P r o o f. By Table 6a and Table 6b, an L2d+6
(k) exists for d even and k ∈
d
[2d+8, 4d+12] . The required values of the extension k are obtained by reversing
these sequences.
829
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 11, 4d + 9] − {3d + 10, 3d + 12}
Table 5b. Sequences hL2d+5
d
5b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
ai
bi
|bi − ai |
1
2
3
4+j
(d + 5)/2
(d + 7)/2
(d + 9)/2 + j
d+6+j
(3d + 15)/2 + j
2d + 6
2d + 7
2d + 8
3d + 10
4d + 11
1
(3d + 13)/2
d+5
2d + 10 + 2j
(3d + 11)/2
2d + 9
3d + 9 + 2j
2d + 11 + 2j or 1
3d + 14 + 2j or 1
3d + 7
4d + 12
3d + 12
4d + 10
4d + 11
0
(3d + 9)/2
d+2
2d + 6 + j
d+3
(3d + 11)/2
(5d + 9)/2 + j
d+5+j
(3d + 13)/2 + j
d+1
2d + 5
d+4
d
0
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
–
–
≤ (d − 5)/2
≤ (d − 1)/2
≤ (d − 3)/2
≤ (d − 5)/2
Table 6a. Sequences L2d+6
(k), d even, k ∈ [2d + 8, 4d + 8] − {3d + 8, 3d + 10}
d
6a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
ai
bi
|bi − ai |
1
2
3
4+j
d/2 + 3
d/2 + 4 + j
d+4
d+5+j
3d/2 + 5
3d/2 + 7 + j
2d + 6
2d + 9
3d + 8
3d + 10
d+3
2
3d/2 + 6
2d + 11 + 2j
2d + 7
3d + 11 + 2j
3d + 9
2d + 8 + 2j or 2
4d + 11
3d + 12 + 2j or 2
4d + 12
4d + 13
4d + 9
4d + 10
d+2
0
3d/2 + 3
2d + 7 + j
3d/2 + 4
5d/2 + 7 + j
2d + 5
d+3+j
5d/2 + 6
3d/2 + 5 + j
2d + 6
2d + 4
d+1
d
–
–
–
0≤j
–
0≤j
–
0≤j
–
0≤j
–
–
–
–
≤ d/2 − 2
≤ d/2 − 2
≤ d/2 − 1
≤ d/2 − 2
6.2
For m = 2d + 6 and odd d ≥ 4, the set [1, 2m + 1] − {k} can be
partitioned into differences [d, d + m − 1] if k is odd.
(k) exists for k ∈ [2d + 7,
P r o o f. By Table 7a, Table 7b, and Table 7c an L2d+6
d
4d + 13] . The other required extensions k are obtained by reversing these sequences.
830
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(k), d even, k ∈ [2d + 12, 4d + 12] − {3d + 12}
Table 6b. Sequences L2d+6
d
6b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
ai
bi
|bi − ai |
1
2
3+j
d/2 + 4
d/2 + 5
d/2 + 6 + j
d+6
d+7+j
3d/2 + 8 + j
3d + 11
3d + 12
d+5
2
2d + 9 + 2j
3d/2 + 7
2d + 10
3d + 13 + 2j
2d + 8
2d + 12 + 2j or 2
3d + 14 + 2j or 2
4d + 11
4d + 13
d+4
0
2d + 6 + j
d+3
3d/2 + 5
5d/2 + 7 + j
d+2
d+5+j
3d/2 + 6 + j
d
d+1
–
–
0≤j
–
–
0≤j
–
0≤j
0≤j
–
–
≤ d/2
≤ d/2 − 2
≤ d/2 − 1
≤ d/2 − 1
Table 7a. A L2d+6
(2d + 7), d odd
d
7a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
ai
bi
|bi − ai |
1
2
3+j
(d + 5)/2
(d + 7)/2 + j
d+5+j
(3d + 11)/2
(3d + 13)/2 + j
2d + 7
2d + 8
3d + 8
3d + 9
d+4
d+3
2d + 9 + 2j
(3d + 9)/2
3d + 10 + 2j
2d + 10 + 2j
4d + 11
3d + 11 + 2j
2d + 7
4d + 13
4d + 12
4d + 9
d+3
d+1
2d + 6 + j
d+2
(5d + 13)/2 + j
d+5+j
(5d + 11)/2
(3d + 9)/2 + j
0
2d + 5
d+4
d
–
–
0≤j
–
0≤j
0≤j
–
0≤j
–
–
–
–
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 1)/2
We now construct some hooked extended Langford sequences for D7 , namely
(k), k ≥ 2d + 8.
hL2d+6
d
6.3 For m = 2d+6 and even d ≥ 4, the set [1, 2m+2]−{k, 2m+1} can
be partitioned into differences [d, d + m − 1] if k is odd and k ≥ m + 2 = 2d + 8.
P r o o f. The required extensions are obtained in Table 8a and Table 8b.
6.4 For m = 2d + 6 and odd d ≥ 4, the set [1, 2m + 2] − {k, 2m + 1} can
be partitioned into differences [d, d + m − 1] if k is even and k ≥ m + 2 = 2d + 8.
P r o o f. The required extensions are obtained in Table 9a and Table 9b.
831
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 9, 4d + 13] − {3d + 12}
Table 7b. Sequences L2d+5
d
7b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
ai
bi
|bi − ai |
1
2
3+j
(d + 5)/2
(d + 7)/2
(d + 9)/2 + j
d+5+j
(3d + 15)/2 + j
3d + 9
3d + 12
1
d+4
2d + 10 + 2j
2d + 8
(3d + 13)/2
3d + 11 + 2j
2d + 9 + 2j or 1
3d + 14 + 2j or 1
4d + 10
4d + 12
0
d+2
2d + 7 + j
(3d + 11)/2
d+3
(5d + 13)/2 + j
d+4+j
(3d + 13)/2 + j
d+1
d
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d + 1)/2
≤ (d − 1)/2
Table 7c. Sequences L2d+5
(k), d odd, k ∈ [2d + 9, 4d + 9] − {3d + 8, 3d + 10}
d
7c
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
ai
bi
|bi − ai |
1
2
3+j
(d + 5)/2
(d + 7)/2
(d + 9)/2 + j
d+5+j
(3d + 11)/2
(3d + 13)/2 + j
2d + 6
2d + 8
3d + 8
3d + 10
1
d+4
2d + 10 + 2j
2d + 7
(3d + 9)/2
3d + 11 + 2j
2d + 9 + 2j or 1
3d + 9
3d + 12 + 2j or 1
4d + 12
4d + 13
4d + 11
4d + 10
0
d+2
2d + 7 + j
(3d + 9)/2
d+1
(5d + 13)/2 + j
d+4+j
(3d + 7)/2
(3d + 11)/2 + j
2d + 6
2d + 5
d+3
d
–
–
0≤j
–
–
0≤j
0≤j
–
0≤j
–
–
–
–
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
7. Difference set D8 , m = 2d + 7
As in the preceding section, we take d ≥ 4 (though many tables are valid for
even smaller defects).
For each d the set D8 = [d, 3d + 6] is a perfect difference set.
7.1 For m = 2d + 7 and d ≥ 4, the set [1, 2m + 1] − {k} can be
partitioned into differences [d, d + m − 1] if k is odd.
P r o o f. When d is even the required extensions are obtained in Table 10a and
Table 10b (and by reversing those sequences).
832
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(k), d even, k ∈ [2d + 9, 4d + 9] − {3d + 7}
Table 8a. Sequences hL2d+6
d
8a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
ai
bi
|bi − ai |
1
2
3+j
d/2 + 3
d/2 + 4
d/2 + 5 + j
d+5+j
3d/2 + 5 + j
2d + 7
2d + 8
3d + 7
3d + 10
4d + 13
1
d+4
2d + 10 + 2j
2d + 6
3d/2 + 4
3d + 12 + 2j
2d + 9 + 2j or 1
3d + 9 + 2j or 1
4d + 12
4d + 14
4d + 10
4d + 11
4d + 13
0
d+2
2d + 7 + j
3d/2 + 3
d
5d/2 + 7 + j
d+4+j
3d/2 + 4 + j
2d + 5
2d + 6
d+3
d+1
0
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
–
–
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 2
≤ d/2
Table 8b. Sequences hL2d+6
(k), d even, k ∈ [2d + 9, 4d + 11] − {3d + 9}
d
8b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
ai
bi
|bi − ai |
1
2
3+j
d/2 + 3
d/2 + 4
d/2 + 5 + j
d+5+j
3d/2 + 6 + j
2d + 8
3d + 9
3d + 10
4d + 13
1
d+4
2d + 10 + 2j
2d + 7
3d/2 + 5
3d + 12 + 2j
2d + 9 + 2j or 1
3d + 11 + 2j or 1
4d + 14
4d + 12
4d + 10
4d + 13
0
d+2
2d + 7 + j
3d/2 + 4
d+1
5d/2 + 7 + j
d+4+j
3d/2 + 5 + j
2d + 6
d+3
d
0
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
–
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 1
≤ d/2
When d ≥ 7 is odd the required extensions are obtained in Table 11a and
Table 11b (and by reversing those sequences). When d = 5 we append the
sequence L17
5 (29) = ljk58bi65ahc86gebfdajlkci97h egdf79.
We now construct some hooked extended Langford sequences for D8 , namely
(k) with k ≥ m + 2 = 2d + 9.
hL2d+7
d
7.2
For m = 2d + 7 and d ≥ 4, the set [1, 2m + 2] − {k, 2m + 1} can
be partitioned into differences [d, d + m − 1] if k is even and k ≥ m + 2.
833
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 12, 4d + 14] − {3d + 11}
Table 9a. Sequences hL2d+6
d
9a
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
ai
bi
|bi − ai |
1
2
3
4+j
(d + 9)/2
(d + 11)/2
(d + 13)/2 + j
d+7+j
(3d + 15)/2 + j
2d + 10
3d + 11
4d + 13
d+5
2
d+6
2d + 11 + 2j
2d + 9
(3d + 13)/2
3d + 14 + 2j
2d + 12 + 2j or 2
3d + 13 + 2j or 2
3d + 12
4d + 11
4d + 13
d+4
0
d+3
2d + 7 + j
(3d + 9)/2
d+1
(5d + 15)/2 + j
d+5+j
(3d + 11)/2 + j
d+2
d
0
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
≤ (d − 1)/2
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d + 1)/2
Table 9b. Sequences hL2d+6
(k), d odd, k ∈ [2d + 8, 4d + 8] − {3d + 7, 3d + 9}
d
9b
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
ai
bi
|bi − ai |
1
2
3
4+j
(d + 7)/2
(d + 9)/2 + j
d+4
d+5+j
(3d + 11)/2
(3d + 13)/2 + j
2d + 7
2d + 9
3d + 7
3d + 9
4d + 13
(3d + 9)/2
2
d+3
2d + 11 + 2j
2d + 6
3d + 12 + 2j
3d + 10
2d + 8 + 2j or 2
4d + 12
3d + 11 + 2j or 2
4d + 11
4d + 14
4d + 9
4d + 10
4d + 13
(3d + 7)/2
0
d
2d + 7 + j
(3d + 5)/2
(5d + 15)/2 + j
2d + 6
d+3+j
(5d + 13)/2
(3d + 9)/2 + j
2d + 4
2d + 5
d+2
d+1
0
–
–
–
0≤j
–
0≤j
–
0≤j
–
0≤j
–
–
–
–
–
≤ (d − 3)/2
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d − 3)/2
P r o o f. For even defects d ≥ 6 the result follows from Table 12a and Table 12b,
but for d = 4 we append hL15
4 (28) = igha468f4bd6ca8e9gihb7fdc95 7e 5.
For odd defects the result follows from Table 13a and Table 13b.
834
Unauthenticated
Download Date | 6/16/17 2:11 PM
HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(k), d even, k ∈ [2d + 11, 4d + 15] − {3d + 11, 3d + 13}
Table 10a. Sequences L2d+7
d
10a
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
1
2
3
4+j
d/2 + 4
d/2 + 5
d/2 + 6 + j
d+6+j
3d/2 + 8 + j
2d + 9
3d + 11
3d + 12
1
3d/2 + 7
d+5
2d + 12 + 2j
2d + 10
3d/2 + 6
3d + 14 + 2j
2d + 11 + 2j or 1
3d + 15 + 2j or 1
3d + 13
4d + 14
4d + 12
0
3d/2 + 5
d+2
2d + 8 + j
3d/2 + 6
d+1
5d/2 + 8 + j
d+5+j
3d/2 + 7 + j
d+4
d+3
d
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 1
≤ d/2
Table 10b. Sequences L2d+7
(k), d even, k ∈ [2d + 9, 4d + 7] − {3d + 9}
d
10b
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
2
3
4+j
d/2 + 2
d/2 + 3 + j
d+5+j
3d/2 + 5
3d/2 + 6 + j
2d + 5
2d + 7
2d + 8
2d + 10
3d + 8
3d + 9
1
d+4
d+3
2d + 12 + 2j
2d + 6
3d + 10 + 2j
2d + 9 + 2j or 1
4d + 11
3d + 11 + 2j or 1
4d + 10
4d + 13
4d + 15
4d + 14
4d + 9
4d + 12
0
d+2
d
2d + 8 + j
3d/2 + 4
5d/2 + 7 + j
d+4+j
5d/2 + 6
3d/2 + 5 + j
2d + 5
2d + 6
2d + 7
2d + 4
d+1
d+3
–
–
–
0≤j
–
0≤j
0≤j
–
0≤j
–
–
–
–
–
–
≤ d/2 − 3
≤ d/2 − 1
≤ d/2 − 1
≤ d/2 − 2
8. Main result
The following is our main result.
8.1 If m ≥ 8d + 3 or m = 2d − 1, 2d or 2d + 1, then the set
[1, 2m + 2] − {k, 2m + 1} can be partitioned into differences [d, d + m − 1] if and
only if (m, k) ≡ (0, 0), (1, d + 1), (2, 1), (3, d) (mod (4, 2)), with the exception of
(d, m, k) = (1, 1, 2), (2, 3, 2).
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 11, 4d + 13] − {3d + 10, 3d + 12, 3d + 14}
Table 11a. Sequences L2d+7
d
11a
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
1
2
3
4+j
(d + 7)/2
(d + 9)/2
(d + 11)/2 + j
d+5
d+6+j
(3d + 13)/2
(3d + 17)/2 + j
3d + 10
3d + 11
3d + 14
1
2d + 9
(3d + 15)/2
2d + 12 + 2j
2d + 10
(3d + 11)/2
3d + 13 + 2j
2d + 8
2d + 11 + 2j or 1
3d + 12
3d + 16 + 2j or 1
4d + 12
4d + 15
4d + 14
0
2d + 7
(3d + 9)/2
2d + 8 + j
(3d + 13)/2
d+1
(5d + 15)/2 + j
d+3
d+5+j
(3d + 11)/2
(3d + 15)/2 + j
d+2
d+4
d
–
–
–
0≤j
–
–
0≤j
–
0≤j
–
0≤j
–
–
–
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
Table 11b. Sequences L2d+7
(k), d odd, k ∈ [2d + 9, 4d + 7] − {3d + 8}
d
11b
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
2
3
4+j
(d + 5)/2
(d + 7)/2 + j
d+5+j
(3d + 9)/2
(3d + 11)/2 + j
2d + 5
2d + 7
2d + 8
2d + 10
3d + 8
3d + 9
1
d+4
d+3
2d + 12 + 2j
2d + 6
3d + 11 + 2j
2d + 9 + 2j or 1
4d + 11
3d + 10 + 2j or 1
4d + 10
4d + 13
4d + 15
4d + 14
4d + 9
4d + 12
0
d+2
d
2d + 8 + j
(3d + 7)/2
(5d + 15)/2 + j
d+4+j
(5d + 13)/2
(3d + 9)/2 + j
2d + 5
2d + 6
2d + 7
2d + 4
d+1
d+3
–
–
–
0≤j
–
0≤j
0≤j
–
0≤j
–
–
–
–
–
–
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 3)/2
P r o o f. If m = 2d−1, 2d or 2d+1, then the result follows from Theorem 3.6. So
we suppose that m ≥ 8d+3 and we proceed to (twice) partition D = [d, d+m−1]
into two intervals.
Write d + m − 1 = 3d − 2 + i, where 0 ≤ i ≤ 2, and let m = d − d so that
[d, d − 1] contains m differences. The first partition of D is
[d, d + m − 1] = [d, d − 1] ∪ [d , d + m − 1].
(10)
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
(k), d even, k ∈ [2d + 10, 4d + 10] − {3d + 10}
Table 12a. Sequences hL2d+7
d
12a
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
2
3
4
5+j
d/2 + 4
d/2 + 5 + j
d+6+j
3d/2 + 6
3d/2 + 7 + j
2d + 7
2d + 11
3d + 10
3d + 11
4d + 15
d+4
2
2d + 9
d+5
2d + 13 + 2j
2d + 8
3d + 13 + 2j
2d + 10 + 2j or 2
4d + 13
3d + 12 + 2j or 2
4d + 14
4d + 16
4d + 12
4d + 11
4d + 15
d+3
0
2d + 6
d+1
2d + 8 + j
3d/2 + 4
5d/2 + 8 + j
d+4+j
5d/2 + 7
3d/2 + 5 + j
2d + 7
2d + 5
d+2
d
0
–
–
–
–
0≤j
–
0≤j
0≤j
–
0≤j
–
–
–
–
–
≤ d/2 − 2
≤ d/2 − 2
≤ d/2 − 1
≤ d/2 − 1
Table 12b. Sequences hL2d+7
(k), d even, k ∈ [2d + 12, 4d + 14] −
d
{3d + 12, 3d + 14, 3d + 16}
12b
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
2
3
4+j
d/2 + 4
d/2 + 5
d/2 + 6
d/2 + 7 + j
d+7+j
3d/2 + 9
3d/2 + 10 + j
2d + 9
3d + 11
3d + 16
4d + 15
3d/2 + 8
2
d+6
2d + 11 + 2j
2d + 10
3d + 12
3d/2 + 7
3d + 15 + 2j
2d + 12 + 2j or 2
3d + 14
3d + 18 + 2j or 2
3d + 13
4d + 13
4d + 16
4d + 15
3d/2 + 7
0
d+3
2d + 7 + j
3d/2 + 6
5d/2 + 7
d+1
5d/2 + 8 + j
d+5+j
3d/2 + 5
3d/2 + 8 + j
d+4
d+2
d
0
–
–
–
0≤j
–
–
–
0≤j
0≤j
–
0≤j
–
–
–
–
≤ d/2 − 1
≤ d/2 − 2
≤ d/2 − 1
≤ d/2 − 2
The second partition of D is
[d, d + m − 1] = [d, d − 3] ∪ [d − 2, d + m − 1].
(11)
Let m = m − 2, so that m is the number of differences in [d, d − 3]. Note
that d + m − 1 = 3(d − 2) − 2 + (6 + i) (with the same i as before) and
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SHAI MOR — VÁCLAV LINEK
(k), d odd, k ∈ [2d + 10, 4d + 10] − {3d + 9}
Table 13a. Sequences hL2d+7
d
13a
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
(14)
(15)
1
2
3
4
5+j
(d + 9)/2
(d + 11)/2 + j
d+6+j
(3d + 11)/2
(3d + 13)/2 + j
2d + 7
2d + 11
3d + 9
3d + 12
4d + 15
d+4
2
2d + 9
d+5
2d + 13 + 2j
2d + 8
3d + 14 + 2j
2d + 10 + 2j or 2
4d + 13
3d + 11 + 2j or 2
4d + 14
4d + 16
4d + 11
4d + 12
4d + 15
d+3
0
2d + 6
d+1
2d + 8 + j
(3d + 7)/2
(5d + 17)/2 + j
d+4+j
(5d + 15)/2
(3d + 9)/2 + j
2d + 7
2d + 5
d+2
d
0
–
–
–
–
0≤j
–
0≤j
0≤j
–
0≤j
–
–
–
–
–
≤ (d − 3)/2
≤ (d − 5)/2
≤ (d − 3)/2
≤ (d − 1)/2
Table 13b. Sequences hL2d+7
(k), d odd, k ∈ [2d + 12, 4d + 14] −
d
{3d + 11, 3d + 13}
13b
ai
bi
|bi − ai |
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)
(11)
(12)
(13)
1
2
3
4+j
(d + 9)/2
(d + 11)/2
(d + 13)/2 + j
d+7+j
(3d + 17)/2 + j
2d + 10
3d + 12
3d + 13
4d + 15
(3d + 13)/2
2
d+6
2d + 11 + 2j
2d + 9
(3d + 15)/2
3d + 14 + 2j
2d + 12 + 2j or 2
3d + 15 + 2j or 2
3d + 11
4d + 16
4d + 13
4d + 15
(3d + 11)/2
0
d+3
2d + 7 + j
(3d + 9)/2
d+2
(5d + 15)/2 + j
d+5+j
(3d + 13)/2 + j
d+1
d+4
d
0
–
–
–
0≤j
–
–
0≤j
0≤j
0≤j
–
–
–
–
≤ (d − 1)/2
≤ (d − 3)/2
≤ (d − 3)/2
≤ (d − 1)/2
that since m ≥ 8d + 3 we have m = d − d − 2 ≥ 2d + 1 (and obviously
m = m + 2 ≥ 2d + 1). The difference interval [d − 2, d + m − 1] of (11) is of the
form Dj for some 6 ≤ j ≤ 8 with j = 6 + i, and Theorem 4.1 and Theorem 4.2
apply to these difference intervals.
We make four cases depending on the perfect/nonperfect status of the difference intervals involved.
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
Case 1.
In this case [d, d + m − 1]P = [d, d − 1]P ∪ [d , d + m − 1]P .
(k )
By Theorem 2.1 a Ldd −d exists, and by Theorem 3.6 there exists a hLd+m−d
d
for all k ∈ [1, 2(m + d − d ) + 2] . Now form the concatenation
d −d
hLm
hLd+m−d
(k ).
d (k) = Ld
d
(12)
Since k = k + 2(d − d), as the extension k in hLd+m−d
(k ) is varied, all
d
m
k-extensions k ∈ [2(d − d) + 1, 2m + 2] are achieved in hLd (k).
Considering the second partition (11), we have [d, d+m−1]P = [d, d −3]N P ∪
+2 (k ) exists for each possible
[d − 2, d + m − 1]N P . By Theorem 4.1 a Ld+m−d
d −2
k , and by Theorem 2.2 there exists a hLdd −d−2 . Now form the concatenation
d+m−d +2 hLm
(k ) hLdd −d−2 .
d (k) = Ld −2
(13)
+2 (k ) is varied all k-extensions
Since k = k , as the extension k in Ld+m−d
d −2
m
k ∈ [1, 2(d + m − d + 2) + 1] in hLd (k) are achieved.
To complete this case we must check that the two intervals for the extension
k overlap, that is, we must check that
2(d − d) + 1 ≤ 2(d + m − d + 2) + 1.
(14)
This inequality is equivalent to 2d ≤ 2d + m + 2, which upon substituting
d + m − 1 = 3d − 2 + i is equivalent to 2d ≤ 3d + d + i + 1, which is true.
Case 1 is complete.
Case 2.
In this case [d, d + m − 1]P = [d, d − 1]N P ∪ [d , d + m − 1]N P .
In this case note that the second partition (11) is of the form [d, d+ m − 1]P =
[d, d − 3]P ∪ [d − 2, d + m − 1]P .
By Theorem 2.1 a Ldd −d−2 exists and by Theorem 2.2 a hLdd −d exists. By
Theorem 3.5 there exists a Ld+m−d
(k ) for all k ∈ [1, 2(d + m − d ) + 1] .
d
d+m−d +2 (k ) exists for all k ∈ [d + m − d + 4, 2(d + m
By Theorem 4.2 a hLd −2
− d + 2) + 2] .
Now form the concatenations
d+m−d
hLm
(k ) hLdd −d ,
d (k) = Ld
(15)
d −d−2
+2 hLd+m−d
(k ).
hLm
d (k) = Ld
d −2
(16)
(k ) is varied all kSince k = k in (15), as the extension k in hLd+m−d
d
m
extensions k ∈ [1, 2(d + m − d ) + 1] are achieved in hLd (k).
+2 Since k = k + 2(d − d − 2) in (15), as the extension k in hLd+m−d
(k ) is
d −2
m
varied all k-extensions k ∈ [m + d − d, 2m + 2] in hLd (k) are achieved.
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To complete this case we check that the two intervals for the extension k
overlap, that is, we check that
m + d − d ≤ 2(d + m − d ) + 1.
(17)
This inequality is equivalent to 3d ≤ 3d + m + 1, which upon substituting
d + m − 1 = 3d − 2 + i is equivalent to 3d ≤ 3d + 2d + i, which is true. Case 2
is complete.
Case 3.
In this case [d, d + m − 1]N P = [d, d − 1]P ∪ [d , d + m − 1]N P .
In this case note that the second partition (11) is of the form [d, d+m−1]N P =
[d, d − 3]N P ∪ [d − 2, d + m − 1]P .
By Theorem 2.1 a Ldd −d exists and by Theorem 2.2 a hLdd −d−2 exists. By
(k ) for all k ∈ [1, 2(d + m − d ) + 2] . By
Theorem 3.6 there exists a hLd+m−d
d
+2 Theorem 4.1 a Ld+m−d
(k ) exists for all k ∈ [1, 2(d + m − d + 2) + 1] .
d −2
Now form the concatenations
d −d
hLm
hLd+m−d
(k ),
d (k) = Ld
d
hLm
d (k)
=
+2 Ld+m−d
(k )
d −2
(18)
hLdd −d−2 .
(19)
(k ) is
Since k = k + 2(d − d) in (18), then as the extension k in hLd+m−d
d
m
varied all k-extensions k ∈ [2(d − d) + 1, 2m + 2] are achieved in hLd (k).
+2 (k ) is varied all
Since k = k in (19), then as the k -extension in Ld+m−d
d −2
k-extensions k ∈ [1, 2(d + m − d + 2) + 1] in hLm
(k)
are
achieved.
d
The two intervals for the extension k will overlap if
2(d − d) + 1 ≤ 2(d + m − d + 2) + 1,
(20)
which earlier was shown to be true (see (14)). Case 3 is complete.
Case 4.
In this case we have [d, d + m − 1]N P = [d, d − 1]N P ∪ [d , d + m − 1]P .
In this case note that the second partition (11) is of the form [d, d+m−1]N P =
[d, d − 3]P ∪ [d − 2, d + m − 1]N P .
By Theorem 2.1 there exists a Ldd −d−2 , and by Theorem 2.2 a hLdd −d exists.
(k ) for all k ∈ [1, 2(d + m − d ) + 1] .
By Theorem 3.5 there exists a Ld+m−d
d
d+m−d +2 (k ) exists for all k ∈ [d + m − d + 4, 2(d + m −
By Theorem 4.2 a hLd −2
d + 2) + 2] .
Now form the concatenations
d+m−d
(k ) hLdd −d ,
hLm
d (k) = Ld
d −d−2
+2 hLd+m−d
(k ).
hLm
d (k) = Ld
d −2
(21)
(22)
(k ) is varied all extensions
Since k = k in (21), as the extension k in Ld+m−d
d
m
k ∈ [1, 2(d + m − d ) + 1] are achieved in hLd (k).
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HOOKED EXTENDED LANGFORD SEQUENCES OF SMALL AND LARGE DEFECTS
+2 Since k = k + 2(d − d − 2) in (22), as the extension k in hLd+m−d
(k ) is
d −2
m
varied all extensions k ∈ [d − d + m, 2m + 2] in hLd (k) are achieved.
The two intervals for the extension k will overlap if
m + d − d ≤ 2(d + m − d ) + 1,
(23)
which earlier was shown to be true (see (17)). Case 4 is complete.
This completes all four cases and proves the result.
9. Conclusion
The results obtained here and in [5], [6] still leave some open cases.
The construction of extended Langford sequences Lm
d (k) is still open for d ≥ 5,
m ∈ [2d + 2, 2d + 4] ∪ [2d + 8, 8d + 5], and k of the appropriate parity.
The construction of hooked extended Langford sequences hLm
d (k) is still open
for d ≥ 3, m ∈ [2d + 2, 2d + 4] ∪ [2d + 8, 8d + 2], and k of the appropriate parity;
and for m ∈ [2d + 5, 2d + 7], 2 ≤ k < m + 2, and k of the appropriate parity.
The degenerate case m = 2d − 2 discussed after Theorem 3.3 is also open, but
such sequences have yet to be used to make combinatorial designs and (so far)
are of limited interest.
Acknowledgement Mr. Mor thanks professor Brett Stevens of Carleton University for proofreading a first draft of this manuscript. Both authors thank the
referees for their comments.
REFERENCES
[1] BAKER, C. A.: Extended Skolem sequences, J. Combin. Des. 3 (1995), 363–379.
[2] BERMOND, J.-C.—BROUWER, A. E.—GERMA, A.: Systèmes de triplets et differences
associées. In: Problèmes combinatoires et théorie des graphes. Colloq. Internat. CNRS,
Univ. Orsay, Orsay, 1976, pp. 35–38.
[3] LANGFORD, C. D.: Problem, Math. Gaz. 42 (1958), 288.
[4] LINEK, V.—JIANG, Z.: Hooked k-extended Skolem sequences, Discrete Math. 196
(1999), 229–238.
[5] LINEK, V.—JIANG, Z.: Extended Langford sequences with small defects, J. Combin.
Theory Ser. A 84 (1998), 38–54.
[6] LINEK, V.—MOR, S.: On partitions of {1, . . . , 2m + 1}{k} into differences d, . . .
. . . , d + m − 1: Extended Langford sequences of large defect, J. Combin. Des. 12 (2004),
421–442.
[7] O’KEEFE, E. S.: Verification of a conjecture of Th. Skolem, Math. Scand. 9 (1961),
80–82.
[8] LINEK, V.—SHALABY, S.: The existence of (p, q)-extended Rosa sequences, Discrete
Math. 308 (2008), 1583–1602.
[9] NORDH, G.: Generalization of Skolem Sequences. M.Sc. Thesis. LITH-MAT-EX2003/05, Department of Mathematics, Linköpings Universitet, 2003.
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SHAI MOR — VÁCLAV LINEK
[10] NICKERSON, R. S.: Problem E1845; Solution, Amer. Math. Monthly 73; 74 (1966;
1967), 81; 591–592.
[11] PRIDAY, C. J.: On Langford’s problem (I), Math. Gaz. 43 (1959), 250–253.
[12] SHALABY, N.: Skolem and Langford Sequences. In: Handbook of Combinatorial Designs
(2nd ed.) (C. J. Colbourn, J. H. Dinitz, eds.) Chapman & Hall/CRC Press, Boca Raton,
2007, pp. 612–616.
[13] SIMPSON, J. E.: Langford sequences: perfect and hooked, Discrete Math. 44 (1983),
97–104.
[14] SKOLEM, TH.: On certain distributions of integers in pairs with given differences, Math.
Scand. 5 (1957), 57–68.
[15] SKOLEM, TH.: Some remarks on the triple systems of Steiner, Math. Scand. 6 (1958),
273–280.
Received 6. 9. 2011
Accepted 12. 4. 2012
* School of Mathematics and Statistics
Carleton University
Ottawa, ON K1S 5B6
CANADA
** Department of Mathematics and Statistics
University of Winnipeg
Winnipeg, MB R3B 2E9
CANADA
E-mail : [email protected]
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