Gas in Semigrous J.L. Ramírez Alfonsín Université Pierre et Marie Curie, Paris 6, Equie Combinatoire - Case 189, 4 Place Jussieu Paris 755 Cedex 05, France. Abstract In this aer we investigate the behaviour of the gas in numerical semigrous. We will give an exlicit formula for the i th ga of a semigrou generated by k+1 consecutive integers (generalizing a result due to Brauer) as well as for a secial semigrou of three generators. It is also roved that the number of gas of the semigrou <, q > in the interval [q (k + 1)( + q),..., q k( + q)] is equals to (k + 1) + kq + k q for each k = 1,..., q +q. We actually give two roofs of the latter result, the first one uses the so-called Aery sets and the second one is an alication of the well-known Pick s theorem. Keywords Semigrous, gas, Frobenius number AMS Mathematics Subject Classification: 11B5, 11B75 1 Introduction Let s 1,..., s n be ositive integers such that their greatest common divisor, denoted by (s 1,..., s n ), is one. Let S =< s 1,..., s n > be the numerical e-mail: ramirez@math.jussieu.fr 1
semigrou 1 generated by s 1,..., s n. The genus of a numerical semigrou S is the number N(S) = #(IN + {0} \ S). The ositive elements of S (res. elements of IN + {0} \ S) are called the non-gas (res. gas) of S. One motivation to study gas comes from the imortant role they lay in the concet of symmetry [3] as well as in the investigations of Weierstrass semigrous [5, 6, 8] and its alications to algebraic codes [9]. In 188, while investigating the artition number function, J.J. Sylvester [15, age 134] (see also [14]) roved that if S q =<, q > with (, q) = 1 then N(S q ) = 1 ( 1)(q 1) (1) The so-called Frobenius number denoted by g(s) = g(s 1,..., s n ), is the largest integer not belonging to S (g(s) is also known as the conductor of S). From equation (1) it can be deduced that g(s q ) = q q. () Finding g(s) is a difficult roblem from the comutational oint of view, in general [11]. There exist numerous bounds (and formulas for articular semigrous S) for g(s) as well as generalizations and alications of it. We refer the reader to [1] for a detailed discussion on the Frobenius number. Let us enumerate the gas of S by increasing order l 1 < < l N(S). So, l N(S) = g(s) is both the largest ga of S as the N(S) th ga. Not surrisingly, comuting the i th ga of a semigrou is also difficult since it comes down to calculate the Frobenius number. Indeed, for calculating the i th ga of a semigrou S we may calculate g N(S) i (S) = g(s {g N(S) (S)} {g N(S) i+1 (S)}) where g N(S) (S) = g(s). There is not much known about the 1 Recall that a semigrou (S, ) consists of a nonemty set S and an associative binary oeration on S. If, in addition, there exists an element, which is usually denoted by 0, in S such that a 0 = 0 a = a for all a S, we say that (S, ) is a monoid. A numerical semigrou is a submonoid of (IN, +) such that the greatest common divisor of its elements is equal to one. In the introduction section of [1], A. Brauer stated that G. Frobenius mentioned, occasionally in his lectures, the roblem of finding the largest natural number that is not reresentable as a nonnegative integer combination of s 1,..., s n.
behaviour of the gas in semigrous. The main intention of this aer is to investigate the distribution of gas in some semigrous. In the next section, we rove the following two theorems. Theorem 1.1 Let a, k 1 be integers and let S =< a, a + 1,..., a + k > be a semigrou with gas l 1 < < l N(S). Let v m = (m+1)(a 1) k ( ) m(m+1), v 1 = 0 and r = a k. Then, N(S) = v r and l i = t i (a + k) + i v ti 1 for each i = 1,..., N(S) where t i is the smallest integer such that v ti i. In [1], Brauer found a formula for the Frobenius number of k + 1 consecutive ositive integers, this is given by ( a g(a, a + 1,..., a + k) = a k ) + 1 1. We remark that Theorem 1.1 contains Brauer s result as a articular case. Indeed, since v r = v r 1 + a (rk + 1) and t N(S) = r then l N(S) = r(a + k) + N(S) v r 1 = r(a + k) + v r (v r + rk + 1 a) = r(a + k) + a 1 kr = a(r + 1) 1 = a ( ) a k + 1 1. Theorem 1.1 yields us to consider the following question. Question: Let S =< a, a + d,..., a + kd > with d 1 and gcd(a, d) = 1. Is there a formula that comutes the i th ga of S for each 1 i N(S)? A ositive answer to the above question is obtained when i = N(S) since, for any integer d, it is known [13] that 3
( ) a l N(S) = g(a, a + d,..., a + kd) = a + d(a 1). (3) k The following Theorem answers this question when k = and d = a 1. Theorem 1. Let a 1 be an integer and let S =< a, a 1, 3a > be a semigrou with gas l 1 < < l N(S). Let r = (r 0, r 1,..., r 3n ) = (0, n 1, n, n 3, n 3, n 4, n 5, n 5, n 6,...,, 1, 1) and r = (r 0, r 1,..., r 3n 1) = (0, n, n 1, n, n, n 3, n 4, n 4, n 5,...,,, 1). We set v m = m r i and v m = m r i. Then, N(S) = { 3 4 n 3 n + 1 if a = n, n 1, 3 4 (n n + 1) if a = n + 1, n 1, and l i = { (ti 1)n + i v ti 1 if a = n, n 1, (t i 1)(n + 1) + i v t i 1 if a = n + 1, n 1, where t i is the smallest integer such that v ti i if a = n (or v t i i if a = n + 1). We notice that Theorem 1. contains equality (3) when k = and d = a 1. Indeed, equation (3) imlies that if a = n, n 1 then 4
g(a, a 1, 3a ) = a ( ) a + (a 1) = n ( n ) + (n 1) = n(n 1) + (n 1) = 6n 6n + 1 = 6n(n 1) + 1 and a = n + 1, n 1 then g(a, a 1, 3a ) = a ( ) a + (a 1) = (n + 1) ( n+1 ) + (n) = (n + 1)(n 1) + 4n = 6n n 1 = n(6n 1) 1. On other hand, by Theorem 1., we have that. (a) If a = n, n 1 then t N(S) = 3n, N(S) = v 3n and v 3n = v 3n 3 + r 3n 1 = v 3n 3 + 1. So, l N(S) = (3n 3)n + N(S) v 3n 1 = (3n 3)n + v 3n (v 3n 1) = 6n 6n + 1 = 6n(n 1) + 1. (b) If a = n + 1, n 1 then t N(S) = 3n 1, N(S) = v 3n 1 and v 3n 1 = v 3n + r 3n 1 = v 3n + 1. So, l N(S) = (3n )(n + 1) + N(S) v 3n 1 1 = (3n )(n + 1) + v 3n 1 (v 3n 1 1) = 6n 4n + 3n + 1 = n(6n + 1) 1. In this aer, we also study the gas of semigrous with two generators. 5
Theorem 1.3 Let, q ositive integers such that g.c.d.(, q) = 1. Let g k (S q ) be the number of gas of S q =<, q > in the interval [q (k + 1)( + q),..., q k( + q)], for each 0 k q +q 1. Then, { 1 if k = 0 g k (S q ) = (k + 1) + kq + k q if 1 k q +q 1. In Section 3, we give two roofs of Theorem 1.3, the first one uses the notions of Aery set and the second one the well-known Pick s theorem. These algebraic and geometric roofs may motivate further investigations on the study of gas of semigrous with n 4. We remark that Theorem 1.3 can be regarded as a generalization, to some extent, of equation () contained when k = 0 (an arithmetical and an algebraic roofs of equation () are given in [7] and [] resectively). Arithmetic sequences We may now rove Theorems 1.1 and 1.. Proof of Theorem 1.1. Let S i = {i(a + k) + 1,..., (i + 1)a 1} for each i = 0,..., r = a k. So, Si = a (ik + 1). We shall rove, by induction on i, that if s S i then s S. It is clear for i = 0 since S 0 = {1,..., a 1}. Let us suose that it is true for i = m r 1 and let s S m+1. Notice that s S if and only if s (a + j) S for some 0 j k. So, if we show that s (a + j) S m for all 0 j k then, by the inductive hyothesis, s S. To this end, it can be easily checked that m(a + k) + 1 s (a + j) = (m + 1)(a + k) + l (a + j) (m + 1)a 1 6
for all 1 l a (k(m + 1) + 1) and all 0 j k with equality on the left-hand side when l = 1 and j = k and equality on the right-hand side when l = a (k(m + 1) + 1) and j = 0. Now, we claim that if t S i for all i = 0,..., r = a k then t S. For this, we consider the sets T i = {(i + 1)a,..., (i + 1)(a + k)} for each i = 0,..., r = a k and Tr+1 = {x x (r + 1)(a + k) + 1}. We show, by induction on i, that if t T i for some i then t S. This is clear for i = 0 since T 0 = {a,..., a + k}. Now, suose that it is true for i = m 1 r 1. Then, by inductive hyothesis, all elements in T m 1 = {ma,..., ma + mk = m(a + k)} belong to S. So, all elements in A = {x + a x T m 1 } also belong to S. Moreover, all elements in B = {m(a + k) + a + 1, m(a + k) + a +,..., m(a + k) + a + k} also belong to S since m(a + k) belong to T m. Hence, all the elements in T m = {(m + 1)a,..., (m + 1)(a + k)} = {ma + a,..., m(a + k) + a + k} = {ma + a,..., ma + a + mk = m(a + k) + a, m(a + k) + a + 1,..., m(a + k) + a + k} = A B belong to S. Finally, note that ( ) a T r = T a k = + 1 k + 1 a. k So, T r contains a consecutive elements forming a comlete system modulo a, that is, T r contain elements t(0),..., t(a 1) such that t(i) i mod a. It is clear if t T r+1 then t = t(i) + la with l 0 integer and where t i mod a. 7
Thus, the set of all gas in S is given by {S i } 0 i r. So, ( ) r r(r + 1) N(S) = S i = a (ik + 1) = (r + 1)(a 1) k and l i = t i (a + k) + i v ti 1 for each i = 1,..., N(S) where t i is the smallest integer such that v ti i. Examle: Let S =< 8, 9, 10, 11 >. Then, k = 3 and r =. S 0 = {1,, 3, 4, 5, 6, 7}, S 1 = {1, 13, 14, 15}, S = {3}, T 0 = {8, 9, 10, 11}, T 1 = {16, 17, 19, 19, 0, 1, }, T = {4,..., 33} and T 3 = {34, 35,...}. Proof of Theorem 1.. We consider two cases according with the arity of a. Case A) If a = n, n 1 then S =< n, 4n 1, 6n >. We consider the sets S i = {(i 1)n + 1,..., (i 1)n + r i } for each i = 1,..., 3n. We shall show, by induction on i, that if s S i then s S. This can be easily checked for i = 1, and 3. Let us suose that it is true for i = m 4 and let s S m+1. Again, as in the above roof, we use the fact that s S if and only if s l S with either l = n or l = 4n 1 or l = 6n. We then show that s l S j, for some 1 j m when either l = n or l = 4n 1 or l = 6n and thus, by the inductive hyothesis, s l S (and therefore s S). Remark.1 (a) If r m+1 = r m then r m 1 = r m+1 + 1 and r m = r m+1 +. (b) If r m+1 < r m then r m = r m+1 + 1 and either r m 1 = r m+1 + 1 and r m = r m+1 + or r m 1 = r m+1 + and r m = r m+1 +. From (a) and (b) we have that r m+1 r m, r m+1 + 1 r m 1 and r m+1 + = r m. We now check three cases. (1) If l = n then s l = m(n) + j n = (m 1)n + j with 1 j r m+1. So, s l S m since, by the above remark, 1 j r m+1 r m. 8
() If l = 4n 1 then s l = m(n) + j (4n 1) = (m )n + j + 1 with 1 j r m+1. So, s l S m 1 since, by the above remark, 1 j + 1 r m+1 + 1 r m 1. (3) If l = 6n then s l = m(n) + j (6n ) = (m 3)n + j + with 1 j r m+1. So, s l S m since, by the above remark, 1 j + r m+1 + = r m. Now, we claim that if t S i for all i = 1,..., 3n then t S. For this, we consider the set T i = {(i 1)n + r i + 1,..., in}) for each i = 1,..., 3n and T 3n 1 = {(3n )n + 1,..., (3n )n + n}. We show, by induction on i, that if t T i for some i then s S. This is clear for i = 1, and 3. Indeed, the elements in T 1 = {n}, T = {4n 1, 4n = (n)} and T 3 = {6n, 6n 1 = 4n 1 + n, 6n = 3(n)} all belong to S. Now, we suose that it is true for i = m with 4 m 3n 3 and notice that T j = (j 1)n + s where r j + 1 s n for each 1 j 3n. Let t T m+1. If s = n then t = (m + 1 1)n + n = (m + 1)n that clearly belongs to S (t is a multile of n) and if s = n 1 then t = (m + 1 1)n + n 1 = (m + 1)n 1 = 4n 1 + (m 1)n that also belongs to S (t is a combination of elements 4n 1 and n). Let us suose now that r m+1 + 1 s < n 1. We have three subcases according to the above remark (we recall that t S if and only if t l S where l = n or 4n 1 or 6n ). (1) If r m+1 = r m then t l = m(n) + s n = (m 1)n + s with r m+1 + 1 s n. But, t l S m since r m + 1 = r m+1 + 1 s n. Thus, by the inductive hyothesis, t l S and therefore t S. () If r m = r m+1 + 1, r m 1 = r m+1 + 1 and r m = r m+1 + then t l = m(n) + s (4n 1) = (m )n + s + 1 with r m+1 + 1 s < n 1. But, t l S m 1 since r m 1 + 1 = r m+1 + s + 1 n. Thus, by the inductive hyothesis, t l S and therefore t S. (3) If r m = r m+1 + 1, r m 1 = r m+1 + and r m = r m+1 + then t l = m(n) + j (6n ) = (m 3)n + s + with with r m+1 + 1 s < n 1. But, t l S m since r m 1 + 1 = r m+1 + 3 s + n. Thus, by the inductive hyothesis, t l S and therefore t S. For the case i = 3n 1 we have that T 3n 1 = (3n )n+j with j = 1,..., n. So, if t T 3n 1 then t = (3n )n + j = (3n 3)n + j + n and since 9
(3n )n + j belongs to T 3n then t S. Since the elements in T 3n 1 are consecutive and T 3n 1 = a then they form a comlete system modulo a, that is, T 3n 1 contain elements t(0),..., t(a 1) such that t(i) i mod a. From this, it is clear that if t > max{t 3n 1 } then t = t(i) + la with l 0 integer and where t i mod a. Thus, the sets S i are the gas in S (increasingly ordered) and N(S) = v 3n = 3n r i = 3 4 n 3 n + 1. In order to find the i th -ga of S we first find out to which set S j the ga l i belongs to. The latter is done by comuting the smallest integer t i such that v ti i obtaining that l i S ti. And thus, l i is given by (t i 1)n + j with j = i v ti 1. Case B) If a = n + 1, n 1 then S =< n + 1, 4n + 1, 6n + 1 >. This case is analogous as the first one by considering the sets S i = {(i 1)(n + 1) + 1,..., (i 1)(n + 1) + r i} for each i = 1,..., 3n 1 and T j = {(j 1)(n + 1) + r j + 1,..., j(n + 1)} with j = 1,..., 3n. Again, in this case, the sets S i are the gas in S (increasingly ordered). So, N(S) = v 3n 1 = 3n 1 r i = 3 4 (n n + 1) and l i = (i 1)(n + 1) + i v t i 1. Examle: Let S =< 7, 13, 19 >. Then, n = 3 and r = (0, 6, 5, 4, 4, 3,,, 1). S 1 = {1,, 3, 4, 5, 6}, S = {8, 9, 10, 11, 1}, S 3 = {15, 16, 17, 18}, S 4 = {, 3, 4, 5}, S 5 = {9, 30, 31}, S 6 = {36, 37}, S 7 = {43, 44}, S 8 = {50}, T 1 = {7}, T = {13, 14}, T 3 = {19, 0, 1}, T 4 = {6, 7, 8}, T 5 = {3, 33, 34, 35}, T 6 = {38, 39, 40, 41, 4} and T 7 = {45, 46, 47, 48, 49}. 10
3 Gas in <, q > The Aéry set of element a, a S \ {0} is defined as A(S, a) = {s S s a S}. It is known that the set A(S, a) is a comlete system modulo a, that is, A(S, a) = {0 = w(0),..., w(a 1)} where w(i) is the least element in S congruent with i modulo a. Let, q ositive integers such that g.c.d.(, q) = 1. The semigrou S q =<, q > is symmetric, that is, for any integer x [0,..., g(s q )], x S q if and only if g(s q ) x S q. Proof of Theorem 1.3. As S q is symmetric then for each k = 0,..., q +q 1 there exists a one-to-one corresondence between the sets and {x IN x S q and q (k + 1)( + q) x q k( + q)} {x S q (k 1)( + q) x k( + q)}. Let A(S q, + q) = {w(0),..., w( + q 1)} where w(i) is the least element in S q congruent with i modulo + q. Then, for each i {0,..., + q 1} we have that (k 1)( + q) + i S q if and only if (k 1)( + q) + i w(i). Hence, {x S q (k 1)(+q) x k(+q)} = {w A(S q, +q) w < k(+q)} +1 (4) Since g.c.d.(, q) = 1 then A(S q, +q) = {0,,,..., (q 1), q, q,..., ( 1)q, q}. Besides, tq < k( + q) if and only if t < k + k/q or equivalently, t k + k/q. Analogously, t < k( + q) if and only if t k + kq/. Therefore, {w A(S q, +q) w < k(+q)} = {0,,..., (k+ kq/ ), q,..., (k+ k/q )q} and {0,,..., (k + kq/ ), q,..., (k + k/q )q} = 1+k + kq/ +k + k/q. Thus, by equation (4), 11
{x S q (k 1)( + q) x k( + q)} = (k + 1) + kq + k q and the result follows by the above bijection. Pick s theorem [10] is considered as one of the gems of elementary mathematics. It asserts that the area of a simlest lattice olygon 3 S, denoted by A(S), is given by I(S) + B(S)/ 1 where I(S) and B(S) are the number of lattice oints in the interior of S and in the boundary of S resectively; see [16] for a short roof of Pick s theorem. Second roof of Theorem 1.3. Let P be the lattice olygon with vertices (q 1, 1), ( 1, 1), (q, 0) and (0, ). Notice that there are no other lattice oints on the boundary of P and that the set of lattice oints inside P, denoted by I(P ), are all in the first quadrant. The equation of the line connecting the the first (res. the last) two oints is given by x + qy = q q (res. by x + qy = q). Let T 1 and T be the triangles formed by oints (q, 0), (0, ), ( 1, 1) and ( 1, 1), (q 1, 1), (q, 0) resectively. Since A(T 1 ) = 1 q 0 1 0 1 1 1 1 = 1(q + ) = 1 1 1 1 q 1 1 1 q 0 1 = A(T ) then A(P ) = A(T 1 ) + A(T ) = + q and, by Pick s theorem, we have that I(P ) = + q 1. We claim that line x + qy = q q + i contains 3 We call a olygon simle if its boundary is a simle closed curve. A lattice olygon is a olygon where its vertices have integer coordinates. 1
exactly one oint in I(P ) for each i = 1,..., + q 1. Suose that there exists 1 j + q 1 such that the line x + qy = q q + j contains two oints of I(P ), that is, x 1 + qy 1 = q q + j = x + qy for some 0 x 1, x < q, x 1 x and 0 y 1, y < q, y 1 y. But then, (x 1 x ) = (y y 1 )q and since (, q) = 1 then (x 1 x ) = sq q which is imossible. So each line x + qy = q q + i contains at most one oint of I(P ). Moreover, each line x + qy = q q + i has at least one oint of I(P ) otherwise, by the igeon hole rincile, it would exists a line x + qy = q q + j for some 1 j + q 1 containing two oints of I(P ), which is a contradiction. Notice that, since all lines x + qy = n q clearly have at least one lattice oint in the first quadrant then q q is the largest value for which x + qy = q q does not have solution on the nonnegative integers. Thus F 0 (S q ) = 1. Let k be the largest integer such that q k ( + q) 0. Let Q k be the olygon formed by the oints (q k, k), (q (k + 1), (k + 1)), ( k, k) and ( (k+1), (k+1)). Notice that Q 0 = P, that Q k is just a translation of Q 0 and that Q k 1 does not contains oints ( x, y), x, y > 0 (by definition of k ). Let rk 1 (res. rk) be the intersection of the line x+qy = q k(+q) with the x-axis (res. with the y-axis) for each k = 0,..., k. Let Q 1 k (res. Q k), k = 0,..., k 1 be the (not necessarily lattice) olygon formed by the oints (rk, 1 0), (rk+1, 1 0), (q k, k) and (q (k + 1), (k + 1)) (res. formed by the oints (0, rk), (0, rk+1), ( k, k) and ( (k + 1), (k + 1))). We notice that Q 1 k (res. Q k) is the iece of Q k that lies below the x-axis (res. on the left-hand side of the y-axis). Since each line x+qy = q (+q)+j, 1 j + q 1 has a unique solution with nonnegative integers (x, y) then the translated line x + qy = q k( + q) + j (lying in Q k ) does not have solution with nonnegative integers if the corresonding translation of (x, y) lies either in Q 1 k or Q k. Hence, F k (S q ) = I(Q 1 k) + I(Q k) + for each k = 1,..., k 1 (the term counts the gas corresonding to the inexistence of solutions for x + qy = q k( + q) and x + qy = q (k + 1)( + q)). We calculate I(Q 1 k) and I(Q k) for each k = 1,..., k 1. To this end, we first observe that the number of integer oints lying on the interval [( rk, 1 0),..., (q, 0)[= [(q k kq, 0),..., (q, 0)[ (res. lying on the interval [(0, ),..., (0, rk )[= [(0, ),..., (0, k k )[) is equals to k + kq q (res. 13
equals to k + k q ). Let 1 k and k the number of integer oints lying on the intervals [(r 1 k+1, 0),..., (r 1 k, 0)[ and [(0, r k+1),..., (0, r k)[ resectively for each k = 1,..., k 1. Then, I(Q 1 k) = k 1 = k 1 1 i = k 1 ( 1 + (i+1)q ( ( )) (i + 1) + (i+1)q i + iq iq ) = k + kq. Similarly, I(Q k) = k + k q, k = 1,..., k 1 and the result follows. We end this section with the following question. Let ρ i (S) = ρ i be the i th non-ga of S and let M(i) be the number of gas smaller than ρ i. Question 3.1 Is M(i) comutable in olynomial time? References [1] A. Brauer, On a roblem of artitions, Amer. Journal of Math. 64 (194), 99-31. [] T.C. Brown and P.J. Shiue, A remark related to the Frobenius roblem, The Fibonacci Quarterly 31(1) (1993), 3-36. [3] R. Frőberg, C. Gottlieb and R. H aggkvist, On numerical semigrous, Semigrou Forum 35 (1987), 63-83. [4] R. Kannan, Lattice translates of a olytoe and the Frobenius roblem, Combinatorica 1() (199), 161-177. [5] J. Komeda, Non-Weierstrass numerical semigrous, Semigrou Forum 57 (1998), 157-185. 14
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