On \(p\)-Frobenius Numbers for the Numerical Semigroups Generated by Three Consecutive Star Numbers

#A106 INTEGERS 25 (2025) ON p-FROBENIUS NUMBERS FOR THE NUMERICAL SEMIGROUPS GENERATED BY THREE CONSECUTIVE STAR NUMBERS Takao Komatsu Institute of Mathematics, Henan Academy of Sciences, Zhengzhou, China; Department of Mathematics, Institute of Science Tokyo, 2-12-1 Ookayama, Meguro-ku, Tokyo, Japan [email protected]; [email protected] Tapas Chatterjee Department of Mathematics, Indian Institute of Technology Ropar, Punjab, India [email protected] Palak Narula Department of Mathematics, Indian Institute of Technology Ropar, Punjab, India [email protected] Received: 12/30/24, Revised: 5/22/25, Accepted: 11/3/25, Published: 11/25/25 Abstract The Frobenius coin problem involves computing the largest integer, known as the Frobenius number, that cannot be expressed as a non-negative integral linear combination of given relatively prime positive integers. A more generalized version of this problem, termed as the p-Frobenius number, aims to find the largest integer that has at most prepresentations in terms of linear combinations, where pis any non-negative integer. In this article, we give the closed-form expressions of the pFrobenius numbers for the numerical semigroups generated by the triplets of the consecutive star numbers for the cases p= 0,1,and 2.Also, we present explicit formulas for their p-Sylvester numbers (which count the positive integers having no more than prepresentations) where p= 0,1,and 2. 1. Introduction The Star numbers, denoted by Sn,are commonly referred to as the centered 12gonal numbers or centered dodecagonal numbers. The formula expressing the nth star number is given by Sn= 6n(n−1)+1,where n≥1. The first few star numbers DOI: 10.5281/zenodo.17711576 INTEGERS: 25 (2025) 2 [21, A.131] are as follows {Sn}n≥1= 1,13,37,73,121,181,253,337,433,541,661,793,937,1093, 1261,1441,1633,1837,2053, . . . . These numbers appear in many number theoretic problems. Some well-known formulas include [27, OEIS A306980] ∞ X n=1 1 Sn =πtan(π/2√3) 2√3, ∞ X n=0 Sn n!= 7e, and ∞ X n=1 Sn 2n= 25. One can also easily prove the first identity above using the Cauchy’s residue theorem. Geometrically, the nth star number consists of a central point along with 12 copies of the (n−1)th triangular number tn−1.A notable observation is that infinitely many star numbers are also triangular numbers and among the initial instances, we have S1=1=t1,S7= 253 = t22,S91 = 49141 = t313, and S1261 = 9533161 = t4366 on the OEIS entry A156712. The star numbers are used for a new set of vector-valued Teichm¨uller modular forms, defined on the Teichm¨uller space, strictly related to the Mumford forms, which are holomorphic global sections of the vector bundle [19]. Let A={a1, a2, . . . , ak}be a set of relatively prime positive integers, where k≥2, and let pbe any non-negative integer. The p-numerical semigroup Sp(A) is defined as the set of integers whose non-negative integral linear combinations in terms of given positive integers a1, a2, . . . , akcan be expressed in more than pways [18]. For some background on the number of representations, refer, e.g., [4,6,10,28]. For the set of non-negative integers N0,the set Gp(A):=N0\Sp(A) is finite if and only if gcd (a1, a2, . . . , ak) = 1. Then, the maximum element of the set Gp(A),denoted by gp(A) is called the p-Frobenius number. The cardinality of the set Gp(A) is called the p-Sylvester number (or the p-genus) and is denoted by np(A).This kind of concept is a generalization of the famous Diophantine problem of Frobenius [2,25], since p= 0 is the classical case, where the original Frobenius number is denoted by g(A) = g0(A) and the genus as n(A) = n0(A).Here, the set Ais called the system of generators of the p-numerical semigroup Sp(A). When k= 2,there exists an explicit closed formula of the p-Frobenius number for any non-negative integer p[3]. However, for k= 3,the p-Frobenius number cannot be given by any set of closed formulas that can be reduced to a finite set of certain polynomials [9]. Since it is very difficult to give a closed explicit formula of any general sequence for three or more variables, many researchers have tried to find the Frobenius numbers for some special cases (see, e.g., [14,22,23,24] for more details). Recently, in [7,8], the Frobenius numbers for the triplets of successive centered triangular, centered square, centered pentagonal, and centered hexagonal numbers were studied. Though it is even more difficult when p>0 (see, e.g., [12,15, 16,17]), in [11], the p-Frobenius numbers of three consecutive triangular numbers INTEGERS: 25 (2025) 3 were studied. In this paper, the p-Frobenius numbers of the three consecutive star numbers are examined. Initially, the focus is on understanding the structure of the p-Ap´ery set for the classical case, followed by other positive integral values of p= 1,2. Additionally, we present explicit formulas for their p-Sylvester numbers for p= 0,1,2. 2. Preliminaries In this section, we recall the notion of the p-Ap´ery set [1] and some results to compute the p-Frobenius number and p-Sylvester number which will play a crucial role in proving the main theorem. Let us define the p-Ap´ery set. Definition 1. Consider a set of positive integers A={a1, a2, . . . , ak}(k≥2) with gcd(A) = 1. Without loss of generality, assume that a1= min(A) and let pbe any non-negative integer. Then, the p-Ap´ery set of A, denoted by App(A),is defined as App(A) = Ap(a1, a2, . . . , ak)={m(p) 0, m(p) 1, . . . , m(p) a1−1}, where m(p) iis the least positive integer of Sp(A),and satisfies the congruence m(p) i≡ i(mod a1), for 0 ≤i≤a1−1. This definition is equivalent to saying that m(p) i∈ Sp(A), m(p) i−a1∈ Sp(A),and m(p) i≡i(mod a1). Note that m(0) 0is defined to be 0. It follows that for given p, App(A)≡ {0,1, . . . , a1−1}(mod a1). In other words, App(A) forms a complete residue system modulo a1. One of the convenient formulas to obtain the p-Frobenius number and the pSylvester number (or p-genus) is via the elements in the corresponding p-Ap´ery set. The lemma given below describes the relationship between the Frobenius number and the Sylvester number with the associated Ap´ery set [13]. Lemma 1. Let gcd(a1, . . . , ak)=1with a1= min{a1, . . . , ak}. Then, we have gp(a1, . . . , ak) = max 0≤j≤a1−1m(p) j−a1, np(a1, . . . , ak) = 1 a1  a1−1 X j=0 m(p) j −a1−1 2. Remark 1. When p= 0 (classical case), the Frobenius number is essentially due to Brauer and Shockley [5], and the classical Sylvester number is due to Selmer [26]. More general formulas, including the p-power sum and the p-weighted sum, can also be seen in [13]. INTEGERS: 25 (2025) 4 In order to discuss the Frobenius number for triples of successive star numbers, we must ensure that they are relatively prime. Here, we give a lemma that articulates this requirement. Lemma 2. For any three consecutive star numbers Sn, Sn+1, and Sn+2, gcd(Sn, Sn+1, Sn+2)=1. Proof. We know that gcd(Sn, Sn+1, Sn+2) = gcd(gcd(Sn, Sn+1),gcd(Sn+1, Sn+2)). Let dn= gcd(Sn, Sn+1) and observe that (n+ 1)Sn−(n−1)Sn+1 = 2. This implies dn|2 and since star numbers are odd, we conclude that dn= 1. Thus, gcd(Sn, Sn+1) = 1 and hence, gcd(Sn, Sn+1, Sn+2) = 1. In addition, in a later section we use a classical identity known as B´ezout’s Lemma, which is stated as follows. Lemma 3 ([20]).Let aand bbe integers with greatest common divisor d. Then there exist integers xand ysuch that ax +by =d. Moreover, the integers of the form az +bt are exactly the multiples of d. 3. Main Results Now, we derive the explicit expressions for the p-Frobenius numbers and the pSylvester numbers for the triples consisting of successive star numbers, discussed in Section 3.1 and Section 3.2, respectively. 3.1. p-Frobenius Numbers The p-Frobenius numbers of the numerical semigroups generated by three consecutive star numbers are given as follows. Theorem 1. For p= 0,1,2, we have gp(Sn, Sn+1, Sn+2) = (2nSn+1 + (p+ 2)nSn+2 −Sn,if 6≤n≤9; (2n−11)Sn+1 + (p+ 3)nSn+2 −Sn,if n≥10. Remark 2. More explicitly, we can write gp(Sn, Sn+1, Sn+2) = (24n3+ 42n2+ 34n−1 + (6n3+ 18n2+ 13)p, if 6 ≤n≤9; 30n3−6n2−19n−12 + (6n3+ 18n2+ 13)p, if n≥10. When p≥3,the situation becomes complex, and no explicit formula has been derived so far. The cases where n= 2,3,4,5 are discussed later. INTEGERS: 25 (2025) 5 Proof. Our main goal is to find the p-Ap´ery set and establish the validity of the theorem case by case for n≥6. Case 1: p= 0. For convenience, substitute ty,z := ySn+1 +zSn+2 for non-negative integers yand z. Then, we can show that the elements of the 0-Ap´ery set are given as in Table 1. t0,0··· t2n−11,0t2n−10,0··· ··· t2n,0 . . .. . .. . .. . . t0,2n··· t2n−11,2nt2n−10,2n··· ··· t2n,2n t0,2n+1 ··· t2n−11,2n+1 . . .. . . . . .. . . t0,3n··· t2n−11,3n Table 1: Ap0(Sn, Sn+1, Sn+2) Firstly, we prove that the elements ty,z in Table 1form a complete residue system modulo Sn.To prove this we show that any such two elements in the table are incongruent modulo Sn.Assume for a contradiction that there exist ordered pairs (y1, z1) and (y2, z2) such that ty1,z1=ty2,z2in Table 1with ty1,z1≡ty2,z2(mod Sn).(1) Substituting the values of ty,z in Equation (1), we have y1Sn+1 +z1Sn+2 ≡y2Sn+1 +z2Sn+2 (mod Sn). Re-arranging the above equation, we get (y1−y2)Sn+1 + (z1−z2)Sn+2 ≡0 (mod Sn). This implies (y1−y2)(Sn+1 −Sn)+(z1−z2)(Sn+2 −Sn)≡0 (mod Sn). Consequently, 12n(y1−y2) + 12(2n+ 1)(z1−z2)≡0 (mod Sn). Furthermore, gcd(12, Sn)=1,and hence n(y1−y2) + (2n+ 1)(z1−z2)≡0 (mod Sn).(2) For y1, y2, z1, z2as in Table 1, we have |y1−y2| ≤ 2nand |z1−z2|≤3n. (3) INTEGERS: 25 (2025) 6 From Equation (2), n(y1−y2) + (2n+ 1)(z1−z2) is a multiple of Sn, and under the constraints of Equation (3), the only possible values are 0, Sn,or −Sn. If n(y1−y2) + (2n+ 1)(z1−z2) = 0, it follows that ty1,z1=ty2,z2which contradicts our initial assumption. Now, consider the case when n(y1−y2) + (2n+ 1)(z1−z2)=Sn= 6n2−6n+ 1.(4) Using B´ezout’s lemma, the extended Euclidean algorithm, and the inequalities in Equation (3), the only integral solution to Equation (4) is (y1−y2, z1−z2) = (2n−10,2n+ 1). As a result, y1= 2n−10 + y2, z1= 2n+1+z2. Since y2and z2are non-negative integers, it follows that ty1,z1lies outside Table 1. Hence, our initial assumption was wrong. On a similar line of reasoning, we can argue that n(y1−y2) + (2n+ 1)(z1−z2)=−Sn.Thus, no two elements in Table 1are congruent modulo Sn. Additionally, note that the count of elements in Table 1is equal to (2n+ 1)2+n(2n−10) = Sn.Therefore, we obtain that the set {ty,z :y, z ∈Table 1}constitutes a complete residue system modulo Sn.In other words, we show that for each i∈ {0,1,...,Sn−1},there exists a unique element ty,z in Table 1such that ty,z ≡i(mod Sn). We now proceed to prove that the elements in Table 1are indeed the smallest ones in their corresponding residue classes. Observe that (2n−10)Sn+1 + (2n+ 1)Sn+2 = (4n+ 3)Sn,(5) (2n+ 1)Sn+1 −nSn+2 = (n+ 1)Sn,(6) (3n+ 1)Sn+2 −11Sn+1 = (3n+ 2)Sn.(7) Consequently, t2n−10+y,z ≡ty,z−2n−1(mod Sn) and t2n−10+y,z > ty,z−2n−1 (2n+ 1 ≤z≤3n, 0≤y≤10), t2n+1+y,z ≡ty,n+z(mod Sn) and t2n+1+y,z > ty,n+z (0 ≤y≤2n−11,0≤z≤2n), t4n−9+y,z ≡t2n−10+y,n+z(mod Sn) and t4n−9+y,z > t2n−10+y,n+z (0 ≤y≤10,0≤z≤n), ty,z ≡ty−11,z+3n+1 (mod Sn) and ty,z > ty−11,z+3n+1 (0 ≤z≤n−1,11 ≤y≤2n). INTEGERS: 25 (2025) 7 Therefore, the elements in Table 1form the 0-Ap´ery set Ap0(Sn, Sn+1, Sn+2). Now, from Table 1, there are two possibilities for the largest value of the set Ap0(Sn, Sn+1, Sn+2): t2n,2nor t2n−11,3n.Note that, t2n,2n< t2n−11,3nif and only if nSn+2 >11Sn+1,which is equivalent to 6n3−48n2−53n−11 >0. Observe that the roots of the equation 6n3−48n2−53n−11 = 0 are −0.7214,−0.2822,and 9.0036. Therefore, for 6 ≤n≤9, t2n,2nis the largest element of the Ap´ery set, so by Lemma 1, we have that g0(Sn, Sn+1, Sn+2)=2nSn+1 + 2nSn+2 −Sn. While for n≥10, we have t2n−11,3nis the maximum among all the elements, so using Lemma 1 g0(Sn, Sn+1, Sn+2) = (2n−11)Sn+1 + 3nSn+2 −Sn. Case 2: p= 1. The elements of the 1-Ap´ery set can be determined from those of the 0-Ap´ery set as follows. t2n+1,0··· ··· t4n+1,0 . . .. . . . . .. . . t2n+1,n ··· ··· t4n+1,n t2n+1,n+1 ··· t4n−10,n+1 . . .. . . t2n+1,2n··· t4n−10,2n t2n−10,2n+1 ··· ··· t2n,2n+1 . . .. . . t2n−10,3n··· ··· t2n,3n t0,3n+1 ··· t2n−11,3n+1 . . .. . . t0,4n··· t2n−11,4n Table 2: Ap1(Sn, Sn+1, Sn+2) From the set of Equations (5), (6), and (7) and the relation (2n+ 12)Sn+1 + (2n+ 1)Sn= (4n+ 1)Sn+2, we have the following one-to-one correspondence between the elements of the 0Ap´ery set (on the left-hand side of congruences) and that of the 1-Ap´ery set (on the right-hand side of congruences): ty,z ≡ty+2n+1,z−n(mod Sn) (0 ≤y≤2n, n ≤z≤2n; 0 ≤y≤2n−11,2n+ 1 ≤z≤3n), ty,z ≡ty+2n−10,z+2n+1 (mod Sn) (0 ≤y≤10,0≤z≤n−1), ty,z ≡ty−11,z+3n+1 (mod Sn) (11 ≤y≤2n, 0≤z≤n−1). INTEGERS: 25 (2025) 8 The elements in the first nrows of Table 1are divided into two parts. One part is simply moved below the 0-Ap´ery set to fill in the gap as shown in Table 2. However, the remaining portion is moved to the lower left of the 0-Ap´ery set. Elements other than the first nrows of Table 1are shifted to the right side of the 0-Ap´ery set by shifting up by nrows. Set τx,y,z := xSn+ySn+1 +zSn+2. We show that all the elements of Table 2 have at least two different representations. In fact, for 0 ≤y≤2n, n ≤z≤2nand 0≤y≤2n−11,2n+ 1 ≤z≤3n, we have τn+1,y,z =τ0,y+2n+1,z−n. Similarly, we get τ4n+3,y,z =τ0,y+2n−10,z+2n+1 (for 0 ≤y≤10,0≤z≤n−1), τ3n+2,y,z =τ0,y−11,z+3n+1 (for 11 ≤y≤2n,0≤z≤n−1). From Table 2, there are four possible choices for the largest value of Ap1(Sn, Sn+1, Sn+2): t4n+1,n, t4n−10,2n, t2n,3n,and t2n−11,4n. Since 18n2+ 18n−1>0,we have t4n+1,n < t2n,3nand t4n−10,2n< t2n−11,4n. Analogous to the case p= 0,we obtain t2n,3n< t2n−11,4nif and only if n>9. Therefore, for 6 ≤n≤9,the maximum element is t2n,3n, and by Lemma 1, g1(Sn, Sn+1, Sn+2) = 2nSn+1 + 3nSn+2 −Sn. When n≥10, t2n−11,4nis the largest of all the elements, and hence using Lemma 1, g1(Sn, Sn+1, Sn+2) = (2n−11)Sn+1 + 4nSn+2 −Sn. Case 3: p= 2. The elements of the 2-Ap´ery set can be derived from those of the 1-Ap´ery set in the following manner (see Table 3). t4n+2,0··· t6n−9,0t6n−8,0··· t6n+2,0 t4n+2,1··· t6n−9,1 . . .. . . t4n+2,n ··· t6n−9,n t4n−9,n+1 ··· t4n+1,n+1 . . .. . . t4n−9,2n··· t4n+1,2n t2n+1,2n+1 ··· t4n−10,2n+1 . . .. . . t2n+1,3n··· t4n−10,3n t2n−10,3n+1 ··· t2n,3n+1 . . .. . . t2n−10,4n··· t2n,4n t0,4n+1 ··· t2n−11,4n+1 . . .. . . t0,5n··· t2n−11,5n Table 3: Ap2(Sn, Sn+1, Sn+2) INTEGERS: 25 (2025) 9 Similar to the case when p= 1,the elements in the first nrows of the main part of the 1-Ap´ery set are subdivided into two parts and then relocated beneath the two staircase sections of the 1-Ap´ery set. Elements other than the first nrows of the main portion undergo a shift to the right side of the 1-Ap´ery set which is achieved by moving up by nrows. The other two staircase parts are shifted to the right by (2n+ 1) steps and upward by nsteps. More precisely, ty,z ≡ty+2n+1,z−n(mod Sn) (2n+ 1 ≤y≤4n+ 1, z =n; 2n+ 1 ≤y≤4n−10, n + 1 ≤z≤2n), ty,z ≡ty−11,z+3n+1 (mod Sn) (2n+ 1 ≤y≤2n+ 11,0≤z≤n−1), ty,z ≡ty−2n−12,z+4n+1 (mod Sn) (2n+ 12 ≤y≤4n+ 1,0≤z≤n−1), ty,z ≡ty+2n+1,z−n(mod Sn) (0 ≤y≤2n−11,3n+ 1 ≤z≤4n; 2n−10 ≤y≤2n, 2n+ 1 ≤z≤3n). Using Equations (5), (6), and (7), we can demonstrate that the elements of the 2-Ap´ery set possess atleast three distinct representations. For 0 ≤y≤2n, z = 2n and 0 ≤y≤2n−11,2n+ 1 ≤z≤3n, we have τ2n+2,y,z =τn+1,y+2n+1,z−n=τ0,y+4n+2,z−2n. Similarly, we have τ4n+3,y,z =τ3n+2,y+2n+1,z−n=τ0,y+2n−10,z+2n+1 (0 ≤y≤10, n ≤z≤2n−1), τ3n+2,y,z =τ2n+1,y+2n+1,z−n=τ0,y−11,z+3n+1 (11 ≤y≤2n, n ≤z≤2n−1), τ5n+4,y,z =τn+1,y+2n−10,z+2n+1 =τ0,y+4n−9,z+n+1 (0 ≤y≤10,0≤z≤n−1), τ4n+3,y,z =τn+1,y−11,z+3n+1 =τ0,y+2n−10,z+2n+1 (11 ≤y≤2n, 0≤z≤n−1) . In Table 3, by comparing the six candidates t2n−11,5n, t2n,4n, t4n−10,3n, t4n+1,2n, t6n−9,n,and t6n+2,0, we find that t2n,4nis the largest element of the 2-Ap´ery set, when 6 ≤n≤9,and t2n−11,5nis the largest element of the 2-Ap´ery set, when n≥10. Therefore, g2(Sn, Sn+1, Sn+2) = (2nSn+1 + 4nSn+2 −Sn,if 6 ≤n≤9; (2n−11)Sn+1 + 5nSn+2 −Sn,if n≥10. This completes the proof.