Hyperharmonic numbers arise by taking repeated partial sums of harmonic numbers. These numbers can be expressed in terms of r-Stirling numbers, leading to combinatorial interpretations of many interesting identities

Benjamin, Arthur T.

Gaebler, David, \u2704

Gaebler, Robert, \u2704

English

Scholarship@Claremont

Claremont CollegesScholarship @ ClaremontAll HMC Faculty Publications and Research HMC Faculty Scholarship10-1-2003A Combinatorial Approach to HyperharmonicNumbersArthur T. BenjaminHarvey Mudd CollegeDavid Gaebler '04Harvey Mudd CollegeRobert Gaebler '04Harvey Mudd CollegeThis Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusionin All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contactscholarship@cuc.claremont.edu.Recommended CitationBenjamin, A.T., Gaebler, D., & Gaebler R. (2003). A combinatorial approach to hyperharmonic numbers. Integers: The ElectronicJournal of Combinatorial Number Theory, 3: 1-9.A COMBINATORIAL APPROACH TO HYPERHARMONIC NUMBERSArthur T. BenjaminDepartment of Mathematics, Harvey Mudd College, Claremont, CA, USAbenjamin@math.hmc.eduDavid GaeblerDepartment of Mathematics, Harvey Mudd College, Claremont, CA, USAdgaebler@hmc.eduRobert GaeblerDepartment of Mathematics, Harvey Mudd College, Claremont, CA, USArgaebler@hmc.eduReceived: 3/26/03, Accepted: 10/21/03, Published: 10/23/03AbstractHyperharmonic numbers arise by taking repeated partial sums of harmonic numbers.These numbers can be expressed in terms of r-Stirling numbers, leading to combinatorialinterpretations of many interesting identities.1. IntroductionThe harmonic numbers, de¯ned by Hn =Pnk=11karise frequently in the solution tocombinatorial problems and in the analysis of algorithms. It is well known [6] thatHn =£n+12¤n!(1)where£nk¤denotes the (unsigned) Stirling number of the ¯rst kind, counting the permu-tations of n elements that are the product of k disjoint cycles. This allows harmonicnumber identities to be viewed combinatorially, as in [2].The quantity Hn can be generalized many ways (see, for instance, [9]). The general-ization we pursue, called hyperharmonic numbers by Conway and Guy [5], are obtainedby taking repeated partial sums of the Harmonic numbers.Formally, we de¯ne Hrn, the hyperharmonic number of order r as follows.De¯nition 1 Let Hrn = 0 for r < 0 or n ≤ 0, H0n = 1n for n ≥ 1, and for r; n ≥ 1, letHrn =nXi=1Hr−1i : (2)INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 2Note that H1n is equal to the ordinary harmonic number Hn.In [5], Conway and Guy express the hyperharmonic numbers in terms of ordinaryharmonic numbers.Hrn =µn+ r − 1r − 1¶(Hn+r−1 −Hr−1): (3)The presence of the binomial coe±cient suggests that a deeper combinatorial rela-tionship exists. As we shall see, the hyperharmonic numbers satisfy several interestingidentities like the ones below.nHrn =µn+ r − 1r¶+ rHr+1n−1 (4)µm− 1r − 1¶Hmn =µn+m− rn¶Hrn+m−r −µn+m− 1n¶Hrm−r for 1 ≤ r ≤ m (5)rXt=1H tn = Hrn+1 −1n+ 1(6)Hrn =nXt=1µn+ r −m− t− 1r −m− 1¶Hmt for 0 ≤ m ≤ r − 1 (7)Hrn =mXt=0µmt¶Hr−tn+t−m for 0 ≤ m ≤ r (8)Hrn =nXs=mµn+ r − l − sr − l¶H l−1s +rXt=lµn+ r −m− tn−m¶H tm−1 (9)for 1 ≤ l ≤ r and 1 ≤ m ≤ n:A generating function for the hyperharmonic numbers is− ln(1− x)(1− x)r =∞Xi=1Hri xisince − ln(1−x) = P xnn, and multiplying by 1(1−x)r has the e®ect of taking partial sums rtimes. This generating function is equivalent by (3) to an identity in [7]. Although manyof the identities in this paper can be proved using algebraic methods, we shall prove allof these by simple combinatorial arguments.INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 32. Basic properties of hyperharmonic numbersThere are two di®erent ways to approach the hyperharmonic numbers combinatorially.One is to investigate the numerator of the unreduced fraction, i.e., n!Hrn. This approachwill be followed in the next section. The other is to write Hrn as the sum of fractions1t,where t ranges from 1 to n, and ask how many times each fraction appears in this sum.The answer is given by the following theorem.Theorem 1Hrn =nXt=1µn+ r − t− 1r − 1¶1t:Proof. Given a particular t, where 1 ≤ t ≤ n, we claim that the fraction 1tis countedexactly¡n+r−t−1r−1¢times. To see this, consider the weighted directed graph with vertexset V = {(x; y)|x ≥ 1; y ≥ 0} and the arcs leaving (x; y) enter vertices (x′; y + 1), wherex′ ≥ x. We give each vertex (x; y) a weight of Hyx . Hence for all t ≥ 1, (t; 0) has weight1tand for x; y ≥ 1, (x; y) has weight Px′≤xHy−1x′ , which is the sum of the weights of thevertices that point to (x; y).Thus the fraction 1tappears in Hrn for every path from (t; 0) to (n; r), i.e., for everysequence t ≤ x1 ≤ · · · ≤ xr−1 ≤ n. This is the same as the number of size r − 1multisubsets of {t; : : : ; n}, which equalsµµn− t+ 1r − 1¶¶=µn+ r − t− 1r − 1¶as desired, where¡¡nk¢¢=¡n+k−1k¢is the number of size k multisubsets of an n-elementset. 2More generally, for 0 ≤ m ≤ r− 1, by counting all paths from (t;m) to (n; r), we seethat for 1 ≤ t ≤ n, Hmt is contributed¡¡n−t+1r−1−m¢¢=¡n+r−t−m−1r−m−1¢times, and Identity (7)follows. Similar path counting arguments can also be used to establish equations (6), (8)and (9), but we shall provide other combinatorial proofs of these in Section 4.We now begin our discussion of the combinatorial nature of the numerator of Hn;r.We can write Hrn as a (typically non-reduced) fraction of the formHrn =an;rn!: (10)Thus a0;r = 0 and for n ≥ 1, an;0 = (n− 1)! . For n; r ≥ 1,Hrn =nXi=1Hr−1i = Hrn−1 +Hr−1n=⇒ an;rn!=an−1;r(n− 1)! +an;r−1n!=⇒ an;r = n an−1;r + an;r−1 (11)INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 4We shall use this recurrence and initial condition to prove a generalization of equation(1).3. r-Stirling numbersThe r-Stirling numbers, which are similar to the weighted Stirling numbers of Carlitz [4]and are a special case of the generalized Stirling numbers of Hsu and Shiue [8], can bede¯ned combinatorially in terms of restricted permutations. Broder [3] gives the followingcombinatorial de¯nition, which we will use throughout this paper:De¯nition 2£nk¤ris the number of permutations of the set {1; 2; : : : ; n} having k disjoint,non-empty cycles, in which the elements 1 through r are restricted to appear in di®erentcycles.Note that£nk¤0and£nk¤1are both equal to the ordinary Stirling number£nk¤.For convenience, we let Tn;k;r denote the set of permutations of {1; 2; : : : ; n} intok cycles, in which the elements 1 through r appear in di®erent cycles. Thus£nk¤risthe number of permutations in Tn;k;r. We can think of£nk¤ras counting the number ofpermutations with r \restricted" and n−r \free" elements. If a cycle contains a restrictedelement, we call it a restricted cycle; otherwise we call it a free cycle.We shall adopt the standard permutation notation of writing the smallest element ina cycle at the beginning of that cycle, and listing the cycles in ascending order accordingto their smallest elements. For example, (1 5 3 7)(2)(4 9 8)(6) is in standard permutationnotation, while (1 5 3 7)(2)(6)(4 9 8) and (1 5 3 7)(2)(9 8 4)(6) are not.Generating functions for the r-Stirling numbers are presented in [3]:nXk=r·nk¸rxk = xrn−1Yi=r(x+ i):∞Xn=k·n+ rk + r¸rxnn!=1k!µ11− x¶r ·lnµ11− x¶¸k:X0≤k≤n·n+ rk + r¸rxnn!tk =µ11− x¶r+t:By de¯nition, we have£00¤r= 1 and for n > 0,£nk¤r= 0 for k < r or k > n. Noticethat for 1 ≤ r ≤ n ·nr¸r=(n− 1)!(r − 1)! (12)INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 5since we are counting permutations of the form (1 · · ·)(2 · · ·) · · · (r · · ·). The elements r+1through n may be entered one at a time. Each element r+1 ≤ i ≤ n can be placed to theright of any of the i−1 elements already in place, so there are r(r+1) · · · (n−1) = (n−1)!(r−1)!such permutations. Notice that r-Stirling numbers also satisfy the traditional Stirlingrecurrence: for 0 ≤ r ≤ k ≤ n,·n+ 1k¸r=·nk − 1¸r+ n·nk¸r(13)which may be proved by conditioning on whether or not element n + 1 is alone in itscycle.Another useful identity we shall rely on is:Identity 1 For r ≥ 1,·n+ rk¸r= n·n+ r − 1k¸r+·n+ r − 1k − 1¸r−1:This may be proved by conditioning on whether there are any free elements in the¯rst cycle. If so, then choose one of the n free elements to appear immediately after 1;the other elements can be arranged£n+r−1k¤rways. Otherwise, element 1 is alone, andelements 2 through n+r can be arranged£n+r−1k−1¤r−1 ways. This brings us to the followingtheorem.Theorem 2Hrn =£n+rr+1¤rn!:Proof 1. Let An;r =£n+rr+1¤r. Hence A0;r =£rr+1¤r= 0 = a0;r and for n ≥ 1, An;0 =£n1¤0= (n − 1)! = a0;r. For n; r ≥ 1, Identity 1 implies An;r = nAn−1;r + An;r−1 just likein equation (11). Since An;r and an;r satisfy the same initial conditions and the samerecurrence, then by equation (10) we have£n+rr+1¤r= An;r = an;r = n!Hrn; as desired. 2It is also possible to combinatorially prove Theorem 2 without relying on a recurrence.We ¯rst prove the following general identity, similar to one proved in [3].Identity 2 For 0 ≤ m ≤ r ≤ n and 0 ≤ l ≤ k − r,µk − rl¶·n+ rk¸r=n−lXt=k−r−lµnt¶·t+ r −mk −m− l¸r−m·n+m− tm+ l¸mINTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 6Proof. The left side counts permutations of Tn+r;k;r where the ¯rst m cycles and l freecycles are colored red, and the other cycles colored green. On the right, we conditionon the number t of free elements in green cycles. There are¡nt¢ways to choose the freeelements,£t+r−mk−m−l¤r−m ways to arrange them in k − (m + l) green cycles, then£n+m−tl+m¤mways to arrange the other (n+ r)− (r−m+ t) elements among the m+ l red cycles. 2Proof 2 of Theorem 2. As a special case of Identity 2 with l = 0, m = r, and k = r + 1,we have·n+ rr + 1¸r=nXt=1n!t!(n− t)! · (t− 1)! ·(n+ r − t− 1)!(r − 1)! =nXt=1µn+ r − t− 1r − 1¶n!t= n!Hrnby Theorem 1. 24. r-Stirling identitiesTheorem 2 expresses Hrn in terms of r-Stirling numbers. This allows us to give combina-torial proofs of hyperharmonic identities by proving the equivalent r-Stirling identities.This will be the focus of the rest of our paper.We make use of the following lemma:Lemma 1 For r+ 1 ≤ t ≤ n+ r, the number of permutations in Tn+r;r+1;r with t as thesmallest element in the right cycle is (n+r−1)!(r−1)!(t−1) :Proof 1. First place elements 1 through t− 1 in the restricted cycles. There are £t−1r¤r=(t−2)!(r−1)! ways to do this. Next, place element t at the beginning of the right cycle. Finally,place each of the elements t+1 through n+r one at a time. Each element can be placed tothe right of any element already placed, so there are t·(t+1)·(t+2) · · · (n+r−1) = (n+r−1)!(t−1)!ways to place them. Altogether, then, the number of permutations is(t− 2)!(r − 1)! ·(n+ r − 1)!(t− 1)! =(n+ r − 1)!(r − 1)!(t− 1) :2Proof 2. Alternatively, we could list the numbers 1 through n+ r in any order with theprovision that 1 is ¯rst and the numbers 1 through r must be in increasing order. Thereare¡n+r−1r−1¢ways to place the elements 1 through r in our list, then n! ways to order thenumbers r + 1 through n + r in the remaining positions. We then convert our list toa restricted permutation by starting a new cycle at each of 1; 2; : : : ; r; t: Element t hasequal probability of having any position among elements 2; 3; : : : ; t, so its probability ofbeing last among them is 1t−1 . Thus the probability that we have a valid permutation isINTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 71t−1 . The number of permutations in Tn+r;r+1;r with t as the smallest element in the rightcycle, then, is µn+ r − 1r − 1¶n!1t− 1 =(n+ r − 1)!(r − 1)!(t− 1) :2Now we are ready to prove equation (3).Proof 1. Since every permutation in Tn+r;r+1;r must have some smallest element t in itsright cycle, with r + 1 ≤ t ≤ n+ r, then by Theorem 2 we haveHrn =1n!·n+ rr + 1¸r=1n!n+rXt=r+1(n+ r − 1)!(r − 1)!(t− 1) =µn+ r − 1r − 1¶(Hn+r−1 −Hr−1):2Proof 2. By Theorem 2, (3) is transformed into the following r-Stirling identity:·n+ r2¸= (r − 1)!·n+ rr + 1¸r+(n+ r − 1)!(r − 1)!·r2¸;which is just a special case of the following identity with r = 1 and k = 2. 2Identity 3 For 0 ≤ r ≤ m ≤ n,·nk¸r=kXt=r·mt¸r·nk +m− t¸m:Proof. The left side counts the permutations in Tn;k;r. On the right, we condition on thenumber t of cycles that contain the elements 1 through m, where r ≤ t ≤ k. We can buildsuch a permutation in two steps. First we arrange elements 1 through n into k + m− tcycles so that elements 1 through m are all in separate cycles. Next, we reduce this to kcycles by converting the ¯rst m cycles into t cycles by treating the m cycles as elements,where the ¯rst r cycles are restricted to stay separate from each other. For example, ifr = 3 and m = 5, then to create the permutation (1 5 8)(2)(3 7 4 6), we would ¯rst createthe permutation (1)(2)(3 7)(4 6)(5 8), and then arrange and combine the ¯rst 5 cycles,keeping 1 through 3 in separate cycles. There are£nk+m−t¤mways to accomplish the ¯rststep and£mt¤rfor the second. Thus, for a given t there are£mt¤r£nk+m−t¤mpermutations.Summing over t, we have the desired result. 2We note that hyperharmonic identities (3) and (4) are special cases of equation (5),which by Theorem 2, is equivalent toIdentity 4 For 0 ≤ r ≤ m ≤ n,·nr + 1¸r=·nm+ 1¸m·mr¸r+·mr + 1¸r·nm¸m:INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 8Proof. This is a special case of Identity 3, with k = r+ 1. We just condition on whetherelements 1 through m are in restricted cycles. 2Next we prove equation (6), which is equivalent to the following identity when k=r+1.Identity 5 ·n+ rk¸r=·nk − r¸+ nrXt=1·n+ r − tk − t+ 1¸r−t+1:Proof. The left side counts the permutations in Tn+r;k;r. On the right, we condition onwhether there are any free elements in restricted cycles. There are£nk−r¤permutationswith no free elements in restricted cycles. Otherwise, condition on the number t of the ¯rstcycle with a free element. There are n ways to choose the free element m to immediatelyfollow restricted element t. Next, there are£n+r−tk−t+1¤r−t+1 ways to place elements 1 throught− 1 in their own cycles, and the other elements except m into all the cycles except the¯rst t− 1. Finally, we insert element m immediately to the right of t. 2Using Theorem 2, equation (7) becomes the following r-Stirling identity:Identity 6 For 0 ≤ m ≤ r,·n+ rk¸r=nXs=k−rµns¶·s+ r −mk −m¸r−m·n+m− sm¸m:Proof. This is just a special case of Identity 2 with l = 0. We condition on the numbers of free elements that do not appear in cycles 1 through m, where 0 ≤ m ≤ r. 2Equation (8) is equivalent to the following r-Stirling identity when k = r + 1:Identity 7 For 0 ≤ m ≤ r,·n+ rk¸r=mXt=0µmt¶n!(n+ t−m)!·n+ r −mk − t¸r−t:Proof. We color the ¯rst m cycles blue, and condition on the number t of blue cyclesthat have only one element. There are¡mt¢ways to choose these cycles and place therestricted elements in them. Next we choose but do not place the leftmost free element togo in each of the other blue cycles; this can be done in¡nm−t¢ways. Now we can arrangethe remaining n+ r−m elements in the k− t allowed cycles in £n+r−mk−t¤r−t ways. Finally,there are (m − t)! ways to place the m − t chosen free elements, each to the right of arestricted element. 2Finally, equation (9), is equivalent to the following r-Stirling identity.INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 9Identity 8 For 1 ≤ d ≤ r and 1 ≤ c ≤ n,·n+ rk¸r=d−1Xi=0cµnc¶·c+ ii+ 1¸i+1·n+ r − c− ik − i¸r−i+c−1Xj=0µnj¶·d+ jd¸d·n+ r − d− jk − d¸r−d:Proof. The left side counts the permutations in Tn+r;k;r. The right side conditions onwhether there are at least c free elements in the ¯rst d restricted cycles. If there are,then condition on i, the last restricted cycle with fewer than c free elements before or init (if there are at least c free elements in the ¯rst cycle, then i = 0). There are¡nc¢waysto choose which c free elements are listed ¯rst in the permutation. Since at least one freeelement is in cycle i+ 1, there are c ways to choose which is the ¯rst free element of thatcycle. Then there are£c+ii+1¤i+1ways to place the other c − 1 of the c ¯rst free elements.Finally, we can place the remaining n− c free elements, but not in the ¯rst i cycles andnot in cycle i + 1 before any elements already placed. There are£n+r−c−ik−i¤r−i ways toplace them.If instead there are fewer than c free elements in the ¯rst d cycles, then condition onthe number j of free elements in the ¯rst d cycles. There are¡nj¢ways to choose these jelements,£d+jd¤dways to place them, and£n+r−d−jk−d¤r−d ways to place the remaining freeelements in the other k − d cycles. 2References[1] M. Aigner, Combinatorial Theory, Ch. 3 (Springer, New York, 1979) 98{99.[2] A.T. Benjamin, G.O. Preston, and J.J. Quinn, A Stirling Encounter with Harmonic Numbers,Math. Mag. 75 (2002), 95{103.[3] A.Z. Broder, The r-Stirling Numbers, Disc. Math. 49 (1984) 241{259.[4] L. Carlitz, Weighted Stirling Numbers of the First and Second Kind{1, Fib. Quart. 18 (1980)147{162.[5] J.H. Conway and R.K. Guy, The Book of Numbers, Copernicus, 1996.[6] Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison Wes-ley, 1993.[7] D.H. Greene and D.E. Knuth, Mathematics for the Analysis of Algorithms, Birkhauser, Boston,1981.[8] L. Hsu and P. Shiue, A Uni¯ed Approach to Generalized Stirling Numbers, Adv. Appl. Math. 20(1998) 366{384.[9] J.M. Santmyer, A Stirling Like Sequence of Rational Numbers, Disc. Math. 171 (1997) 229{235.

A Combinatorial Approach to Hyperharmonic Numbers

Keck Graduate Institute

Claremont Colleges
Scholarship @ Claremont
All HMC Faculty Publications and Research HMC Faculty Scholarship
10-1-2003
A Combinatorial Approach to Hyperharmonic
Numbers
Arthur T. Benjamin
Harvey Mudd College
David Gaebler '04
Harvey Mudd College
Robert Gaebler '04
Harvey Mudd College
This Article is brought to you for free and open access by the HMC Faculty Scholarship at Scholarship @ Claremont. It has been accepted for inclusion
in All HMC Faculty Publications and Research by an authorized administrator of Scholarship @ Claremont. For more information, please contact
scholarship@cuc.claremont.edu.
Recommended Citation
Benjamin, A.T., Gaebler, D., & Gaebler R. (2003). A combinatorial approach to hyperharmonic numbers. Integers: The Electronic
Journal of Combinatorial Number Theory, 3: 1-9.
A COMBINATORIAL APPROACH TO HYPERHARMONIC NUMBERS
Arthur T. Benjamin
Department of Mathematics, Harvey Mudd College, Claremont, CA, USA
benjamin@math.hmc.edu
David Gaebler
Department of Mathematics, Harvey Mudd College, Claremont, CA, USA
dgaebler@hmc.edu
Robert Gaebler
Department of Mathematics, Harvey Mudd College, Claremont, CA, USA
rgaebler@hmc.edu
Received: 3/26/03, Accepted: 10/21/03, Published: 10/23/03
Abstract
Hyperharmonic numbers arise by taking repeated partial sums of harmonic numbers.
These numbers can be expressed in terms of r-Stirling numbers, leading to combinatorial
interpretations of many interesting identities.
1. Introduction
The harmonic numbers, de¯ned by Hn =
Pn
k=1
1
k
arise frequently in the solution to
combinatorial problems and in the analysis of algorithms. It is well known [6] that
Hn =
£
n+1
2
¤
n!
(1)
where
£
n
k
¤
denotes the (unsigned) Stirling number of the ¯rst kind, counting the permu-
tations of n elements that are the product of k disjoint cycles. This allows harmonic
number identities to be viewed combinatorially, as in [2].
The quantity Hn can be generalized many ways (see, for instance, [9]). The general-
ization we pursue, called hyperharmonic numbers by Conway and Guy [5], are obtained
by taking repeated partial sums of the Harmonic numbers.
Formally, we de¯ne Hrn, the hyperharmonic number of order r as follows.
De¯nition 1 Let Hrn = 0 for r < 0 or n ≤ 0, H0n = 1n for n ≥ 1, and for r; n ≥ 1, let
Hrn =
nX
i=1
Hr−1i : (2)
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 2
Note that H1n is equal to the ordinary harmonic number Hn.
In [5], Conway and Guy express the hyperharmonic numbers in terms of ordinary
harmonic numbers.
Hrn =
µ
n+ r − 1
r − 1
¶
(Hn+r−1 −Hr−1): (3)
The presence of the binomial coe±cient suggests that a deeper combinatorial rela-
tionship exists. As we shall see, the hyperharmonic numbers satisfy several interesting
identities like the ones below.
nHrn =
µ
n+ r − 1
r
¶
+ rHr+1n−1 (4)µ
m− 1
r − 1
¶
Hmn =
µ
n+m− r
n
¶
Hrn+m−r −
µ
n+m− 1
n
¶
Hrm−r for 1 ≤ r ≤ m (5)
rX
t=1
H tn = H
r
n+1 −
1
n+ 1
(6)
Hrn =
nX
t=1
µ
n+ r −m− t− 1
r −m− 1
¶
Hmt for 0 ≤ m ≤ r − 1 (7)
Hrn =
mX
t=0
µ
m
t
¶
Hr−tn+t−m for 0 ≤ m ≤ r (8)
Hrn =
nX
s=m
µ
n+ r − l − s
r − l
¶
H l−1s +
rX
t=l
µ
n+ r −m− t
n−m
¶
H tm−1 (9)
for 1 ≤ l ≤ r and 1 ≤ m ≤ n:
A generating function for the hyperharmonic numbers is
− ln(1− x)
(1− x)r =
∞X
i=1
Hri x
i
since − ln(1−x) = P xn
n
, and multiplying by 1
(1−x)r has the e®ect of taking partial sums r
times. This generating function is equivalent by (3) to an identity in [7]. Although many
of the identities in this paper can be proved using algebraic methods, we shall prove all
of these by simple combinatorial arguments.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 3
2. Basic properties of hyperharmonic numbers
There are two di®erent ways to approach the hyperharmonic numbers combinatorially.
One is to investigate the numerator of the unreduced fraction, i.e., n!Hrn. This approach
will be followed in the next section. The other is to write Hrn as the sum of fractions
1
t
,
where t ranges from 1 to n, and ask how many times each fraction appears in this sum.
The answer is given by the following theorem.
Theorem 1
Hrn =
nX
t=1
µ
n+ r − t− 1
r − 1
¶
1
t
:
Proof. Given a particular t, where 1 ≤ t ≤ n, we claim that the fraction 1
t
is counted
exactly
¡
n+r−t−1
r−1
¢
times. To see this, consider the weighted directed graph with vertex
set V = {(x; y)|x ≥ 1; y ≥ 0} and the arcs leaving (x; y) enter vertices (x′; y + 1), where
x′ ≥ x. We give each vertex (x; y) a weight of Hyx . Hence for all t ≥ 1, (t; 0) has weight
1
t
and for x; y ≥ 1, (x; y) has weight Px′≤xHy−1x′ , which is the sum of the weights of the
vertices that point to (x; y).
Thus the fraction 1
t
appears in Hrn for every path from (t; 0) to (n; r), i.e., for every
sequence t ≤ x1 ≤ · · · ≤ xr−1 ≤ n. This is the same as the number of size r − 1
multisubsets of {t; : : : ; n}, which equalsµµ
n− t+ 1
r − 1
¶¶
=
µ
n+ r − t− 1
r − 1
¶
as desired, where
¡¡
n
k
¢¢
=
¡
n+k−1
k
¢
is the number of size k multisubsets of an n-element
set. 2
More generally, for 0 ≤ m ≤ r− 1, by counting all paths from (t;m) to (n; r), we see
that for 1 ≤ t ≤ n, Hmt is contributed
¡¡
n−t+1
r−1−m
¢¢
=
¡
n+r−t−m−1
r−m−1
¢
times, and Identity (7)
follows. Similar path counting arguments can also be used to establish equations (6), (8)
and (9), but we shall provide other combinatorial proofs of these in Section 4.
We now begin our discussion of the combinatorial nature of the numerator of Hn;r.
We can write Hrn as a (typically non-reduced) fraction of the form
Hrn =
an;r
n!
: (10)
Thus a0;r = 0 and for n ≥ 1, an;0 = (n− 1)! . For n; r ≥ 1,
Hrn =
nX
i=1
Hr−1i = H
r
n−1 +H
r−1
n
=⇒ an;r
n!
=
an−1;r
(n− 1)! +
an;r−1
n!
=⇒ an;r = n an−1;r + an;r−1 (11)
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 4
We shall use this recurrence and initial condition to prove a generalization of equation
(1).
3. r-Stirling numbers
The r-Stirling numbers, which are similar to the weighted Stirling numbers of Carlitz [4]
and are a special case of the generalized Stirling numbers of Hsu and Shiue [8], can be
de¯ned combinatorially in terms of restricted permutations. Broder [3] gives the following
combinatorial de¯nition, which we will use throughout this paper:
De¯nition 2
£
n
k
¤
r
is the number of permutations of the set {1; 2; : : : ; n} having k disjoint,
non-empty cycles, in which the elements 1 through r are restricted to appear in di®erent
cycles.
Note that
£
n
k
¤
0
and
£
n
k
¤
1
are both equal to the ordinary Stirling number
£
n
k
¤
.
For convenience, we let Tn;k;r denote the set of permutations of {1; 2; : : : ; n} into
k cycles, in which the elements 1 through r appear in di®erent cycles. Thus
£
n
k
¤
r
is
the number of permutations in Tn;k;r. We can think of
£
n
k
¤
r
as counting the number of
permutations with r \restricted" and n−r \free" elements. If a cycle contains a restricted
element, we call it a restricted cycle; otherwise we call it a free cycle.
We shall adopt the standard permutation notation of writing the smallest element in
a cycle at the beginning of that cycle, and listing the cycles in ascending order according
to their smallest elements. For example, (1 5 3 7)(2)(4 9 8)(6) is in standard permutation
notation, while (1 5 3 7)(2)(6)(4 9 8) and (1 5 3 7)(2)(9 8 4)(6) are not.
Generating functions for the r-Stirling numbers are presented in [3]:
nX
k=r
·
n
k
¸
r
xk = xr
n−1Y
i=r
(x+ i):
∞X
n=k
·
n+ r
k + r
¸
r
xn
n!
=
1
k!
µ
1
1− x
¶r ·
ln
µ
1
1− x
¶¸k
:
X
0≤k≤n
·
n+ r
k + r
¸
r
xn
n!
tk =
µ
1
1− x
¶r+t
:
By de¯nition, we have
£
0
0
¤
r
= 1 and for n > 0,
£
n
k
¤
r
= 0 for k < r or k > n. Notice
that for 1 ≤ r ≤ n ·
n
r
¸
r
=
(n− 1)!
(r − 1)! (12)
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 5
since we are counting permutations of the form (1 · · ·)(2 · · ·) · · · (r · · ·). The elements r+1
through n may be entered one at a time. Each element r+1 ≤ i ≤ n can be placed to the
right of any of the i−1 elements already in place, so there are r(r+1) · · · (n−1) = (n−1)!
(r−1)!
such permutations. Notice that r-Stirling numbers also satisfy the traditional Stirling
recurrence: for 0 ≤ r ≤ k ≤ n,·
n+ 1
k
¸
r
=
·
n
k − 1
¸
r
+ n
·
n
k
¸
r
(13)
which may be proved by conditioning on whether or not element n + 1 is alone in its
cycle.
Another useful identity we shall rely on is:
Identity 1 For r ≥ 1,·
n+ r
k
¸
r
= n
·
n+ r − 1
k
¸
r
+
·
n+ r − 1
k − 1
¸
r−1
:
This may be proved by conditioning on whether there are any free elements in the
¯rst cycle. If so, then choose one of the n free elements to appear immediately after 1;
the other elements can be arranged
£
n+r−1
k
¤
r
ways. Otherwise, element 1 is alone, and
elements 2 through n+r can be arranged
£
n+r−1
k−1
¤
r−1 ways. This brings us to the following
theorem.
Theorem 2
Hrn =
£
n+r
r+1
¤
r
n!
:
Proof 1. Let An;r =
£
n+r
r+1
¤
r
. Hence A0;r =
£
r
r+1
¤
r
= 0 = a0;r and for n ≥ 1, An;0 =£
n
1
¤
0
= (n − 1)! = a0;r. For n; r ≥ 1, Identity 1 implies An;r = nAn−1;r + An;r−1 just like
in equation (11). Since An;r and an;r satisfy the same initial conditions and the same
recurrence, then by equation (10) we have
£
n+r
r+1
¤
r
= An;r = an;r = n!H
r
n; as desired. 2
It is also possible to combinatorially prove Theorem 2 without relying on a recurrence.
We ¯rst prove the following general identity, similar to one proved in [3].
Identity 2 For 0 ≤ m ≤ r ≤ n and 0 ≤ l ≤ k − r,µ
k − r
l
¶·
n+ r
k
¸
r
=
n−lX
t=k−r−l
µ
n
t
¶·
t+ r −m
k −m− l
¸
r−m
·
n+m− t
m+ l
¸
m
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 6
Proof. The left side counts permutations of Tn+r;k;r where the ¯rst m cycles and l free
cycles are colored red, and the other cycles colored green. On the right, we condition
on the number t of free elements in green cycles. There are
¡
n
t
¢
ways to choose the free
elements,
£
t+r−m
k−m−l
¤
r−m ways to arrange them in k − (m + l) green cycles, then
£
n+m−t
l+m
¤
m
ways to arrange the other (n+ r)− (r−m+ t) elements among the m+ l red cycles. 2
Proof 2 of Theorem 2. As a special case of Identity 2 with l = 0, m = r, and k = r + 1,
we have·
n+ r
r + 1
¸
r
=
nX
t=1
n!
t!(n− t)! · (t− 1)! ·
(n+ r − t− 1)!
(r − 1)! =
nX
t=1
µ
n+ r − t− 1
r − 1
¶
n!
t
= n!Hrn
by Theorem 1. 2
4. r-Stirling identities
Theorem 2 expresses Hrn in terms of r-Stirling numbers. This allows us to give combina-
torial proofs of hyperharmonic identities by proving the equivalent r-Stirling identities.
This will be the focus of the rest of our paper.
We make use of the following lemma:
Lemma 1 For r+ 1 ≤ t ≤ n+ r, the number of permutations in Tn+r;r+1;r with t as the
smallest element in the right cycle is (n+r−1)!
(r−1)!(t−1) :
Proof 1. First place elements 1 through t− 1 in the restricted cycles. There are £t−1
r
¤
r
=
(t−2)!
(r−1)! ways to do this. Next, place element t at the beginning of the right cycle. Finally,
place each of the elements t+1 through n+r one at a time. Each element can be placed to
the right of any element already placed, so there are t·(t+1)·(t+2) · · · (n+r−1) = (n+r−1)!
(t−1)!
ways to place them. Altogether, then, the number of permutations is
(t− 2)!
(r − 1)! ·
(n+ r − 1)!
(t− 1)! =
(n+ r − 1)!
(r − 1)!(t− 1) :
2
Proof 2. Alternatively, we could list the numbers 1 through n+ r in any order with the
provision that 1 is ¯rst and the numbers 1 through r must be in increasing order. There
are
¡
n+r−1
r−1
¢
ways to place the elements 1 through r in our list, then n! ways to order the
numbers r + 1 through n + r in the remaining positions. We then convert our list to
a restricted permutation by starting a new cycle at each of 1; 2; : : : ; r; t: Element t has
equal probability of having any position among elements 2; 3; : : : ; t, so its probability of
being last among them is 1
t−1 . Thus the probability that we have a valid permutation is
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 7
1
t−1 . The number of permutations in Tn+r;r+1;r with t as the smallest element in the right
cycle, then, is µ
n+ r − 1
r − 1
¶
n!
1
t− 1 =
(n+ r − 1)!
(r − 1)!(t− 1) :
2
Now we are ready to prove equation (3).
Proof 1. Since every permutation in Tn+r;r+1;r must have some smallest element t in its
right cycle, with r + 1 ≤ t ≤ n+ r, then by Theorem 2 we have
Hrn =
1
n!
·
n+ r
r + 1
¸
r
=
1
n!
n+rX
t=r+1
(n+ r − 1)!
(r − 1)!(t− 1) =
µ
n+ r − 1
r − 1
¶
(Hn+r−1 −Hr−1):
2
Proof 2. By Theorem 2, (3) is transformed into the following r-Stirling identity:·
n+ r
2
¸
= (r − 1)!
·
n+ r
r + 1
¸
r
+
(n+ r − 1)!
(r − 1)!
·
r
2
¸
;
which is just a special case of the following identity with r = 1 and k = 2. 2
Identity 3 For 0 ≤ r ≤ m ≤ n,·
n
k
¸
r
=
kX
t=r
·
m
t
¸
r
·
n
k +m− t
¸
m
:
Proof. The left side counts the permutations in Tn;k;r. On the right, we condition on the
number t of cycles that contain the elements 1 through m, where r ≤ t ≤ k. We can build
such a permutation in two steps. First we arrange elements 1 through n into k + m− t
cycles so that elements 1 through m are all in separate cycles. Next, we reduce this to k
cycles by converting the ¯rst m cycles into t cycles by treating the m cycles as elements,
where the ¯rst r cycles are restricted to stay separate from each other. For example, if
r = 3 and m = 5, then to create the permutation (1 5 8)(2)(3 7 4 6), we would ¯rst create
the permutation (1)(2)(3 7)(4 6)(5 8), and then arrange and combine the ¯rst 5 cycles,
keeping 1 through 3 in separate cycles. There are
£
n
k+m−t
¤
m
ways to accomplish the ¯rst
step and
£
m
t
¤
r
for the second. Thus, for a given t there are
£
m
t
¤
r
£
n
k+m−t
¤
m
permutations.
Summing over t, we have the desired result. 2
We note that hyperharmonic identities (3) and (4) are special cases of equation (5),
which by Theorem 2, is equivalent to
Identity 4 For 0 ≤ r ≤ m ≤ n,·
n
r + 1
¸
r
=
·
n
m+ 1
¸
m
·
m
r
¸
r
+
·
m
r + 1
¸
r
·
n
m
¸
m
:
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 8
Proof. This is a special case of Identity 3, with k = r+ 1. We just condition on whether
elements 1 through m are in restricted cycles. 2
Next we prove equation (6), which is equivalent to the following identity when k=r+1.
Identity 5 ·
n+ r
k
¸
r
=
·
n
k − r
¸
+ n
rX
t=1
·
n+ r − t
k − t+ 1
¸
r−t+1
:
Proof. The left side counts the permutations in Tn+r;k;r. On the right, we condition on
whether there are any free elements in restricted cycles. There are
£
n
k−r
¤
permutations
with no free elements in restricted cycles. Otherwise, condition on the number t of the ¯rst
cycle with a free element. There are n ways to choose the free element m to immediately
follow restricted element t. Next, there are
£
n+r−t
k−t+1
¤
r−t+1 ways to place elements 1 through
t− 1 in their own cycles, and the other elements except m into all the cycles except the
¯rst t− 1. Finally, we insert element m immediately to the right of t. 2
Using Theorem 2, equation (7) becomes the following r-Stirling identity:
Identity 6 For 0 ≤ m ≤ r,·
n+ r
k
¸
r
=
nX
s=k−r
µ
n
s
¶·
s+ r −m
k −m
¸
r−m
·
n+m− s
m
¸
m
:
Proof. This is just a special case of Identity 2 with l = 0. We condition on the number
s of free elements that do not appear in cycles 1 through m, where 0 ≤ m ≤ r. 2
Equation (8) is equivalent to the following r-Stirling identity when k = r + 1:
Identity 7 For 0 ≤ m ≤ r,·
n+ r
k
¸
r
=
mX
t=0
µ
m
t
¶
n!
(n+ t−m)!
·
n+ r −m
k − t
¸
r−t
:
Proof. We color the ¯rst m cycles blue, and condition on the number t of blue cycles
that have only one element. There are
¡
m
t
¢
ways to choose these cycles and place the
restricted elements in them. Next we choose but do not place the leftmost free element to
go in each of the other blue cycles; this can be done in
¡
n
m−t
¢
ways. Now we can arrange
the remaining n+ r−m elements in the k− t allowed cycles in £n+r−m
k−t
¤
r−t ways. Finally,
there are (m − t)! ways to place the m − t chosen free elements, each to the right of a
restricted element. 2
Finally, equation (9), is equivalent to the following r-Stirling identity.
INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 3 (2003), #A15 9
Identity 8 For 1 ≤ d ≤ r and 1 ≤ c ≤ n,·
n+ r
k
¸
r
=
d−1X
i=0
c
µ
n
c
¶·
c+ i
i+ 1
¸
i+1
·
n+ r − c− i
k − i
¸
r−i
+
c−1X
j=0
µ
n
j
¶·
d+ j
d
¸
d
·
n+ r − d− j
k − d
¸
r−d
:
Proof. The left side counts the permutations in Tn+r;k;r. The right side conditions on
whether there are at least c free elements in the ¯rst d restricted cycles. If there are,
then condition on i, the last restricted cycle with fewer than c free elements before or in
it (if there are at least c free elements in the ¯rst cycle, then i = 0). There are
¡
n
c
¢
ways
to choose which c free elements are listed ¯rst in the permutation. Since at least one free
element is in cycle i+ 1, there are c ways to choose which is the ¯rst free element of that
cycle. Then there are
£
c+i
i+1
¤
i+1
ways to place the other c − 1 of the c ¯rst free elements.
Finally, we can place the remaining n− c free elements, but not in the ¯rst i cycles and
not in cycle i + 1 before any elements already placed. There are
£
n+r−c−i
k−i
¤
r−i ways to
place them.
If instead there are fewer than c free elements in the ¯rst d cycles, then condition on
the number j of free elements in the ¯rst d cycles. There are
¡
n
j
¢
ways to choose these j
elements,
£
d+j
d
¤
d
ways to place them, and
£
n+r−d−j
k−d
¤
r−d ways to place the remaining free
elements in the other k − d cycles. 2
References
[1] M. Aigner, Combinatorial Theory, Ch. 3 (Springer, New York, 1979) 98{99.
[2] A.T. Benjamin, G.O. Preston, and J.J. Quinn, A Stirling Encounter with Harmonic Numbers,
Math. Mag. 75 (2002), 95{103.
[3] A.Z. Broder, The r-Stirling Numbers, Disc. Math. 49 (1984) 241{259.
[4] L. Carlitz, Weighted Stirling Numbers of the First and Second Kind{1, Fib. Quart. 18 (1980)
147{162.
[5] J.H. Conway and R.K. Guy, The Book of Numbers, Copernicus, 1996.
[6] Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison Wes-
ley, 1993.
[7] D.H. Greene and D.E. Knuth, Mathematics for the Analysis of Algorithms, Birkhauser, Boston,
1981.
[8] L. Hsu and P. Shiue, A Uni¯ed Approach to Generalized Stirling Numbers, Adv. Appl. Math. 20
(1998) 366{384.
[9] J.M. Santmyer, A Stirling Like Sequence of Rational Numbers, Disc. Math. 171 (1997) 229{235.


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A Combinatorial Approach to Hyperharmonic Numbers

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