# Two-generator numerical semigroups and Fermat and Mersenne numbers

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1 Two-generator numerical semigroups and Fermat and Mersenne numbers Shalom Eliahou and Jorge Ramírez Alfonsín Abstract Given g N, what is the number of numerical semigroups S = a,b in N of genus N\S = g? After settling the case g = 2 k for all k, we show that attempting to extend the result to g = p k for all odd primes p is linked, quite surprisingly, to the factorization of Fermat and Mersenne numbers. Keywords: Semigroups, factorization, Fermat, Mersenne. MSC 2010: 05A15, 11A05, 11A51, 20M14 1 Introduction Anumerical semigroup isasubsets ofncontaining0, stableunderaddition, and with finite complement G(S) = N\S. The elements of G(S) are called the gaps of S, and their number is denoted g(s) and called the genus of S. The Frobenius number of S is its largest gap. See [6] for more details. If a 1,...,a r are positive integers with gcd(a 1,...,a r ) = 1, then they generate a numerical semigroup S = Na 1 + +Na r, denoted S = a 1,...,a r. For example, the numerical semigroup S = 4,5,7 has gaps G(S) = {1,2,3,6}, whence genus g(s) = 4 and Frobenius number 6. It is well known that every numerical semigroup admits a unique finite minimal generating set [3]. Given g N, what is the number n g of numerical semigroups S of genus g? Maria Bras-Amorós recently determined n g for all g 50 by computer. On this basis, she made three conjectures suggesting that the numbers n g behave closely like the Fibonacci numbers [1, 2]. For instance, the inequality n g n g 1 +n g 2, valid for g 50, is conjectured to hold for all g. 1

2 We propose here the refined problem of counting numerical semigroups S of genus g with a specified number of generators. Notation 1.1 Given g,r 1, let n(g,r) denote the number of numerical semigroups S of genus g having a minimal generating set of cardinality r. Of course n g = r 1n(g,r). Is there an explicit formula for n(g,r), and might it be true that n(g,r) n(g 1,r)+n(g 2,r)? In this paper, we focus on the case r = 2, i.e. on numerical semigroups S = a,b with gcd(a,b) = 1. The genus of S is equal to (a 1)(b 1)/2, by a classical theorem of Sylvester. This allows us to show in Section 2 that n(g,2) depends on the factorizations of both 2g and 2g 1, and to determine n(g,2) when 2g 1 is prime. In Section 3, we determine n(g,2) for g = 2 k and all k 1. We then tackle the case g = p k for odd primes p in Section 4. On the one hand, we provide explicit formulas for n(p k,2) when k 6. On the other hand, we show that obtaining similar formulas for all k 1 is linked to the factorization of Fermat and Mersenne numbers. We conclude with a few open questions about n(g,2). 2 Basic properties of n(g, 2) We will show that n(g,2) is linked with the factorizations of 2g and 2g 1. For this, we need the following theorem of Sylvester [7]. Theorem 2.1 Let a,b be coprime positive integers, and let S = a,b. Then maxg(s) = ab a b, and for all x {0,1,...,ab a b}, one has x G(S) ab a b x S. In particular, g(s) = (a 1)(b 1)/ Link with factorizations of 2g and 2g 1 We first derive that n(g, 2) is the counting function of certain particular factorizations of 2g. As usual, the cardinality of a set X will be denoted X. Proposition 2.2 Let g 1 be a positive integer. Then we have n(g,2) = {(u,v) N 2 1 u v, uv = 2g, gcd(u+1,v +1) = 1}. 2

3 Proof. Indeed, let S = a,b with 1 a b and gcd(a,b) = 1, and assume that g(s) = g. By Theorem 2.1, we have g = (a 1)(b 1)/2, i.e. 2g = (a 1)(b 1). The claim follows by setting u = a 1,v = b 1. A first consequence is that every g 1 is the genus of an appropriate 2-generator numerical semigroup. Corollary 2.3 n(g,2) 1 for all g 1. Proof. This follows from the factorization 2g = uv with u = 1,v = 2g. Concretely, the numerical semigroup S = 2,2g +1 has genus g. Ournextremarkshowsthatn(g,2)isalsolinkedwiththefactorsof2g 1. Lemma 2.4 Let g 1 be a positive integer, and let 2g = uv with u,v positive integers. Then gcd(u+1,v +1) divides 2g 1. Proof. Set δ = gcd(u+1,v+1). Then u v 1 mod δ, and therefore 2g = uv 1 mod δ. 2.2 The case where 2g 1 is prime We can now determine n(g,2) when 2g 1 is prime. As customary, for n N we denote by d(n) the number of divisors of n in N. Proposition 2.5 Let g 3, and assume that 2g 1 is prime. Then n(g,2) = d(2g)/2. In particular, n(g,2) = d(g) if g is odd. Proof. Let 2g = uv be any factorization of 2g in N. We claim that gcd(u+1,v+1) = 1. Indeed, by Lemma 2.4 we know that gcd(u+1,v+1) divides 2g 1. Assume for a contradiction that gcd(u+1,v+1) 1. Then gcd(u+1,v +1) = 2g 1, since 2g 1 is assumed to be prime. It follows that u,v 2g 2, implying 2g = uv 4(g 1) 2. 3

4 However, the inequality 2g 4(g 1) 2, while true at g = 2, definitely fails for g 3 as assumed here. Thus gcd(u+1,v +1) = 1, as claimed. Hence, by Proposition 2.2, we have n(g,2) = {(u,v) N 2 u v,2g = uv}. (1) Clearly 2g counts as many divisors u < 2g as divisors v > 2g. Moreover 2g is not a perfect square. This is clear for g = 3 or 4. If g 5 and 2g = a 2 with a N, then a 3 and 2g 1 = a 2 1 = (a 1)(a+1), contradicting the primality of 2g 1. We conclude from (1) that n(g,2) = d(2g)/2. Finally, if g is further assumed to be odd, then clearly d(2g)/2 = d(g). Proposition 2.5 cannot be extended to g = 2, even though 2g 1 is prime. Indeed n(2,2) = 1 as easily seen, whereas d(4)/2 is not even an integer. Since n(g,2) is controlled by the factorizations of both 2g and 2g 1, its determination is expected to be hard in general, even if the factors of g are known. Nevertheless, below we determine n(g,2) when g = 2 k for all k N, despite the fact that the prime factors of 2 k+1 1 are generally unknown. 3 The case g = 2 k Let g = 2 p 1 with p an odd prime, and assume that 2g 1 is prime. 1 Proposition 2.5 then applies, and gives n(2 p 1,2) = d(2 p )/2 = (p+1)/2. But we shall now determine n(2 k,2) for all k N, and show that its value only depends on the largest odd factor s of k +1. Theorem 3.1 Let g = 2 k with k N. Write k +1 = 2 µ s with µ N and s odd. Then n(2 k,2) = (s+1)/2. Proof. Since 2g = 2 k+1, the only integer factorizations 2g = uv with 1 u v are given by u = 2 i, v = 2 k+1 i 1 In fact a Mersenne prime, since 2g 1 = 2 p 1. See also Section

5 with 0 i (k +1)/2. In order to determine n(2 k,2) with Proposition 2.2, we must count those i in this range for which gcd(2 i + 1,2 k+1 i + 1) = 1. This condition is taken care of by the following claim. Claim. We have gcd(2 i +1,2 k+1 i +1) = 1 if and only if 2 µ divides i. The claim is proved by examining separately the cases where 2 µ divides i or not. Case 1: 2 µ divides i. Assume for a contradiction that there is a prime p dividing gcd(2 i +1,2 k+1 i +1). Then p is odd, and we have 2 i 2 k+1 i 1 mod p. (2) It follows that 2 2i 2 k+1 1 mod p. (3) Let m denote the multiplicative order of 2 mod p. It follows from (3) that m divides gcd(2i,k +1). Now, in the present case, we have gcd(2i,k +1) = gcd(i,k +1), since 2 µ divides i and k + 1, while 2 µ+1 divides 2i without dividing k + 1. Consequently m divides i, not only 2i. Hence 2 i 1 mod p, in contradiction with (2). Therefore gcd(2 i +1,2 k+1 i +1) = 1, as desired. Case 2: 2 µ does not divide i. We may then write i = 2 ν j with j odd and ν < µ. Set q = 2 2ν +1, and note that q 3. We claim that q divides gcd(2 i +1,2 k+1 i +1). Indeed, observe that 2 2ν 1 mod q, by definition of q. Since i = 2 ν j with j odd, we have 2 i +1 = (2 2ν ) j +1 ( 1) j +1 0 mod q. Similarly, we have k + 1 i = 2 ν j where j = 2 µ ν s j. Then j is odd, since µ ν > 0 and j is odd. As above, this implies that 2 k+1 i = (2 2ν ) j +1 0 mod q. It follows that q divides gcd(2 i +1,2 k+1 i +1), thereby settling the claim. 5

6 We may now conclude the proof. Indeed, the above claim yields n(2 k,2) = {i 0 i (k +1)/2, i 0 mod 2 µ } = {j 0 j (k +1)/2 µ+1 } = (k +1)/2 µ+1 +1 = (s+1)/2. Corollary 3.2 For every N 1, there are infinitely many g 1 such that n(g,2) = N. Proof. Let s = 2N 1. Then s is odd, and for all k = 2 µ s 1 with µ N, we have n(2 k,2) = (s+1)/2 = N by Theorem 3.1. In particular, there are infinitely many g 1 for which n(g,2) = 1. Since n(h,2) 1 for all h 1, the inequality n(g,2) n(g 1,2) + n(g 2,2) fails to hold infinitely often. This says nothing, of course, about the original conjecture n g n g 1 +n g 2 of Bras-Amorós. 4 The case g = p k for odd primes p Wehavedeterminedn(2 k,2) for allk 1. Attemptingtosimilarly determine n(p k,2) for odd primes p leads to a somewhat paradoxical situation. Indeed, while the case where k is small is relatively straightforward, formidable difficulties arise when k grows. This Jekyll-and-Hyde behavior is shown below. 4.1 When k is small Given positive integers q 1,...,q t, we denote by ρ q1,...,q t : Z Z/q 1 Z Z/q t Z the canonical reduction morphism ρ q1,...,q t (n) = (n mod q 1,...,n mod q t ), and shall write n a mod q instead of n a mod q. For example, the condition ρ 3,5,17 (p) = (2, 3, 8) means that p 2 mod 3, p 3 mod 5 and p 8 mod 17. 6

7 Proposition 4.1 Let p be an odd prime number. Then we have: { 1 if ρ3 (p) = 2 1. n(p,2) = 2 if ρ 3 (p) = 2, 2. n(p 2,2) = 3, 1 if ρ 3,5 (p) = (2,2) 3. n(p 3 2 if ρ,2) = 3,5 (p) = (2, 2) 3 if ρ 3,5 (p) = ( 2,2) 4 if ρ 3,5 (p) = ( 2, 2), { 4 if 4. n(p 4 ρ7 (p) = 3,2) = 5 if ρ 7 (p) = 3, 1 if ρ 3,5,17 (p) = (2,3,8) 2 if ρ 3,5,17 (p) = (2,3, 8) or (2, 3,8) 5. n(p 5 3 if ρ,2) = 3,5,17 (p) = (2, 3, 8) 4 if ρ 3,5,17 (p) = ( 2,3,8) 5 if ρ 3,5,17 (p) = ( 2,3, 8) or ( 2, 3,8) 6 if ρ 3,5,17 (p) = ( 2, 3, 8), { 6 if 6. n(p 6 ρ31 (p) = 15,2) = 7 if ρ 31 (p) = 15. With the Chinese Remainder Theorem, the above result implies that n(p 3,2) depends on the class of p modulo 15, and that n(p 4,2), n(p 5,2) and n(p 6,2) depend on the class of p modulo 7, 255 and 31, respectively. Proof. Let k 6. We determine n(p k,2) using Proposition 2.2. As p is an odd prime, counting the factorizations 2p k = uv with u v and gcd(u + 1,v + 1) = 1 amounts to count the number of exponents i in the range 0 i k satisfying the condition gcd(p i +1,2p k i +1) = 1. A convenient way to ease the computation of this gcd is to replace p by a variable x and to reduce, in the polynomial ring Z[x], the greatest common divisor of x i +1 and 2x j +1 to the simpler form gcd(x i +1,2x j +1) = gcd(f,g), (4) 7

8 where either polynomial f or g is constant. Polynomials are used here precisely for allowing such degree considerations. Since gcd(p i + 1,2p j + 1) is odd, we may equivalently work in the rings Z[2 1 ] or Z[2 1,x], where 2 is made invertible. Note that these rings are still unique factorization domains. We obtain the following table, with a method explained below. For simplicity, we write (f,g) rather than gcd(f,g), and 1 whenever either f or g is invertible in the ring Z[2 1,x]. The cases i = 0 and j = 0 are not included, since then x i +1 and 2x j +1 are already constant, respectively. gcd 2x+1 2x x x x x 6 +1 x+1 1 (x+1,3) 1 (x+1,3) 1 (x+1,3) x 2 +1 (2x+1,5) 1 (2x 1,5) (x 2 +1,3) (2x+1,5) 1 x 3 +1 (2x+1,7) (x 2,9) 1 (2x 1,9) (x+2,7) (x 3 +1,3) x 4 +1 (2x+1,17) (2x 2 +1,5) (x 2,17) 1 (2x 1,17) (2x 2 1,5) x 5 +1 (2x+1,31) (x+4,33) (4x+1,31) (x 2,33) 1 (2x 1,33) x 6 +1 (2x+1,65) (2x 2 +1,7) (2x 3 +1,5) (x 2 2,9) (x 2,65) 1 Table 1: Reduction of gcd(x i +1,2x j +1) for 1 i,j 6. In order to construct this table, we use the most basic trick for computing gcd s in a unique factorization domain A, namely: g 1 g 2 mod f gcd(f,g 1 ) = gcd(f,g 2 ) (5) for all f,g 1,g 2 A. As an illustration, let us reduce gcd(x 2 +1,2x 3 +1) to the form (4) in the ring Z[2 1,x]. We have gcd(x 2 +1,2x 3 +1) = gcd(x 2 +1, 2x+1) (6) = gcd(2 2 +1, 2x+1) (7) = gcd(1+2 2,2x 1), (8) where steps (6) and (7) follow from (5) and the respective congruences x 2 1 mod (x 2 +1), x 2 1 mod ( 2x+1). Hence gcd(x 2 +1,2x 3 +1) = gcd(2x 1,5), as displayed in Table

9 Now, from that table, it is straightforward to determine those pairs of exponents i,j with i+j 6 and those odd primes p for which gcd(p i +1,2p j +1) = 1, and hence to obtain the stated formulas for n(p k,2). Consider, for instance, the case k = 3. We shall count those exponents i {0,1,2,3} for which i = 0: Condition (9) is always satisfied. gcd(p i +1,2p 3 i +1) = 1. (9) i = 1: Table 4.1 gives gcd(p+1,2p 2 +1) = gcd(p+1,3), which equals 1 exactly when p 2 mod 3. i = 2: Table 4.1 gives gcd(p 2 + 1,2p + 1) = gcd(2p + 1,5), which equals 1 exactly when p 2 mod 5. i = 3: Finally, we have gcd(p 3 +1,3) = 1 exactly when p 2 mod 3. It follows that n(p 3,2) is entirely determined by the classes of p mod 3 and 5, with a value ranging from 1 to 4 depending on whether ρ 3,5 (p) equals (2,2),(2, 2),( 2,2) or ( 2, 2), as stated. The cases k = 1,2,4,5,6 are similar and left to the reader. We leave the determination of n(p 7,2) as an exercise to the reader. Let us just mention that the value of this function depends on the class of the prime p mod , and that its range is equal to {1,2,...,8}. The case k = 8 is much simpler. We state the result without proof. Proposition 4.2 Let p be an odd prime number. Then we have: 6 if ρ 7,31,127 (p) = (5,23,63) n(p 8 7 if ρ,2) = 7,31,127 (p) = ( 5,23,63), (5, 23,63) or (5,23, 63) 8 if ρ 7,31,127 (p) = ( 5, 23,63), ( 5,23, 63) or (5, 23, 63) 9 if ρ 7,31,127 (p) = ( 5, 23, 63). That was the gentle side of the story. Here comes the harder one. 9

10 4.2 When k grows When k grows arbitrarily, the task of determining n(p k,2) for all odd primes p using Proposition 2.2 becomes much more complicated, and turns out to be linked to hard problems. Let us focus on one specific factorization of 2p k, namely 2p k = uv with u = p k 1, v = 2p. In order to find when this factorization contributes 1 to n(p k,2), we need to decide when gcd(p k 1 +1,2p+1) is equal to 1. Here is the key reduction. Lemma 4.3 Let p be an odd prime and let m N. Then { gcd(2 gcd(p m +1,2p+1) = m +1,2p+1) if m is even, gcd(2 m 1,2p+1) if m is odd. Proof. As earlier, we will reduce gcd(x m + 1,2x + 1) in Z[2 1,x] to the form gcd(f,g), where either f or g is a constant polynomial. Since x 2 1 mod (2x+1), trick (5) yields gcd(x m +1,2x+1) = gcd(( 2) m +1,2x+1) Substituting x = p gives the stated formula. = gcd(2 m +( 1) m,2x+1). Hence, in order to determine when gcd(p m +1,2p+1) equals 1, we need to know the prime factors of 2 m + 1 for m even, and of 2 m 1 for m odd. This is an ancient open problem. It is not even known at present whether there are finitely or infinitely many Fermat or Mersenne primes, i.e. primes of the form F t = 2 2t +1 or M q = 2 q 1 with t 0 and q prime, respectively. Assume for instance that k = 2 t +1 for some t 1. Then k 1 is even, and thus Lemma 4.3 yields gcd(p k 1 +1,2p+1) = gcd(f t,2p+1). (10) Therefore, as long as the prime factors of the Fermat number F t remain unknown, we cannot determine those primes p for which the gcd in (10) equals 1, and hence write down an exact formula for n(p k,2) in the spirit of Proposition 4.1. For the record, as of 2010, the prime factorization of F t is completely known for t 11 only [5]. 10

11 Assume now that k = q +1 for some large prime q. Then k 1 = q is odd, and Lemma 4.3 yields gcd(p k 1 +1,2p+1) = gcd(2 q 1,2p+1). (11) Here again, we do not know the prime factors of M q = 2 q 1 in general; it may even happen that 2 q 1 hits some unknown Mersenne prime. Thus, we will not know for which primes p the gcd in (11) equals 1, i.e. when the specific factorization 2p k = p k 1 2p contributes 1 to n(p k,2). For the record, the largest prime currently known is the Mersenne prime p = 2 43,112,609 1, found in August 2008 [4]. The above difficulties concern the specific factorization 2p k = p k 1 2p. However, most other ones will also lead to trouble for some exponents k. For instance, consider the factorization 2p k = p k 2 2p 2, and let k = 2 t Then, a computation as in the proof of Lemma 4.3 yields gcd(p k 2 +1,2p 2 +1) = gcd(f t,2p 2 +1). Once again, not knowing the prime factors of F t = 2 2t + 1 prevents us to know for which primes p this gcd equals 1. 5 Concluding remarks and open questions We have determined n(g,2) when 2g 1 is prime, for g = 2 k for all k 1, and for g = p k for all odd primes p and k 6. The general case is probably out of reach. However, here are a few questions which might be more tractable, yet which we cannot answer at present. 1. Is there an explicit formula for n(3 k,2) as a function of k? Is it true that n(3 k,2) goes to infinity as k does? Here are the values of this function for k = 1,2,...,20: 2,3,4,4,5,7,8,9,8,9,11,13,11,15,16,14,14,18,20, Can one characterize those integers g 1 for which n(g,2) = 1? In special cases, we know enough to get a complete answer, for instance when g is prime using Proposition 4.1, or when g = 2 k using 11

12 Theorem 3.1. However, the general case seems to be very hard. As an appetizer, let us mention that a prime p satisfies n(p 21,2) = 1 if and only if p 8 mod ; the smallest such prime is p = 12,884,901, Letr N, r 1. Doesn(p 1 p r,2) attaineveryvalue i {1,2,...,2 r } for suitable distinct primes p 1,...,p r, and infinitely often so? 4. Let l N, l 1. Does n(p 2l 1,2) attain every value i {1,2,...,2l} for suitable odd primes p, and infinitely often so? 5. In contrast, is it true that min{n(p 2l,2) p odd prime} goes to infinity with l? Using the above methods, and a classical theorem of Dirichlet, it is fairly easy to show that, independently of the parity of k, the function n(p k,2) attains its maximal value k +1 infinitely often. References [1] M. Bras-Amorós, Fibonacci-like behavior of the number of numerical semigroups of a given genus, Semigroup Forum 76 (2008) [2] M. Bras-Amorós, Bounds on the number of numerical semigroups of a given genus, J. Pure and Applied Algebra 213 (2009) [3] R. Fröberg, C. Gottlieb, R. Häggkvist, On numerical semigroups, Semigroup Forum 35 (1987) [4] GIMPS, Great Internet Mersenne Prime Search, [5] W. Keller, Prime factors k 2 n + 1 of Fermat numbers F m and complete factoring status, web page available at [6] J.L. Ramírez Alfonsín, The Diophantine Frobenius problem, Oxford Lecture Series in Mathematics and its Applications 30, Oxford University Press, Oxford, (2005). 12

13 [7] J.J. Sylvester, On subinvariants, i.e. semi-invariants to binary quantities of an unlimited order, Amer. J. Math. 5 (1882) Authors addresses: Shalom Eliahou a,b,c, a Univ Lille Nord de France, F Lille, France b ULCO, LMPA J. Liouville, B.P. 699, F Calais, France c CNRS, FR 2956, France Jorge Ramírez Alfonsín, Institut de Mathématiques et de Modélisation de Montpellier Université Montpellier 2 Case Courrier 051 Place Eugène Bataillon Montpellier, France UMR 5149 CNRS 13

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