The Frobenius problem, sums of powers of integers, and recurrences for the Bernoulli numbers

Transcription

1 Journal of Nuber Theory 117 ( The Frobenius proble, sus of powers of integers, and recurrences for the Bernoulli nubers Hans J.H. Tuenter Schulich School of Business, York University, Toronto, Canada M3J 1P3 Received 14 February 2000; revised 7 March 2005 Available online 10 October 2005 Counicated by D. Goss Dedicated to Prof. R.A. Cuninghae-Green Abstract In the Frobenius proble with two variables, one is given two positive integers a and b that are relative prie, and is concerned with the set of positive nubers NR that have no representation by the linear for ax + by in nonnegative integers x and y. We give a coplete characterization of the set NR, and use it to establish a relation between the power sus over its eleents and the power sus over the natural nubers. This relation is used to derive new recurrences for the Bernoulli nubers Published by Elsevier Inc. Keywords: Frobenius proble; Sus of powers of integers; Bernoulli nubers 1. Introduction Given two positive integers a and b that are relative prie, one can define the set of positive nubers NR that have no representation by the linear for ax + by in nonnegative integers x and y. It is well known that NR is finite and that its largest eleent is ab a b, and thus that every integer larger than or equal to (a 1(b 1 is representable as a nonnegative linear cobination of a and b. An early version was published as working paper of the Schulich School of Business in July E-ail address: htuenter@schulich.yorku.ca X/$ - see front atter 2005 Published by Elsevier Inc. doi: /j.jnt

2 H.J.H. Tuenter / Journal of Nuber Theory 117 ( Alfred Brauer ( referred in [2,3] to the proble of deterining the largest integer that cannot be represented as a nonnegative, linear cobination of a set of positive, coprie integers as the proble of Frobenius. It bears the nae of Brauer s copatriot, the Geran atheatician, Ferdinand Georg Frobenius ( , who would ention it occasionally in his lectures [2]. Nowadays, the scope of the Frobenius proble is usually extended to include questions relating to the entire set of integers that are not representable. As an illustration of the Frobenius proble in two variables, consider the exaple that was given by Jaes Joseph Sylvester ( as part of a question posed in the June 1883 issue of The Educational Ties, and Journal of the College of Preceptors [18]. For a = 4 and b = 7, one has 1, 2, 3, 5, 6, 9, 10, 13 and 17 as the only integers that are neither ultiples of 4 or 7, nor can be ade up by adding together ultiples of 4 and 7, and thus every integer larger than or equal to (a 1(b 1 = 18 is representable. Several questions of interest, such as the cardinality of the set NR, are related to the power sus S (a, b = n, where is a nonnegative integer. In recognition of the initiating role played by Sylvester, we shall refer to these sus as Sylvester sus. Sylvester asked to show that there are exactly S 0 (a, b = 2 1 (a 1(b 1 integers not representable as a nonnegative linear cobination of a and b. In response, a solution was given by W. J. Curran Sharp, in the Noveber 1883 issue of the said journal. This solution is based upon enuerating ters in the product of two finite geoetric series, and is of historical interest only as ore concise solutions are now available. It was not until a good century later that Brown and Shiue [4] published a closed-for expression for S 1 (a, b by deterining the generating function of the characteristic function of the set NR, and coputing its first derivative. The following year, Rödseth [16] deterined the exponential generating function of the Sylvester sus, and settled the general case, by deriving the expression S 1 (a, b = 1 (+1 i i=0 j=0 ( ( i B i j i B j a j b i 1 B, (1 where B 0 = 1, B 1 = 1/2, B 2 = 1/6, B 3 = 0, B 4 = 1/30,... are the Bernoulli nubers defined by the iplicit recurrence relation j=0 ( + 1 B j j = 0 for 1.

3 378 H.J.H. Tuenter / Journal of Nuber Theory 117 ( The approaches used in [4,16] are based upon an enueration of the eleents of the set NR. In this paper, we take a different approach, and derive an identity that copletely characterizes the set NR, and allows one to deterine expressions for the Sylvester sus in a succinct and elegant anner. Using this characterization the previous results follow easily, and a new identity is derived in the process. More iportant, however, it shows a deeper connection between the Sylvester sus, the sus of powers over the natural nubers, and the Bernoulli nubers. 2. A characterization The ain idea on which this paper rests is very siple and intuitive; it is perhaps best illustrated by eans of an exaple, such as the one used by Sylvester. Write down all the nubers that are not representable as a linear cobination of 4 and 7. Below this write down the sae list, but with all eleents increented by a = 4. Now cross out the nubers appearing in both lists to give the following: //// 5 //// 6 //// 9 ///// 10 ///// 13 ///// 17 //// 5 //// 6 7 //// 9 ///// 10 ///// ///// This leaves the nubers 1, 2, 3 of the first list and 7, 14, 21 of the second list. Now note that the second residual list is the sae as the first residual list, but with all eleents ultiplied by b = 7. The reader is encouraged to discover the general structure by repeating the exercise with the second list constructed fro the first by adding b = 7, instead of a = 4. The structure that one observes is a consequence of the following characterization of the set of nubers that are not representable. Theore 2.1. For every function f, the following identity holds: [f(n+ a f (n] = [f (nb f (n], (2 where a and b are interchangeable. Proof. Rewrite the su on the left-hand side as {n n a NR} f (n f (n, and cross out every index appearing in both sets (just as in the exaple above. The residual index set of the first suation consists of the nubers satisfying n a NR and n/ NR. The condition n/ NR iplies either n<0, or n = ax + by for soe nonnegative integers x and y. The first iplication cannot hold since n a NR ust be positive. The latter iplies x = 0, as otherwise n a is representable, and thus n = by ust hold. A necessary condition on y is a/b < y < a; the first inequality holds because n a ust be positive, and the second inequality because otherwise n a = b(y a + (b 1a is representable. This condition is also sufficient (as a 1 n=1

4 H.J.H. Tuenter / Journal of Nuber Theory 117 ( shown in the next few lines and gives n = by with y { a/b +1,...,a 1} as those n that satisfy n/ NR and n a NR. The first condition is obviously satisfied, since n = by is representable. The second is proved by contradiction; suppose that by a = λa + μb, for soe nonnegative λ and μ, then λ = b 1 + kb with k 0, but this iplies y = a + ak + μ a, and gives the desired contradiction. The residual index set of the second suation consists of the nubers that satisfy n NR and n a / NR. The condition n a / NR iplies either n a <0, or n a = ax + by for soe nonnegative integers x and y. The latter obviously cannot hold since this iplies n / NR. It is easily seen that the integers n<afor which n NR is obtained by deleting the ultiples of b; this gives {1,...,a 1}/{b, 2b,..., a/b b} as the integers n that satisfy n NR and n a/ NR. The suation that we started out with at the beginning of the proof, now reduces to y { a/b +1,...,a 1} f(yb n {1,...,a 1} n/ {b,2b,..., a/b b} a 1 f (n = [f (nb f (n] n=1 and proves the identity. Since the only assuption ade 1 is that a and b are coprie, they are interchangeable, and this observation copletes the proof. We will now show that equality (2 copletely characterizes the set NR. Suppose that it holds for every function f, with the set NR replaced by soe other set of positive integers S. First, by taking f (n = n, it follows that both sets have the sae cardinality: S = NR = 1 2 (a 1(b 1. Now consider f (n = n, where is a positive integer. Expanding (n + a, using the binoial theore, it is easily shown by induction that n = n for = 0, 1, 2,... (3 n S Taking sufficiently large will show that, for this relation to hold, S and NR ust have their largest eleent in coon. One can now eliinate this largest eleent fro both sets, and apply the sae arguent to the next largest eleent, and so on, until all eleents of the two sets are shown to be equal, and thus prove the assertion. This shows that (2 characterizes the set NR copletely, and therefore all properties of this set can be derived (at least theoretically fro (2. 1 It should be noted that the proof siplifies considerably if we ake the assuption a<b. This has the effect that a/b =0, and siplifies and shortens the proof accordingly. However, this does not allow one to draw the conclusion that a and b are interchangeable, a syetry that is of iportance.

5 380 H.J.H. Tuenter / Journal of Nuber Theory 117 ( Applications As an application of Theore 2.1, take f (n as a threshold function: f (n = 1, for n τ, and 0, otherwise, setting the threshold at τ = (a 1b. The right-hand side of (2 evaluates to one, so that there ust be an n NR such that n + a τ. It is easily seen that increasing τ renders the right-hand side of (2 as zero, so that there cannot be an n NR such that n + a>τ. This iplies that L, the largest eleent of NR, satisfies L + a = (a 1b, and thus that every integer larger than or equal to (a 1(b 1 is representable. Sylvester s observation that there are exactly 2 1 (a 1(b 1 integers not representable is readily obtained by taking f (n = n, followed by an application of the su forula for an arithetic progression. As another application, take f (n = e nz and extend the suation on the right-hand side of (2 to include n = 0. Now use the su forula for a finite geoetric series, and divide by (e az 1 to give the exponential generating function of the sequence {S } as =0 S (a, b z! = e nz = e abz 1 (e az 1(e bz 1 1 e z 1. This is the forula derived by Rödseth; he expands its right-hand side, uses the exponential generating function of the Bernoulli nubers, given by z/(e z 1 = n 0 B nz n /n!, and equates the coefficients of the powers of z to derive expression (1. It is useful to observe the syetry in the forula. 2 Fro these applications we see that all known results on the Frobenius proble in two variables, as listed in the introduction, are contained in (2 and easily derived fro it. An identity that sees not to have been noticed hitherto is obtained by choosing f (n = ln n, and gives (1 + a/n = b a 1, where a and b are interchangeable. As a last application, take f (n = n/a, where x denotes the greatest integer that is no larger than x. This gives the identity a 1 nb/a = n=1 1 = 2 1 (a 1(b 1, 2 Indeed, one could have proceeded by proving Theore 2.1 for a<b, then derive the exponential generating function, and given its uniqueness and the syetry in a and b, draw the conclusion that a and b are interchangeable, and thus that (2 also holds for a>b.

6 H.J.H. Tuenter / Journal of Nuber Theory 117 ( where a and b are coprie, positive integers, and a new interpretation of this su as the nuber of integers that are not representable as a nonnegative linear cobination of a and b. 4. Sus of powers of integers For every nonnegative integer r, define σ r (n = 0 r + 1 r + +n r. As perhaps one of the ost frequent occurring sus in atheatics, it has been studied extensively, dating back as far as 1631, when Johann Faulhaber ( of Ul published his Acadeia Algebra [8]; ore accessible accounts of Faulhaber s work can be found in [7,12,17]. Nowadays, the treatent of sus of powers of the integers is a standard occurrence in eleentary textbooks on nuber theory, see for instance [9]. It is instruental to review soe of the ain results. Recall the derivation by Blaise Pascal ( of the recursive forula for the su-of-powers polynoials. For positive integers r one has the following: (n + 1 r = n [(i + 1 r i r ]= i=0 n r 1 ( r i j = j i=0 j=0 r 1 j=0 ( r j σ j (n. (4 The structure shows ore clearly in atrix forat, as illustrated for r = 4 ( ( ( 2 2 σ 0 (n (n ( ( ( σ 1 (n σ 2 (n = (n (n (5 ( σ 3 (n (n ( ( ( This gives the failiar σ 0 (n = n + 1, σ 1 (n = 2 1 n(n + 1, σ 2(n = 6 1 n(n + 1(2n + 1, and σ 3 (n = 4 1 n2 (n An explicit expression for σ j (n, in ters of the Bernoulli nubers, is given by σ j (n 1 = 1 j + 1 j i=0 ( j + 1 B i i n j+1 i. (6 This expression and ore details on the sus of powers and the Bernoulli nubers can be found in [9, p. 269] and [15, p. 159]. As one would expect, the Sylvester sus are closely related to the sus of powers of the natural nubers. In the next two sections we will show their precise relationship, and how one can express one in ters of the other.

7 382 H.J.H. Tuenter / Journal of Nuber Theory 117 ( Recurrences for the Sylvester sus Following the idea of Pascal, for positive integers r one has r 1 [(n + a r n r ]= while (2 gives =0 ( r r 1 n a r = =0 ( r a r S (a, b, [(n + a r n r ]= [(nb r n r ]=(b r 1σ r (a 1. a 1 n=0 Cobining both yields the recurrence relation r 1 =0 ( r a r S (a, b = (b r 1σ r (a 1 (7 for the Sylvester sus. Its structure is ore transparent in atrix forat, as illustrated for r = 4 ( ( ( 2 2 S 0 (a, b (b 1σ 1 (a 1/a ( ( ( S 1 (a, b/a S 2 (a, b/a 2 = (b 2 1σ 2 (a 1/a 2 (b 3 1σ 3 (a 1/a 3. (8 ( S 3 (a, b/a (b 4 1σ 4 (a 1/a 4 ( ( ( Note the siilarity to (5. The first few sus are readily deterined as S 0 (a, b = 2 1 (a 1(b 1, S 1 (a, b = 12 1 (a 1(b 1(2ab a b 1, S 2 (a, b = 12 1 (a 1(b 1ab(ab a b. This, of course, agrees with the results in [4,16]. If one prefers a recurrence without the su-of-powers polynoials; take (7, divide by (b r 1, ultiply by the binoial coefficient ( p r, su both sides fro r = 1top 1, siplify the right-hand side using

8 H.J.H. Tuenter / Journal of Nuber Theory 117 ( (4, and change the order of suation on the left-hand side to give [ p p =0 r= ( p + 2 r + 1 ( ] r + 1 a r b r+1 S (a, b = a p Note that, because of Theore 2.1, this recurrence and recurrence (7 also hold upon the interchanging of a and b. 6. Explicit expression for the Sylvester sus To obtain an explicit expression for the Sylvester sus, we consider the function f(x= a 1 B (x/a, with the Bernoulli polynoials defined as B (x = r B r x r. By the difference forula for the Bernoulli polynoials B (x + 1 B (x = x 1, the function f (n inherits the property f(n+ a f (n = n 1. A direct application of (2 results in a 1 S 1 (a, b = a 1 n=0 a 1 B (nb/a a 1 n=0 B (n/a. (9 The last su is seen to be equal to the th Bernoulli nuber B by Raabe s ultiplication theore for the Bernoulli polynoials [14] a 1 B (ax = a 1 n=0 B (x + n/a, evaluated at x = 0. Making this siplification, expanding the reaining Bernoulli polynoial and replacing the suation over n by the corresponding power-su polynoial gives S 1 (a, b = r a r 1 B r b r σ r (a 1 B. (10

9 384 H.J.H. Tuenter / Journal of Nuber Theory 117 ( Upon substitution of expression (6 for σ r (a 1, and a change of variables we recover Rödseth s forula (1. 7. Recurrences for the Bernoulli nubers In the previous section, we have seen that the Sylvester sus can be expressed in ters of the Bernoulli nubers. It should not be a surprise that one can also reverse the process, and derive recurrences for the Bernoulli nubers in ters of the Sylvester sus and the power su polynoials. One such recurrence is based upon the identity r a r 1 B r σ r (a 1 = B, (11 where is a nonnegative integer and a a positive integer. Apparently, this identity did not appear in the open literature until It was posed, in a slightly different forat, as a proble in The Aerican Matheatical Monthly by Belinfante [1], generalizing a forula by Naias [13]. Proofs can be found in [1,5], but these do not use representation (6 of the Bernoulli nubers; they are based upon the representation by the exponential generating function. Howard [11, p. 159] notes that (11 is a consequence of Raabe s ultiplication theore for the Bernoulli polynoials, and therefore not really a new identity. However, he shows that it is an excellent vehicle to prove several theores (including one by Frobenius describing divisibility properties of the Bernoulli nubers. It is easily seen that (10 represents a generalization of (11; for b = 1 the set NR is epty, S 1 (a, 1 = 0, and consequently (10 reduces to (11. In order to use (10 to generate recurrences for the Bernoulli nubers, one has to deterine the eleents of the set NR explicitly, so that one can deterine S 1 (a, b. For instance, taking a = 2 and b = 3givesNR ={1}, and thus S 1 (2, 3 = 1. This results in the recurrence B = (1 2 1 r 3 r 2 r B r. Taking a = 3 and b = 2 results in B = (1 3 1 r [2 r + 4 r ]3 r B r. The exaple of Sylvester, given in the introduction, can be used to illustrate the cornucopia of weird and wonderful recurrences for the Bernoulli nubers that (10

10 H.J.H. Tuenter / Journal of Nuber Theory 117 ( represents. For a = 4 and b = 7 this gives B = 4 1 ( (1 4 r 7 r (1 + 2 r + 3 r 4 r B r. Taking a = 7 and b = 4 results in a copanion recursion. 8. Bibliographical issues Sylvester s question was first published as ite 7382 in the Questions for Solution section of the June 1883 issue of The Educational Ties, and Journal of the College of Preceptors, vol. XXXVI, New Series, No. 266, p Later that year, a solution was given by W. J. Curran Sharp, and published, together with the original question, in the Matheatics section of the Noveber issue. The Educational Ties is a onthly journal, first published in 1847, with a long and rich history, and its influence is still felt today, (see [6,10] for a ore detailed account of its history and the historical context. As noted in [19], the journal s proble section proved to be very popular, and as space for it was restricted by advertising pressure, probles and solutions that had appeared in the journal were bundled with additional aterial, and, fro 1864 to 1918, republished in seiannual volues as Matheatical Questions, with their solutions, fro the Educational Ties. Sylvester s question, proble 7382, was republished in 1884 in vol. 41, p. 21 of the Matheatical Questions, and it is this reference that has traditionally been given in the acadeic literature as the origin of Sylvester s question. However, as noted, the pedigree of Sylvester s question goes back slightly further. Acknowledgents I thank the anonyous referee for the suggestion to copleent Theore 2.1 by proving that equality (2 copletely characterizes the set NR. References [1] J.G.F. Belinfante, Proble E 3237, Aer. Math. Monthly 94 (10 ( ; J.G.F. Belinfante, Solution by Ira Gessel, Aer. Math. Monthly 96 (4 ( [2] A. Brauer, On a proble of partitions, Aer. J. Math. 64 (1/4 ( [3] A. Brauer, B.M. Seelbinder, On a proble of partitions II, Aer. J. Math. 76 (2 ( [4] T.C. Brown, P.J.-S. Shiue, A reark related to the Frobenius proble, Fibonacci Quart. 31 (1 ( [5] E.Y. Deeba, D.M. Rodriguez, Stirling s series and Bernoulli nubers, Aer. Math. Monthly 98 (5 (

11 386 H.J.H. Tuenter / Journal of Nuber Theory 117 ( [6] J. Delve, The College of Preceptors and the Educational Ties: changes for British atheatics education in the id-nineteenth century, Historia Math. 30 (2 ( [7] A.W.F. Edwards, A quick route to sus of powers, Aer. Math. Monthly 93 (6 ( [8] J. Faulhaber, Acadeia Algebrae, Verlag Johann Reelins, Augsburg, 1631 A copy on icrofil resides under Call Nuber MN in the special collections of the Butler library at Colubia University, New York. [9] R.L. Graha, D.E. Knuth, O. Patashnik, Concrete Matheatics, Addison-Wesley, Reading, MA, [10] I. Grattan-Guinness, A note on The Educational Ties and Matheatical Questions, Historia Math. 19 (1 ( [11] F.T. Howard, Applications of a recurrence for the Bernoulli nubers, J. Nuber Theory 52 (1 ( [12] D.E. Knuth, Johann Faulhaber and sus of powers, Math. Coput. 61 (203 ( [13] V. Naias, A siple derivation of Stirling s asyptotic series, Aer. Math. Monthly 93 (1 ( [14] J.L. Raabe, Zurückführung einiger suen und bestiten integrale auf die Jakob Bernoullische function, J. Reine Angew. Math. 42 (4 ( [15] J. Riordan, Cobinatorial Identities, Wiley, New York, [16] ].J. RZdseth, A note on T.C. Brown and P.J.-S. Shiue s paper: A reark related to the Frobenius proble, Fibonacci Quart. 32 (5 ( [17] I. Schneider, Potenzsuenforeln i 17. Jahrhundert, Historia Math. 10 (3 ( [18] J.J. Sylvester, Proble 7382, The Educational Ties, and J. College Preceptors, New Ser. 36 (266 ( ; J.J. Sylvester, Solution by W.J. Curran Sharp, The Educational Ties, and J. College Preceptors, New Ser. 36 (271 ( (Republished as J.J. Sylvester, Proble 7382, in: W.J.C. Miller (Ed., Matheatical Questions, with their Solutions, fro the Educational Ties, vol. 41, Francis, Hodgson, London, 1884, p. 21.J.J. Sylvester, Solution by W.J. Curran Sharp, The Educational Ties, and J. College Preceptors, New Ser. 36 (271 ( [19] J.J. Tattersall, S.L. McMurran, Woen and the educational ties, in: Proceedings of the History and Pedagogy of Matheatics Conference, July 2004, Uppsala University, Sweden, pp