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Subsequence sums: Direct and inverse problems
Journal of Number Theory 148 (2015) 235–256
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Journal of Number Theory www.elsevier.com/locate/jnt
Subsequence sums: Direct and inverse problems Raj Kumar Mistri a,1 , Ram Krishna Pandey b,∗ , Om Praksh a a
Department of Mathematics, Indian Institute of Technology Patna, Patna 800013, India b Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee 247667, India
a r t i c l e
i n f o
Article history: Received 7 July 2014 Received in revised form 30 August 2014 Accepted 2 September 2014 Available online 4 November 2014 Communicated by David Goss Keywords: Subsequence sums Subset sums Direct problems Inverse problems
a b s t r a c t Let A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) be a r0 copies
r1 copies
rk−1 copies
finite sequence of integers with k distinct terms, denoted alternatively by (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < r = (r0 , r1 , . . . , rk−1 ), ri ≥ 1 for i = 0, 1, . . . , k − 1. ak−1 , ¯ The sum of all the terms of a subsequence of length at least one of the sequence A is said to be a subsequence sum of A. The set of all subsequence sums of A is denoted by S(¯ r, A). The direct problem for subsequence sums is to find the lower bound for |S(¯ r, A)| in terms of the number of distinct terms in the sequence A. The inverse problem for subsequence sums is to determine the structure of the finite sequence A of integers for which |S(¯ r, A)| is minimal. In this paper, both the problems are solved and some well-known results for subset sum problem are obtained as corollaries of the results for subsequence sum problem. © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. E-mail addresses: [email protected] (R.K. Mistri), [email protected] (R.K. Pandey), [email protected] (O. Praksh). 1 Research supported by the University Grants Commission of India (No. F. 2-10/2011 (SA-I)). http://dx.doi.org/10.1016/j.jnt.2014.09.010 0022-314X/© 2014 Elsevier Inc. All rights reserved.
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0. Notation and terminology For integers a, b, c and for a set S of integers, let [a, b] = {n ∈ Z : a ≤ n ≤ b}, c ∗ S = {ca : a ∈ S} and c + S = {c + a : a ∈ S}. Let A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) be a finite sequence of integers, where a0 < r0 copies
r1 copies
rk−1 copies
a1 < · · · < ak−1 and ri ≥ 1 for i = 0, 1, . . . , k−1. Let ¯r = (r0 , r1 , . . . , rk−1 ) be the ordered k-tuple. Then we denote the sequence A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) by r0 copies
r1 copies
rk−1 copies
(a0 , a1 , . . . , ak−1 )¯r . In this paper, by a finite sequence of integers, we always mean a finite sequence of integers whose terms are in nondecreasing order. This does not cause any harm, because the results in this paper are related to subsequence sums which remain unaltered by changing the order of the terms of the sequence. For an integer c, let c + (a0 , a1 , . . . , ak−1 )¯r = (c + a0 , c + a1 , . . . , c + ak−1 )¯r and for a nonzero integer c, let c ∗ A = c ∗ (a0 , a1 , . . . , ak−1 )¯r =
(ca0 , ca1 , . . . , cak−1 )¯r , (cak−1 , cak−2 , . . . , ca0 )¯s ,
if c > 0, if c < 0,
where ¯s = (rk−1 , . . . , r0 ). For integers a, b and c such that b = a + k − 1, let [a, b]¯r = (a, a + 1, . . . , b)¯r , and c + [a, b]¯r = [c + a, c + b]¯r . We shall call [a, b]¯r a sequence-interval. In particular, if ¯r = (r, r, . . . , r), then we simply write (a0 , a1 , . . . , ak−1 )r in place of (a0 , a1 , . . . , ak−1 )¯r and [a, b]r in place of [a, b]¯r . Let [a, b]r \(0)r = (a, a + 1, . . . , −1, 1, 2, . . . , b)r . If A is a sequence, then we write a ∈ A to mean that a is a term of the sequence A. If A is a set, then the notation a ∈ A has the usual meaning. We say that two finite sequences of integers are disjoint if there is no common term in these sequences. In this case, we write A ∪ B to denote the sequence obtained by merging these sequences and reordering the terms in nondecreasing order. For example, if A = (1, 2, 4)u¯ , where ¯ = (2, 1, 3) and B = (3, 5, 6)v¯ , where v ¯ = (1, 2, 2) are two sequences, then A ∪ B = u (1, 1, 2, 4, 4, 4) ∪ (3, 5, 5, 6, 6) = (1, 2, 3, 4, 5, 6)¯r , where ¯r = (2, 1, 1, 3, 2, 2). If A and B are two sets, then A ∪ B has usual meaning. Finally, we agree with the convention that a b = 0 if a and b are two positive integers such that a < b.
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1. Introduction Let A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of integers, where a0 < a1 < · · · < ak−1 , ¯r = (r0 , r1 , . . . , rk−1 ) and ri ≥ 1 for i = 0, 1, . . . , k − 1. Let B be a subsequence of A of length greater than or equal to one. We define the subsequence sum s(B) of B by s(B) :=
b,
b∈B
and
S(¯r, A) := s(B) : B is a subsequence of A of length ≥ 1 . In particular, if ¯r = (r, r, . . . , r), then we write S(r, A) instead of S(¯r, A). The direct problem for the subsequence sums is to find a lower bound for |S(¯r, A)| in terms of k, the number of distinct terms in the sequence A. The inverse problem for the subsequence sums is to determine the structure of the finite sequence A of integers for which |S(¯r, A)| is minimal. The similar direct and inverse problems for h-fold sumset, restricted h-fold sumset and subset sums of a finite set of integers have been studied in the past. One can refer for example [8–11]. Very recently, the h-fold sumset and the restricted h-fold sumset have been generalized as a generalized sumset and the corresponding direct and inverse problems have been studied (see [7,12]). There is a closely related problem for subsequence sums in additive abelian groups. Let a = (a1 , a2 , . . . , am ), where m ≥ 1, be a sequence of elements of an additive abelian group G. Let (a) denote the set of all group elements representable as a sum of a subsequence of a of length . Then the problems such as the direct problem for (a) and the properties of (a) (for example ai ∈ (a) for some i ∈ {1, 2, . . . , m}) etc. have been studied [1,3–6]. DeVos, Goddyn and Mohar [2] further generalize the sequence a = (a1 , a2 , . . . , am ) of elements of G to the sequence A = (A1 , A2 , . . . , Am ) of finite subsets of G and determine the direct theorem for (A), where
(A) = {ai1 + · · · + ail : 1 ≤ i1 < · · · < i ≤ m and aij ∈ Aij ∀1 ≤ j ≤ }.
In this paper, we solve both the direct and the inverse problems for the sumset S(r, A) of subsequence sums. We also solve the direct and the inverse problems for the general sumset S(¯r, A) of subsequence sums where A contains either only nonnegative terms or only nonpositive terms. These direct and inverse problems of subsequence sums naturally generalize the direct and the inverse problems of subset sums studied by Nathanson [8] by taking r = 1 in S(r, A). Thus the subset sum problem may be considered as a particular type of subsequence sum problem.
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Since the proofs of the results for general sequence are somewhat complicated, first we restrict ourselves to the case of particular sequence of the type (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 and r ≥ 1. This is discussed in Section 2. By this generalization, we obtain the direct and the inverse theorems of Nathanson [8] as corollaries of these results. Furthermore, the direct and the inverse theorems for the general sumset S(¯r, A) of subsequence sums, where A contain either only nonnegative terms or only nonpositive terms, are discussed in Section 3. It would be interesting if one studies the direct and the inverse problems for the general sumset S(¯r, A) of subsequence sums, where A contains positive integers, negative integers and/or zero. 2. Subsequence sum problem for the sequence (a0 , a1 , . . . , ak−1 )r 2.1. Direct problem For the finite sequence of integers A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 and r ≥ 1, we shall say that A is a finite sequence of integers with k distinct terms each with repetitions r. Theorem 2.1. Let k ≥ 1 and r ≥ 1. Let A be a finite sequence of positive integers with k distinct terms each with repetitions r. Then
S(r, A) ≥ r k + 1 . 2
(2.1)
Let A be a finite sequence of nonnegative integers with k distinct terms each with repetitions r and 0 ∈ A. Then
S(r, A) ≥ 1 + r k . 2
(2.2)
The lower bounds in (2.1) and (2.2) are best possible. Proof. First assume that k = 1. Then A = (a)r for some integer a. If a > 0, then 1 |S(r, A)| = r = r 1+1 2 . If a = 0, then |S(r, A)| = 1 = 1 + r 2 . Therefore, the theorem is true for k = 1. Hence we may assume that k ≥ 2. Let A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 , k ≥ 2 and r ≥ 1 be a finite sequence of positive integers. For = 0, 1, . . . , k − 1 and j = 1, 2, . . . , r, we define Sr+j :=
ai + (j − 1)ak−−1 +
m=1
It is easy to see that
rak−m : i = 0, 1, . . . , k − − 1 .
(2.3)
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|Sr+j | = k −
239
for = 0, 1, . . . , k − 1 and j = 1, 2, . . . , r,
and max(Si ) < min(Si+1 ) for i = 1, 2, . . . , rk − 1. It follows that the sets S1 , S2 , . . . , Srk are pairwise disjoint, and so
r rk rk k−1 k−1
k+1
S(r, A) ≥ Si
= |Si | = |Sr+j | = r(k − ) = r .
2 i=1
i=1
=0 j=1
=0
Therefore,
S(r, A) ≥ r k + 1 . 2 Thus (2.1) is established. Now, let A = (a0 , a1 , . . . , ak−1 )r be a finite sequence of nonnegative integers, where 0 = a0 < a1 < · · · < ak−1 , k ≥ 2 and r ≥ 1 so that 0 ∈ A. Let A = (a1 , a2 , . . . , ak−1 )r . Then S(r, A) = {0} ∪ S r, A . Therefore, by the previous case, we have
S(r, A) = 1 + S r, A ≥ 1 + r k . 2 This establishes (2.2). Next we show that the lower bounds in (2.1) and (2.2) are best possible. Let k ≥ 2 and r ≥ 1. Let A0 = [1, k]r . Then the smallest and the largest integers in S(r, A0 ) are 1 and r k+1 2 , respectively. Therefore, k+1 S(r, A0 ) ⊆ 1, r , 2 and so
S(r, A0 ) ≤ r k + 1 . 2 This inequality together with the inequality (2.1) implies
S(r, A0 ) = r k + 1 . 2 Thus the lower bound in (2.1) is best possible.
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Similarly, by considering the sequence A1 = [0, k − 1]r , where k ≥ 2 and r ≥ 1, it can be verified that the lower bound in (2.2) is best possible. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.1. (See [8], Theorem 3.) Let k ≥ 2. If A is a set of k positive integers, then
S(A) ≥
k+1 . 2
(2.4)
If A is a set of k nonnegative integers and 0 ∈ A, then
S(A) ≥ 1 + k . 2
(2.5)
The lower bounds in (2.4) and (2.5) are best possible. Remark 2.1. If in the above theorem, A is a finite sequence of negative integers (nonpositive integers), then also the inequality (2.1) ((2.2)) follows. In fact, if A is a finite sequence of negative integers (nonpositive integers), then A := −1 ∗ A is a sequence of positive integers (nonnegative integers). Moreover, S(r, A) = −1 ∗ S(r, A ), and so |S(r, A)| = |S(r, A )|. Theorem 2.2. Let k ≥ 2 and r ≥ 1. Let A be a finite sequence of integers with k distinct terms each with repetitions r. If 0 ∈ A, then
S(r, A) ≥
r(k2 −1) + 4 rk2 4 +1
1
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2).
(2.6)
If 0 ∈ / A, then
S(r, A) ≥
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2).
+1
(2.7)
The lower bounds in (2.6) and (2.7) are best possible. Proof. First assume that k = 2. If 0 ∈ A, then the sequence A contains a nonzero integer a. Let A = (a)r . Then S(r, A) = {0} ∪ S r, A , and so
S(r, A) = 1 + S r, A .
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It follows that
S(r, A) = 1 + r, and so 2
S(r, A) ≥ r.2 + 1. 4
If 0 ∈ / A, then A = (a0 , a1 )r , where a0 and a1 are nonzero integers and a0 < a1 . If a0 and a1 are both positive integers or both negative integers, then by Theorem 2.1, 2
S(r, A) ≥ r 3 = 3r ≥ 2r + 1 = r (2 + 1) − 1 + 1. 2 4 If a0 < 0 and a1 > 0, then let A = (a0 )r and A = (a1 )r . Now, a0 + a1 ∈ S(r, A) and S(r, A ) ∩ S(r, A ) = ∅. Therefore,
S(r, A) ≥ S r, A + S r, A + 1, and so 2
S(r, A) ≥ r + r + 1 = 2r + 1 = r (2 + 1) − 1 + 1. 4
Thus the theorem is true for k = 2. Hence we may assume that k ≥ 3. It is easy to see that for k ≥ 3, r(k2 −1) + 1 if k ≡ 1 (mod 2), k 4 1+r ≥ rk2 2 if k ≡ 0 (mod 2), 4 +1 and
k+1 r 2
≥
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2).
+1
Therefore, we may assume that A contains both positive and negative integers. Let A contains p distinct positive integers and n distinct negative integers. By Theorem 2.1, the set S(r, A) contains at least r p+1 positive integers and at least r n+1 negative 2 2 integers.
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Case 1 (0 ∈ A). In this case k = p + n + 1 and 0 ∈ S(r, A). Therefore,
S(r, A) ≥ r p + 1 + r n + 1 + 1 2 2 p+1 k−p =r + +1 2 2 k2 − k 2 +1 = r p − p(k − 1) + 2 2 k−1 k2 − 1 =r p− + 1. + 2 4 Now, if k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m. Therefore, 2
S(r, A) ≥ r (p − m)2 + k − 1 + 1, 4 and so 2
S(r, A) ≥ r(k − 1) + 1. 4
If k ≡ 0 (mod 2), then k = 2m for some positive integer m. Therefore, 2 2m − 1 k2 − 1 p− +1 + 2 4 2 1 k2 − 1 =r p−m+ +1 + 2 4 k2 2 = r (p − m) + (p − m) + + 1, 4
S(r, A) ≥ r
and so 2
S(r, A) ≥ rk + 1. 4
Thus (2.6) is established. / A). In this case k = p + n. Let p0 be the smallest positive integer in A, and Case 2 (0 ∈ let −n0 be the largest negative integer in A. The set S(r, A) contains at least r p+1 2 n+1 positive integers, each greater than or equal to p0 and at least r 2 negative integers each less than or equal to −n0 . The set S(r, A) also contains the integer p0 − n0 . Since −n0 < p0 − n0 < p0 , it follows that
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S(r, A) ≥ r p + 1 + r n + 1 + 1 2 2 p+1 k−p+1 =r + +1 2 2 k2 + k 2 +1 = r p − kp + 2 2 k k2 + 2k =r p− + 1. + 2 4 Now, if k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m. Therefore,
1 (k + 1)2 − +1 + 4 4 2 k+1 2 = r (p − m) − (p − m) + + 1, 2
S(r, A) ≥ r
1 p−m− 2
2
and so 2
S(r, A) ≥ r k + 1 + 1. 2 A similar argument holds for the case k ≡ 0 (mod 2). Thus (2.7) is established. Next we show that the lower bounds in (2.6) and (2.7) are best possible. Let k ≡ 1 (mod 2). Then k = 2m + 1 for some positive integer m. Let A0 = [−m, m]r . Then 0 ∈ A0 , A0 contains 2m + 1 distinct integers, and m+1 m+1 S(r, A0 ) ⊆ −r ,r . 2 2 Therefore,
S(r, A0 ) ≤ 2r m + 1 + 1 = r m2 + m + 1, 2 and so
S(r, A0 ) ≤ r
k−1 2
2
r(k2 − 1) k−1 +1= + 1. + 2 4
This inequality together with the inequality (2.6) implies 2
S(r, A0 ) = r(k − 1) + 1. 4
(2.8)
Now, let A1 = [−m, m + 1]r \(0)r . Then 0 ∈ / A1 , A1 contains 2m + 1 distinct integers, and
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m+1 m+2 S(r, A1 ) ⊆ −r ,r . 2 2 Therefore,
S(r, A1 ) ≤ r m + 2 + r m + 1 + 1 = r(m + 1)2 + 1, 2 2 and so 2 2
S(r, A1 ) ≤ r k − 1 + 1 + 1 = r k + 1 + 1. 2 2 This inequality together with the inequality (2.7) implies 2
S(r, A1 ) = r k + 1 + 1. 2
(2.9)
By (2.8) and (2.9), it follows that the lower bounds in (2.6) and (2.7) are best possible for k ≡ 1 (mod 2). Similarly, if k ≡ 0 (mod 2), then by considering the sequences A2 = [−m, m − 1]r and A3 = [−m, m]r \(0)r , it can be verified that the lower bounds in (2.6) and (2.7) are best possible for k ≡ 0 (mod 2). This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.2. (See [8], Theorem 4.) Let k ≥ 2 and let A be a set of k integers. If 0 ∈ A, then 2
S(A) ≥ k + 1. (2.10) 4 If 0 ∈ / A, then 2
S(A) ≥ (k + 1) + 1. 4
(2.11)
The lower bounds in (2.10) and (2.11) are best possible. 2.2. Inverse problem Let k ≥ 3 and let A0 = [0, k − 1]r and A1 = 1 + A0 = [1, k]r , where r ≥ 1. Since
S(r, A0 ) = 1 + r k , 2 and
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S(r, 1 + A0 ) = S(r, A1 ) = r k + 1 , 2 we have
S(r, A0 ) < S(r, A1 ) . Therefore,
S(r, A0 ) = S(r, A1 ) . Thus we see that |S(r, A)| is not translational invariant. However, if d is a nonzero integer, then S(r, d ∗ A) = d ∗ S(r, A), and so
S(r, d ∗ A) = S(r, A) . Thus |S(r, A)| is invariant under scalar multiplication. The inverse problem for S(r, A) is to find the finite sequence A for which |S(r, A)| is minimal in Theorem 2.1 and Theorem 2.2. We shall prove that every such sequence is a scalar multiple of a sequence-interval. Theorem 2.3. Let k ≥ 3 and r ≥ 1. If A is a finite sequence of positive integers with k distinct terms each with repetitions r such that
S(r, A) = r k + 1 , 2 then A = d ∗ [1, k]r for some positive integer d. If A is a finite sequence of nonnegative integers with k distinct terms each with repetitions r such that 0 ∈ A and
S(r, A) = 1 + r k , 2 then A = d ∗ [0, k − 1]r for some positive integer d. Proof. Let A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 . If a0 ≥ 1, then it follows from the proof of Theorem 2.1 that S(r, A) =
r k−1 j=1 =0
Sr+j ,
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where Sr+j is the set defined in (2.3). For = 0, 1, . . . , k − 2, we have max(S(+1)r ) = ak−−1 + (r − 1)ak−−1 +
rak−m =
m=1
+1
rak−m ,
m=1
and min(S(+1)r+1 ) = a0 +
+1
rak−m .
m=1
Now, for = 0, 1, . . . , k − 2, we have ak−−2 + (r − 1)ak−−1 +
rak−m < max(S(+1)r ) < min(S(+1)r+1 ),
(2.12)
m=1
and ak−−2 + (r − 1)ak−−1 +
rak−m
m=1
< a0 + ak−−2 + (r − 1)ak−−1 +
rak−m
m=1
< min(S(+1)r+1 ). Since ak−−2 + (r − 1)ak−−1 + that
m=1
(2.13)
rak−m ∈ S(+1)r , it follows from (2.12) and (2.13)
max(S(+1)r ) = a0 + ak−−2 + (r − 1)ak−−1 +
rak−m ,
m=1
and so +1
rak−m = a0 + ak−−2 + (r − 1)ak−−1 +
m=1
rak−m .
m=1
Therefore, ak−−1 = a0 + ak−−2
for = 0, 1, . . . , k − 2,
and so A = a0 ∗ [1, k]r . Thus we have proved that A = d ∗ [1, k]r with d = a0 .
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If a0 = 0, then A1 = (a1 , a2 , . . . , ak−1 )r is a sequence of positive integers containing k − 1 distinct integers. Since S(r, A) = {0} ∪ S(r, A1 ), it follows that
S(r, A) = 1 + S(r, A1 ) , and so
S(r, A1 ) = S(r, A) − 1 = r k . 2 Therefore, by the previous arguments, it follows that A1 = a1 ∗ [1, k − 1]r , and so A = a1 ∗ [0, k − 1]r . Thus we have proved that A = d ∗ [0, k − 1]r with d = a1 ≥ 1. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.3. (See [8], Theorem 5.) Let k ≥ 3. If A is a set of k positive integers such that |S(A)| = k+1 2 , then A = d ∗ [1, k] for some positive integer d. If A is a set of k nonnegative integers such that 0 ∈ A and |S(A)| = 1 + k2 , then A = d ∗ [0, k − 1] for some positive integer d. Theorem 2.4. Let k ≥ 3 and r ≥ 1. Let A be a finite sequence of integers with k distinct terms each with repetitions r. If 0 ∈ A and |S(r, A)| =
r(k2 −1) + 4 rk2 4 +1
1
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2),
(2.14)
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then there is a nonzero integer d such that A=
k−1 d ∗ [− k−1 2 , 2 ]r
if k ≡ 1 (mod 2),
d ∗ [− k2 , k2 − 1]r
if k ≡ 0 (mod 2).
(2.15)
If 0 ∈ / A and |S(r, A)| =
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2),
+1
(2.16)
then there is a nonzero integer d such that A=
k+1 d ∗ [− k−1 2 , 2 ]r \(0)r
d∗
[− k2 , k2 ]r \(0)r
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2).
(2.17)
Proof. The conditions on |S(r, A)| imply that A contains p distinct positive integers and n distinct negative integers, where p ≥ 1 and n ≥ 1. Case 1 (0 ∈ A). In this case k = p + n + 1 and by the proof of Theorem 2.2,
S(r, A) ≥ r
2 k−1 k2 − 1 p− + 1. + 2 4
If k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m, and 2
S(r, A) ≥ r (p − m)2 + k − 1 + 1. 4
(2.18)
By (2.14), we have 2
S(r, A) = r(k − 1) + 1. 4
(2.19)
From (2.18) and (2.19), it follows that (p − m)2 = 0, and so p = m. Therefore n = k − p − 1 = m. Thus p = n = m, and so S(r, A) contains exactly r m+1 positive integers 2 m+1 and exactly r 2 negative integers. Therefore, it follows from Theorem 2.3 that there exist positive integers p0 and n0 such that the positive part of A, say, A+ is p0 ∗ [1, m]r and the negative part of A, say, A− is −n0 ∗ [1, m]r . Since 0, p0 − n0 ∈ S(r, A) and −n0 < p0 − n0 < p0 , it follows that p0 − n0 = 0. Let d = p0 . Then A = A− ∪ (0)r ∪ A+ , and so A = d ∗ [−m, m]r .
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Therefore, k−1 k−1 , A=d∗ − . 2 2 r
(2.20)
Similarly, if k ≡ 0 (mod 2), then it can be shown that k k A=d∗ − , −1 2 2 r
(2.21)
for some nonzero integer d. Thus by (2.20) and (2.21), we have proved the first part of the theorem. Case 2 (0 ∈ / A). In this case k = p + n and by the proof of Theorem 2.2,
S(r, A) ≥ r p + 1 + r n + 1 + 1, 2 2 and so
S(r, A) ≥ r
p−
k 2
2 +
k2 + 2k + 1. 4
If k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m, and
S(r, A) ≥ r
p−m−
1 2
2 +
1 (k + 1)2 − + 1, 4 4
and so 2
S(r, A) ≥ r (p − m)2 − (p − m) + k + 1 + 1. 2
(2.22)
By (2.16), we have 2
S(r, A) = r k + 1 + 1. 2
(2.23)
Since (p −m)2 −(p −m) ≥ 0, it follows from (2.22) and (2.23) that (p −m)2 −(p −m) = 0, and so either p = m or p = m + 1. Therefore, either n = m + 1 or n = m. Thus either p = m and n = m + 1 or p = m + 1 and n = m. If p = m and n = m + 1, then the set positive integers and exactly r m+2 negative integers. S(r, A) contains exactly r m+1 2 2 Therefore, it follows from Theorem 2.3 that there exist positive integers p0 and n0 such that the positive part of A, say, A+ is p0 ∗ [1, m]r and the negative part of A, say, A− is −n0 ∗ [1, m + 1]r . The set S(r, A) also contains p0 − n0 . Now, if p0 − n0 = 0, then the number of positive or negative integers in S(r, A) will exceed. Hence p0 − n0 = 0, and so p0 = n0 . Therefore,
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A = A− ∪ A+ = p0 ∗ −(m + 1), m r \(0)r . Let d = −p0 . Then A = d ∗ [−m, m + 1]r \(0)r , and so k−1 k+1 , A=d∗ − (0)r . 2 2 r If p = m + 1 and n = m, then by a similar argument, it can be shown that there exists a nonzero integer d such that k−1 k+1 , A=d∗ − (0)r . 2 2 r Thus in all cases, we have shown that there exists a nonzero integer d such that k−1 k+1 , A=d∗ − (0)r 2 2 r
if k ≡ 1 (mod 2).
(2.24)
Similarly, if k ≡ 0 (mod 2), then it can be shown that k k A=d∗ − , (0)r 2 2 r
(2.25)
for some nonzero integer d. By (2.24) and (2.25), we have proved the second part of the theorem. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.4. (See [8], Theorem 6.) Let k ≥ 3 and let A be a set of k integers. If 0 ∈ A 2 and |S(A)| = k4 + 1, then there is a nonzero integer d such that A=
d ∗ [−m, m] d ∗ [−m, m − 1]
if k = 2m + 1, if k = 2m.
2
If 0 ∈ / A and |S(A)| = (k+1) + 1, then there is a nonzero integer d such that 4 A=
d ∗ [−m, m]\{0} d ∗ [−m, m + 1]\{0}
if k = 2m, if k = 2m + 1.
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251
3. Subsequence sum problem for the general sequence (a0 , a1 , . . . , ak−1 )¯r 3.1. Direct problem Theorem 3.1. Let k ≥ 1. Let A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of integers, where a0 < a1 < · · · < ak−1 and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1. If a0 > 0, then
k−1
S(¯r, A) ≥ (i + 1)ri .
(3.26)
i=0
If a0 = 0, then k−1
S(¯r, A) ≥ 1 + iri .
(3.27)
i=1
The lower bounds in (3.26) and (3.27) are best possible. Proof. For k = 1, the result is obvious. Hence we may assume that k ≥ 2. Let A = (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < ak−1 , and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1. First assume that a0 > 0. For = 0, 1, . . . , k − 1 and t = 1, 2, . . . , rk−−1 , we define S
j=1
rk−j +t
:=
ai + (t − 1)ak−−1 +
rk−m ak−m : i = 0, 1, . . . , k − − 1 .
m=1
(3.28) It is easy to see that |S
j=1
rk−j +t |
=k−
for = 0, 1, . . . , k − 1 and t = 1, 2, . . . , rk−−1 ,
and max(Si ) < min(Si+1 ) for i = 1, 2, . . . ,
k−1
rk−−1 − 1.
=0
It follows that the sets S1 , S2 , . . . , S k−1 rk−−1 are pairwise disjoint, and so =0
k−1 r
k−1 rk−−1 =0
=0k−−1
S(¯r, A) ≥ Si = |Si |
i=1
i=1
=
k−1 rk−−1
|S
j=1
=0 t =1
rk−j +t |
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=
k−1 rk−−1
(k − )
=0 t =1
=
k−1
k−1
1
t =1
=0
=
rk−−1
(k − )
(k − )rk−−1
=0
=
k−1
(i + 1)ri .
i=0
Therefore,
k−1
S(¯r, A) ≥ (i + 1)ri , i=0
and thus (3.26) is established. Now, assume that a0 = 0. Let ¯r = (r1 , r2 , . . . , rk−1 ) and let A = (a1 , a2 , . . . , ak−1 )¯r . Then S(¯r, A) = {0} ∪ S ¯r , A . Therefore, by the previous case, we have
S(¯r, A) = 1 + S ¯r , A , and so k−1
S(¯r, A) ≥ 1 + iri . i=1
This establishes (3.27). Next we show that the lower bounds in (3.26) and (3.27) are best possible. Let k ≥ 2 and ¯r = (r0 , r1 , . . . , rk−1 ), where ri ≥ 1 for i = 0, 1, . . . , k − 1. Let A0 = [1, k]¯r . Then k−1 the smallest and the largest integers in S(¯r, A0 ) are 1 and i=0 (i + 1)ri , respectively. Therefore, S(¯r, A0 ) ⊆ 1,
k−1 i=0
and so
(i + 1)ri ,
R.K. Mistri et al. / Journal of Number Theory 148 (2015) 235–256
253
k−1
S(¯r, A0 ) ≤ (i + 1)ri . i=0
This inequality together with the inequality (3.26) implies
k−1
S(¯r, A0 ) = (i + 1)ri . i=0
Thus the lower bound in (3.26) is best possible. Similarly, by considering the sequence A1 = [0, k − 1]¯r , where k ≥ 2 and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1, it can be verified that the lower bound in (3.27) is best possible. This completes the proof. 2 Remark 3.1. Theorem 3.1 implies Theorem 2.1 by taking ¯r = (r, r, . . . , r). Remark 3.2. Let ¯r = (r0 , r1 , . . . , rk−1 ) and ¯s = (rk−1 , . . . , r0 ) where ri ≥ 1 for i = 0, 1, . . . , k − 1. Let A0 = [1, k]¯r and A1 = [0, k − 1]¯r , where k ≥ 2, so that A0 = 1 + A1 . By the above proof, it follows that |S(¯r, A0 )| = |S(¯r, A1 )|. Thus |S(¯r, A)| is not translational invariant. However, |S(¯r, A)| is invariant under scalar multiplication: Let d is a nonzero integer. For d > 0, |S(¯r, d ∗ A)| = |d ∗ S(¯r, A)|, and for d < 0, |S(¯s, d ∗ A)| = |d ∗ S(¯r, A)|. 3.2. Inverse problem Theorem 3.2. Let k ≥ 3 and let ¯r = (r0 , r1 , . . . , rk−1 ), where ri ≥ 1 for i = 0, 1, . . . , k − 1. If A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of positive integers, where a0 < a1 < · · · < ak−1 , such that
k−1
S(¯r, A) = (i + 1)ri ,
(3.29)
i=0
then A = d ∗ [1, k]¯r for some positive integer d. If A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of nonnegative integers, where a0 < a1 < · · · < ak−1 , such that a0 = 0 and k−1
S(¯r, A) = 1 + iri ,
(3.30)
i=1
then A = d ∗ [0, k − 1]¯r for some positive integer d. Proof. Let A = (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < ak−1 . If a0 ≥ 1, then it follows from the proof of Theorem 3.1 that
R.K. Mistri et al. / Journal of Number Theory 148 (2015) 235–256
254
S(¯r, A) =
k−1 rk−−1
S
j=1
rk−j +t ,
=0 t =1
where S
j=1
rk−j +t
is the set defined in (3.28). For = 0, 1, . . . , k − 2, we have
max(S +1 rk−j ) = ak−−1 + (rk−−1 − 1)ak−−1 +
j=1
rk−m ak−m
m=1 +1
=
rk−m ak−m ,
m=1
and min(S +1 rk−j +1 ) = a0 +
+1
j=1
rk−m ak−m .
m=1
Now, for = 0, 1, . . . , k − 2, we have ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m < max(S +1 rk−j ) j=1
m=1
< min(S +1 rk−j +1 ), j=1
(3.31)
and ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m
m=1
< a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m
m=1
< min(S +1 rk−j +1 ).
(3.32)
j=1
It follows from (3.31) and (3.32) that max(S +1 rk−j ) = a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m ,
j=1
m=1
and so +1 m=1
Therefore,
rk−m ak−m = a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
m=1
rk−m ak−m .
R.K. Mistri et al. / Journal of Number Theory 148 (2015) 235–256
ak−−1 = a0 + ak−−2
255
for = 0, 1, . . . , k − 2,
and so A = a0 ∗ [1, k]¯r . Thus we have proved that A = d ∗ [1, k]¯r with d = a0 . If a0 = 0, then A1 = (a1 , a2 , . . . , ak−1 )¯r is a sequence of positive integers containing k − 1 distinct integers, where ¯r = (r1 , . . . , rk−1 ). Since S(¯r, A) = {0} ∪ S ¯r , A1 , it follows that
S(¯r, A) = 1 + S ¯r , A1 , and so k−1
S ¯r , A1 = S(¯r, A) − 1 = iri . i=1
Therefore, by the previous arguments, it follows that A1 = a1 ∗ [1, k − 1]¯r , and so A = a1 ∗ [0, k − 1]¯r . Thus we have proved that A = d ∗ [0, k − 1]¯r with d = a1 ≥ 1. This completes the proof. 2 Remark 3.3. Theorem 3.2 implies Theorem 2.3 by taking ¯r = (r, r, . . . , r). Remark 3.4. Theorem 3.1 and Theorem 3.2 are also true if the sequence A is considered to be a finite sequence of negative integers (nonpositive integers). In fact, if A is a finite sequence of negative integers (nonpositive integers), then A := −1 ∗ A is a sequence of positive integers (nonnegative integers). Moreover, S(¯r, A) = −1 ∗ S(¯s, A ), and so |S(¯r, A)| = |S(¯s, A )|. Here ¯s is same as in Remark 3.2
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References [1] B. Bollobás, I. Leader, The number of k-sums modulo k, J. Number Theory 78 (1999) 27–35. [2] M. DeVos, L. Goddyn, B. Mohar, A generalization of Kneser’s addition theorem, Adv. Math. 220 (2009) 1531–1548. [3] P. Erdős, A. Ginzburg, A. Ziv, Theorem in the additive number theory, Bull. Res. Counc. 10F (1961) 41–43. [4] D. Grynkiewicz, On a partition analog of the Cauchy–Davenport theorem, Acta Math. Hungar. 107 (2005) 167–181. [5] D. Grynkiewicz, On a conjecture of Hamidoune for subsequence sums, Integers 5 (2) (2005) A07. [6] Y.O. Hamidoune, Subsequence sums, Combin. Probab. Comput. 12 (2003) 413–425. [7] R.K. Mistri, R.K. Pandey, A generalization of sumsets of set of integers, J. Number Theory 143 (2014) 334–356. [8] M.B. Nathanson, Inverse theorems for subset sums, Trans. Amer. Math. Soc. 347 (1995) 1409–1418. [9] M.B. Nathanson, Additive Number Theory: Inverse Problems and the Geometry of Sumsets, Springer, 1996. [10] E. Szemerédi, V.H. Vu, Finite and infinite arithmetic progressions in sumsets, Ann. of Math. (2) 163 (1) (2006) 1–35. [11] V.H. Vu, Some new results on subset sums, J. Number Theory 124 (2007) 229–233. [12] Q.-H. Yang, Y.-G. Chen, On the cardinality of general h-fold sumsets, arXiv preprint, arXiv:1406.7346, 2014.
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Subsequence sums: Direct and inverse problems Raj Kumar Mistri a,1 , Ram Krishna Pandey b,∗ , Om Praksh a a
Department of Mathematics, Indian Institute of Technology Patna, Patna 800013, India b Department of Mathematics, Indian Institute of Technology Roorkee, Roorkee 247667, India
a r t i c l e
i n f o
Article history: Received 7 July 2014 Received in revised form 30 August 2014 Accepted 2 September 2014 Available online 4 November 2014 Communicated by David Goss Keywords: Subsequence sums Subset sums Direct problems Inverse problems
a b s t r a c t Let A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) be a r0 copies
r1 copies
rk−1 copies
finite sequence of integers with k distinct terms, denoted alternatively by (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < r = (r0 , r1 , . . . , rk−1 ), ri ≥ 1 for i = 0, 1, . . . , k − 1. ak−1 , ¯ The sum of all the terms of a subsequence of length at least one of the sequence A is said to be a subsequence sum of A. The set of all subsequence sums of A is denoted by S(¯ r, A). The direct problem for subsequence sums is to find the lower bound for |S(¯ r, A)| in terms of the number of distinct terms in the sequence A. The inverse problem for subsequence sums is to determine the structure of the finite sequence A of integers for which |S(¯ r, A)| is minimal. In this paper, both the problems are solved and some well-known results for subset sum problem are obtained as corollaries of the results for subsequence sum problem. © 2014 Elsevier Inc. All rights reserved.
* Corresponding author. E-mail addresses: [email protected] (R.K. Mistri), [email protected] (R.K. Pandey), [email protected] (O. Praksh). 1 Research supported by the University Grants Commission of India (No. F. 2-10/2011 (SA-I)). http://dx.doi.org/10.1016/j.jnt.2014.09.010 0022-314X/© 2014 Elsevier Inc. All rights reserved.
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236
0. Notation and terminology For integers a, b, c and for a set S of integers, let [a, b] = {n ∈ Z : a ≤ n ≤ b}, c ∗ S = {ca : a ∈ S} and c + S = {c + a : a ∈ S}. Let A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) be a finite sequence of integers, where a0 < r0 copies
r1 copies
rk−1 copies
a1 < · · · < ak−1 and ri ≥ 1 for i = 0, 1, . . . , k−1. Let ¯r = (r0 , r1 , . . . , rk−1 ) be the ordered k-tuple. Then we denote the sequence A = (a0 , . . . , a0 , a1 , . . . , a1 , . . . , ak−1 , . . . , ak−1 ) by r0 copies
r1 copies
rk−1 copies
(a0 , a1 , . . . , ak−1 )¯r . In this paper, by a finite sequence of integers, we always mean a finite sequence of integers whose terms are in nondecreasing order. This does not cause any harm, because the results in this paper are related to subsequence sums which remain unaltered by changing the order of the terms of the sequence. For an integer c, let c + (a0 , a1 , . . . , ak−1 )¯r = (c + a0 , c + a1 , . . . , c + ak−1 )¯r and for a nonzero integer c, let c ∗ A = c ∗ (a0 , a1 , . . . , ak−1 )¯r =
(ca0 , ca1 , . . . , cak−1 )¯r , (cak−1 , cak−2 , . . . , ca0 )¯s ,
if c > 0, if c < 0,
where ¯s = (rk−1 , . . . , r0 ). For integers a, b and c such that b = a + k − 1, let [a, b]¯r = (a, a + 1, . . . , b)¯r , and c + [a, b]¯r = [c + a, c + b]¯r . We shall call [a, b]¯r a sequence-interval. In particular, if ¯r = (r, r, . . . , r), then we simply write (a0 , a1 , . . . , ak−1 )r in place of (a0 , a1 , . . . , ak−1 )¯r and [a, b]r in place of [a, b]¯r . Let [a, b]r \(0)r = (a, a + 1, . . . , −1, 1, 2, . . . , b)r . If A is a sequence, then we write a ∈ A to mean that a is a term of the sequence A. If A is a set, then the notation a ∈ A has the usual meaning. We say that two finite sequences of integers are disjoint if there is no common term in these sequences. In this case, we write A ∪ B to denote the sequence obtained by merging these sequences and reordering the terms in nondecreasing order. For example, if A = (1, 2, 4)u¯ , where ¯ = (2, 1, 3) and B = (3, 5, 6)v¯ , where v ¯ = (1, 2, 2) are two sequences, then A ∪ B = u (1, 1, 2, 4, 4, 4) ∪ (3, 5, 5, 6, 6) = (1, 2, 3, 4, 5, 6)¯r , where ¯r = (2, 1, 1, 3, 2, 2). If A and B are two sets, then A ∪ B has usual meaning. Finally, we agree with the convention that a b = 0 if a and b are two positive integers such that a < b.
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237
1. Introduction Let A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of integers, where a0 < a1 < · · · < ak−1 , ¯r = (r0 , r1 , . . . , rk−1 ) and ri ≥ 1 for i = 0, 1, . . . , k − 1. Let B be a subsequence of A of length greater than or equal to one. We define the subsequence sum s(B) of B by s(B) :=
b,
b∈B
and
S(¯r, A) := s(B) : B is a subsequence of A of length ≥ 1 . In particular, if ¯r = (r, r, . . . , r), then we write S(r, A) instead of S(¯r, A). The direct problem for the subsequence sums is to find a lower bound for |S(¯r, A)| in terms of k, the number of distinct terms in the sequence A. The inverse problem for the subsequence sums is to determine the structure of the finite sequence A of integers for which |S(¯r, A)| is minimal. The similar direct and inverse problems for h-fold sumset, restricted h-fold sumset and subset sums of a finite set of integers have been studied in the past. One can refer for example [8–11]. Very recently, the h-fold sumset and the restricted h-fold sumset have been generalized as a generalized sumset and the corresponding direct and inverse problems have been studied (see [7,12]). There is a closely related problem for subsequence sums in additive abelian groups. Let a = (a1 , a2 , . . . , am ), where m ≥ 1, be a sequence of elements of an additive abelian group G. Let (a) denote the set of all group elements representable as a sum of a subsequence of a of length . Then the problems such as the direct problem for (a) and the properties of (a) (for example ai ∈ (a) for some i ∈ {1, 2, . . . , m}) etc. have been studied [1,3–6]. DeVos, Goddyn and Mohar [2] further generalize the sequence a = (a1 , a2 , . . . , am ) of elements of G to the sequence A = (A1 , A2 , . . . , Am ) of finite subsets of G and determine the direct theorem for (A), where
(A) = {ai1 + · · · + ail : 1 ≤ i1 < · · · < i ≤ m and aij ∈ Aij ∀1 ≤ j ≤ }.
In this paper, we solve both the direct and the inverse problems for the sumset S(r, A) of subsequence sums. We also solve the direct and the inverse problems for the general sumset S(¯r, A) of subsequence sums where A contains either only nonnegative terms or only nonpositive terms. These direct and inverse problems of subsequence sums naturally generalize the direct and the inverse problems of subset sums studied by Nathanson [8] by taking r = 1 in S(r, A). Thus the subset sum problem may be considered as a particular type of subsequence sum problem.
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Since the proofs of the results for general sequence are somewhat complicated, first we restrict ourselves to the case of particular sequence of the type (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 and r ≥ 1. This is discussed in Section 2. By this generalization, we obtain the direct and the inverse theorems of Nathanson [8] as corollaries of these results. Furthermore, the direct and the inverse theorems for the general sumset S(¯r, A) of subsequence sums, where A contain either only nonnegative terms or only nonpositive terms, are discussed in Section 3. It would be interesting if one studies the direct and the inverse problems for the general sumset S(¯r, A) of subsequence sums, where A contains positive integers, negative integers and/or zero. 2. Subsequence sum problem for the sequence (a0 , a1 , . . . , ak−1 )r 2.1. Direct problem For the finite sequence of integers A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 and r ≥ 1, we shall say that A is a finite sequence of integers with k distinct terms each with repetitions r. Theorem 2.1. Let k ≥ 1 and r ≥ 1. Let A be a finite sequence of positive integers with k distinct terms each with repetitions r. Then
S(r, A) ≥ r k + 1 . 2
(2.1)
Let A be a finite sequence of nonnegative integers with k distinct terms each with repetitions r and 0 ∈ A. Then
S(r, A) ≥ 1 + r k . 2
(2.2)
The lower bounds in (2.1) and (2.2) are best possible. Proof. First assume that k = 1. Then A = (a)r for some integer a. If a > 0, then 1 |S(r, A)| = r = r 1+1 2 . If a = 0, then |S(r, A)| = 1 = 1 + r 2 . Therefore, the theorem is true for k = 1. Hence we may assume that k ≥ 2. Let A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 , k ≥ 2 and r ≥ 1 be a finite sequence of positive integers. For = 0, 1, . . . , k − 1 and j = 1, 2, . . . , r, we define Sr+j :=
ai + (j − 1)ak−−1 +
m=1
It is easy to see that
rak−m : i = 0, 1, . . . , k − − 1 .
(2.3)
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|Sr+j | = k −
239
for = 0, 1, . . . , k − 1 and j = 1, 2, . . . , r,
and max(Si ) < min(Si+1 ) for i = 1, 2, . . . , rk − 1. It follows that the sets S1 , S2 , . . . , Srk are pairwise disjoint, and so
r rk rk k−1 k−1
k+1
S(r, A) ≥ Si
= |Si | = |Sr+j | = r(k − ) = r .
2 i=1
i=1
=0 j=1
=0
Therefore,
S(r, A) ≥ r k + 1 . 2 Thus (2.1) is established. Now, let A = (a0 , a1 , . . . , ak−1 )r be a finite sequence of nonnegative integers, where 0 = a0 < a1 < · · · < ak−1 , k ≥ 2 and r ≥ 1 so that 0 ∈ A. Let A = (a1 , a2 , . . . , ak−1 )r . Then S(r, A) = {0} ∪ S r, A . Therefore, by the previous case, we have
S(r, A) = 1 + S r, A ≥ 1 + r k . 2 This establishes (2.2). Next we show that the lower bounds in (2.1) and (2.2) are best possible. Let k ≥ 2 and r ≥ 1. Let A0 = [1, k]r . Then the smallest and the largest integers in S(r, A0 ) are 1 and r k+1 2 , respectively. Therefore, k+1 S(r, A0 ) ⊆ 1, r , 2 and so
S(r, A0 ) ≤ r k + 1 . 2 This inequality together with the inequality (2.1) implies
S(r, A0 ) = r k + 1 . 2 Thus the lower bound in (2.1) is best possible.
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Similarly, by considering the sequence A1 = [0, k − 1]r , where k ≥ 2 and r ≥ 1, it can be verified that the lower bound in (2.2) is best possible. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.1. (See [8], Theorem 3.) Let k ≥ 2. If A is a set of k positive integers, then
S(A) ≥
k+1 . 2
(2.4)
If A is a set of k nonnegative integers and 0 ∈ A, then
S(A) ≥ 1 + k . 2
(2.5)
The lower bounds in (2.4) and (2.5) are best possible. Remark 2.1. If in the above theorem, A is a finite sequence of negative integers (nonpositive integers), then also the inequality (2.1) ((2.2)) follows. In fact, if A is a finite sequence of negative integers (nonpositive integers), then A := −1 ∗ A is a sequence of positive integers (nonnegative integers). Moreover, S(r, A) = −1 ∗ S(r, A ), and so |S(r, A)| = |S(r, A )|. Theorem 2.2. Let k ≥ 2 and r ≥ 1. Let A be a finite sequence of integers with k distinct terms each with repetitions r. If 0 ∈ A, then
S(r, A) ≥
r(k2 −1) + 4 rk2 4 +1
1
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2).
(2.6)
If 0 ∈ / A, then
S(r, A) ≥
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2).
+1
(2.7)
The lower bounds in (2.6) and (2.7) are best possible. Proof. First assume that k = 2. If 0 ∈ A, then the sequence A contains a nonzero integer a. Let A = (a)r . Then S(r, A) = {0} ∪ S r, A , and so
S(r, A) = 1 + S r, A .
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It follows that
S(r, A) = 1 + r, and so 2
S(r, A) ≥ r.2 + 1. 4
If 0 ∈ / A, then A = (a0 , a1 )r , where a0 and a1 are nonzero integers and a0 < a1 . If a0 and a1 are both positive integers or both negative integers, then by Theorem 2.1, 2
S(r, A) ≥ r 3 = 3r ≥ 2r + 1 = r (2 + 1) − 1 + 1. 2 4 If a0 < 0 and a1 > 0, then let A = (a0 )r and A = (a1 )r . Now, a0 + a1 ∈ S(r, A) and S(r, A ) ∩ S(r, A ) = ∅. Therefore,
S(r, A) ≥ S r, A + S r, A + 1, and so 2
S(r, A) ≥ r + r + 1 = 2r + 1 = r (2 + 1) − 1 + 1. 4
Thus the theorem is true for k = 2. Hence we may assume that k ≥ 3. It is easy to see that for k ≥ 3, r(k2 −1) + 1 if k ≡ 1 (mod 2), k 4 1+r ≥ rk2 2 if k ≡ 0 (mod 2), 4 +1 and
k+1 r 2
≥
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2).
+1
Therefore, we may assume that A contains both positive and negative integers. Let A contains p distinct positive integers and n distinct negative integers. By Theorem 2.1, the set S(r, A) contains at least r p+1 positive integers and at least r n+1 negative 2 2 integers.
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Case 1 (0 ∈ A). In this case k = p + n + 1 and 0 ∈ S(r, A). Therefore,
S(r, A) ≥ r p + 1 + r n + 1 + 1 2 2 p+1 k−p =r + +1 2 2 k2 − k 2 +1 = r p − p(k − 1) + 2 2 k−1 k2 − 1 =r p− + 1. + 2 4 Now, if k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m. Therefore, 2
S(r, A) ≥ r (p − m)2 + k − 1 + 1, 4 and so 2
S(r, A) ≥ r(k − 1) + 1. 4
If k ≡ 0 (mod 2), then k = 2m for some positive integer m. Therefore, 2 2m − 1 k2 − 1 p− +1 + 2 4 2 1 k2 − 1 =r p−m+ +1 + 2 4 k2 2 = r (p − m) + (p − m) + + 1, 4
S(r, A) ≥ r
and so 2
S(r, A) ≥ rk + 1. 4
Thus (2.6) is established. / A). In this case k = p + n. Let p0 be the smallest positive integer in A, and Case 2 (0 ∈ let −n0 be the largest negative integer in A. The set S(r, A) contains at least r p+1 2 n+1 positive integers, each greater than or equal to p0 and at least r 2 negative integers each less than or equal to −n0 . The set S(r, A) also contains the integer p0 − n0 . Since −n0 < p0 − n0 < p0 , it follows that
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S(r, A) ≥ r p + 1 + r n + 1 + 1 2 2 p+1 k−p+1 =r + +1 2 2 k2 + k 2 +1 = r p − kp + 2 2 k k2 + 2k =r p− + 1. + 2 4 Now, if k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m. Therefore,
1 (k + 1)2 − +1 + 4 4 2 k+1 2 = r (p − m) − (p − m) + + 1, 2
S(r, A) ≥ r
1 p−m− 2
2
and so 2
S(r, A) ≥ r k + 1 + 1. 2 A similar argument holds for the case k ≡ 0 (mod 2). Thus (2.7) is established. Next we show that the lower bounds in (2.6) and (2.7) are best possible. Let k ≡ 1 (mod 2). Then k = 2m + 1 for some positive integer m. Let A0 = [−m, m]r . Then 0 ∈ A0 , A0 contains 2m + 1 distinct integers, and m+1 m+1 S(r, A0 ) ⊆ −r ,r . 2 2 Therefore,
S(r, A0 ) ≤ 2r m + 1 + 1 = r m2 + m + 1, 2 and so
S(r, A0 ) ≤ r
k−1 2
2
r(k2 − 1) k−1 +1= + 1. + 2 4
This inequality together with the inequality (2.6) implies 2
S(r, A0 ) = r(k − 1) + 1. 4
(2.8)
Now, let A1 = [−m, m + 1]r \(0)r . Then 0 ∈ / A1 , A1 contains 2m + 1 distinct integers, and
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m+1 m+2 S(r, A1 ) ⊆ −r ,r . 2 2 Therefore,
S(r, A1 ) ≤ r m + 2 + r m + 1 + 1 = r(m + 1)2 + 1, 2 2 and so 2 2
S(r, A1 ) ≤ r k − 1 + 1 + 1 = r k + 1 + 1. 2 2 This inequality together with the inequality (2.7) implies 2
S(r, A1 ) = r k + 1 + 1. 2
(2.9)
By (2.8) and (2.9), it follows that the lower bounds in (2.6) and (2.7) are best possible for k ≡ 1 (mod 2). Similarly, if k ≡ 0 (mod 2), then by considering the sequences A2 = [−m, m − 1]r and A3 = [−m, m]r \(0)r , it can be verified that the lower bounds in (2.6) and (2.7) are best possible for k ≡ 0 (mod 2). This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.2. (See [8], Theorem 4.) Let k ≥ 2 and let A be a set of k integers. If 0 ∈ A, then 2
S(A) ≥ k + 1. (2.10) 4 If 0 ∈ / A, then 2
S(A) ≥ (k + 1) + 1. 4
(2.11)
The lower bounds in (2.10) and (2.11) are best possible. 2.2. Inverse problem Let k ≥ 3 and let A0 = [0, k − 1]r and A1 = 1 + A0 = [1, k]r , where r ≥ 1. Since
S(r, A0 ) = 1 + r k , 2 and
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S(r, 1 + A0 ) = S(r, A1 ) = r k + 1 , 2 we have
S(r, A0 ) < S(r, A1 ) . Therefore,
S(r, A0 ) = S(r, A1 ) . Thus we see that |S(r, A)| is not translational invariant. However, if d is a nonzero integer, then S(r, d ∗ A) = d ∗ S(r, A), and so
S(r, d ∗ A) = S(r, A) . Thus |S(r, A)| is invariant under scalar multiplication. The inverse problem for S(r, A) is to find the finite sequence A for which |S(r, A)| is minimal in Theorem 2.1 and Theorem 2.2. We shall prove that every such sequence is a scalar multiple of a sequence-interval. Theorem 2.3. Let k ≥ 3 and r ≥ 1. If A is a finite sequence of positive integers with k distinct terms each with repetitions r such that
S(r, A) = r k + 1 , 2 then A = d ∗ [1, k]r for some positive integer d. If A is a finite sequence of nonnegative integers with k distinct terms each with repetitions r such that 0 ∈ A and
S(r, A) = 1 + r k , 2 then A = d ∗ [0, k − 1]r for some positive integer d. Proof. Let A = (a0 , a1 , . . . , ak−1 )r , where a0 < a1 < · · · < ak−1 . If a0 ≥ 1, then it follows from the proof of Theorem 2.1 that S(r, A) =
r k−1 j=1 =0
Sr+j ,
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where Sr+j is the set defined in (2.3). For = 0, 1, . . . , k − 2, we have max(S(+1)r ) = ak−−1 + (r − 1)ak−−1 +
rak−m =
m=1
+1
rak−m ,
m=1
and min(S(+1)r+1 ) = a0 +
+1
rak−m .
m=1
Now, for = 0, 1, . . . , k − 2, we have ak−−2 + (r − 1)ak−−1 +
rak−m < max(S(+1)r ) < min(S(+1)r+1 ),
(2.12)
m=1
and ak−−2 + (r − 1)ak−−1 +
rak−m
m=1
< a0 + ak−−2 + (r − 1)ak−−1 +
rak−m
m=1
< min(S(+1)r+1 ). Since ak−−2 + (r − 1)ak−−1 + that
m=1
(2.13)
rak−m ∈ S(+1)r , it follows from (2.12) and (2.13)
max(S(+1)r ) = a0 + ak−−2 + (r − 1)ak−−1 +
rak−m ,
m=1
and so +1
rak−m = a0 + ak−−2 + (r − 1)ak−−1 +
m=1
rak−m .
m=1
Therefore, ak−−1 = a0 + ak−−2
for = 0, 1, . . . , k − 2,
and so A = a0 ∗ [1, k]r . Thus we have proved that A = d ∗ [1, k]r with d = a0 .
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If a0 = 0, then A1 = (a1 , a2 , . . . , ak−1 )r is a sequence of positive integers containing k − 1 distinct integers. Since S(r, A) = {0} ∪ S(r, A1 ), it follows that
S(r, A) = 1 + S(r, A1 ) , and so
S(r, A1 ) = S(r, A) − 1 = r k . 2 Therefore, by the previous arguments, it follows that A1 = a1 ∗ [1, k − 1]r , and so A = a1 ∗ [0, k − 1]r . Thus we have proved that A = d ∗ [0, k − 1]r with d = a1 ≥ 1. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.3. (See [8], Theorem 5.) Let k ≥ 3. If A is a set of k positive integers such that |S(A)| = k+1 2 , then A = d ∗ [1, k] for some positive integer d. If A is a set of k nonnegative integers such that 0 ∈ A and |S(A)| = 1 + k2 , then A = d ∗ [0, k − 1] for some positive integer d. Theorem 2.4. Let k ≥ 3 and r ≥ 1. Let A be a finite sequence of integers with k distinct terms each with repetitions r. If 0 ∈ A and |S(r, A)| =
r(k2 −1) + 4 rk2 4 +1
1
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2),
(2.14)
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then there is a nonzero integer d such that A=
k−1 d ∗ [− k−1 2 , 2 ]r
if k ≡ 1 (mod 2),
d ∗ [− k2 , k2 − 1]r
if k ≡ 0 (mod 2).
(2.15)
If 0 ∈ / A and |S(r, A)| =
2 r( k+1 2 ) +1
if k ≡ 1 (mod 2),
2 −1 r (k+1) 4
if k ≡ 0 (mod 2),
+1
(2.16)
then there is a nonzero integer d such that A=
k+1 d ∗ [− k−1 2 , 2 ]r \(0)r
d∗
[− k2 , k2 ]r \(0)r
if k ≡ 1 (mod 2), if k ≡ 0 (mod 2).
(2.17)
Proof. The conditions on |S(r, A)| imply that A contains p distinct positive integers and n distinct negative integers, where p ≥ 1 and n ≥ 1. Case 1 (0 ∈ A). In this case k = p + n + 1 and by the proof of Theorem 2.2,
S(r, A) ≥ r
2 k−1 k2 − 1 p− + 1. + 2 4
If k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m, and 2
S(r, A) ≥ r (p − m)2 + k − 1 + 1. 4
(2.18)
By (2.14), we have 2
S(r, A) = r(k − 1) + 1. 4
(2.19)
From (2.18) and (2.19), it follows that (p − m)2 = 0, and so p = m. Therefore n = k − p − 1 = m. Thus p = n = m, and so S(r, A) contains exactly r m+1 positive integers 2 m+1 and exactly r 2 negative integers. Therefore, it follows from Theorem 2.3 that there exist positive integers p0 and n0 such that the positive part of A, say, A+ is p0 ∗ [1, m]r and the negative part of A, say, A− is −n0 ∗ [1, m]r . Since 0, p0 − n0 ∈ S(r, A) and −n0 < p0 − n0 < p0 , it follows that p0 − n0 = 0. Let d = p0 . Then A = A− ∪ (0)r ∪ A+ , and so A = d ∗ [−m, m]r .
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Therefore, k−1 k−1 , A=d∗ − . 2 2 r
(2.20)
Similarly, if k ≡ 0 (mod 2), then it can be shown that k k A=d∗ − , −1 2 2 r
(2.21)
for some nonzero integer d. Thus by (2.20) and (2.21), we have proved the first part of the theorem. Case 2 (0 ∈ / A). In this case k = p + n and by the proof of Theorem 2.2,
S(r, A) ≥ r p + 1 + r n + 1 + 1, 2 2 and so
S(r, A) ≥ r
p−
k 2
2 +
k2 + 2k + 1. 4
If k ≡ 1 (mod 2), then k = 2m + 1 for some positive integer m, and
S(r, A) ≥ r
p−m−
1 2
2 +
1 (k + 1)2 − + 1, 4 4
and so 2
S(r, A) ≥ r (p − m)2 − (p − m) + k + 1 + 1. 2
(2.22)
By (2.16), we have 2
S(r, A) = r k + 1 + 1. 2
(2.23)
Since (p −m)2 −(p −m) ≥ 0, it follows from (2.22) and (2.23) that (p −m)2 −(p −m) = 0, and so either p = m or p = m + 1. Therefore, either n = m + 1 or n = m. Thus either p = m and n = m + 1 or p = m + 1 and n = m. If p = m and n = m + 1, then the set positive integers and exactly r m+2 negative integers. S(r, A) contains exactly r m+1 2 2 Therefore, it follows from Theorem 2.3 that there exist positive integers p0 and n0 such that the positive part of A, say, A+ is p0 ∗ [1, m]r and the negative part of A, say, A− is −n0 ∗ [1, m + 1]r . The set S(r, A) also contains p0 − n0 . Now, if p0 − n0 = 0, then the number of positive or negative integers in S(r, A) will exceed. Hence p0 − n0 = 0, and so p0 = n0 . Therefore,
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A = A− ∪ A+ = p0 ∗ −(m + 1), m r \(0)r . Let d = −p0 . Then A = d ∗ [−m, m + 1]r \(0)r , and so k−1 k+1 , A=d∗ − (0)r . 2 2 r If p = m + 1 and n = m, then by a similar argument, it can be shown that there exists a nonzero integer d such that k−1 k+1 , A=d∗ − (0)r . 2 2 r Thus in all cases, we have shown that there exists a nonzero integer d such that k−1 k+1 , A=d∗ − (0)r 2 2 r
if k ≡ 1 (mod 2).
(2.24)
Similarly, if k ≡ 0 (mod 2), then it can be shown that k k A=d∗ − , (0)r 2 2 r
(2.25)
for some nonzero integer d. By (2.24) and (2.25), we have proved the second part of the theorem. This completes the proof. 2 As a consequence, for r = 1, we get the following corollary. Corollary 2.4. (See [8], Theorem 6.) Let k ≥ 3 and let A be a set of k integers. If 0 ∈ A 2 and |S(A)| = k4 + 1, then there is a nonzero integer d such that A=
d ∗ [−m, m] d ∗ [−m, m − 1]
if k = 2m + 1, if k = 2m.
2
If 0 ∈ / A and |S(A)| = (k+1) + 1, then there is a nonzero integer d such that 4 A=
d ∗ [−m, m]\{0} d ∗ [−m, m + 1]\{0}
if k = 2m, if k = 2m + 1.
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3. Subsequence sum problem for the general sequence (a0 , a1 , . . . , ak−1 )¯r 3.1. Direct problem Theorem 3.1. Let k ≥ 1. Let A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of integers, where a0 < a1 < · · · < ak−1 and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1. If a0 > 0, then
k−1
S(¯r, A) ≥ (i + 1)ri .
(3.26)
i=0
If a0 = 0, then k−1
S(¯r, A) ≥ 1 + iri .
(3.27)
i=1
The lower bounds in (3.26) and (3.27) are best possible. Proof. For k = 1, the result is obvious. Hence we may assume that k ≥ 2. Let A = (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < ak−1 , and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1. First assume that a0 > 0. For = 0, 1, . . . , k − 1 and t = 1, 2, . . . , rk−−1 , we define S
j=1
rk−j +t
:=
ai + (t − 1)ak−−1 +
rk−m ak−m : i = 0, 1, . . . , k − − 1 .
m=1
(3.28) It is easy to see that |S
j=1
rk−j +t |
=k−
for = 0, 1, . . . , k − 1 and t = 1, 2, . . . , rk−−1 ,
and max(Si ) < min(Si+1 ) for i = 1, 2, . . . ,
k−1
rk−−1 − 1.
=0
It follows that the sets S1 , S2 , . . . , S k−1 rk−−1 are pairwise disjoint, and so =0
k−1 r
k−1 rk−−1 =0
=0k−−1
S(¯r, A) ≥ Si = |Si |
i=1
i=1
=
k−1 rk−−1
|S
j=1
=0 t =1
rk−j +t |
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=
k−1 rk−−1
(k − )
=0 t =1
=
k−1
k−1
1
t =1
=0
=
rk−−1
(k − )
(k − )rk−−1
=0
=
k−1
(i + 1)ri .
i=0
Therefore,
k−1
S(¯r, A) ≥ (i + 1)ri , i=0
and thus (3.26) is established. Now, assume that a0 = 0. Let ¯r = (r1 , r2 , . . . , rk−1 ) and let A = (a1 , a2 , . . . , ak−1 )¯r . Then S(¯r, A) = {0} ∪ S ¯r , A . Therefore, by the previous case, we have
S(¯r, A) = 1 + S ¯r , A , and so k−1
S(¯r, A) ≥ 1 + iri . i=1
This establishes (3.27). Next we show that the lower bounds in (3.26) and (3.27) are best possible. Let k ≥ 2 and ¯r = (r0 , r1 , . . . , rk−1 ), where ri ≥ 1 for i = 0, 1, . . . , k − 1. Let A0 = [1, k]¯r . Then k−1 the smallest and the largest integers in S(¯r, A0 ) are 1 and i=0 (i + 1)ri , respectively. Therefore, S(¯r, A0 ) ⊆ 1,
k−1 i=0
and so
(i + 1)ri ,
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k−1
S(¯r, A0 ) ≤ (i + 1)ri . i=0
This inequality together with the inequality (3.26) implies
k−1
S(¯r, A0 ) = (i + 1)ri . i=0
Thus the lower bound in (3.26) is best possible. Similarly, by considering the sequence A1 = [0, k − 1]¯r , where k ≥ 2 and ¯r = (r0 , r1 , . . . , rk−1 ) with ri ≥ 1 for i = 0, 1, . . . , k − 1, it can be verified that the lower bound in (3.27) is best possible. This completes the proof. 2 Remark 3.1. Theorem 3.1 implies Theorem 2.1 by taking ¯r = (r, r, . . . , r). Remark 3.2. Let ¯r = (r0 , r1 , . . . , rk−1 ) and ¯s = (rk−1 , . . . , r0 ) where ri ≥ 1 for i = 0, 1, . . . , k − 1. Let A0 = [1, k]¯r and A1 = [0, k − 1]¯r , where k ≥ 2, so that A0 = 1 + A1 . By the above proof, it follows that |S(¯r, A0 )| = |S(¯r, A1 )|. Thus |S(¯r, A)| is not translational invariant. However, |S(¯r, A)| is invariant under scalar multiplication: Let d is a nonzero integer. For d > 0, |S(¯r, d ∗ A)| = |d ∗ S(¯r, A)|, and for d < 0, |S(¯s, d ∗ A)| = |d ∗ S(¯r, A)|. 3.2. Inverse problem Theorem 3.2. Let k ≥ 3 and let ¯r = (r0 , r1 , . . . , rk−1 ), where ri ≥ 1 for i = 0, 1, . . . , k − 1. If A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of positive integers, where a0 < a1 < · · · < ak−1 , such that
k−1
S(¯r, A) = (i + 1)ri ,
(3.29)
i=0
then A = d ∗ [1, k]¯r for some positive integer d. If A = (a0 , a1 , . . . , ak−1 )¯r be a finite sequence of nonnegative integers, where a0 < a1 < · · · < ak−1 , such that a0 = 0 and k−1
S(¯r, A) = 1 + iri ,
(3.30)
i=1
then A = d ∗ [0, k − 1]¯r for some positive integer d. Proof. Let A = (a0 , a1 , . . . , ak−1 )¯r , where a0 < a1 < · · · < ak−1 . If a0 ≥ 1, then it follows from the proof of Theorem 3.1 that
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S(¯r, A) =
k−1 rk−−1
S
j=1
rk−j +t ,
=0 t =1
where S
j=1
rk−j +t
is the set defined in (3.28). For = 0, 1, . . . , k − 2, we have
max(S +1 rk−j ) = ak−−1 + (rk−−1 − 1)ak−−1 +
j=1
rk−m ak−m
m=1 +1
=
rk−m ak−m ,
m=1
and min(S +1 rk−j +1 ) = a0 +
+1
j=1
rk−m ak−m .
m=1
Now, for = 0, 1, . . . , k − 2, we have ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m < max(S +1 rk−j ) j=1
m=1
< min(S +1 rk−j +1 ), j=1
(3.31)
and ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m
m=1
< a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m
m=1
< min(S +1 rk−j +1 ).
(3.32)
j=1
It follows from (3.31) and (3.32) that max(S +1 rk−j ) = a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
rk−m ak−m ,
j=1
m=1
and so +1 m=1
Therefore,
rk−m ak−m = a0 + ak−−2 + (rk−−1 − 1)ak−−1 +
m=1
rk−m ak−m .
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ak−−1 = a0 + ak−−2
255
for = 0, 1, . . . , k − 2,
and so A = a0 ∗ [1, k]¯r . Thus we have proved that A = d ∗ [1, k]¯r with d = a0 . If a0 = 0, then A1 = (a1 , a2 , . . . , ak−1 )¯r is a sequence of positive integers containing k − 1 distinct integers, where ¯r = (r1 , . . . , rk−1 ). Since S(¯r, A) = {0} ∪ S ¯r , A1 , it follows that
S(¯r, A) = 1 + S ¯r , A1 , and so k−1
S ¯r , A1 = S(¯r, A) − 1 = iri . i=1
Therefore, by the previous arguments, it follows that A1 = a1 ∗ [1, k − 1]¯r , and so A = a1 ∗ [0, k − 1]¯r . Thus we have proved that A = d ∗ [0, k − 1]¯r with d = a1 ≥ 1. This completes the proof. 2 Remark 3.3. Theorem 3.2 implies Theorem 2.3 by taking ¯r = (r, r, . . . , r). Remark 3.4. Theorem 3.1 and Theorem 3.2 are also true if the sequence A is considered to be a finite sequence of negative integers (nonpositive integers). In fact, if A is a finite sequence of negative integers (nonpositive integers), then A := −1 ∗ A is a sequence of positive integers (nonnegative integers). Moreover, S(¯r, A) = −1 ∗ S(¯s, A ), and so |S(¯r, A)| = |S(¯s, A )|. Here ¯s is same as in Remark 3.2
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