ARS MATHEMATICA CONTEMPORANEA
ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.)
ARS MATHEMATICA CONTEMPORANEA 14 (2018) 415-432 https://doi.org/10.26493/1855-3974.1098.ca0
(Also available at http://amc-journal.eu)
On the number of additive permutations and Skolem-type sequences*
*
Diane M. Donovan
Centre for Discrete Mathematics and Computing, University of Queensland, St. Lucia 4072, Australia
Michael J. Grannell *
School of Mathematics and Statistics, The Open University, Walton Hall, Milton Keynes MK7 6AA, United Kingdom
Received 3 May 2016, accepted 31 May 2017, published online 9 November 2017
Cavenagh and Wanless recently proved that, for sufficiently large odd n, the number of transversals in the Latin square formed from the addition table for integers modulo n is greater than (3.246)n. We adapt their proof to show that for sufficiently large t the number of additive permutations on [—t, t] is greater than (3.246)2i+1 and we go on to derive some much improved lower bounds on the numbers of Skolem-type sequences. For example, it is shown that for sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order n = 7t+3 is greater than (3.246)6t+3. This compares with the previous best bound of 2 Ln/3J.
Keywords: Additive permutation, Skolem sequence, transversal. Math. Subj. Class.: 05B07, 05B10
1 Introduction
This paper is concerned with counting additive permutations and Skolem-type sequences. Additive permutations are related to certain kinds of transversals in Latin squares. A Latin square of order n may be envisaged as an n x n array having n distinct entries, each of which appears once in any one row and once in any one column. We adopt a slightly wider
* We thank the referees for helpful comments and for drawing to our attention the papers [6, 7, 8]. t Corresponding author.
E-mail addresses: dmd@maths.uq.edu.au (Diane M. Donovan), m.j.grannell@open.ac.uk (Michael J.
Abstract
Grannell)
©® This work is licensed under http://creativecommons.org/licenses/by/3.0/
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than usual definition of a transversal in an n x n array (not necessarily a Latin square) as a set of n (row, column, entry) triples that cover every row, every column and n distinct entries from the array. Consequently, a transversal in a Latin square will necessarily cover every entry precisely once.
In a recent paper, Cavenagh and Wanless [3] proved that, for all sufficiently large odd n, the number of transversals in the Latin square formed from the addition table for integers modulo n is greater than (3.246)n. This result was a substantial improvement on previous results in this area, although Vardi [10] conjectured that this number exceeds cnn! for some constant c G (0,1). A proof of this conjecture is claimed in the arXiv paper [6]. However, it is important to note that not all transversals are suitable for our purposes.
For integers a and b with a < b we use the notation [a, b] to denote the set of integers i such that a < i < b. An additive permutation n on [—t, t] is a permutation of these integers such that {i + n(i) : i G [—t, t]} is also a permutation on the same set of integers. This definition of an additive permutation is the one employed by Abram [1] and others in connection with Skolem sequences, but the reader is cautioned that it differs from that used in [6] where pointwise addition of permutations on [1, n] is taken modulo n. An examination of the proof given in [3] shows that it is possible to adapt the proof to show that the number of additive permutations on [—t, t] is greater than (3.246)2i+1 for all sufficiently large t, and this is done in Theorem 2.1 below. There are strong connections between additive permutations and Skolem-type sequences. We investigate some of these connections and obtain much improved lower bounds on the numbers of some Skolem-type sequences.
A pure Skolem sequence, sometimes simply called a Skolem sequence, of order n is a sequence (s1, s2,..., s2n) of 2n integers satisfying the following conditions.
(C1) For each k G {1,2,..., n} there are precisely two elements of the sequence, say s¿ and sj, such that s¿ = Sj = k.
(C2) If Sj = Sj = k and i < j then j — i = k.
For example, (4,1,1, 5,4,2,3, 2, 5, 3) is a pure Skolem sequence of order 5. It is well known that a pure Skolem sequence of order n exists if and only if n = 0 or 1 (mod 4). For this and other existence results mentioned below see, for example, [4, 5].
An extended Skolem sequence of order n is a sequence (s1, s2,..., s2n+1) of 2n +1 integers satisfying (C1) and (C2) above and such that precisely one element of the sequence is zero. An extended Skolem sequence of order n exists for every positive integer n. If the zero element of an extended Skolem sequence of order n appears in the 2n-th position, i.e. s2n = 0, then the sequence is called a hooked Skolem sequence. A hooked Skolem sequence of order n exists if and only if n = 2 or 3 (mod 4). If the zero element of an extended Skolem sequence of order n appears in the (n + 1)-th position, i.e. sn+1 = 0, then the sequence is called a split Skolem sequence or a Rosa sequence. A split Skolem sequence of order n exists if and only if n = 0 or 3 (mod 4).
A split-hooked Skolem sequence (also known as a hooked Rosa sequence) of order n is a sequence (s1, s2,..., s2n+2) of 2n + 2 integers satisfying (C1) and (C2) above and such that sn+1 = s2n+1 = 0. A split-hooked Skolem sequence of order n exists if and only if n = 1 or 2 (mod 4) and n =1.
The various types of Skolem sequence described above may be used to construct solutions to Heffter's first and second difference problems. These, in turn, may be used to construct cyclic Steiner triple systems. We will refer to the sequences just described, and
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to some near relatives, somewhat loosely as Skolem-type sequences. It was shown in [1, 2] that for many of these Skolem-type sequences (in particular, pure, hooked, split and split-hooked) the number of them is essentially bounded below by 2 Lf J, where n is the order of the sequence.
2 Additive permutations
Define the Latin square An of odd order n = 2t +1 to have its rows and columns indexed by the integers in [—t, t] and the (i, j) entry k G [—t, t] given by k = i + j (mod n). The array An gives the addition table on Zn. A Z-transversal T in An is a transversal in which every (row, column, entry) triple (i, j, k) G T has k = i + j in Z, so that no triples of a Z-transversal have i + j < -t or i + j > t. Not all transversals in An are Z-transversals, for example the transversal formed by the leading diagonal in An is not a Z-transversal. We will only count Z-transversals: in effect the (i, j) cells in An where i + j < -t or i + j > t are ignored. The entries in these cells are therefore irrelevant to our discussions and it will sometimes be helpful to take i + j as the entry, rather than i + j reduced modulo n. This has the advantage that the "ignored" cells are easily identified, particularly when considering subarrays of An - such cells are then precisely those with entries outside the range [—t, t].
Figure 1 shows the array A19 with the "ignored" entries greyed-out, and with a Z-transversal having its entries marked in boxes. For the present, disregard the highlighting of the subarrays.
If the (row, column, entry) triples of a Z-transversal in An are listed as a 3 x n array T* with row numbers of An forming the first row of T*, column numbers the second, and entries the third, then each row of this array contains the integers [—t,t], and the entries in the third row are the sums of the corresponding entries in the other two rows. Taking the first row of T* as [—t, t], the second row as a permutation n on [—t, t], and the third row as the vector sum of the first two rows, then n is an additive permutation on [—t, t]. Conversely, if n is an additive permutation on [—t, t] then the (row, column, entry) triples (i, n(i), i + n(i)) for i G [—t, t] form a Z-transversal in An. If the entries in the third row of such a 3 x n array T* are multiplied by ( — 1), then each column of the resulting array sums to zero, and the new array is called a zero-sum array. Thus, there is an equivalence between Z-transversals, additive permutations, and zero-sum arrays. Table 1 (taken from [3]) gives the number of these, here denoted by zn, for n = 2t +1 < 23. These numbers form the sequence A002047 in Sloane's encyclopaedia [9] and have been independently checked by ourselves using our own computer program. The table also gives a rounded down value for (zn)f which will be used subsequently. We remark that z2i+1 is also the number of extremal Langford sequences with defect t + 1 (i.e. starting with t + 1) - see [1, 4] for definitions.
Theorem 2.1. Suppose that b and n are odd and that n > 3b > 9. Then zn > (zb)2 L f—J.
Proof. Our proof is a re-working of that of Cavenagh and Wanless [3], ensuring that for general b and n the subarrays R and S (defined below) have appropriate (sub-)transversals and that the transversals constructed in An are indeed Z-transversals. We take b = 2a +1 (so that a > 1) and n = 2t + 1. Put k = |_n— J = |_^J, s = t — a — bk, and r = s + b. Then 0 < s < b, b < r < 2b, and one of r, s is odd and the other is even.
Next define the subarray M(¿,j),c of An to be the c x c block whose top left entry is in
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	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7	8	9
-9	1	2	3	4	5	6	7	8		0	-8	-7	-6	-5	-4	-3	-2	-1	0
-8	2	3	4	5	6	7	8	9	-9	-8	0	-6	-5	-4	-3	-2	-1	0	1
-7	3	4	5	6	7	8	9	-9	H	-7	-6	-5	-4	-3	-2	-1	0	1	2
-6	4	5	6	7	8	9	-9	-8	-7	-6	-5	-4	-3	0	-1	0	1	2	3
-5	5	6	7	8	9	-9	-8	-7	-6	-5	-4	0	-2	-1	0	1	2	3	4
-4	6	7	8	9	-9	-8	-7	-6	-5	-4	-3	-2	0	0	1	2	3	4	5
-3	7	8	9	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	0	4	5	6
-2	8	9	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	0	7
-1	9	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	0	5	6	7	8
0	-9	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7	8	0
1	-8	-7	-6	-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7	0	9	
2	-7		-5	-4	-3	-2	-1	0	1	2	3	4	5	6	7	8	9	-9	-8
3	-6	-5	H	-3	-2	-1	0	1	2	3	4	5	6	7	8	9	-9	-8	-7
4		-4	-3	-2	-1	0	1	2	3	4	5	6	7	8	9	-9	-8	-7	-6
5	-4	-3	-2	-1	0	□	2	3	4	5	6	7	8	9	-9	-8	-7	-6	-5
6	-3	-2	-1		1	2	3	4	5	6	7	8	9	-9	-8	-7	-6	-5	-4
7	-2	-1	0	1	H	3	4	5	6	7	8	9	-9	-8	-7	-6	-5	-4	-3
8	-1	0	1	2	3	4	H	6	7	8	9	-9	-8	-7	-6	-5	-4	-3	-2
9	0	1	2	3	4	5	6	0	8	9	-9	-8	-7	-6	-5	-4	-3	-2	-1
Figure 1: The Latin square A19.
Table 1: The number of Z-transversals in An
3 5 7 9 11 13 15 17 19 21 23
2 6 28 244 2544 35600 659632 15 106128 425 802 176 14409 526080 577 386122880
(Zn) n >
1.259 1.430 1.609 1.841 2.039 2.239 2.443 2.644 2.845 3.046 3.246
n
z
n
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position (i, j) of the array An. For example, if n =19 (see Figure 1) then
/-4 -3 -2> M(_6,2),3 = ( -3 -2 -1
V-2 -1
As previously mentioned, it is convenient to take the entries of subarrays unreduced modulo n. Allowing this, M(ij)c has entries from i + j to i + j + 2(c - 1) and, if c is odd, there is a central entry i + j + c - 1. We use the following subarrays:
Type 1:	A = M{_t+ib_a+ib)ib for i = 0,1,..., (k - 1),
Type 2:	E = M(-t+r+(k+i)b,-t+ib),b for i = 0,1,..., (k - 1),
Type 3:	R = M(-t+kb, -a+kb),r and
Type 4:	S = M(t_s+i,_a-s),s.
In Figure 1, n = 19 and b = 3, so that t = 9, a = 1, k = 2, s = 2 and r = 5. The subarrays of types 1 and 2 are lightly shaded and the subarrays R and S are shaded more heavily.
Altogether there are 2k + 2 subarrays of the four types. No two of these have a common row and the total number of rows covered is 2bk + r + s = n, so each row of An is covered by precisely one of these subarrays. Similarly, each column of An is covered by precisely one of these subarrays. Consequently we may attempt to construct Z-transversals in An from transversals in the subarrays.
The type 1 subarray Di has a central entry 2ib -1 + a. If this value is subtracted from every entry in Di, the resulting array is a copy of Ab. Since Ab has zb Z-transversals, each Di has zb transversals that are symmetric about its central entry, that is to say transversals each covering the entries from 2ib - t to 2ib - t + (b - 1) inclusive. Note that these transversals avoid "ignored" cells of An. Similarly, the type 2 subarray Ei has zb transversals symmetric about its central entry 2ib -1 + 3a + 1, and each of these covers the entries from 2ib -1 + b to 2ib -1 + 2b - 1 inclusive. Collectively the type 1 and type 2 subarrays have (zb)2k transversals covering the entries from -t to t - r - s inclusive. All of these are partial Z-transversals of An.
To complete the proof for all b > 3 and n > 3b we must show that the remaining subarrays R and S have appropriate transversals (i.e. avoiding "ignored" cells of An) that cover the entry values from t - r - s + 1 to t inclusive. It will then follow that An has at least (zb)2k = (zb)2LZ-transversals.
The subarray R has entries ranging from t - r - s - a + 1 to t + a inclusive, and so R always contains some "ignored" cells of An, namely those with entries exceeding t. The subarray S has entries ranging from t - r - s + 2 + a to t - a - 1 inclusive, so S does not contain any "ignored" cells of An. If we subtract t - s - a from all the entries in R and S we obtain equivalent arrays R' and S', where R' has entries ranging from -(s + 2a) to s + 2a, while S' has entries ranging from - (s -1) to s -1, and we are seeking transversals in these arrays that cover the entry values from -(s + a) to s + a inclusive. Cells in R' that contain entries greater than s + a correspond to the "ignored" cells of R. Our proof that such transversals always exist falls into a number of cases, the details of which are lengthy, and so are postponed until Section 4.	□
Corollary 2.2. If n is odd and sufficiently large, then zn > (3.246)n.
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Proof. Theorem 2.1 gives zn > (zb)2LtitJ > (zb)n_3 for b odd and all sufficiently large n. From Table 1 we have (z23) 23 > 3.246, so taking b = 23 we obtain zn > (3.246)n for all sufficiently large n.	□
Putting the corollary into words, the number of additive permutations on [—t, t] is greater than (3.246)2i+1 for all sufficiently large t.
3 Skolem-type sequences
A connection between additive permutations and Skolem-type sequences is formed by so-called (m, 3, c)-systems. A set D = {D1, D2,..., Dm}, where each Dj is a triple of positive integers (a^ bj, aj + bj) with aj < b and |Jm=1 Dj = {c, c + 1,..., c + 3m - 1} is called an (m, 3, c)-system. As remarked in [1], such a system exists if and only if
(i)	m > 2c - 1 , and
(ii)	m = 0 or 1 (mod 4) if c is odd, or m = 0 or 3 (mod 4) if c is even.
Given an (m, 3, c)-system D = {D1, D2,..., Dm}, where Dj = (aj, bj, aj + bj), and putting r = c + 3m - 1, a (Skolem-type) sequence (x_r, x_r+1,..., xr-1, xr) may be constructed by putting x_(ai+bi) = aj, x_bi = aj, x_a = aj + bj, xa = bj, xbi = aj + bj, xai+6i = bj for i = 1, 2,..., m, and xj = 0 for — c < j < c. For example, if c =2 and m = 3, and if D = {D1,D2,D3} where D1 = (2,6, 8), D2 = (3, 7,10), D3 = (4, 5,9) then the constructed sequence is
(3,4, 2, 3, 2,4, 9,10, 8, 0, 0,0, 6, 7, 5, 9, 8,10, 6, 5, 7).
Observe that in such a sequence, for each k e {c, c + 1,..., r} the two positions occupied by k are precisely k apart. Further observe that, independently for each i e{1,2,..., m}, wemayreplace xj for j e {—aj — bj, —bj, —aj ,aj,bj,aj + bj} by xj where x_(a) = bj, x'_^ = aj + bj, x_a. = bj, x^. = aj + bj, x^. = aj, x^= aj. Thus we may obtain 2m distinct sequences of length 2r + 1 each of which has the property that for each k e {c, c + 1,..., r}, the two positions occupied by k are precisely k apart. Each such sequence has zeros in the central 2c — 1 positions.
If n is an additive permutation on [—t, t] then a (2t + 1,3, t + 1)-system is formed by the set of triples {Dj : i e [—t, t]} where
Dj = (i + 2t + 1, n(i) + 4t + 2, i + n(i) + 6t + 3).
Note that the first entries in these triples cover the interval [t + 1,3t +1], the second entries cover [3t + 2,5t + 2], and the third entries cover [5t + 3, 7t + 3]. If n1 and n2 are two different additive permutations on [—t, t], then the two resulting (2t + 1, 3, t + 1)-systems contain different triples. Consequently the number of different (2t + 1, 3, t + 1)-systems is bounded below by z2t+1, and hence by (3.246)2i+1 for all sufficiently large t. Combining this with the previous observation that each such sequence gives rise to 22i+1 Skolem-type sequences, we obtain the following result.
Theorem 3.1. For all sufficiently large t, there are more than (6.492)2i+1 Skolem-type sequences of length 14t + 7 having the following properties:
(a) there are zeros in the central 2t +1 positions, and
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(b) for k G {t + 1, t + 2,..., 7t + 3}, the two positions occupied by k are precisely k apart.
If the central 2t + 1 zero entries in such a Skolem-type sequence are replaced by a split Skolem sequence of order t (which exists for t = 0 or 3 (mod 4)) then a split Skolem sequence of order 7t + 3 is obtained. Hence we have the corollary:
Corollary 3.2. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 split Skolem sequences of order 7t + 3.
In fact we can achieve slightly better than this because the number of split Skolem sequences of order t is at least 2L 3 J for all t = 0 or 3 (mod 4) [1, 2], so we have at least that number of choices for replacing the central zeros. The bound (6.492)2i+1 > 25-3973t is (for large t) substantially better than the previous best bound of 2L 73+3j .
Given a split Skolem sequence of order n, we can form a pure Skolem sequence of order n +1 by replacing the central zero with n + 1 and placing a further entry n +1 at either the start or the end of the sequence. Hence we obtain:
Corollary 3.3. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 pure Skolem sequences of order 7t + 4.
In the next two corollaries, the lower bound of (6.492)2i+1 is extended to hooked and split-hooked Skolem sequences of orders 7t + 4 and 7t + 5 respectively. In each case the basic approach is as follows. For given t, choose a small positive integer c such that t — c = 2 or 11 (mod 12) if c is odd, or t — c = 8 or 11 (mod 12) if c is even. Put m = (t — c +1)/3 and assume that t is large enough to ensure that m > 2c — 1 (i.e. t > 7c — 4). Then there exists an (m, 3, c)-system from which a Skolem-type sequence T may be constructed that has length 2t + 1, has zeros in the central 2c — 1 positions and, for each k G {c, c + 1,..., t}, the two positions occupied by k are precisely k apart. The sequence T is used to replace the central 2t + 1 zeros in each sequence S of length 14t + 7 given by Theorem 3.1. We denote the resulting sequence as T ^ S (T into S), and this sequence has zeros in its central 2c — 1 positions. These are then replaced by a sequence Q of length 2c — 1 to form Q ^ (T ^ S), and a short sequence R of further entries is appended at the right-hand end of this sequence to form a sequence S' = (Q ^ (T ^ S)) A R (where A denotes appending). By choosing c, Q and R appropriately, it is possible to form hooked and split-hooked Skolem sequences S'.
To illustrate the procedure, we explain how to convert a Skolem-type sequence S of length 147 of the form described in Theorem 3.1, to a hooked Skolem sequence S' of order 74. Note that for k G {11,12,..., 73}, the two positions in S occupied by k are precisely k apart, and that S has zeros in the central 21 positions. Next take a (3,3,2)-system (for example, the one previously described) and from it form a Skolem-type sequence T of length 21 that has zeros in the central three positions and, for each k G {2,3,..., 10}, the two positions in T occupied by k are precisely k apart. Replace the central 21 zeros of S by the sequence T to form T ^ S. Then T ^ S has length 147, zeros in the central three positions and, for each k G {2,3,..., 73}, the two positions occupied by k are precisely k apart. Finally, replace the central three zeros in T ^ S by the sequence Q = (1,1, 74), and append the sequence R = (0, 74) to the right-hand end of Q ^ (T ^ S). The resulting sequence S' has length 149, has a zero in the penultimate position and, for each k g {1,, 2,..., 74}, the two positions occupied by k are precisely k apart. Hence S' is a
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hooked Skolem sequence of order 74. Clearly different sequences S give rise to different sequences S'.
We now return to the general cases. For given t it is obvious that different sequences S will result in different sequences S' = (Q ^ (T ^ S)) A R. So, to extend the bound, it suffices to specify c (and hence T), Q and R, and to check the parity conditions for t - c. In the next two corollaries to Theorem 3.1, we establish the bound by tabulating appropriate c, Q and R. We leave the reader to check the parity conditions for t - c and that the sequence S' is of the required type.
Corollary 3.4. For sufficiently large t = 1 or 2 (mod 4), there are more than (6.492)2i+1 hooked Skolem sequences of order 7t + 4.
Proof. Table 2 covers the possible values of t modulo 12.
Table 2: Construction of hooked Skolem sequences.
t (mod 12)	c	Q	R	t > 7c - 4
1,10	2	(1,1, 7t + 4)	(0, 7t + 4)	t > 10
2, 5	3	(1,1, 2, 7t + 4, 2)	(0, 7t + 4)	t > 17
6, 9	7	(2,4, 2, 5, 6,4, 3, 7t + 4, 5, 3, 6,1,1)	(0, 7t + 4)	t > 45
For each Skolem-type sequence S of the form described in Theorem 3.1, the resulting sequence S' = (Q ^ (T ^ S)) A R is a hooked Skolem sequence of order 7t + 4. □
Corollary 3.5. For sufficiently large t = 0 or 3 (mod 4), there are more than (6.492)2i+1 split-hooked Skolem sequences of order 7t + 5.
Proof. Table 3 covers the possible values of t modulo 12.
Table 3: Construction of split-hooked Skolem sequences.
t (mod 12)	c	Q	R	t > 7c - 4
0, 3	4	(1,1, 7t + 5, 3, 7t + 4, 0, 3)	(7t + 5, 7t + 4, 2,0, 2)	t > 24
4, 7	5	(2, 3, 2, 4, 3, 7t + 5, 0, 4, 7t + 4)	(1,1, 7t + 5,0, 7t + 4)	t > 31
8,11	9	(4, 5, 6, 8, 4, 7, 5, 7t + 5, 6, 7t + 4, 0, 8, 7, 3,1,1, 3)	(7t + 5, 7t + 4, 2,0, 2)	t > 59
For each Skolem-type sequence S of the form described in Theorem 3.1, the resulting sequence S' = (Q ^ (T ^ S))AR is a split-hooked Skolem sequence of order 7t+5. □
The bound obtained in each of the preceding four corollaries only applies to restricted parts of the appropriate residue classes. We believe that it is possible to extend the bound to all possible residue classes in each case. We do not give a proof of this because our argument breaks into a considerable number of subcases. However, to support our contention, we give one example for split Skolem sequences. Corollary 3.2 gives the bound
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(6.492)2t+1 when n = 7t + 3 and t = 0 or 3 (mod 4), thereby dealing with n = 3 or 24 (mod 28). The necessary and sufficient conditions on n for the existence of a split Skolem sequence of order n may be written as n = 0, 3,4,7, 8,11,12,15,16,19, 20, 23,24 or 27 (mod 28). For our example, we show how the bound may be extended to n = 0 or 7
(mod 28).
Put n = 7t + 7 where t = 0 or 3 (mod 4). Take S to be a Skolem-type sequence as described in Theorem 3.1. Depending on the residue of t modulo 12, take c as in Table 4 and put m = (t - c +1)/3. For t > 7c - 4, use an (m, 3, c)-system to construct a Skolem-type sequence T of length 2t + 1 having zeros in the central 2c - 1 positions and such that for each k G {c, c + 1,..., t}, the two positions occupied by k are precisely k apart. Take Q and R as specified in the table and form S' = (Q ^ (T ^ S)) A R. Then S' is a split Skolem sequence of length 14t + 15 (i.e. of order n = 7t + 7). Hence, for all sufficiently large t, there are more than (6.492)2i+1 split Skolem sequences of order n = 7t + 7 where t = 0 or 3 (mod 4), so that n = 0 or 7 (mod 28).
Table 4: Further split Skolem sequences.
t (mod 12)	c	Q	R	t > 7c - 4
4, 7	5	(4, 7t + 7, 3, 7t + 6,4, 3, 7t + 4, 7t + 5,0)	(7t + 7, 7t + 6, 7t + 4, 2, 7t + 5, 2,1,1)	t > 31
8,11	9	(1,1, 3, 8,4, 3, 7t + 7, 7,4, 7t + 6, 6, 8,0, 7t + 5, 7, 7t + 4, 6)	(5, 7t + 7, 2, 7t + 6, 2, 5, 7t + 5, 7t + 4)	t > 59
0, 3	16	(11, 8,4, 7,13, 9, 4,14,10, 8, 7,11,12, 7t + 7, 9,15, 7t + 6,13,10,0, 7t + 5,14, 7t + 4, 6,12, 3,1,1, 3, 6,15)	(5, 7t + 7, 2, 7t + 6, 2, 5, 7t + 5, 7t + 4)	t > 108
There are methods other than the one described above for generating Skolem-type sequences from additive permutations. For certain orders the construction technique given below can give improved bounds.
Suppose that s > i > 0 and that S is a Skolem-type sequence of length 2s + 1 having zeros in the central 2i-1 positions, and such that for k G {i, i+1,..., s} the two positions where k appears in S are precisely k apart. If i =1 then S is a split Skolem sequence of order s (and such a sequence exists if s = 0 or 3 (mod 4)), otherwise the earlier discussion following Corollary 3.3 shows that such a sequence exists when s -i = 2 or 11 (mod 12) if i is odd, or s - i = 8 or 11 (mod 12) if i is even, provided that s > 7i - 4. Let S be indexed by [—s, s].
Construction 3.6.
•	For j = i, i + 1,..., s, denote by aj , bj (with aj < bj) the positions in S occupied by the entry j, so that bj - aj = j.
•	For each j = i, i + 1,..., s, let nj be an additive permutation on [-1, t], and denote by n the ordered (s - i + 1)-tuple (n^, ..., ns).
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•	Form a new sequence Sn indexed by [—((2t + 1)s + t), (2t + 1)s + t] by placing the entry (2t + 1)j + nj (i) at positions (2t + 1)aj + i and (2t + 1)bj + i + nj (i) for j = t, t + 1,..., s and i G [—t, t]. For each j these entries cover the interval [(2t + 1)j — t, (2t + 1)j +1], and so they collectively cover [(2t + 1)t—t, (2t + 1)s+1]. These entries cover the positions
[—((2t + 1)s +1), (2t + 1)s + t] \ [—(2t + 1)t + t +1, (2t + 1)t — t — 1].
Place zeros in the vacant positions. Then Sn is a Skolem-type sequence of length 2(2st + s + t) + 1 having zeros in the central 2((2t + 1)t — t) — 1 positions, and such that for k G [(2t + 1)t — t, (2t + 1)s +1] the two positions where k appears in Sn are precisely k apart.
•	Now suppose that t = 0 or 3 (mod 4) and let P be a split Skolem sequence of order t, indexed by [—t, t]. Apply the previous three steps to P using additive permutations oj on [—(t — 1), t — 1] for j = 1, 2,..., t to form a new sequence Ps, where £ is the ordered t-tuple (oi, o2,..., ot). Then Ps is a Skolem-type sequence of length 2(2t(t — 1) + t + (t — 1)) + 1 = 2((2t + 1)t — t) — 1 having zeros in the central 2((2(t — 1) + 1) — (t — 1))) — 1 = 2t — 1 positions, and such that for k G [t, (2t + 1)t — t — 1] the two positions where k appears in Ps are precisely k apart. Replace the central zeros of Sn by Ps to form the sequence Ps ^ Sn.
Then Ps ^ Sn is a Skolem-type sequence of length 2(2st + s + t) + 1 having zeros in the central 2t — 1 positions, and such that for k G {t, t + 1,..., 2st + s + t} the two positions where k appears in Ps ^ Sn are precisely k apart.
Given a sequence Ps ^ Sn constructed in this fashion for given t, s and t, the ingredients S, n, P and £ may be recovered by considering entries and positions. Suppose that entry e > 0 occupies positions a and b with a < b. If e > (2t +1)t — t then e = (2t + 1)j+nj (i) for some i and j, and j = |_(e+t)/(2t +1)_|, while a = (2t +1)aj + i, so aj = |_(a + t)/(2t + 1)J. Similarly, bj = |(b + t)/(2t +1)J, while i,nj(i) G [—t,t] are given by i = a (mod 2t + 1) and nj(i) = e (mod 2t + 1). This process recovers S and n. If e < (2t + 1)t — t — 1 = (2(t — 1) + 1)t + (t — 1), then P and £ may be recovered in the same way from |(e + (t — 1))/(2t — 1)J, etc.
Hence, varying any of S, n, P and £ will yield different Skolem-type sequences Ps ^ Sn. Disregarding variation due to selection of S, P, and £, the following result is obtained.
Theorem 3.7. Suppose that there exists a Skolem-type sequence S of length 2s + 1 having zeros in the central 2t—1 positions, and such that for k G {t, t+1,..., s} the two positions where k appears in S are precisely k apart. Then for t = 0 or 3 (mod 4) there are at least (z2t+i)s-e+i Skolem-type sequences Ps ^ Sn of length 2(2st + s +1) + 1 having zeros in the central 2t — 1 positions, and such that for k G {t, t + 1,..., 2st + s + t} the two positions where k appears in Ps ^ Sn are precisely k apart.
This result generates lower bounds for the numbers of pure, split, hooked and split-hooked Skolem sequences of various orders. Recall from Corollary 2.2 that for sufficiently large t, z2t+i > (3.246)2t+i.
Corollary 3.8. If s = 0 or 3 (mod 4), then for all sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order 2st + s +1 is greater than (3.246)2st+s.
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Proof. Invoke the above construction with i = 1.	□
As an example, taking s = 3 gives that for all sufficiently large t = 0 or 3 (mod 4), the number of split Skolem sequences of order n = 7t + 3 is greater than (3.246)6t+3. This is substantially better than the bound given earlier in Corollary 3.2, although this approach does not appear to offer the generalization to n = 0 or 3 (mod 4) mentioned in connection with the previous method.
If the central zero in a split Skolem sequence of order n is replaced by n + 1 and an additional entry n + 1 is placed at either the start or the end of the sequence, then a pure Skolem sequence of order n + 1 is formed. Applying this adaptation to the split Skolem sequence constructed in Corollary 3.8 gives the following result.
Corollary 3.9. If s = 0 or 3 (mod 4), then for all sufficiently large t = 0 or 3 (mod 4), the number of pure Skolem sequences of order 2st + s +1 + 1 is greater than (3.246)2st+s.
To deal with hooked and split-hooked Skolem sequences, we may replace the central 2i - 1 zeros of Vs ^ Sn with an appropriate sequence Q, and append a short sequence R to the right hand end to form (Q ^ (Vs ^ Sn)) A R.
Corollary 3.10. If s = 1 or 2 (mod 4) and s > 45, then for all sufficiently large t = 0 or 3 (mod 4), the number of hooked Skolem sequences of order 2st + s +1 +1 is greater than (3.246)(s-6)(2i+1).
Proof. Given s and t, put m = 2st + s +1. The sequences Q and R for possible values of s modulo 12 are covered by using Table 2 with t replaced by s, c replaced by i and 7t + 3 replaced by m. Note that in Table 2, c < 7 and so we may assume that i < 7. In each case, the resulting sequence (Q ^ (Vs ^ Sn))A R is a hooked Skolem sequence of order m + 1 = 2st + s +1 +1, and the result follows.	□
Corollary 3.11. If s = 0 or 3 (mod 4) and s > 59, then for all sufficiently large t = 0 or 3 (mod 4), the number of split-hooked Skolem sequences of order 2st + s +1 + 2 is greater than (3.246)(s-8)(2i+1).
Proof. Given s and t, put m = 2st + s +1. The sequences Q and R for possible values of s modulo 12 are covered by using Table 3 with t replaced by s, c replaced by i and 7t + 3 replaced by m. Note that in Table 3, c < 9 and so we may assume that i < 9. In each case, the resulting sequence (Q ^ (Vs ^ Sn))A R is a split-hooked Skolem sequence of order
m + 2 = 2st + s +1 + 2, and the result follows.	□
4 Completing the proof of Theorem 2.1
As previously described, the arithmetic is simplified by subtracting t - s - a from all the entries in R and S to obtain equivalent arrays R' and S', where R' has entries ranging from — (s + 2a) to s + 2a, and S' has entries ranging from — (s - 1) to s - 1. Transversals are sought in these arrays that cover the entry values from -(s + a) to s + a inclusive. We renumber the rows and columns so that for R' the row and column numbers run from 0 to r - 1 = s + 2a and for S' they run from 0 to s - 1. The entry in cell (i, j) of R' is then i + j - (s + 2a), and that in cell (i, j) of S' is i + j - (s - 1). The identification of transversals falls into several cases.
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Table 5: Case 1 (r even), transversal in S'.
	Range	Row	Column	Entry
(a)	j =0,..., 2—1	j	2—1 + j	- 2 — 1 +2j
(b) f	j 2—1 - 1	2—1 + 1+ j	j	- 2—1 + 1 + 2 j
Case 1: r even. Since r is even, s must be odd. Table 5 identifies a suitable transversal in S'. Line (b) of the table, marked with f, is omitted when s = 1. In S', and subject to f, line (a) of the table covers rows 0 to 2—1, and line (b) covers rows 2—1 + 1 to s - 1. For columns, line (b) covers 0 to 2—1 - 1, and line (a) covers 2—1 to s - 1. As regards entries, lines (a) and (b) together cover - 2—1 to 2—1.
Subcase 1.1: r = 0 (mod 4). Table 6 identifies a suitable transversal in R'. The line of the table marked with * is omitted when s = b - 2, and the line marked f is omitted when s = 1. Subject to * and f, rows, columns and entries of R' are covered by lines of the
Table 6: Subcase 1.1 (r = 0 (mod 4)), transversal in R'.
	Range	Row	Column	Entry
(a)	j = 0,..., 4 -1	j	a + j	-a - s + 2j
(b)	j = °..., a - 4	4 + j	a + 4 + s + j	2—1 + 1 + 2 j
(c)	j = 0,..., 4 -1	a + 1 + j		-a - s + 1 + 2j
(d)	j =0,...,2—1	a+4+1+j	a + 4 + 2!T1 + j	a + 1 + 2 j
(e) f	j =0,..., 2—1 - 1	a + 4 + TT + 2 + j	a + 4 + j	a + 2 + 2j
(f) *	j = 0,. .., a - 4 - 1	a + 4 + s + 1+ j	4 + j	2—1 + 2 + 2 j
table in the following orders, where notation such as (a&c) means that entries from lines (a) and (c) are interleaved and taken together. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f) in that order. Columns 0 to 2a + s by lines (c)(f)(a)(e)(d)(b) in that order. Entries -a - s to - 2—1 - 1 by lines (a&c), and entries 2—1 + 1 to a + s by lines (b&f)(d&e) in that order; the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
Subcase 1.2: r = 2 (mod 4). Table 7 identifies a suitable transversal in R'. The line of the table marked with f is omitted when s = 1. Subject to f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f) in that order. Columns 0 to 2a + s by lines (c)(f)(a)(e)(d)(b) in that order. Entries -a - s to -^—r1 - 1 by lines (a&c), and entries ^—r1 + 1 to a + s by lines (b&f)(d&e) in that order; the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
Case 2: r odd. If r is odd then s must be even. If s = 0 then r = b and R is a copy of Ab
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Table 7: Subcase 1.2 (r = 2 (mod 4)), transversal in R'.
	Range	Row	Column	Entry
(a)	j = 0,...,^ -1	j	a + j	-a - s + 2j
(b)	j = 0,. .., a -	-+§ + j	a + + s + j	+ 2 + 2 j
(c)	j = 0,..., r-+§ - 2	a + 1 + j	j	-a - s + 1 + 2j
(d)	j ^	a + -+§ + j	a + + ¥ + j	a + 1 + 2j
(e) f	j =0,..., ^ - 1	a + -+§ + + 1+j	a + + j	a + 2 + 2j
(f)	j = 0,. .., a - -+§	a + + s + j	- 1+ j	+ 1 + 2 j
(with rows and columns appropriately renumbered) and any one of the transversals already identified in Ab provides a suitable transversal in R. So throughout Case 2, we may assume that s > 0, and then Table 8 identifies a suitable transversal in S'. Line (b) of the table, marked with f, is omitted when s = 2. Subject to f, rows, columns and entries of S' are
Table 8: Case 2 (r odd), transversal in S'.
	Range	Row	Column	Entry
(a)	j § - 1	j	§ - 1+ j	- § +2j
(b) f	j = 0,..., § - 2	§ + j	j	- § + 1 + 2j
(c)	single cell	s - 1	s - 1	s - 1
covered by lines of the table in the following orders. Rows 0 to s - 1 by lines (a)(b)(c) in that order. Columns 0 to s - 1 by lines (b)(a)(c) in that order. Entries -§ to § - 2 by lines (a&b), and entry s - 1 by line (c).
To deal with R', we consider four subcases depending on the values of r and s modulo 4.
Subcase 2.1: r = 1, s = 0 (mod 4). These conditions imply that b = 1 (mod 4) and we may assume that s > 4. Table 9 identifies a suitable transversal in R'. Lines of the table marked with * are omitted when s = b - 1, and lines marked with f are omitted when s = 4. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
Subcase 2.2: r = 1, s = 2 (mod 4). These conditions imply that b = 3 (mod 4) and a is odd. Table 10 identifies a suitable transversal in R' when s > 6. The line of the table
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Table 9: Subcase 2.1 (r = 1, s = 0 (mod 4)), transversal in fi'.
	Range	Row	Column	Entry
(a) *	j = o,..., a - 4 -1	j	a + s + j	-a + 2j
(b)	j = 0,..., 4 -1	a - s + j 2 4 + j	a + 4 + j	-a - s + 2j
(c)	single cell	a 2	a + s - 1	-a - 1
(d) t	j = 0,..., 4 - 2	a + 1+j	a + 22 + j	-a - f + 1 + 2j
(e)	single cell	a i s 2 + 4	a + f - 1 2 + 4 1	-a - 2 - 1
(f) t	j = 0,..., 4 - 2	a+4+1+j	a - 4 + j	-a - s + 1 + 2j
(g) t	j = 0,..., 4 - 2	a + 2 + j	a + j	-a - 2 + 2j
(h)	j = 0,..., 4 -1	a + 3s -1+j	3a + 3s + j 2 + 4 +j	2 -1 + 2 j
(i)	single cell	a + s -1	a -1	-a - 2
(j)	j = 0,..., a	a + s + j	3a + s + j	s + 2 j
(k) *	j = 0,._a - 4 -1	a + s + 1 + j	j	-a + 1 + 2j
(l)	j = 0,..., 4 -1	3a + 3f + 1+ j	a + 3f - 1+ j	2 + 2 j
(m)	j = 0,..., a -1	+ s + 1+ j	a + s + j	s + 1 + 2 j
marked with * is omitted when s = b - 1, and the line of the table marked with f is omitted when s = 6. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a+s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a+s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(c)(i)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
The case s = 2 may be obtained from the table by omitting lines (b), (d), (e), (f), (g) and (l). Subject to * (i.e. when b = 3), rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + 2 by lines (a)(c)(h)(i)(j)(k)(m) in that order. Columns 0 to 2a + 2 by lines (k)(i)(c)(m)(a)(h)(j) in that order. Entries -a - 2 to -2 by lines (c)(i)(a&k) in that order, entry 0 by line (h), and entries 2 to a + 2 by lines (j&m); the remaining entries required to complete the transversal values from -(a + 2) to (a + 2) come from the transversal in S'.
Subcase 2.3: r = 3, s = 0 (mod 4). These conditions imply that b = 3 (mod 4) and a is odd. Table 11 identifies a suitable transversal in R'. The line of the table marked with * is omitted when s = b - 3, and the lines of the table marked with f are omitted when s = 4. Subject to * and f, rows, columns and entries of R' are covered by lines of the table in the following orders. Rows 0 to 2a + s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to - § - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries § - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
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Table 10: Subcase 2.2 (r = 1, s = 2 (mod 4)), transversal in fi'.
	Range	Row	Column	Entry
(a) *	j =0,..., ^ -s-2 1 4 1	j	a + s + j	-a + 2j
(b)	j = 0,..., ^ - 1	a—1 s-2 i j 2 4 +j	a—1 + s —2 + 2 + 4 + 1+ j	—a — s + 2j
(c)	single cell	a —1 2	^ + s - 1	-a — 2
(d)	j = 0,..., ^ - 1	s-r + 1+ j	^ + 2 + j	—a — f + 2j
(e)	single cell	a—1 + s-2 + 1 2 + 4 + 1	a—1 + s —2 2 + 4	—a — 2 — 1
(f)	j = 0,..., ^ - 1	a—1 + s —2 + 2 + 4 + 2 + j	a —1 s—2 + j 2 4 +j	—a — s + 1 + 2 j
(g) t	j = 0,..., ^ - 2	a—T + 2 + 1+ j	^ + 1+ j	—a — 2 + 1 + 2 j
(h)	j =0,..., ^	a—1 i 3s —2 i j 2 + 4 +j	3a+1 i 3s —2 i j 2 1 4 +j	2 — 1 + 2j
(i)	single cell	^ + s	a—1 2	—a — 1
(j)	j =0,..., ^	a—T + s +1+ j	^ + s + j	s + 1 + 2 j
(k) *	j =0,..., ^ -s-2 i 4 1	a + s + 1 + j	j	—a + 1 + 2j
(l)	j = 0,..., ^ - 1	3a+1 i 3s —2 i 2 1 4 + 1+ j	a—1 i 3s—2 i j 2 + 4 +j	2 + 2j
(m)	j = 0,..., ^	3a+1 + s + j	^ + s + j	s + 2j
Subcase 2.4: r = 3, s = 2 (mod 4). These conditions imply that b = 1 (mod 4) and a is even. Table 12 identifies a suitable transversal in ñ' when s > 6. The line of the table marked with * is omitted when s = b - 3, and the line of the table marked with f is omitted when s = 6. Subject to * and f, rows, columns and entries of ñ' are covered by lines of the table in the following orders. Rows 0 to 2a+s by lines (a)(b)(c)(d)(e)(f)(g)(h)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a+s by lines (k)(f)(i)(g)(e)(b)(d)(l)(c)(m)(a)(h)(j) in that order. Entries -a - s to -1 - 1 by lines (b&f)(e)(d&g)(i)(c)(a&k) in that order, entries | - 1 to s - 2 by lines (h&l), and entries s to a + s by lines (j&m); the remaining entries required to complete the transversal values from -(s + a) to (s + a) come from the transversal in S'.
The case s = 2 may be obtained from the table by omitting lines (b), (d), (e), (f), (g) and (h). Subject to * (i.e. when b = 5), rows, columns and entries of ñ' are covered by lines of the table in the following orders. Rows 0 to 2a + 2 by lines (a)(c)(i)(j)(k)(l)(m) in that order. Columns 0 to 2a + 2 by lines (k)(i)(l)(c)(m)(a)(j) in that order. Entries -a - 2 to -2 by lines (i)(c)(a&k) in that order, entry 0 by line (l), and entries 2 to a + 2 by lines (j&m); the remaining entries required to complete the transversal values from -(a + 2) to (a + 2) come from the transversal in S'.
This concludes the proof of Theorem 2.1.
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Table 11: Subcase 2.3 (r = 3, s = 0 (mod 4)), transversal in R'.
	Range	Row	Column	Entry
(a)	j = o,..., ^ - 4		a + s + j	-a + 2j
(b)	j = 0,..., s -1	o±l _ s + j 2 4 + j	a-T + 4 + j	-a - s + 2j
(c)	single cell	a+1 2	a-T + s - 1	-a - 1
(d) t	j = 0,..., s - 2	a+1 + 1+ j	a—1 + 22 + j	-a - § + 1 + 2j
(e)	single cell	a+1 + s 2 + 4	a — 1 I s i ~ + 4 - 1	-a - 2 - 1
(f) t	j = 0,..., s - 2	^+4+1+j	TT - 4 + j	-a - s + 1 + 2j
(g) t	j = 0,..., s - 2	a+1 + 22 + j	a-i+j	-a - 2 + 2j
(h)	j = 0,..., s -1	^ + 3s -1 + j	3a+1 | 3s | j 2 + 4 +j	2 +2j
(i)	single cell	at-1 + s - 1	a — 1 i ~2--1	-a - 2
(j)	j =0,..., ^	^ + s + j	^ + s + j	s + 1 + 2j
(k) *	j 1 -s -1	a + s + 1 + j	j'	-a + 1 + 2j
(l)	j = 0,..., s -1	3a+1 , 3s , j 2 + 4 + j	a—1 + 3s 2 + 4 1+ j	2 - 1 + 2 j
(m)	j = 0,..., 1	^ + s + j	a—1 + s + j	s + 2j
5 Concluding remarks
Our results improve the known lower bounds for the number of additive permutations, zero-sum arrays, some Skolem-type sequences, and some extremal Langford sequences. It seems highly likely that the bounds obtained in this paper apply to all pure, split, hooked and split-hooked Skolem sequences of sufficiently large orders. The recent paper [8] combines such bounds with graph labellings to generate Langford sequences. It seems likely that our new bounds can be combined with these techniques to generate improved estimates for the numbers of Langford sequences.
For small orders, the numbers of (pure) Skolem sequences and hooked Skolem sequences (and other related sequences) are tabulated in [4], while Table 8 of [7] gives the numbers of split Skolem (Rosa) sequences of orders n < 12. These numerical results strongly suggest that further improvements to our lower bounds are possible.
Since Skolem sequences may be used to construct solutions to Heffter's first and second difference problems, the bounds inform the numbers of these and of resulting cyclic Steiner triple systems. If improved bounds for z2i+1 are obtained in the future, these methods will lead to improved bounds for many related sequences. From Table 1, it will be seen that the ratio z2i+1/z2i-1 appears to increase with t, and to exceed 2t for t > 6, strongly suggesting that z2i+1 > 24t! for all sufficiently large t. This is a weaker bound than might be suggested by Vardi's conjecture, but it is strongly supported by the computational evidence, and one might expect that most transversals are not Z-transversals.
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Table 12: Subcase 2.4 (r = 3, s = 2 (mod 4)), transversal in R'.
	Range	Row	Column	Entry
(a)	3 = o,..., a - s-2 i		a + s + j	-a + 2j
(b)	j = 0,..., ^ - 1	a _ s-2 + j 2 4 +j	a + +j	-a - s + 2j
(c)	single cell	a 2	a+s -1	-a - 1
(d)	j = o,..., -1	a + 1+ j	a + 2-1+j	-a - § + 2j
(e)	single cell	a 1 s-2 1 1 2 + 4 + 1	a I s — 2 -| 2 + ~ 1	-a - f - 1
(f)	j = o,..., -1	a + +2+j	a - t2 -1+j	-a - s + 1 + 2j
(g) t	j = 0,..., - 2	a+2+1+j	a+j	-a - 2 + 1 + 2j
(h)	j = o,..., ^ - 1	a 1 3s-2 | j 2 + 4 +j	3a | 3s+2 | j 2 + 4 +j	2 +2j
(i)	single cell	a + s - 1	a -1	-a - 2
(j)	j = o,..., a	a + s + j	3a + s+j	s + 2j
(k) *	j = o,..., a -s-1 - 2	a + s + 1 + j		-a + 1 + 2j
(l)	j =o,..., ^	3a , 3s+2 , j 2 + 4 +j	a I 3s-6 , j 2 + 4 +j	2 -1 + 2 j
(m)	j = o,..., a -1	3a + s +1 + j	a+s+j	s + 1 + 2 j
References
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