a $q$-analog of the hyperharmonic numbers

Afr. Mat.DOI 10.1007/s13370-012-0106-6

A q-analog of the hyperharmonic numbers

Toufik Mansour · Mark Shattuck

Received: 10 April 2012 / Accepted: 27 August 2012© African Mathematical Union and Springer-Verlag 2012

Abstract Recently, the q-analog of the harmonic numbers obtained by replacing eachpositive integer n with nq has been shown to satisfy congruences which generalize Wolsten-holme’s theorem. Here, we wish to consider further algebraic properties of these numbers.Recall that the r-harmonic, or hyperharmonic, numbers arise by taking repeated partial sumsof harmonic numbers. In this paper, we introduce and study properties of a q-analog of the r -harmonic numbers, which reduces to the aforementioned q-harmonic numbers when r = 1.It is defined in terms of a statistic on the set of permutations of length n in which the elements1, 2, . . . , r belong to distinct cycles (which is enumerated by the r -Stirling number of thefirst kind).

Keywords Harmonic number · Hyperharmonic number · q-Generalization

Mathematics Subject Classification (2000) 11B65, 05A19

1 Introduction

The harmonic numbers H(n), defined by H(n) = ∑nk=1

1k , have important applications in

combinatorics, number theory, and the analysis of algorithms. The quantity H(n) has beengeneralized in many ways (see, e.g., [11]). Among these generalizations are the hyperhar-monic numbers, so termed by Conway and Guy [6], which are obtained by taking repeatedpartial sums of the harmonic numbers. Let H(n, r) denote the hyperharmonic number oforder r (also called the r-harmonic number), which is defined by letting H(n, r) = 0 if

T. Mansour (B)Department of Mathematics, University of Haifa, 31905 Haifa, Israele-mail: [email protected]

M. ShattuckDepartment of Mathematics, University of Tennessee, Knoxville, TN 37996, USAe-mail: [email protected]

123

T. Mansour, M. Shattuck

n ≤ 0 or r < 0, H(n, 0) = 1n if n ≥ 1, and if n, r ≥ 1, letting

H(n, r) =n∑

t=1

H(t, r − 1). (1)

Since − log(1 − x) = ∑n≥1

xn

n and since multiplying by 1(1−x)r has the effect of taking

partial sums r times, the hyperharmonic numbers have generating function

∑

n≥1

H(n, r)xn = − log(1 − x)

(1 − x)r, r ≥ 0. (2)

Two q-analogs of H(n) are given by

Hn(q) :=n∑

j=1

1

jq, n ≥ 0, (3)

and

H̃n(q) :=n∑

j=1

q j

jq, n ≥ 0, (4)

where q is an indeterminate and jq = 1 + q + · · · + q j−1. A noteworthy property of theharmonic numbers is that they satisfy the congruence of Wolstenholme, which states thatHp−1 ≡ 0 (mod p2) for all primes p ≥ 5. Andrews [2] proved q-analogs of a weakerversion (mod p) of the Wolstenholme congruence for Hp−1(q) and H̃p−1(q), taken in thepolynomial ring Z[q], which was extended, more recently, by Shi and Pan [13] to a q-analogof the full version. See also the related paper by Dilcher [8] for a different generalization.

Our aim here is to study combinatorial properties of the numbers H̃n(q) (and hence alsoof Hn(q) = q H̃n(1/q)). Indeed, we consider a q-analog of the r -harmonic numbers whichreduces to H̃n(q) when r = 1. First recall that the r -Stirling numbers of the first kind (see[4]), denoted here by A(n, k, r), satisfy

A(n, k, r) = A(n − 1, k − 1, r) + (n − 1)A(n − 1, k, r) (5)

for all n, k ≥ 1 and n > r ≥ 0, where A(n, k, r) = 0 for all n, k ∈ N if n < k or k < r ,with A(n, 0, 0) = δn,0 and A(r, r, r) = 1 for all n, r ≥ 0. They count the permutations of[n] = {1, 2, . . . , n} having k cycles in which the elements of [r ] occupy distinct cycles andreduce to the signless Stirling numbers of the first kind when r = 0, 1. The following result(see, e.g., [3]) allows the hyperharmonic numbers to be viewed combinatorially.

Theorem 1.1 If n, r ∈ Z+, then

H(n, r) = A(n + r, r + 1, r)

n! . (6)

In this paper, we consider the q-generalization of H(n, r), which we will denote byHq(n, r), obtained by replacing the numerator on the right-hand side of (6) with a distributionpolynomial for a certain statistic on the set of permutations enumerated by A(n + r, r + 1, r)

and replacing n! with qn+(r2)nq !, where nq ! = 1q2q · · · nq if n ≥ 1 with 0q ! = 1. Starting

with this definition, we will derive several properties of Hq(n, r), including generalizationsof (1) and (2). When r = 1, it will be seen that H̃n(q) = q Hq(n, 1). A further generalizationof Hq(n, r) involving a second parameter may also be given.

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2 q-Hyperharmonic numbers

Let An,k,r denote the set of all permutations of [n] having exactly k cycles where the elements1, 2, . . . , r lie in distinct cycles. The cardinality of An,k,r is the r -Stirling number introducedby Broder [4], which we denote by A(n, k, r). We consider the following statistic on the setAn,k,r .

Definition 2.1 Suppose σ ∈ An,k,r is represented in the standard cycle form, namely, cycleswritten with the smallest element as the first entry and arranged in ascending order accordingto first entries. Erase parentheses and, in the resulting permutation α = α1α2 · · · αn , countthe number of permanences, i.e., the number of ordered pairs (i, j) with 1 ≤ i < j ≤ n andαi < α j . Denote the value so obtained by perm∗(σ ).

For example, if σ = 3452176 ∈ A7,3,2, then the standard form is (135) (24) (67) andperm∗(σ ) = 6 + 4 + 2 + 3 + 2 + 1 = 18, the number of permanences in the word 1352467.Note that perm∗ is a variant of the usual perm statistic on Sn , which records the number ofpermanences when a permutation is expressed as a word.

Remark The analogous statistic, denoted inv∗, which counts the number of inversionsbetween elements within cycles was first introduced by Carlitz [5] on Sn . See also sequenceA129178 in [14] as well as section 2.2 of [12]. However, the restriction of either the perm∗or inv∗ statistics to An,k,r seems not to have been previously considered.

Furthermore, following [3], let us call elements 1 through r within σ ∈ An,k,r restrictedand the cycles in which they occur restricted cycles. Any member of [r +1, n] = {r +1, r +2, . . . , n} or any additional cycle will be described as free.

We define the polynomial Aq(n, k, r) by

Aq(n, k, r) =∑

σ∈An,k,r

q perm∗(σ ).

For example, if n = 4, k = 3, and r = 2, then

A4,3,2 = {(14)(2)(3), (1)(24)(3), (1)(2)(34), (13)(2)(4), (1)(23)(4)},and Aq(4, 3, 2) = q4(1 + 2q + 2q2). The Aq(n, k, r) satisfy the two-term recurrence

Aq(n, k, r) = qn−1 Aq(n − 1, k − 1, r) + q(n − 1)q Aq(n − 1, k, r) (7)

for all n, k ≥ 1 and n > r ≥ 0, where Aq(n, k, r) = 0 for all n, k ∈ N if n < k or k < r ,

with Aq(n, 0, 0) = δn,0 and Aq(r, r, r) = q(r2) for all n, r ≥ 0. Note that (7) reduces to (5)

when q = 1. To see (7), note that there are n − 1 permanences caused by the element n if thecycle (n) occurs within some member of An,k,r and anywhere from 1 to n − 1 permanencescaused by n otherwise depending on its position. The boundary conditions are clear. TheAq(n, k, r) may be expressed explicitly as follows.

Proposition 2.2 If n ≥ k ≥ r ≥ 0, then

Aq(n, k, r) = q(n2)Sn−k{q1−i iq}r≤i≤n−1, (8)

where Sn−k denotes the (n − k)-th symmetric function.

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Proof Let Bq(n, k, r) := Aq (n,k,r)

q(n2)

. Dividing both sides of (7) by q(n2) implies

Bq(n, k, r) = Bq(n − 1, k − 1, r) + q2−n(n − 1)q Bq(n − 1, k, r) (9)

for all n, k ≥ 1 and n > r ≥ 0. If Bn,r,q(x) := ∑nk=r Bq(n, k, r)xk , then multiplying both

sides of (9) by xk and summing over r ≤ k ≤ n implies

Bn,r,q(x) = (x + q2−n(n − 1)q)Bn−1,r,q(x), n ≥ r + 1,

which we iterate to obtain

Bn,r,q(x) = xrn−1∏

i=r

(x + q1−i iq), n ≥ r, (10)

upon noting Br,r,q(x) = Bq(r, r, r)xr = Aq (r,r,r)

q(r2)

xr = xr . Collecting the coefficient of xk

on the right-hand side of (10), and multiplying by q(n2), gives (8). ��

We note the following special cases.

Corollary 2.3 If n, r ≥ 1, then

Aq(n + r, r, r) = qn+(r2)

(n + r − 1)q !(r − 1)q ! (11)

and

Aq(n + r, r + 1, r) =n+r∑

t=r+1

qn+t−2+(r2)(n + r − 1)q !

(r − 1)q !(t − 1)q. (12)

Proof Formulas (11) and (12) follow from taking k = r and k = r + 1 in (8) and replacingn with n + r . Alternatively, one may provide combinatorial proofs as follows. To prove(11), note first that the elements 1, 2, . . . , r must all belong to distinct cycles, accounting for1 + 2 + · · · + (r − 1) = (r

2

)permanences. Then there are r + i − 1 choices regarding the

position of the element r + i for each i ∈ [n], with each choice creating anywhere from 1to r + i − 1 additional permanences. Thus, Aq(n + r, r, r) = q(r

2)∏n−1

i=0 q(r + i)q , whichgives (11).

To prove (12), suppose that the first element of the (r + 1)-st cycle is t , where r + 1 ≤t ≤ n + r . Note that the placement of the elements 1 through t − 1 in the restricted cycles

contributes qt−r−1+(r2) (t−2)q !

(r−1)q ! , by the first part. Also, the placement of the element t + i ,where i ∈ [n + r − t], contributes q(t + i − 1)q for each i and there are t − 1 additionalpermanences between t and the members of [t − 1]. Thus, we get

qt−r−1+(r2)

(t − 2)q !(r − 1)q !qt−1

n+r−t∏

i=1

q(t + i − 1)q = qn+t−2+(r2)(n + r − 1)q !

(r − 1)q !(t − 1)q

in this case. Summing over t completes the proof. ��We now define a q-analog of H(n, r) which reduces to H(n, r) when q = 1 using (6).

Definition 2.4 Let Hq(n, r) be given by

Hq(n, r) = Aq(n + r, r + 1, r)

qn+(r2)nq ! , n, r ≥ 1, (13)

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Table 1 Values of Aq (n, k, 3) for n = 3, 4, 5, 6 and 3 ≤ k ≤ n

n\k 3 4 5 6

3 q3

4 q4(3)q q6

5 q5(3)q (4)q q7 ∑1i=0 qi (4 − i)q q10

6 q6(3)q (4)q (5)q q10(3)q (4)q + q8(5)q∑1

i=0 qi (4 − i)q q11 ∑2i=0 qi (5 − i)q q15

Table 2 Values of Hq (n, 3) for n = 1, 2, 3, 4, 5

n Hq (n, 3)

1 q2

2 q2(2q3+2q2+2q+1)q+1

3 q2(3q7+6q6+9q5+10q4+9q3+6q2+3q+1)

(q2+q+1)(q+1)

4 q2(4q9+4q8+8q7+9q6+10q5+8q4+7q3+4q2+2q+1)

q3+q2+q+1

5 q2(5q15+10q14+20q13+31q12+43q11+52q10+59q9+58q8+54q7+45q6+34q5+23q4+14q3+7q2+3q+1)

(q4+q3+q2+q+1)(q3+q2+q+1)

with Hq(n, 0) = 1qnq

if n ≥ 1 and Hq(n, r) = 0 if n ≤ 0 or r < 0.

For example, if n = r = 2, then

Hq(2, 2) = Aq(4, 3, 2)

q32q ! = q(1 + 2q + 2q2)

1 + q.

Note that formula (13) continues to hold when r = 0 and n ≥ 1 since it is seen thatAq(n, 1, 0) = qn−1(n − 1)q ! (note (12) does not apply in the case when r = 0). By dividing

by qn+(r2) in (13), we are essentially subtracting the permanences caused by the restricted

elements 1 through r always occurring in the natural order within σ ∈ An+r,r+1,r as well asthose caused by the members of [r + 1, n + r ] always lying to the right of the element 1;alternatively, one could have considered the statistic perm∗ − n − (r

2

)on An+r,r+1,r . Above

are tables for the values of Aq(n, k, r) and Hq(n, r) in the case when r = 3 (Tables 1, 2).Let Hq(n) = Hq(n, 1). Taking r = 1 in (13), and using (12), yields a combinatorial

interpretation for the q-harmonic number H̃n(q) defined by (4) above.

Proposition 2.5 If n ∈ Z+, then H̃n(q) = q Hq(n).

It turns out that the Hq(n, r) defined by (13) above satisfy the following q-version of thehyperharmonic recurrence.

Theorem 2.6 If n, r ∈ Z+, then

Hq(n, r) =n∑

t=1

qt Hq(t, r − 1). (14)

Proof We first show the identity

Aq(n + r, k, r) =n−k+r∑

t=0

qn+r−1tq !(

n

t

)

qAq(n + r − t − 1, k − 1, r − 1), (15)

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where n, r ≥ 1, k ≥ r +1, and(n

t

)q = nq !

tq !(n−t)q ! denotes the q-binomial coefficient. To show(15), we consider the number t of free elements belonging to the first cycle of σ ∈ An+r,k,r ,where 0 ≤ t ≤ n −k +r (note t ≤ n −k +r since σ has k −r free cycles). Observe that

(nt

)q

accounts for both the choice of the subset S of free elements to go in the first cycle as well asthe permanences of σ between members of S and free elements occurring in later cycles (tosee this, encode members of S with 0 and members of [r + 1, n + r ] − S with 1 and countpermanences in the resulting binary word). Furthermore, qt tq ! accounts for the ordering ofelements within the first cycle of σ as well as all permanences between two members bothbelonging to the first cycle. Finally, there are Aq(n+r −t −1, k−1, r −1) ways to arrange theremaining free elements within the last k − 1 cycles and there are n + r − t − 1 permanencesbetween 1 and elements of [n + r ] occurring past the first cycle. Thus, we get a contributionof

(n

t

)

qtq !Aq(n + r − t − 1, k − 1, r − 1)qn+r−1

in this case and summing over t yields (15).Taking k = r + 1 in (15), we get by (13),

Hq(n, r) = Aq (n+r,r+1,r)

qn+(r2)nq ! = q(

r−12 )

q(r2)

∑n−1t=0 qn+r−t−1 Aq (n−t+r−1,r,r−1)

qn−t+(r−12 )(n−t)q !

= ∑n−1t=0 qn−t Hq(n − t, r − 1) = ∑n

t=1 qt Hq(t, r − 1),

as required. ��Remark Using (15), one can establish a formula for Aq(n + r, r + d, r) involving d sumswhich extends (12).

3 A generating function for Hq(n, r)

The next result generalizes the well-known expansion, see [3]:

H(n, r) =n∑

t=1

(n + r − t − 1

r − 1

)1

t, n, r ∈ Z

+.

Proposition 3.1 If n, r ∈ Z+, then

Hq(n, r) =n∑

t=1

(n + r − t − 1

r − 1

)

q

qrt−1

tq. (16)

Proof We establish the identity

Aq(n + r, r + 1, r) = nq !n∑

t=1

(n + r − t − 1

r − 1

)

q

qrt+n−1+(r2)

tq, n, r ≥ 1, (17)

from which (16) follows, via (13), upon dividing both sides by qn+(r2)nq !. To show (17), we

consider the number t of elements occurring in the (r + 1)-st cycle within σ ∈ An+r,r+1,r .Upon encoding the free elements lying in one of the first r cycles of σ by 0 and encodingthose that lie in the last cycle by 1, we see that

(nt

)q accounts for both the choice of subset

S for the elements (necessarily of [r + 1, n + r ]) comprising the last cycle as well as all of

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the permanences between elements in the final cycle and free elements in restricted cycles.Furthermore, there are r t permanences in which the left entry corresponds to a restrictedelement and the right entry to an element of S and

(r2

)permanences in which both entries are

restricted elements. There are also qt−1(t −1)q ! ways to arrange the elements in the last cycle.Finally, we consider the positions of the elements of [r + 1, n + r ] − S within the restrictedcycles. Place each of these elements one at a time, starting with the smallest. Each elementmay be placed to the right of any restricted element or any free element already placed,

which may be done in∏n+r−t−1

i=r qiq = qn−t (n+r−t−1)q !(r−1)q ! ways. Thus, the total number of

possibilities in this case is

(n

t

)

qq(r+1)t−1+(r

2)(t − 1)q !qn−t (n + r − t − 1)q !(r − 1)q ! = nq !

(n + r − t − 1

r − 1

)

q

qrt+n−1+(r2)

tq.

Summing over t gives (17) and completes the proof of (16). ��If n ∈ Z

+, let (x : q)n denote the product (1 − x)(1 − qx) · · · (1 − qn−1x), with (x :q)0 = 1. Recall the well-known fact

∑j≥0

( j+m−1m−1

)

qx j = 1

(x :q)m, which is equivalent to a

special case of the q-binomial theorem (see [1], for example). Using this, we obtain from(16) the following formula for the ordinary generating function which generalizes (2).

Theorem 3.2 If r ∈ Z+ is fixed, then

∑

n≥1

Hq(n, r)xn = − logq(1 − qr x)

q(x : q)r, (18)

where − logq(1 − y) = ∑t≥1

yt

tq.

See, e.g., [10,16,17] for other definitions of q-logarithm functions. Using (18), one maygeneralize (14) and (16).

Theorem 3.3 Let n, r ∈ Z+. If 0 ≤ m ≤ r − 1, then

Hq(n, r) =n∑

i=1

qi(r−m)

(n + r − m − i − 1

r − m − 1

)

qHq(i, m). (19)

If 1 ≤ m ≤ r , then

Hq(n, r) =n−1∑

i=0

qm(n−i)(

i + m − 1

m − 1

)

qHq(n − i, r − m). (20)

Proof Let Hr,q(x) = ∑n≥1 Hq(n, r)xn . To prove (19), first note that by Theorem 3.2, we

have

Hm,q(x) = − logq(1 − qm x)

q∏m−1

i=0 (1 − qi x)

and

Hr,q

(x

qr−m

)

= − logq(1 − qm x)

q∏r−1

i=0 (1 − qi−r+m x)

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so that division yields

Hr,q

(x

qr−m

)

Hm,q(x)= Fq(x),

where

Fq(x) = 1∏r−m−1

i=0 (1 − qi−r+m x).

Note that

F1/q

(x

q

)

= 1∏r−m−1

i=0 (1 − qi x)=

∑

j≥r−m−1

(j

r − m − 1

)

qx j−(r−m−1),

whence

Fq(x) =∑

j≥r−m−1

(j

r − m − 1

)

1/q

(x

q

) j−(r−m−1)

. (21)

Then Hr,q

(x

qr−m

)= Hm,q(x)Fq(x), with (21), implies

∑

n≥1

Hq(n, r)

(x

qr−m

)n

=∑

i≥1

Hq(i, m)xi∑

j≥r−m−1

(j

r −m−1

)

1/q

(x

q

) j−(r−m−1)

. (22)

Extracting the coefficient of xn on both sides of (22), where n = i + j − (r − m − 1) on theright, yields

Hq(n, r) =n∑

i=1

q(r−m−1)n+i(

n + r − m − i − 1

r − m − 1

)

1/qHq(i, m).

Formula (19) now follows from the fact(n

k

)q = qk(n−k)

(nk

)1/q .

The proof of (20) is similar, upon noting

Hr,q(x)

Hr−m,q(qm x)= 1

(1 − x)(1 − qx) · · · (1 − qm−1x)

and thus∑

n≥1

Hq(n, r)xn =∑

i≥1

Hq(i, r − m)(qm x)i∑

j≥m−1

(j

m − 1

)

qx j−(m−1).

��Remark The q = 1 case of (19) occurs as equation 7 in [3], while we were unable to findthe q = 1 case of (20). Note that (19) and (20) reduce to (16) when m = 0 and m = r ,respectively.

Using (18), it is also possible to q-generalize equation 8 in [3].

Theorem 3.4 If n, r ∈ Z+ and 0 ≤ m ≤ r , then

Hq(n, r) =m∑

t=0

qt (n−m+t)(

m

t

)

qHq(n − m + t, r − t). (23)

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Proof Let Hq,r (x) = ∑n≥0 Hq(n, r)xn . We show, equivalently,

Hr,q(x) =m∑

t=0

(m

t

)

qxm−t Hr−t,q(qt x), (24)

where n, r ∈ Z+. By (18), identity (24) is equivalent to

logq(1 − qr x)

q(x : q)r=

m∑

t=0

(m

t

)

q

xm−t logq(1 − qr x)

q(1 − qt x) · · · (1 − qr−1x),

i.e.,m∑

t=0

(m

t

)

qxm−t (x : q)t = 1. (25)

To show (25), note that by the q-binomial theorem (see, e.g., [1]), we have

m∑

t=0

(m

t

)

qxm−t (x : q)t =

m∑

t=0

(m

t

)

qxm−t

t∑

k=0

(−1)kq(k2)

(t

k

)

qxk

=m∑

s=0

xs

(m∑

t=m−s

(−1)s+t−mq(s+t−m2 )

(m

t

)

q

(t

m − s

)

q

)

=m∑

s=0

xs · δs,0 = 1,

upon letting s = m − t + k and using the fact

n∑

i=m

(−1)i−mq(i−m2 )

(n

i

)

q

(i

m

)

q= δn,m, 0 ≤ m ≤ n;

see, e.g., Theorem 4 in [7]. ��

4 Some further identities

We first provide q-generalizations of the following formulas for H(n, r) which occur in [3]where n, r ∈ Z

+:

H(n, r) =(

n + r − 1

r − 1

)

(H(n + r − 1) − H(r − 1)),

H(n + 1, r) − 1

n + 1=

r∑

t=1

H(n, t),

and

H(n, r) =d−1∑

i=0

(c + i − 1

c − 1

)

H(n − c, r − i) +c−1∑

j=0

(d + j − 1

d − 1

)

H(n − j, r − d),

where 1 ≤ c ≤ n and 1 ≤ d ≤ r in the last identity; recall H(n, r) = 0 if n ≤ 0 or r < 0and H(n, 0) = 1

n if n ≥ 1.

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Identity 4.1 If n, r ∈ Z+, then

Hq(n, r) =(

n + r − 1

r − 1

)

q

(Hq(n + r − 1) − Hq(r − 1)

). (26)

Proof By (13), (12) and (16), we have

Hq(n, r) = Aq(n + r, r + 1, r)

qn+(r2)nq !

= 1

qn+(r2)nq !

n+r∑

t=r+1

qn+t−2+(r2)(n + r − 1)q !

(r − 1)q !(t − 1)q

=(

n + r − 1

r − 1

)

q

(Hq(n + r − 1) − Hq(r − 1)

).

��

The q = 1 case of (26) occurs on p. 258 of [6]. Using (26), one can show the followingequalities which hold for all n, r ∈ Z

+, the q = 1 cases of which are equations 4 and 5,respectively, in [3]:

nq Hq(n, r) = qr−1(

n + r − 1

r

)

q+ rq Hq(n − 1, r + 1) (27)

and, for m ≥ r ,(

m − 1

r − 1

)

qHq(n, m) =

(n + m − r

n

)

qHq(n + m − r, r)

−(

n + m − 1

n

)

qHq(m − r, r). (28)

For example, to show (27), substitute (26) on both sides and use the facts

Hq(r) = qr−1

rq+ Hq(r − 1)

and nq(n+r−1

r−1

)

q= rq

(n+r−1r

)q .

We now provide a q-generalization of the second hyperharmonic identity above.

Identity 4.2 If n, r ∈ Z+, then

Hq(n, r) − qnr−1

nq=

r∑

t=1

qn(t−1) Hq(n − 1, r − t + 1). (29)

Proof We will first prove the identity

Aq(n + r, k, r) = qnr+(r2) Aq(n, k − r, 0)

+nq

r∑

t=1

q(n+r−t)(t−1)+(t2)+1 Aq(n + r − t, k − t + 1, r − t + 1), (30)

where n, k, r ∈ Z+. To show (30), we q-generalize the argument given for Identity 5 in [3].

Note that the first term on the right side counts (according to perm∗) the members of An+r,k,r

123


in which each restricted cycle has length one. Otherwise, we consider the position t of thefirst cycle having length at least two. Let m ∈ [r + 1, n + r ] denote the number directly tothe right of t in its cycle. Then elements 1 through t − 1 must go in their own cycles andthere are Aq(n +r − t, k − t +1, r − t +1) possibilities regarding placement of the elementsof [t, n + r ] − {m}. The nq factor accounts for the choice of m as it can create anywherefrom 0 to n − 1 permanences with the members of [r + 1, n + r ] − {m}. There are also(n + r − t)(t − 1) additional permanences caused by the members of [t + 1, n + r ] comingto the right of the members of [t − 1] as well as

(t2

)permanences caused by the members

of [t] since they occur in the natural order and a single permanence between the elements tand m. Summing over t establishes (30). Taking k = r + 1 in (30), dividing both sides byqn+(r

2)nq !, and noting Aq(n, 1, 0) = qn−1(n − 1)q ! then gives

Hq(n, r) = qnr−1

nq+

r∑

t=1

q(n+r−t)(t−1)+(t2)+(r−t+1

2 )−(r2)

Aq(n+r −t, r −t +2, r −t+1)

qn−1+(r−t+12 )(n − 1)q !

,

from which (29) follows by using (13). ��Identity 4.3 If n, r ∈ Z

+, with 1 ≤ c ≤ n and 1 ≤ d ≤ r , then

Hq(n, r) =d−1∑

i=0

qi(n−c+1)

(c + i − 1

c − 1

)

qHq(n − c, r − i)

+c−1∑

j=0

qd(n− j)(

d + j − 1

d − 1

)

qHq(n − j, r − d). (31)

Proof We first establish the identity

Aq(n + r, k, r)

=d−1∑

i=0

qi(n+r−c−i)+1cq

(n

c

)

qAq(c + i, i + 1, i + 1)Aq(n + r − c − i, k − i, r − i)

+c−1∑

j=0

qd(n+r−d− j)(

n

j

)

qAq(d + j, d, d)Aq(n + r − d − j, k − d, r − d). (32)

For this, we q-generalize (and slightly modify) the argument given for Identity 8 in [3]. Therethey considered whether there were at least c free elements in the first d cycles of σ ∈ An+r,k,r

or less than c. In the former case, consider i , the last restricted cycle with fewer than c freeelements before or in it (if there are at least c free elements in the first cycle, then i = 0).Furthermore, consider the value of the c-th free element from the left (which occurs in cyclei + 1), instead of the value of the first element in the (i + 1)-st cycle (as was done in [3]).Note that the cq factor in the first sum in (32) reflects the choice of this free element. Observefurther that Aq(c+i, i +1, i +1) accounts for the positions of the c−1 left-most free elementsalong with the elements of [i + 1], while Aq(n + r − c − i, k − i, r − i) accounts for thepositions of the elements of the set [i +1, n +r ]− S in the final k − i cycles, where S denotesthe set comprising the leftmost c free elements. Summing over all possible i gives the firstsum in (32) above. Note that there are i + 1 permanences between the c-th free element andrestricted elements as well as i(n + r − c − i − 1) permanences between members of [i] and[i +2, n+r ]−S, which implies that there are i +1+i(n+r −c−i −1) = i(n+r −c−i)+1additional permanences to be accounted for, whence the factor of q appearing.

123


If there are less than c free elements in the first d cycles, then consider the number j offree elements in these cycles. Note that there are d(n + r − d − j) permanences caused bythe members of [d] and elements occurring in the final k − d cycles, which gives the secondsum in (32) and completes its proof. Taking k = r + 1 in (32), dividing by qn+(r

2)nq !, andusing (13) and (11) yields (31). ��

We conclude this section with the following identity which relates a positive sum involvingthe hyperharmonic numbers to one that alternates.

Identity 4.4 If 1 ≤ m ≤ n and r ≥ 0, then

n−1∑

k=m

qk−m(

k

m

)

qHq(k, r) = (−1)r q(r

2)+r(m+1)−1

(m + r)q

(n − 1

m + r

)

q

+r−1∑

j=0

(−1) j q( j2)+ j (m+1)

(n

m + j + 1

)

qHq(n − 1, r − j).

(33)

Proof Let an,m,r = ∑n−1k=m qk−m

( km

)q Hq(k, r). From the summation-by-parts formula, see,

for example, p. 172 of [15], and the fact

n−1∑

k=m

qk−m(

k

m

)

q=

(n

m + 1

)

q,

see [7], we have, by (14),

an,m,r = Hq(n − 1, r)

n−1∑

k=m

qk−m(

k

m

)

q−

n−2∑

k=m

(Hq(k + 1, r) − Hq(k, r))

k∑

i=m

qi−m(

i

m

)

q

= Hq(n − 1, r)

(n

m + 1

)

q−

n−2∑

k=m

qk+1 Hq(k + 1, r − 1)

(k + 1

m + 1

)

q

= Hq(n − 1, r)

(n

m + 1

)

q− qm+1an,m+1,r−1. (34)

Since Hq(k, 0) = 1qkq

if k ≥ 1, equation (34) when r = 1 gives

n−1∑

k=m

qk−m(

k

m

)

qHq(k, 1) = Hq(n − 1, 1)

(n

m + 1

)

q− qm+1

n−1∑

k=m+1

qk−m−1 1

qkq

(k

m + 1

)

q

= Hq(n−1, 1)

(n

m+1

)

q− qm

(m + 1)q

n−1∑

k=m+1

qk−1−m(

k − 1

m

)

q

= Hq(n−1, 1)

(n

m+1

)

q− qm

(m + 1)q

(n − 1

m + 1

)

q. (35)

When r ≤ n − m, equation (33) follows from iterating (34) and using (35) with m replacedby m + r − 1. Equation (33) is also seen to hold in the case when r > n − m, upon iterating(34) and noting an,n, j = 0 for j ≥ 0, which completes the proof. ��

123


Remark Letting q = 1 in (35) gives

n−1∑

k=m

(k

m

)

H(k) = H(n − 1)

(n

m + 1

)

− 1

m + 1

(n − 1

m + 1

)

= H(n)

(n

m + 1

)

− 1

n

(n

m + 1

)

− 1

m + 1

(n − 1

m + 1

)

=(

n

m + 1

) (

H(n) − 1

m + 1

)

. (36)

Formula (36) occurs as equation 6.70 on p. 280 of [9].

5 A further generalization

By considering a second statistic on An,k,r , one may generalize Hq(n, r) as follows. Recallthat if p and q are indeterminates, then n p,q := pn−1 + pn−2q + · · · + pqn−2 + qn−1 ifn ∈ Z

+, with 0p,q := 0. Let n p,q ! := 1p,q2p,q · · · n p,q for n ∈ Z+, with 0p,q ! := 1, and let

(nk

)p,q := n p,q !

kp,q !(n−k)p,q ! for n ∈ N and 0 ≤ k ≤ n, with(n

k

)p,q = 0 if 0 ≤ n < k or if k < 0

(see, e.g., [7]).

Definition 5.1 Suppose σ ∈ An,k,r is written in the standard cycle form. Erase parenthesesand, in the resulting permutation α = α1α2 · · · αn , count the number of inversions, i.e., thenumber of ordered pairs (i, j) with 1 ≤ i < j ≤ n and αi > α j . Denote the value so obtainedby inv∗(σ ).

The statistic inv∗ was first considered by Carlitz [5] on the set of all permutations of sizen as well as those having a prescribed number of cycles. Define the polynomial Ap,q(n, k, r)

by

Ap,q(n, k, r) =∑

σ∈An,k,r

pinv∗(σ )q perm∗(σ ).

Note that Ap,q(n, k, r) reduces to Aq(n, k, r) when p = 1. Considering the position of theelement n yields the recurrence

Ap,q(n, k, r) = qn−1 Ap,q(n − 1, k − 1, r) + q(n − 1)p,q Ap,q(n − 1, k, r), (37)

subject to the same boundary conditions as Aq(n, k, r). We now define a p, q-analog of thehyperharmonic numbers.

Definition 5.2 Let Hp,q(n, r) be given by

Hp,q(n, r) = Ap,q(n + r, r + 1, r)

qn+(r2)n p,q ! , n, r ≥ 0, (38)

with Hp,q(n) = Hp,q(n, 1).

123


For example, we have Ap,q(4, 3, 2) = q4(p2 + 2pq + 2q2) and

Hp,q(2, 2) = Ap,q(4, 3, 2)

q32p,q ! = q(p2 + 2pq + 2q2)

p + q.

Reasoning as in the second section above, one may establish the following p, q-versionof the hyperharmonic recurrence for n, r ∈ Z

+:

Hp,q(n, r) =n∑

t=1

p(r−1)(n−t)qt Hp,q(t, r − 1), (39)

where Hp,q(t, 0) = 1qtp,q

if t ∈ Z+. Most of the other properties discussed earlier may be

generalized as well. For example, formulas (18) and (26) may be extended, respectively, to

∑

n≥1

Hp,q(n, r)xn = − logp,q(1 − qr x)

q∏r−1

i=0 (1 − pi qr−1−i x), (40)

where r ∈ Z+ and − logp,q(1 − y) = ∑

t≥1yt

tp,q, and to

Hp,q(n, r) =(

n + r − 1

r − 1

)

p,q(Hp,q(n + r − 1) − Hp,q(r − 1)), (41)

where n, r ∈ Z+.

References

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2. Andrews, G.E.: q-analogs of the binomial coefficient congruences of Babbage, Wolstenholme andGlaisher. Discret. Math. 204, 15–25 (1999)

3. Benjamin, A., Gaebler, D., Gaebler, R.: A combinatorial approach to hyperharmonic numbers. Integers3, #A15 (2003)

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Electron. J. Combin. 15, #R63 (2008)9. Graham, R., Knuth, D., Patashnik, O.: Concrete Mathematics: A Foundation for Computer Science, 2nd

edn. Addison-Wesley, Menlo Park (1989)10. Metala-Aho, T., Vaananen, K.: On approximation measures of q-logarithms. Bull. Aus. Math. Soc. 58,

15–31 (1998)11. Santmyer, J.M.: A Stirling like sequence of rational numbers. Discret. Math. 171, 229–235 (1997)12. Shattuck, M.: Parity theorems for statistics on permutations and Catalan words. Integers 5, #A07 (2005)13. Shi, L.L., Pan, H.: A q-analogue of Wolstenholme’s harmonic series congruence. Amer. Math. Monthly

114, 529–531 (2007)14. Sloane, N.J.: The On-Line Encyclopedia of Integer Sequences, http://oeis.org15. Wade, W.: An Introduction to Analysis, 2nd edn. Prentice-Hall, New Jersey (2000)16. Yamano, T.: Some properties of the q-logarithm and q-exponential functions in Tsallis statistics. Phys.

A 305, 486–496 (2002)17. Zudilin, W.: Approximations to q-logarithms and q-dilogarithms, with applications to q-zeta values.

J. Math. Sci. (New York) 137, 4673–4683 (2006)

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a $q$-analog of the hyperharmonic numbers

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