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2

Symmetric functions and a natural framework for combinatorial and

number theoretic sequences

Cormac O’Sullivan

Abstract

Certain triples of power series, considered by I. Macdonald, give a natural framework for many

combinatorial and number theoretic sequences, such as the Stirling, Bernoulli and harmonic numbers

and partitions of different kinds. The power series in such a triple are closely linked by identities coming

from the theory of symmetric functions. We extend the work of Z-H. Sun, who developed similar ideas,

and Macdonald, revealing more of the structure of these triples. De Moivre polynomials play a key role

in this study.

1 Introduction

We begin with an example due to E.T. Bell from [Bel28] that will help introduce the main ideas. It is based

on Euler’s pentagonal number theorem, which states that

∞∏

j=1

(1− qj) =∞∑

m=0

c(m)qm for c(m) =

{

(−1)r if m = r(3r−1)2 for r ∈ Z,

0 otherwise.(1.1)

Expanding the logarithm of the product shows

log

∞∏

j=1

(1− qj) =

∞∑

j=1

log(1− qj) = −∞∑

j=1

∞∑

k=1

qkj

k= −

∞∑

m=1

σ(m)

mqm, (1.2)

where σ(m) indicates the sum of the divisors of m. Then exponentiating (1.2) and comparing coefficients of

powers of q gives the equalities

c(1) = −σ(1), c(2) =σ(1)2

2− σ(2)

2, c(3) = −σ(1)

3

6+σ(1)σ(2)

2− σ(3)

3, (1.3)

and, in general, c(n) is a degree n polynomial in σ(1), . . . , σ(n). Bell called these partition polynomials,

presumably since they have one term for each of the p(n) partitions of the integer n. See (6.18) for his

general definition.

One aspect of this is quite ancient and arises whenever polynomials or power series are raised to powers.

Definition 1.1. For integers n, k with k > 0, the De Moivre polynomial An,k(a1, a2, . . . ) is defined by

(

a1x+ a2x2 + a3x

3 + · · ·)k

=∑

n∈Z

An,k(a1, a2, a3, . . . )xn (k ∈ Z>0). (1.4)

Date: Mar 6, 2022.

2010 Mathematics Subject Classification: 05E05, 05A15, 11B68, 11B73, 05A17

Support for this project was provided by a PSC-CUNY Award, jointly funded by The Professional Staff Congress and The City

University of New York.

1

These polynomial coefficients An,k may be traced back to De Moivre’s paper [DM97] from 1697, and

their properties are gathered in [O’S]. If n < k then clearly An,k(a1, a2, a3, . . . ) = 0. If n > k then

An,k(a1, a2, a3, . . . ) =∑

1j1+2j2+···+mjm=nj1+j2+···+jm=k

(

k

j1, j2, . . . , jm

)

aj11 aj22 · · · ajmm , (1.5)

where m = n − k + 1, the sum is over all possible j1, j2, . . . , jm ∈ Z>0 and 00 always means 1. It is a

polynomial in a1, a2, . . . , am of homogeneous degree k with positive integer coefficients.

In this notation the general form of (1.3) is

c(n) =

n∑

k=0

(−1)k

k!An,k

(

σ(1)

1,σ(2)

2,σ(3)

3, · · ·

)

. (1.6)

This is equivalent to Bell’s formulation (see the first example in [Bel28, p. 44]) and we will see that it follows

from (2.7). Besides c(n) and σ(n), there is an obvious third sequence to consider here since

F (q) :=

∞∏

j=1

1

1− qj=

∞∑

n=0

p(n)qn (1.7)

is the generating function for the partitions. A similar formula to (1.6) with p(n) on the left can be found, as

well as the well-known recurrences,

p(n) = −n−1∑

k=0

p(k)c(n − k), (1.8)

p(n) =1

n

n−1∑

k=0

p(k)σ(n − k), (1.9)

for n > 1. Here (1.8) is due to Euler and comes from multiplying (1.1) by (1.7) and equating coefficients,

while (1.9) comes from equating −F ′(q)/F (q) and the derivative of (1.2).

I. Macdonald related the sequences c(n), p(n) and σ(n) to symmetric functions in [Mac95, Ex. 6, p. 27]

and the identities we have seen follow easily from the known relations among these functions. He gave many

other examples in [Mac95, pp. 26 – 36] and they lead naturally to the following simple abstract definition.

Definition 1.2. Let E(t), H(t) and P (t) be three formal power series with E(t) and H(t) having constant

term 1. We may call them a symmetric triple if they satisfy

P (−t) = d

dtlogE(t), P (t) =

d

dtlogH(t). (1.10)

Independently, Z-H. Sun studied Newton-Euler pairs in [Sun05] – these are the (H(t), P (t)) part. We

will look at almost all of the symmetric triples contained in Macdonald and Sun’s work, find new cases, and

reveal more of their overall structure; these triples form a group for instance.

Following Macdonald, it is convenient to write the coefficients of the three series as

E(t) =∞∑

n=0

entn, H(t) =

∞∑

n=0

hntn, P (t) =

∞∑

n=0

pn+1tn, (1.11)

noting that the constant term of P (t) is labelled p1, and p0 is left undefined. In this language our Bell/Euler

example is the following symmetric triple. (All Examples 1.3 to 9.7 are symmetric triples so we will not

always say it explicitly. Also it is understood that en and hn are given for n > 0 while pn is given for n > 1.)

Example 1.3 (Partitions, divisor sums).

E(t) =∞∏

j=1

(1− (−t)j), H(t) =∞∏

j=1

1

1− tj, P (t) =

∞∑

j=1

j · tj−1

1− tj,

with en = (−1)nc(n), hn = p(n), pn = σ(n).

2

As we will see, the coefficients en, hn and pn of a symmetric triple are tightly linked, with five convo-

lution identities connecting them, similar to (1.8), (1.9), and six transition formulas, similar to (1.6), which

correspond to changing the symmetric function basis.

To give a second example based on [Bel28], define Nr(n) to be the number of ways to express n as a

sum of squares of r integers (including the order of the integers and their signs), and let ω2(n) be the sum

of the odd divisors of n. The details of the next symmetric triple, including the pr(n) term, are explained in

section 6.3.

Example 1.4 (Representations of sums of squares). For a positive integer r,

en = Nr(n), hn = pr(n), pn = r(σ(n) + ω2(n)).

Discussion. Bell’s formula for the number of representations of n as a sum of r squares is equivalent to

Nr(n) =n∑

k=0

(−1)n−k rk

k!An,k

(

σ(1) + ω2(1)

1,σ(2) + ω2(2)

2, · · ·

)

. (1.12)

See [Bel28, p. 46] for this and generalizations. We used the following simplification of [Bel28, Eq. (13)]:

(2(−1)n − 1)ω2(n) + σ(n) = (−1)n(σ(n) + ω2(n)).

The identity (1.12) may be compared with Jacobi’s evaluations ofN2(n) andN4(n) in terms of divisor sums,

contained in [AAR99, Thm. 10.6.1]. The case r = 3 of (1.12) gives class numbers of imaginary quadratic

fields; see [Mor17], for instance, for the connection between N3(n) and class numbers.

The plan of this paper is as follows. Section 2 reviews material we will need on De Moivre polynomials.

Sections 3 and 4 then describe the relevant aspects of symmetric polynomials and functions, and explain the

properties of symmetric triples, including how to multiply, divide and take arbitrary powers of them. The

remaining sections look at a wide variety of examples, including those relating to Bernoulli, Catalan, Cauchy

and Stirling numbers, q-series, partitions and various polynomial families. We find this setup gives a natural

organizing structure that allows these sequences to be studied systematically. New triples also often suggest

themselves.

Recall that the Stirling subset numbers{nk

}

count the number of ways to partition n elements into knonempty subsets. The Stirling cycle numbers

[nk

]

count the number of ways to arrange n elements into kcycles. Of particular interest is a similar array of numbers that naturally appears when pk in (1.11) is set to

equal

H(k)n :=

1

1k+

1

2k+ · · ·+ 1

nk, (1.13)

the harmonic number of order k with n summands, (with Hn also meaning H(1)n ). Then hk can be thought of

as a type of harmonic multiset coefficient and we introduce the notation∥

∥

nk

∥

∥ for it. Unexpectedly, it includes

both kinds of Stirling numbers as special cases; see Example 5.6.

In section 8 we show that repeatedly taking powers and compositional inverses of a power series results

in a two-parameter family. This allows us to generalize some of the earlier symmetric triple examples.

2 De Moivre polynomials

2.1 Basic properties

As in [O’S, Sect. 2], for n, k ∈ Z with k > 0 we have the relations

An,0(a1, a2, a3, . . . ) = δn,0, (2.1)

An,1(a1, a2, a3, . . . ) = an (n > 1), (2.2)

An,k(ca1, ca2, ca3, . . . ) = ckAn,k(a1, a2, a3, . . . ), (2.3)

An,k(ca1, c2a2, c

3a3, . . . ) = cnAn,k(a1, a2, a3, . . . ). (2.4)

3

By a power series, in this paper, we mean a formal power series a0 + a1x + a2x2 + . . . with coefficients

aj in an integral domain R that contains Q. These series form another integral domain denoted R[[x]]. The

next basic result is [O’S, Prop. 3.1].

Proposition 2.1. Suppose that f(x) = a1x+ a2x2 + · · · and g(x) = b0 + b1x+ b2x

2 + · · · are two power

series in R[[x]]. Then g(f(x)) = c0 + c1x+ c2x2 + · · · is in R[[x]] with

cn =n∑

k=0

bk · An,k(a1, a2, . . . ). (2.5)

Applying Proposition 2.1 with g(x) = (1 + x)α, eαx and log(1 + αx) gives

[xn](1 + f(x))α =

n∑

k=0

(

α

k

)

An,k(a1, a2, . . . ), (2.6)

[xn]eαf(x) =

n∑

k=0

αk

k!An,k(a1, a2, . . . ), (2.7)

[xn] log(1 + αf(x)) =

n∑

k=1

(−1)k−1αk

kAn,k(a1, a2, . . . ), (2.8)

where [xn] extracts the coefficient of xn in each series. Recall that the general binomial coefficients satisfy(α0

)

:= 1 and

(

α

k

)

:=α(α − 1) · · · (α− k + 1)

k!,

(−αk

)

= (−1)k(

α+ k − 1

k

)

(2.9)

for positive integers k and variable α. So the right sides of (2.6), (2.7) and (2.8) are also degree n polynomials

in α. Therefore the series (1+f(x))α, eαf(x) and log(1+αf(x)) make sense for arbitrary α, giving elements

of R[α][[x]].De Moivre polynomials were introduced in [DM97]. They were rediscovered and appear explicitly in

Arbogast’s work from 1800 on higher derivatives of compositions, as discussed in [O’S, Sect. 3], and in

several of Stern’s papers including [Ste43] from 1843; see (5.46), (5.47). De Moivre polynomials also

appear implicitly in formulas of Waring (3.4), Kramp (5.8), (5.9) and Bell (6.18). In section 3.3 of Comtet’s

influential book [Com74] they are briefly described as ‘partial ordinary Bell polynomials’. The focus there

is on the related ‘partial Bell polynomials’ which include extra factorial terms that are usually not needed.

2.2 Determinant formulas

Define the n× n matrix

Mn(t) :=

a1t 1a2t a1t 1

.... . .

ant an−1t . . . a1t

,

with ones above the main diagonal and zeros above those. Also set

Nn(t) :=

a1t 1a2t a1t 2a3t a2t a1t 3

.... . .

ant an−1t . . . a2t a1t

, On(t) :=

a1t 12a2t a1t 13a3t a2t a1t 1

.... . .

nant an−1t . . . a2t a1t

,

which are the same as Mn(t) except for multiplying by a factor j on row j, above the main diagonal in

Nn(t) and in the first column for On(t).We will need the next identities in section 3. They are inspired by [Mac95, Ex. 8] with (2.10) also giving

a version of [EJ, Thm. 3.1]. See [O’S, Sect. 5] for their proofs.

4

Proposition 2.2. We have

n∑

k=0

(−1)n+ktkAn,k(a1, a2, . . . ) = detMn(t) (2.10)

n∑

k=0

(−1)n+k tk

k!An,k

(a11,a22, . . .

)

=1

n!detNn(t), (2.11)

n∑

k=0

(−1)n+k tk

kAn,k(a1, a2, . . . ) =

1

ndetOn(t). (2.12)

2.3 Stirling numbers and further results

As in [O’S, Sect. 2], generating function arguments show, for n > k > 0,

An,k

((

α

0

)

,

(

α

1

)

,

(

α

2

)

, . . .

)

=

(

kα

n− k

)

, (2.13)

and

(et − 1)k =

∞∑

n=0

k!

n!

{

n

k

}

tn =⇒ An,k

(

1

1!,1

2!,1

3!, . . .

)

=k!

n!

{

n

k

}

, (2.14)

(− log(1− t))k =∞∑

n=0

k!

n!

[

n

k

]

tn =⇒ An,k

(

1

1,1

2,1

3, . . .

)

=k!

n!

[

n

k

]

. (2.15)

We may take the left identities of (2.14) and (2.15) to be our definitions of the Stirling numbers, as in [Com74,

p. 51], and develop all their properties from this starting point. If we add or remove the first coefficient a1 in

An,k there is a simple effect, by the binomial theorem:

Lemma 2.3. For k > 0,

An,k(a2, a3, . . . ) =

k∑

j=0

(−a1)k−j

(

k

j

)

An+j,j(a1, a2, . . . ), (2.16)

An,k(a1, a1, . . . ) =k∑

j=0

ak−j1

(

k

j

)

An−k,j(a2, a3, . . . ). (2.17)

In the remainder of this section some required identities are developed, allowing us to keep later proofs

relatively self contained. Consult [Com74, Chap. 5] and [GKP94, Sect. 6.1] for further information on

Stirling numbers. Using (2.16) to remove the first coefficients in (2.13), (2.14) and (2.15) produces

An,k

((

α

1

)

,

(

α

2

)

,

(

α

3

)

, . . .

)

=

k∑

j=0

(−1)k−j

(

k

j

)(

αj

n

)

, (2.18)

An,k

(

1

2!,1

3!,1

4!, . . .

)

=k!

(n+ k)!

k∑

j=0

(−1)k−j

(

n+ k

n+ j

){

n+ j

j

}

, (2.19)

An,k

(

1

2,1

3,1

4, . . .

)

=k!

(n+ k)!

k∑

j=0

(−1)k−j

(

n+ k

n+ j

)[

n+ j

j

]

. (2.20)

Another application of (2.16) shows

An,k

(

1

1!,1

2!,1

3!, . . .

)

=k∑

j=0

(−1)k−j

(

k

j

)

An+j,j

(

1

0!,1

1!,1

2!, . . .

)

,

5

with the De Moivre polynomial on the right evaluating to jn/n!. Hence by (2.14), we obtain the well-known

{

n

k

}

=1

k!

k∑

j=0

(−1)k−j

(

k

j

)

jn (n, k > 0). (2.21)

The right sides above may be expressed using the difference operator ∆:

∆f(t) := f(t+ 1)− f(t) with ∆kf(t) =k∑

j=0

(−1)k−j

(

k

j

)

f(t+ j). (2.22)

Also ∆k applied to 1/t gives a useful identity, which may be verified by induction, and we obtain

∆k 1

t=

(−1)kk!

t(t+ 1) · · · (t+ k)=

k∑

j=0

(−1)k−j

(

k

j

)

1

t+ j, (2.23)

as in [GKP94, Eqs. (5.40), (5.41)]. We next claim

tk

(1− t)(1 − 2t) · · · (1− kt)=

∞∑

n=k

{

n

k

}

tn. (2.24)

Using (2.23) with 1/t for t, the left side of (2.24) equals

1

(1/t− 1)(1/t − 2) · · · (1/t − k)=

(−1)k

k!

k∑

j=0

(

k

j

)

(−1)j

1− jt=

∞∑

n=0

tn1

k!

k∑

j=0

(

k

j

)

(−1)k−jjn,

and the claim is proved with (2.21).

3 Symmetric polynomials

A polynomial is called symmetric if it remains unchanged under any permutation of its variables. The

elementary symmetric polynomials ek naturally appear when the coefficients of a polynomial are expressed

in terms of its roots, and so have a long history. In n variables they may be defined as

ek = ek(x1, x2, . . . , xn) :=∑

16j1<j2<···<jk6n

xj1xj2 · · · xjk , (3.1)

where e0 = 1 and ek = 0 for k > n+ 1. The power-sum symmetric polynomials are

pk = pk(x1, x2, . . . , xn) := xk1 + xk2 + · · ·+ xkn. (3.2)

Girard in 1625 expressed pk in terms of the elementary symmetric polynomials for k = 1, 2, 3, 4. For

example, if k = 3 then p3 = e31 − 3e1e2 + 3e3 is true for all n, giving

x31 + x32 + x33 = (x1 + x2 + x3)3 − 3(x1 + x2 + x3)(x1x2 + x2x3 + x1x3) + 3(x1x2x3)

when there are n = 3 variables. Newton’s identity of 1707 is

k · ek =

k∑

j=1

(−1)j−1pjek−j (k > 1), (3.3)

and Waring in 1762 used it to extend Girard’s work to all k; see [Fun30, Gou99]. Waring’s formula may be

written conveniently with De Moivre polynomials as

pk = kk∑

j=1

(−1)k−j

jAk,j(e1, e2, e3, . . . ). (3.4)

6

This gives a natural grouping on the right since Ak,j(e1, e2, e3, · · · ) has homogeneous degree j in e1, e2, . . . .The complete homogeneous symmetric polynomials make a third family and they may be defined as

hk = hk(x1, x2, . . . , xn) :=∑

16j16j26···6jk6n

xj1xj2 · · · xjk , (3.5)

with h0 = 1. All three types form bases for the symmetric polynomials:

Theorem 3.1. If f in Z[x1, . . . , xn] is symmetric then there exist unique Φe,Φh in Z[x1, . . . , xn] and Φp in

Q[x1, . . . , xn] so that

f(x1, . . . , xn) = Φe(e1, . . . , en) = Φh(h1, . . . , hn) = Φp(p1, . . . , pn).

We are following [Mac95, Sect. I.2], which can be consulted for detailed explanations. The generating

functions for ek and hk are

E(t) :=

∞∑

k=0

ektk =

n∏

j=1

(1 + xjt), H(t) :=

∞∑

k=0

hktk =

n∏

j=1

1

1− xjt, (3.6)

and the power-sums generating function is

P (t) :=∞∑

k=0

pk+1tk =

n∑

j=1

xj1− xjt

. (3.7)

Differentiating the logs of E(t) and H(t) shows the identities

P (−t) = E′(t)

E(t), P (t) =

H ′(t)

H(t). (3.8)

In particular, E′(t) = E(t)P (−t) implies Newton’s identity (3.3).

Theorem 3.2 (Transition formulas). For r > 0, ℓ > 1,

hr =

r∑

k=0

(−1)r−kAr,k(e1, e2, e3, . . . ), er =

r∑

k=0

(−1)r−kAr,k(h1, h2, h3, . . . ), (3.9)

er =

r∑

k=0

(−1)r−k

k!Ar,k

(p11,p22,p33, . . .

)

,pℓℓ

=

ℓ∑

k=1

(−1)ℓ−k

kAℓ,k(e1, e2, e3, . . . ), (3.10)

hr =

r∑

k=0

1

k!Ar,k

(p11,p22,p33, . . .

)

,pℓℓ

=

ℓ∑

k=1

(−1)k−1

kAℓ,k(h1, h2, h3, . . . ). (3.11)

Proof. Clearly H(t)E(−t) = 1 and hence, by (2.6) with α = −1, we obtain (3.9). Writing

logE(−t) =n∑

j=1

log(1− xjt) = −n∑

j=1

∞∑

r=1

(xjt)r

r= −

∞∑

r=1

prrtr, (3.12)

and similarly for logH(t), shows:

E(−t) = exp

(

−∞∑

r=1

prrtr

)

, logH(t) =

∞∑

r=1

prrtr, H(t) = exp

(

∞∑

r=1

prrtr

)

. (3.13)

By (2.7) and (2.8) we obtain (3.10) and (3.11) from (3.12) and (3.13).

7

Theorem 3.2 has been known for a long time, with the right equality in (3.10) the Girard-Waring formula

again. Expressing these transition formulas with De Moivre polynomials has the advantage of providing

a clear and uniform description. See [Mac95, Eq. (2.14)’, Ex. 20] for equivalent formulas as sums over

partitions and also [Sun05, Thm. 2.2]. With Proposition 2.2 we may give determinant versions, as in [Mac95,

Ex. 8]. For example, using (2.11) makes the left identity in (3.11) into

hn =1

n!det

p1 −1p2 p1 −2p3 p2 p1 −3...

. . .

pn pn−1 . . . p2 p1

.

See [Gou99] for historical references to such determinant formulas, including Salmon’s for ek and pk in

1876.

Note that the transition formulas are finite sums and independent of the xjs and n. Using ek :=(−1)k−1ek instead of ek (and simplifying with (2.3), (2.4)) makes the signs appearing in Theorem 3.2 more

uniform; see also (7.4). By Theorem 3.1 we expect the polynomials expressing pℓ in terms of ek and hk on

the right of (3.10), (3.11) to have integral coefficients. This may be verified since the gcd of the coefficients

of Aℓ,k is k/ gcd(ℓ, k) by [O’S, Thm. 4.1].

Finally in this section we mention a family of symmetric polynomials that were studied by MacMahon

and only include the terms of hk with r distinct indices:

Sk,r = Sk,r(x1, x2, . . . , xn) :=∑

α1+α2+···+αr=k16j1<j2<···<jr6n

xα1

j1xα2

j2· · · xαr

jr. (3.14)

They link ek, hk and pk together with

ek = Sk,k, hk =k∑

r=0

Sk,r, pk = Sk,1,

and based on his work in [Mac60, Vol. 1, p. 5], (see also [Mac95, Ex. 19], [Hof17, Lemma 2]), we have

Sk,r =k∑

j=r

(−1)j−r

(

j

r

)

ejhk−j, ejhk−j =k∑

r=j

(

r

j

)

Sk,r. (3.15)

4 Symmetric triples

4.1 Fundamental properties

If the variables x1, x2, . . . , xn are assigned certain values, then the symmetric polynomials ek, hk and pkalso become numbers. For a simple example, set each xj to 1. Then by (3.6), (3.7) we easily find

E(t) = (1 + t)n, H(t) = (1− t)−n, P (t) =n

1− t,

so that

ek =

(

n

k

)

, hk =

(

n+ k − 1

k

)

, pk = n. (4.1)

These numbers are related by the transition formulas (3.9), (3.10), (3.11), as well as the other identities we

have seen, such as Newton’s identity (3.3). Focussing on the relationships between ek, hk and pk, we may

discard the original xjs and n and work directly with the symmetric triples in Definition 1.2. The following

two propositions develop their basic properties and the proofs are straightforward exercises.

8

Proposition 4.1. Let (E(t),H(t), P (t)) be a symmetric triple. Then we have

H(t)E(−t) = 1, E(t)P (−t) = E′(t), H(t)P (t) = H ′(t), (4.2)

as well as the variants

P (t) = E(−t)H ′(t) = E′(−t)H(t). (4.3)

The transition formulas of Theorem 3.2 for their coefficients ek, hk and pk all hold, and it follows that a

symmetric triple is uniquely defined by specifying one of the formal series E(t), H(t) or P (t). Also,

e1 = h1 = p1. (4.4)

Proposition 4.2. If (E(t),H(t), P (t)) is a symmetric triple then so are

(H(t), E(t), P (−t)), (4.5)

(E(t)α,H(t)α, αP (t)), for α any constant, (4.6)

(E(−φ(−t)),H(φ(t)), φ′(t)P (φ(t))), for φ(t) any power series with φ(0) = 0. (4.7)

Two symmetric triples (E1,H1, P1) and (E2,H2, P2) may be combined to get a third:

(E1 · E2,H1 ·H2, P1 + P2). (4.8)

With (4.6) for α = −1 and (4.8), these triples clearly form an abelian group with identity (1, 1, 0). From

the effect on H(t) we will just call this group operation multiplication. Proposition 4.2 does not seem to

have appeared before, though (4.6) is the first part of [Sun05, Thm. 2.4].

Example 4.3 (Binomial coefficients).

(

(1 + t)α, (1− t)−α,α

1− t

)

with ek =

(

α

k

)

, hk =

(

k + α− 1

k

)

, pk = α. (4.9)

Discussion. This is [Mac95, Ex. 1]. Here α is an arbitrary complex number or variable and it is easy to verify

that the conditions of Definition 1.2 are satisfied. Each of the equalities in (4.2), (4.3) lead to convolution

identities similar to Newton’s identity (3.3), which in this case gives (as in [GKP94, Eq. (5.16)])

k

(

α

k

)

= (−1)k−1k−1∑

j=0

(−1)jα

(

α

j

)

.

The transition formula expressing pk in terms of ek (Girard-Waring) yields

α

r=

r∑

k=1

(−1)k+r

kAr,k

((

α

1

)

,

(

α

2

)

,

(

α

3

)

, . . .

)

. (4.10)

It follows from (2.18), after some simplifications, that (4.10) is equivalent to

r∑

k=0

(−1)k(

r

k

)(

αk − 1

r − 1

)

= 0.

This can be seen directly using (2.22). Since each difference lowers the degree, the rth difference of the

degree r − 1 polynomial(αx−1

r−1

)

in x must be zero. The transition formula for ek in terms of pk also shows

(

α

r

)

=r∑

k=0

(−1)r+k

k!Ar,k

(α

1,α

2,α

3, · · ·

)

=r∑

k=0

(−1)r+kαk

k!Ar,k

(

1

1,1

2,1

3, · · ·

)

=1

r!

r∑

k=0

(−1)r−k

[

r

k

]

αk, (4.11)

9

by (2.3) and (2.15), giving the well-known expansion of the falling factorial as in [GKP94, Eq. (6.13)].

Equivalently, for the rising factorial,

α(α + 1) · · · (α+ r − 1) =r∑

k=0

[

r

k

]

αk. (4.12)

MacMahon’s polynomial (3.14) takes the elegant form

Sk,r =

(

k − 1

r − 1

)(

α

r

)

(k > 1),

after simplification. See also [Egg19, Sect. 3.2] for further identities in this binomial case.

As an example of the manipulations that are possible with Proposition 4.2, consider (4.9) with α = 1/2.

Also apply (4.5) and we obtain the two symmetric triples

(

(1 + t)1/2, (1− t)−1/2,1

2(1− t)

)

,

(

(1− t)−1/2, (1 + t)1/2,1

2(1 + t)

)

. (4.13)

Multiplying them produces

(E(t),H(t), P (t)) :=

(

(

1 + t

1− t

)1/2

,

(

1 + t

1− t

)1/2

,1

1− t2

)

. (4.14)

Dividing the first by the second in (4.13) produces

(E†(t),H†(t), P †(t)) :=

(

(1 + t)1/2(1− t)1/2, (1 + t)−1/2(1− t)−1/2,t

1− t2

)

. (4.15)

Then we easily find p2n−1 = 1, p2n = 0 and p†2n−1 = 0, p†2n = 1. Also

H†(t) = (1− t2)−1/2 implies h†2n = (−1)n(−1/2

n

)

=

(

n− 1/2

n

)

=1

22n

(

2n

n

)

,

as in [GKP94, (5.36), (5.37)], with h†2n−1 = 0. Similarly h2n−1 = h2n = 2−2n(2nn

)

. By the transition

formula between hk and pk we obtain the following identities, equivalent to those given in [Mac95, Ex. 15]:

2n∑

k=0

1

k!A2n,k

(

1

1, 0,

1

3, 0,

1

5, 0, . . .

)

=

2n∑

k=0

1

k!A2n,k

(

0,1

2, 0,

1

4, 0,

1

6, . . .

)

=1

22n

(

2n

n

)

. (4.16)

Symmetric triples correspond to homomorphisms φ from the ring Λ of symmetric functions to a simpler

ring, such as Q[α] in Example 4.3. As described in [Mac95, Sect. I.2] and [MR15, Chap. 2], symmetric

functions are formal power series in infinitely many variables x1, x2, . . . with coefficients in Z (or sometimes

Q). They have bounded degree and are unchanged under any finite permutation of these variables. Symmetric

polynomials appear as the special case when xn+1, xn+2, . . . are all set to 0. The elements ek, hk and pkdefined by (3.1), (3.2) and (3.5) with n = ∞ each form a basis of Λ, as in Theorem 3.1, so that φ may be

defined uniquely by specifying it on one of these bases. The homomorphism φ is called a specialization, see

[Sta99, Sect. 7.8], and the coefficients of the series in a symmetric triple (E(t),H(t), P (t)) correspond to

φ(ek), φ(hk) and φ(pk) in this picture. For each of our triple examples, the images of the other bases of Λ,

such as the Schur functions, can be considered. See for example [Mac95, Ex. 23, p. 58].

If E(t) is a polynomial of degree n, then x1, x2, . . . , xn may be recovered as the negative reciprocals of

the zeros of E(t), as in (3.6). If E(t) is not a polynomial then it can be instructive to look for zeros of E(t),or equivalently poles of H(t). However, E(t) is a formal series and not necessarily a well-defined function.

10

4.2 Changing variables in a triple

The following properties are based on (4.7) and will be useful, especially in section 5.3. If (E,H,P ) is a

symmetric triple then the change of variable t→ ct gives the new triple

(E, H, P ) = (E(ct),H(ct), cP (ct)) with ek = ckek, hk = ckhk, pk = ckpk. (4.17)

Similarly, t→ tm for a positive integer m produces

(E, H, P ) =(

E((−1)m−1tm), H(tm), m · tm−1P (tm))

with new coefficients that are 0 unless m divides the index, in which case

emk =

{

(−1)kek if m is even;

ek if m is odd,hmk = hk, pmk = m · pk.

This is reversed in the next easily verified result.

Proposition 4.4. Let m be a positive integer and (E,H,P ) a symmetric triple. Suppose the coefficients hkare zero whenever k is not a multiple of m. Then we can construct the new symmetric triple

(E, H, P ) =

(

E((−1)(m−1)/mt1/m), H(t1/m),t1/m−1

mP (t1/m)

)

with coefficients

ek =

{

(−1)kemk if m is even;

emk if m is odd,hk = hmk, pk =

pmk

m.

Theorem 2.5 in [Sun05] uses roots of unity to make pairs (H,P ) where H(t) only contains powers that

are multiples of m. We give another version of this result next. Recall that, as usual, R is an integral domain

containing Q.

Theorem 4.5. Let m be a positive integer and (E,H,P ) a symmetric triple with coefficients in R. Then, for

ω = e2πi/m, we can make the new symmetric triple

(E, H, P ) =

m−1∏

j=0

E(ωjt),m−1∏

j=0

H(ωjt),m−1∑

j=0

ωjP (ωjt)

(4.18)

with coefficients in R[ω]. These new coefficients ek, hk and pk are all zero unless m | k and in fact are all in

R. We have pk = m · pk when m | k.

Proof. Use (4.17) and the group operation (4.8) to check that (4.18) is a symmetric triple. Since E(ωt) =E(t), it follows that ekω

k = ek and hence that ek must be 0 unless m|k. Similarly for hk and pk. Also

∞∑

k=1

pktk−1 =

m−1∑

j=0

ωj∞∑

r=1

pr(ωjt)r−1 =

∞∑

r=1

prtr−1

m−1∑

j=0

ωrj

and hence pk = m · pk when m | k. In particular, pk ∈ R and the transition formulas now show that ek and

hk are also in R.

5 Symmetric triple examples

A simple way to build a symmetric triple is to start with any formal power series that has constant term 1.

Letting this be H(t), we produce(

1

H(−t) , H(t),H ′(t)

H(t)

)

, (5.1)

and this lets us study the reciprocal and the derivative of H(t). We will also include compositional inverses

in section 7.

11

5.1 Examples with equal coefficients

Proposition 5.1. We have ek = hk for all k > 1 if and only if P (t) is even.

Proof. Recall that e0 = h0 = 1. Then

E(t) = H(t) ⇐⇒ logE(t) = logH(t)

⇐⇒ d

dtlogE(t) =

d

dtlogH(t) ⇐⇒ P (−t) = P (t).

Proposition 5.2. Suppose hk = pk for all k > 1. Then for some constant c,

(E(t),H(t), P (t)) =

(

1 + ct,1

1− ct,

c

1− ct

)

.

Proof. We have

H(t) = 1 + tP (t) ⇐⇒ 1

E(−t) = 1 + tE′(−t)E(−t) ⇐⇒ 1 = E(t)− tE′(t).

Differentiating shows E′′(t) = 0 and hence E(t) = 1 + ct.

There is a similar result if hk = α · pk for all k > 1 and any fixed α. The next case is stated in [Mac95,

Ex. 16]. The Bernoulli numbers are defined with Bk := k![tk]t/(et − 1).

Proposition 5.3. Suppose ek = pk for all k > 1. Then for some constant c,

(E(t),H(t), P (t)) =

( −cte−ct − 1

,ect − 1

ct,1

t

(

−1 +−ct

e−ct − 1

))

,

with

ek = (−1)kckBk

k!, hk =

ck

(k + 1)!, pk = (−1)k

ckBk

k!. (5.2)

Proof. Expressing E(t) = 1 + tP (t) in terms of H implies H(−t) ddt(tH(t)) = H(t). Set F (t) := tH(t)

and we find

F (−t)F ′(t) = −F (t). (5.3)

This implies the symmetry

F ′(−t) = −F (−t)F (t)

=1

F ′(t). (5.4)

Differentiating (5.3) and simplifying with (5.4) shows

1 + F (t)F ′′(t)/F ′(t) = −F ′(t).

Differentiating this once more finds ddt(F

′′(t)/F ′(t)) = 0. Therefore

d2

dt2logF ′(t) = 0 =⇒ log F ′(t) = ct+ b =⇒ F ′(t) = exp(ct+ b)

for some constants c and b. But F ′(t) = H(t) + tH ′(t), making F ′(0) = 1 and so b = 0. Next, for some a,

F (t) = exp(ct)/c + a.

Finally, F (0) = 0 means a = −1/c and the result follows.

12

5.2 Stirling and harmonic numbers

Example 5.4 (Stirling numbers, power sums). For n > 1,

ek =

[

n+ 1

n+ 1− k

]

, hk =

{

n+ k

n

}

, pk = 1k + 2k + · · ·+ nk. (5.5)

Discussion. This example comes from setting xj = j in (3.6), (3.7) to get the generating functions

E(t) =

n∏

j=1

(1 + jt), H(t) =

n∏

j=1

1

1− jt, P (t) =

n∑

j=1

j

1− jt. (5.6)

Their coefficients in (5.5) follow from (4.12) with α = 1/t and (2.24). If k > n + 1 then ek must be 0 and

we may extend the definition of[nk

]

to be 0 when k 6 0. We now have many identities connecting these

numbers. For example P (t) = E′(−t)H(t) from (4.3) implies

pk =k∑

j=1

(−1)j−1jejhk−j (5.7)

so that, for n, k > 1,

1k + 2k + · · ·+ nk =k∑

j=1

(−1)j−1j

[

n+ 1

n+ 1− j

]{

n+ k − j

n

}

,

as in [Egg19, Prop. 3.17], (including the missing j factor). The transition formula pk ↔ hk also shows that

1k + 2k + · · · + nk =k∑

j=1

(−1)j−1 k

jAk,j

({

n+ 1

n

}

,

{

n+ 2

n

}

, · · ·)

.

By (4.4) we can see[

n+ 1

n

]

=

{

n+ 1

n

}

=

(

n+ 1

2

)

.

In 1796 Kramp found formulas for the symmetric polynomials ek and hk in (5.5). This is discussed by Knuth

[Knu92, p. 413], and in our notation they give, for m,k > 0,

{

m+ k

k

}

= m!

(

m+ k

m

) m∑

j=0

(

k

j

)

Am,j

(

1

2!,1

3!,1

4!, . . .

)

, (5.8)

[

m+ k

k

]

= m!

(

m+ k

m

) m∑

j=0

(

k

j

)

Am,j

(

1

2,1

3,1

4, . . .

)

. (5.9)

We derive them after (9.23). Recall the simpler identities (2.14), (2.15).

In general, as described in [Kon00], there is a natural combinatorial interpretation of ek and hk when

x1, x2, . . . , xn are nonnegative integers. We have this situation in (4.1) with xj = 1, Example 5.4 with

xj = j and Example 6.2 with xj = qj−1 if q ∈ Z>2. Suppose there are n boxes with xj different balls

in the box with index j. Then ek = ek(x1, . . . , xn) is the number of ways to choose k balls from these

boxes, where we mean one ball from each of k different boxes and the order of the boxes is unimportant.

The number hk = hk(x1, . . . , xn) gives a similar count, but the boxes do not have to be different. We may

extend this to pk = pk(x1, . . . , xn), giving the number of ways to choose k balls, with replacement, from a

single box. Also in this setup, An,k(x1, x2, . . . ) counts the number of ways to choose one ball from each of

k (not necessarily different) boxes with indices summing to n, and where the order of the boxes is important.

13

Proposition 5.5 (The generalized Pascal identities). For n, k > 2 and arbitrary x1, . . . , xn,

ek(x1, . . . , xn) = ek(x1, . . . , xn−1) + xn · ek−1(x1, . . . , xn−1), (5.10)

hk(x1, . . . , xn) = hk(x1, . . . , xn−1) + xn · hk−1(x1, . . . , xn), (5.11)

pk(x1, . . . , xn) = pk(x1, . . . , xn−1) + xn · pk−1(0, . . . , 0, xn). (5.12)

Equation (5.12) is clear and (5.10), (5.11) are easily demonstrated using the generating functions (3.6).

In the case of the Stirling numbers in Example 5.4, (5.10) and (5.11) show that, for n > 3, k > 2,

[

n+ 1

k

]

= n

[

n

k

]

+

[

n

k − 1

]

,

{

n+ 1

k

}

= k

[

n

k

]

+

[

n

k − 1

]

. (5.13)

These relations allow the Stirling numbers to be extended to all integers n and k. In fact (5.13) combined

with the initial conditions[

n0

]

={

n0

}

= δn,0 and[

0k

]

={

0k

}

= δk,0 gives an elegant alternate definition for

them. The remarkable duality[nk

]

={−k−n

}

can also be observed; see [GKP94, pp. 266, 267] and [Knu92].

Example 5.6 (Harmonic numbers). For n > 1,

ek =

[

n+ 1

k + 1

]

/n!, hk =

∥

∥

∥

∥

n

k

∥

∥

∥

∥

, pk = H(k)n . (5.14)

Discussion. Recall the definition (1.13) for H(k)n . Here we are setting xj = 1/j for 1 6 j 6 n. The new

notation∥

∥

nk

∥

∥ is meant to suggest multiset coefficients, as in((

nk

))

:=(

n+k−1k

)

, but of an unusual harmonic

kind. By analogy with the above discussion,∥

∥

nk

∥

∥ represents the number of ways to choose k balls, with

replacement, from n boxes where the jth box contains 1/j balls. The generating functions are

E(t) =1

n!

n∏

j=1

(j + t), H(t) = n!

n∏

j=1

1

j − t, P (t) =

n∑

j=1

1

j − t, (5.15)

with the formula for ek in (5.14) coming from (4.12).

Proposition 5.7. We have

∥

∥

∥

∥

n

k

∥

∥

∥

∥

=

n∑

j=1

(−1)j−1

(

n

j

)

1

jk(n, k ∈ Z>1). (5.16)

Proof. By (2.23),

H(t)

−t =

n∑

j=0

(

n

j

)

(−1)j

j − t= −1

t+

n∑

j=1

(−1)j

j

(

n

j

) ∞∑

r=0

tr

jr,

and rearranging and comparing powers of t gives (5.16).

See [Bat17, Thm. 2.1] for another proof and a discussion of the history of this result. By (5.11), for

n, k > 2, we have (after multiplying through by n) the Pascal identity

n

∥

∥

∥

∥

n

k

∥

∥

∥

∥

= n

∥

∥

∥

∥

n− 1

k

∥

∥

∥

∥

+

∥

∥

∥

∥

n

k − 1

∥

∥

∥

∥

. (5.17)

Theorem 5.8. Suppose we define the harmonic multiset numbers∥

∥

nk

∥

∥ using the recursion (5.17) and the

initial conditions∥

∥

∥

∥

n

−1

∥

∥

∥

∥

= δn,1,

∥

∥

∥

∥

−1

k

∥

∥

∥

∥

= δk,1. (5.18)

Then∥

∥

nk

∥

∥ is well-defined for all n, k ∈ Z and agrees with our previous definition. The Stirling numbers are

special cases:

{

n

k

}

=(−1)k+1

k!

∥

∥

∥

∥

k

−n

∥

∥

∥

∥

,

[

n

k

]

= (n− 1)!

∥

∥

∥

∥

−nk

∥

∥

∥

∥

(n > 1, k > 0). (5.19)

14

Proof. For clarity, let S(n, k) denote the numbers defined recursively by (5.17) and (5.18). We first note

that (5.17) implies S(0, k) = 0 for all k and this does not conflict with (5.18). The recursion then gives

S(1, k) = 1 for all k, and next that S(n, 0) = 1 for all n > 1.

Let T (n, k) denote the right side of (5.16). Check that T (1, k) = T (n, 0) = 1 for n > 1 and all k,

agreeing with S(n, k). It follows from the usual Pascal identity that T (n, k) satisfies the same recursive

relation as S(n, k) for n > 2. Hence S(n, k) = T (n, k) for all n, k with n > 1. That means S(n, k) =∥

∥

nk

∥

∥

for all n, k > 1 and the new definition agrees with the old one. The formula (2.21) now shows the left

identity in (5.19) is true.

n

−4

−2

0

2

4

k0 2 4−2−4

∥

∥

∥

∥

n

k

∥

∥

∥

∥

(−1)n+1n!

{−kn

}

1

(−n− 1)!

[−nk

]

Figure 1: The harmonic multiset numbers∥

∥

nk

∥

∥ for −5 6 n, k 6 5

The recursion (5.17) gives no information about S(−1, k), so we may impose the initial condition

S(−1, k) = δk,1 from (5.18). Then recursively we obtain S(n, k) = 0 for all n, k 6 0, (and there is no

conflict with S(n,−1) = δn,1). Now set U(n, k) to be (n − 1)!S(−n, k) for n > 1, k > 0. We have

U(n, 0) = 0 =[n0

]

for n > 1 and U(1, k) = δk,1 =[1k

]

for k > 0. Also U(n, k) satisfies the same recursion

as[nk

]

in (5.13). The right identity in (5.19) follows.

The initial conditions (5.18) at n = ±1, k = ∓1 are highlighted in Figure 1. They are the natural seeds

to obtain the Stirling numbers in quadrants 2 and 4 since the recursions satisfied by the Stirling numbers

and the harmonic multiset numbers are essentially just rotated versions of the same relation. The reason that

quadrant 1 is full of positive numbers instead of zeros, (compare with the top right of [GKP94, Table 267]

in the usual Stirling case), is the slight difference in (5.17) at n = 0 from the Stirling relations, and the fact

that∥

∥

00

∥

∥ is 0 instead of 1.

With our extended definition of∥

∥

nk

∥

∥, the symmetric triple (5.14) becomes

ek =

∥

∥

∥

∥

−n− 1

k + 1

∥

∥

∥

∥

, hk =

∥

∥

∥

∥

n

k

∥

∥

∥

∥

, pk = H(k)n . (5.20)

Also by (4.4),[

n+ 1

2

]

/n! =

∥

∥

∥

∥

−n− 1

2

∥

∥

∥

∥

=

∥

∥

∥

∥

n

1

∥

∥

∥

∥

= Hn (n > 1). (5.21)

With the proof of Theorem 5.8 we see that (5.16) is valid for all n > 0 and k ∈ Z, so that

∥

∥

∥

∥

0

k

∥

∥

∥

∥

= 0,

∥

∥

∥

∥

1

k

∥

∥

∥

∥

= 1,

∥

∥

∥

∥

2

k

∥

∥

∥

∥

= 2− 1

2k,

∥

∥

∥

∥

3

k

∥

∥

∥

∥

= 3− 3

2k+

1

3k(k ∈ Z). (5.22)

With (5.14) we may express∥

∥

nk

∥

∥ for n, k > 1 in terms of the Stirling cycle numbers or the higher order

harmonic numbers using the transition formulas. This last was also done in [FS95, p. 7], [Bat17, Sect. 3]

15

and [Ses17, Sect. 3.1]. We obtain

∥

∥

∥

∥

n

k

∥

∥

∥

∥

=k∑

j=0

1

j!Ak,j

(

H(1)n

1,H

(2)n

2,H

(3)n

3, · · ·

)

(n > 1, k > 0), (5.23)

with, for n > 1,

∥

∥

∥

∥

n

0

∥

∥

∥

∥

= 1,

∥

∥

∥

∥

n

1

∥

∥

∥

∥

= Hn,

∥

∥

∥

∥

n

2

∥

∥

∥

∥

=H2

n

2+H

(2)n

2,

∥

∥

∥

∥

n

3

∥

∥

∥

∥

=H3

n

6+HnH

(2)n

2+H

(3)n

3. (5.24)

We also find

H(k)n = k

k∑

j=1

(−1)k−j

j · (n!)j Ak,j

([

n+ 1

2

]

,

[

n+ 1

3

]

, · · ·)

, (5.25)

[

n+ 1

k + 1

]

= n!

k∑

j=0

(−1)k−j

j!Ak,j

(

H(1)n

1,H

(2)n

2,H

(3)n

3, · · ·

)

, (5.26)

with, for instance,[

n+ 1

3

]

=n!

2

(

H2n −H(2)

n

)

,

[

n+ 1

4

]

=n!

6

(

H3n − 3HnH

(2)n + 2H(3)

n

)

. (5.27)

For further interesting properties of∥

∥

nk

∥

∥when n, k > 1, including connections to polylogarithms, see [Bat17]

where they are denoted Sn(k) and [Ses17] where they are called Roman harmonic numbers with notation

c(k)n , based on earlier work by Loeb, Rota and Roman. Up to a sign and factorial, they are also the negative-

positive Stirling numbers in [Bra06]. The case k < 0 is briefly mentioned in [Bat17] with the connection

to the Stirling subset numbers{nk

}

noted; n < 0 is not discussed. The negative k case is not considered

in [Ses17], though c(k)n with negative n is studied and the Stirling cycle numbers

[nk

]

are found. The initial

conditions and recursion relation for the Roman numbers are not quite the same as (5.18) and (5.17), differing

when n = 0. The result is that c(0)0 = 1, differing from

∥

∥

00

∥

∥ = 0, and for n 6 −1, k > 0 we have

c(k)n = −

∥

∥

nk

∥

∥ 6 0.

In summary, the array of numbers∥

∥

nk

∥

∥ appear naturally in the symmetric triple (5.14), and Theorem 5.8

gives a very convenient way to describe them while also including both kinds of Stirling numbers.

5.3 Examples of exponential type

The first example in this section is [Mac95, Ex. 2]. The standard definitions of Hermite, Bernoulli and

Eulerian polynomials can be seen from the succeeding examples.

Example 5.9 (The exponential function).

(

ext, ext, x)

with ek =xk

k!, hk =

xk

k!, pk = x · δk,1. (5.28)

Example 5.10 (Hermite polynomials Hk(x)).

(

e2xt+t2 , e2xt−t2 , 2x− 2t)

with ek =Hk(ix)

ikk!, hk =

Hk(x)

k!, (5.29)

and pk = 2x · δk,1 − 2 · δk,2.

Example 5.11 (Bernoulli numbers Bk).

ek =(−1)k

(k + 1)!, hk =

Bk

k!, pk = (−1)k−1Bk

k!, (5.30)

coming from(

e−t − 1

−t ,t

et − 1,1

t

(

1− −te−t − 1

))

. (5.31)

16

Discussion. This example corresponds to the Newton-Euler pairs in [Sun05, Exs. 16, 21] and matches (5.2)

with c = −1 using (4.5). The transition formula hk ↔ ek and (2.10) give the simple identities

Bk

k!=

k∑

j=0

(−1)jAk,j

(

1

2!,1

3!,1

4!, · · ·

)

= (−1)k

∣

∣

∣

∣

∣

∣

∣

∣

∣

1/2! 1/1!1/3! 1/2! 1/1!

.... . .

1/(k+1)! 1/k! . . . 1/2!

∣

∣

∣

∣

∣

∣

∣

∣

∣

. (5.32)

Many variations of (5.30) are possible. For instance, let t → −t in (5.31). Then divide that triple by the

same but with t→ 2t to get

ek = (2− 2k+2)Bk+1

(k + 1)!, hk =

(−1)k

2 · k! +δk,02, pk = (2k − 1)

Bk

k!, (5.33)

corresponding to [Sun05, Ex. 20].

Multiplying Example 5.9 by Example 5.11, in the sense of (4.8), produces:

Example 5.12 (Bernoulli polynomials Bk(x)).

ek =xk+1 − (x− 1)k+1

(k + 1)!, hk =

Bk(x)

k!, pk = (−1)k−1Bk

k!+ x · δk,1. (5.34)

Example 5.13 (Eulerian polynomials Ak(x)).

ek =(1− x)k−1

k!+ δk,0

x

x− 1, hk =

Ak(x)

k!, pk =

xAk−1(x)

(k − 1)!+ δk,1(1− x), (5.35)

coming from(

x− e(1−x)t

x− 1,

x− 1

x− e(x−1)t, x

x− 1

x− e(x−1)t− x+ 1

)

. (5.36)

Discussion. This example is based on [Bre93, Sect. 4]. See [Com74, Sect. 6.5] for details about the Eulerian

polynomials (with a slightly different normalization there). They satisfy

∞∑

n=1

nkxn =xAk(x)

(1− x)k+1(k ∈ Z>0), (5.37)

with A0(x) = A1(x) = 1, A2(x) = 1 + x and A3(x) = 1 + 4x + x2. Euler used them to compute the

Riemann zeta values ζ(−k) formally, but correctly. With the hk ↔ ek transition formula and (2.3), (2.4),

Ak(x) = k!

k∑

j=0

Ak,j

(

1,x− 1

2!,(x− 1)2

3!,(x− 1)3

4!, · · ·

)

= k!

k∑

j=0

(x− 1)k−jAk,j

(

1

1!,1

2!,1

3!,1

4!, · · ·

)

=

k∑

j=0

(x− 1)k−jj!

{

k

j

}

,

using (2.14) for the last equality. This proves Frobenius’s 1910 formula for Ak(x); see [Com74, p. 244].

Example 5.13 and (5.35) are just the starting point in [MR15, Chap. 3] for an in-depth study of permutation

statistics.

Example 5.14 (Cosine).

(sec t, cos t,− tan t) with e2k =U2k

(2k)!, h2k =

(−1)k

(2k)!, p2k = − U2k−1

(2k − 1)!. (5.38)

17

Discussion. See also [Sun05, Ex. 19]. The odd indexed elements are zero in this example and the next two.

Here Un counts the number of alternating permutations of n objects. In terms of Bernoulli polynomials and

numbers we have

e2k = (−1)k−124k+2B2k+1(1/4)

(2k + 1)!, p2k = (−1)k

(

24k − 22k) B2k

(2k)!.

The U2k are called secant numbers and the U2k−1 are tangent numbers. See [Rad73, Eq. (14.3)], [GKP94,

p. 287], [MR15, Thm. 3.5] for more information. By the ek ↔ hk transition formula,

U2k

(2k)!=

2k∑

j=0

(−1)k−jA2k,j

(

0,1

2!, 0,

1

4!, 0, · · ·

)

.

Example 5.15 (Sine).

e2k = (−1)k−1

(

22k − 2)

B2k

(2k)!, h2k =

(−1)k

(2k + 1)!, p2k = (−1)k

22kB2k

(2k)!. (5.39)

coming from (t csc t, (sin t)/t,−1/t+ cot t).

Example 5.15 relates to [Sun05, Exs. 18, 22]. Euler’s formula also shows the Riemann zeta connection

p2k = −2π−2kζ(2k). Dividing the sine triple by the cosine triple gives:

Example 5.16 (Tangent).(

t cot t,tan t

t, −1

t+ tan t+ cot t

)

(5.40)

and

e2k = (−1)k22kB2k

(2k)!, h2k =

U2k+1

(2k + 1)!, p2k = (−1)k(22k+1 − 24k)

B2k

(2k)!. (5.41)

The odd indexed coefficients are zero in Examples 5.14, 5.15 and 5.16. By Proposition 4.4 such triples

(E,H,P ) may be rewritten as

(E(t), H(t), P (t)) =

(

E(i√t), H(

√t),

1

2√tP (

√t)

)

, (5.42)

with

ek = (−1)ke2k, hk = h2k, pk = p2k/2.

Example 5.17 (Hyperbolic sine, squared).

(

t

4sin−2

(√t

2

)

,4

tsinh2

(√t

2

)

, −1

t+

1

2√tcoth

(√t

2

))

(5.43)

with

ek = (−1)k(1− 2k)B2k

(2k)!, hk =

2

(2k + 2)!, pk =

B2k

(2k)!. (5.44)

Discussion. This example extends the work in [Saa93, p. 101] and [Sun05, Ex. 17]. To obtain pk =B2k/(2k)! we start with Example 5.15, (alternatively Example 5.12 with x = 1/2 could be used). As in

section 4.2, let t→ t/2 and apply (5.42). Then let t→ −t so that√t→ i

√t and the hyperbolic versions of

sin and cot are needed. Squaring the resulting triple yields (5.43). The formula in (5.44) for ek comes from

the coefficients we have already seen for cot(t) since cot′(t) = − sin−2(t). With sinh2(t) = (cosh(t)−1)/2,

the well-known expansions of cosh(t) and coth(t) give hk and pk.

The pk ↔ hk transition formula implies the identity, for k > 1,

B2k

k · (2k)! = −k∑

j=1

(−2)j

jAk,j

(

1

4!,1

6!,1

8!, · · ·

)

. (5.45)

18

This is equivalent to [Saa93, Eq. LXXVI, p. 101] after a correction. In the same way, from Examples 5.14,

5.15 after applying (5.42),

(22k − 1)22k−1 B2k

k · (2k)! = −k∑

j=1

(−1)j

jAk,j

(

1

2!,1

4!,1

6!, · · ·

)

, (5.46)

22k−1 B2k

k · (2k)! = −k∑

j=1

(−1)j

jAk,j

(

1

3!,1

5!,1

7!, · · ·

)

. (5.47)

These identities (5.46) and (5.47) are due to Stern [Ste43, Eqs. (5), (9)].

5.4 Hypergeometric summation identities

We would like to let n → ∞ in the harmonic number Example 5.6. The convergence issue at k = 1 can be

fixed by dividing by the triple (nt, nt, log n) first. In fact it is simpler to start with the answer, given next,

where the digamma function ψ(t) is defined as Γ′(t)/Γ(t) and pk comes from the expansion [AS64, Eq.

(6.3.14)] for ψ(1 − t), valid for |t| < 1. We also use the Euler reflection formula for the gamma function in

[AAR99, Thm. 1.2.1].

Example 5.18 (Gamma function, zeta values).

(

sinπt

πtΓ(1− t), Γ(1− t), −ψ(1− t)

)

with pk = ζ(k) (5.48)

for k > 2 and p1 = γ, Euler’s constant.

Discussion. The transition formulas can be used to find ek and hk with, for example,

hk =

k∑

j=0

1

j!Ak,j

(

γ,ζ(2)

2,ζ(3)

3, · · ·

)

.

Examples 5.6 and 5.18 give the relations needed to develop the hypergeometric summation identities of

Chu [Chu97]. The idea is to expand hypergeometric summation formulas into power series and compare

coefficients on both sides. We may illustrate this with Gauss’s summation formula, [AAR99, Thm. 2.2.2]:

∞∑

n=0

(x)n(y)nn!(1− z)n

=: 2F1(x, y; 1 − z; 1) =Γ(1− z)Γ(1− x− y − z)

Γ(1− x− z)Γ(1 − y − z), (5.49)

where (x)n means x(x+ 1) · · · (x+ n− 1). With pk as in Example 5.18, the right side of (5.49) is

exp

(

∞∑

k=1

pkk

[

zk + (x+ y + z)k − (x+ z)k − (y + z)k]

)

. (5.50)

The left side is

1 + xy∞∑

n=0

(

1 + x1

)

· · ·(

1 + xn

)

·(

1 + y1

)

· · ·(

1 + yn

)

(n+ 1)2(

1− z1

)

· · ·(

1− zn+1

) (5.51)

and we recognize E and H from Example 5.6.

To make the example simpler, let y = x and z = 0. The ek ↔ pk transition formula for the square of the

triple in Example 5.6 gives

[xm]((

1 +x

1

)

· · ·(

1 +x

n

))2=

m∑

ℓ=0

(−1)m−ℓ 1

ℓ!Am,ℓ

(

2H(1)n

1,2H

(2)n

2,2H

(3)n

3, · · ·

)

.

Comparing this with the expansion of (5.50) by (2.7) proves the following.

19

Proposition 5.19. For m > 0,

m∑

ℓ=0

(−1)m−ℓ 2ℓ

ℓ!

∞∑

n=0

1

(n+ 1)2Am,ℓ

(

H(1)n

1,H

(2)n

2,H

(3)n

3, · · ·

)

=

m+2∑

j=0

1

j!Am+2,j

(

0, (22 − 2)ζ(2)

2, (23 − 2)

ζ(3)

3, (24 − 2)

ζ(4)

4, · · ·

)

.

This generates a harmonic number identity for each m. When m = 1, 2, 3 we find

∞∑

n=0

Hn

(n+ 1)2= ζ(3), (5.52)

∞∑

n=0

2H2n −H

(2)n

(n+ 1)2=

1

2ζ(2) +

7

2ζ(4) =

19π4

360, (5.53)

∞∑

n=0

2H3n − 3HnH

(2)n +H

(3)n

(n+ 1)2= 3ζ(2)ζ(3) + ζ(5). (5.54)

The identity (5.52) is [Chu97, Eq. (1.1a)], and due to Euler, with the left side being the multiple zeta value

ζ(2, 1). See [Chu97], and the many subsequent papers on this topic, for examples of the wide variety of

identities that can be produced in this way. Similar identities appear in [Bat17]. The novelty of Proposition

5.19 is that it shows the structure of all the identities produced at once, at least in this one-variable example.

The paper [Hof17] has more on the connections between symmetric functions and multiple zeta values.

The already very elegant identity [Chu97, Eq. (5.5)] has an even more succinct statement with our

notation from Theorem 5.8:

Proposition 5.20. For all integers p > 0 and q > 1,

∞∑

n=1

1

n2

∥

∥

∥

∥

n

p

∥

∥

∥

∥

·∥

∥

∥

∥

−nq

∥

∥

∥

∥

=

(

p+ q

p

)

ζ(p+ q + 1). (5.55)

A version of the proof from [Chu97] is included next, since we have already done the groundwork. This

result is also Theorem 4 of [Hof17], where it has a different proof.

Proof. Subtracting 1 from (5.51), dividing by x and letting x→ 0 makes

y

∞∑

n=0

(

1 + y1

)

· · ·(

1 + yn

)

(n+ 1)2(

1− z1

)

· · ·(

1− zn+1

) =

n∑

u=0

∞∑

v=0

yu+1zv∞∑

n=0

eu(n)hv(n+ 1)

(n+ 1)2(5.56)

with ek(n) and hk(n) from (5.14) or equivalently (5.20). Doing the same on the right of (5.49) yields

−Γ′(1− y − z)

Γ(1− y − z)+

Γ′(1− z)

Γ(1− z)= ψ(1− y − z) + ψ(1− z)

=

∞∑

k=1

pk

(

(y + z)k−1 − zk−1)

, (5.57)

with pk from Example 5.18. Comparing powers of y and z in (5.56) and (5.57) finishes the argument.

6 q-Series

6.1 q-Binomial coefficients

Define

(a; q)n := (1− a)(1− aq)(1− aq2) · · · (1− aqn−1).

The next result is fundamental and due to Cauchy.

20

Theorem 6.1 (The q-binomial theorem). For |q|, |t| < 1 and all a,

∞∑

m=0

(a; q)m(q; q)m

tm =(at; q)∞(t; q)∞

. (6.1)

See [And98, Thm. 2.1] or [AAR99, Thm. 10.2.1] for the short proof. The q-binomial coefficient is

defined to be(

n

k

)

q

:=(q; q)n

(q; q)k(q; q)n−k.

Then, as in [AAR99, Cor. 10.2.2], the special cases a = q−n and a = qn of Theorem 6.1 imply

(t; q)n =

n∑

k=0

(−1)kq(k

2)(

n

k

)

q

tk,1

(t; q)n=

n∑

k=0

(

n+ k − 1

k

)

q

tk. (6.2)

Therefore, choosing xj = qj−1 in (3.6) and (3.7) yields the next triple.

Example 6.2 (q-Binomial coefficients). For n > 1,

ek = q(k

2)(

n

k

)

q

, hk =

(

n+ k − 1

k

)

q

, pk =1− qnk

1− qk. (6.3)

Discussion. The convolution and transition formulas give relations between ek, hk and pk as rational func-

tions of q. For example, when n = 2 we find the polynomial identity

m∑

k=0

1

k!Am,k

(

1 + q

1,1 + q2

2,1 + q3

3, . . .

)

= 1 + q + q2 + · · · + qm, (6.4)

from the hk ↔ pk transition formula. Taking the limit as q → 1− in Example 6.2 gives (4.1), a special case

of Example 4.3.

We may also take the limit as n→ ∞ of Example 6.2 when |q| < 1.

Example 6.3 (Limit of q-binomial coefficients).

ek = q(k

2) 1

(q; q)k, hk =

1

(q; q)k, pk =

1

1− qk. (6.5)

Examples 6.2 and 6.3 are given in [Mac95, Ex. 3, Ex. 4]. They are special cases of the next symmetric

triple, which is [Mac95, Ex. 5].

Example 6.4 (q-Series with two-parameters).

ek =(−1)k

(q; q)k

k−1∏

j=0

(b− aqj), hk =1

(q; q)k

k−1∏

j=0

(a− bqj), pk =ak − bk

1− qk. (6.6)

Discussion. This follows from setting

H(t) :=(bt; q)∞(at; q)∞

=

∞∑

m=0

am(b/a; q)m(q; q)m

tm (6.7)

where the equality in (6.7) is obtained by letting t→ at, a→ b/a in Theorem 6.1. A calculation finds

log(t; q)∞ =

∞∑

j=0

log(1− tqj) = −∞∑

j=0

∞∑

m=1

(tqj)m

m

= −∞∑

m=1

tm

m

∞∑

j=0

qjm = −∞∑

m=1

tm

m(1− qm). (6.8)

21

Hence

P (t) =d

dtlogH(t) =

∞∑

m=1

am − bm

1− qmtm−1,

and this verifies the formula for pk in (6.6).

For future use, the Jacobi triple product identity is

(t; q)∞(q/t; q)∞(q; q)∞ =∑

k∈Z

(−1)kq(k

2)tk, (6.9)

and can be made to follow from the left identity in (6.2); see [AAR99, Thm. 10.4.1].

6.2 Partitions

For S ⊆ Z>1 and any r, the generating function

PS,r(q) :=∏

j∈S

1

(1− qj)r=

∞∑

n=0

pS,r(n)qn, (6.10)

defines the numbers pS,r(n). In the common case r = 1, pS,1(n) counts the number of partitions of n with

parts in S. When r ∈ Z>1, pS,r(n) is the number of r-color partitions of n with parts in S. These are

partitions of n where each part is colored and the order of the colored parts does not matter. Equivalently,

pS,r(n) counts the number of r-component multipartitions of n with parts in S. These are r-tuples of usual

partitions that sum to n in total.

We have

logPS,r(q) = −r∑

j∈S

log(1− qj) = r∑

j∈S

∞∑

n=1

qnj

n

= r

∞∑

m=1

qm∑

j∈S, j|m

j

m= r

∞∑

m=1

σS(m)qm

mfor σS(m) :=

∑

j∈S, j|m

j. (6.11)

Therefored

dqlog PS,r(q) = r

∞∑

m=1

σS(m)qm−1

and we obtain from PS,r(q) the following symmetric triple.

Example 6.5 (Partitions, divisors).

en = (−1)npS,−r(n), hn = pS,r(n), pn = r · σS(n). (6.12)

Discussion. Some special cases of this may be examined more closely; see also [Mac95, Ex. 6], [Sun05,

Exs. 5 - 9]. These cases all have S = Z>1 and S is omitted from the notation. When r = 1 we find the

symmetric triple

en = (−1)nc(n), hn = p(n), pn = σ(n) (6.13)

from the introduction. Recall Jacobi’s identity, [AAR99, Eq. (10.4.9)]:

∞∏

j=1

(1− qj)3 =∞∑

m=0

d(m)qm, with d(m) :=

{

(−1)r(2r + 1) if m = r(r+1)2 for r ∈ Z>0,

0 otherwise.

(6.14)

Hence, for r = 3 we obtain the symmetric triple

en = (−1)nd(n), hn = p3(n), pn = 3σ(n). (6.15)

22

For r = 24 and Ramanujan’s tau function τ(n), see [O’S, Eq. (5.3)], we find

en = (−1)nτ(n+ 1), hn = p24(n), pn = 24σ(n). (6.16)

Some of the many identities for these last triples appear in section 5 of [O’S].

The transition formulas expressing en in terms of pn above are a special cases of Bell’s following result.

Theorem 6.6 (Partition polynomials). [Bel28] Fix a positive integer s and sets of positive integers C1,

C2, . . . , Cs. Also fix complex numbers a1, . . . , as and z1, . . . , zs. Then we have the formal power series

expansions∏

j=1

∏

nj∈Cj

(1− zj · qnj)aj =

∞∑

n=0

Ψ(n)qn (6.17)

where

Ψ(n) =

n∑

k=0

1

k!An,k

(

ψ(1)

1,ψ(2)

2, . . .

)

, ψ(n) = −s∑

j=1

aj∑

d|n, d∈Cj

d · zn/dj . (6.18)

Proof. Take the logarithm of the product in (6.17) and expand as in (6.11). Exponentiating with (2.7) then

completes the proof.

Bell termed Ψ(n), given in (6.18), the partition polynomial of rank n in the variables z1, . . . , zs and

associated toC1, . . . , Cs and a1, . . . , as. From (1.5) we see that Ψ(n) is of degree at most nwhen considered

as a polynomial in each of z1, . . . , zs or a1, . . . , as or ψ(1), . . . , ψ(n). Theorem 6.6 shows that, for any

identity of the form (6.17), the coefficients Ψ(n) can be expressed as a polynomial of divisor sums.

As an example, take Ramanujan’s famous identity [Rad73, Eq. (105.1)]

5(q5; q5)5∞(q; q)6∞

=

∞∑

n=0

p(5n+ 4)qn, (6.19)

showing that 5 divides p(5n + 4). Applying the theorem with C1 = {5, 10, . . . }, C2 = {1, 2, . . . }, a1 = 5,

a2 = −6 and z1 = z2 = 1 means

ψ(n) = −5∑

d|n, 5|d

d+ 6∑

d|n

d = σ(n) + 5ω5(n) (6.20)

where, in general, ωr(n) is defined to be the sum of all the divisors of n that are not multiples of r. Then

p(5n + 4) = 5

n∑

k=0

1

k!An,k

(

σ(1) + 5ω5(1)

1,σ(2) + 5ω5(2)

2, . . .

)

. (6.21)

An equivalent formula to (6.21) is the main result of [BBT09].

For the corresponding symmetric triple, factor the reciprocal of the product in (6.19) into

(q; q)6∞(q5; q5)5∞

= (q; q)3∞ · (q; q)3∞ · 1

(q5; q5)5∞. (6.22)

The coefficients do not seem to have been studied, but we may express them with (6.22), (6.10) and (6.14)

as

f(n) := [qn](q; q)6∞(q5; q5)5∞

=∑

i+j+5k=n

d(i) · d(j) · p5(k). (6.23)

Example 6.7 (Partitions modulo 5).

en = (−1)nf(n), hn = p(5n+ 4)/5, pn = σ(n) + 5ω5(n).

23

6.3 Sums of squares or triangular numbers

With the theta function definition

ϕ(q) :=∑

n∈Z

qn2

, then ϕ(−q) =∑

n∈Z

(−1)nqn2

=

∞∏

j=1

1− qj

1 + qj(6.24)

by (6.9) when t → q, q → q2. We may set E(q) = ϕ(q) and H(q) = 1/ϕ(−q). Directly as in (6.11), or by

Theorem 6.6,

logH(q) =

∞∑

m=1

(σ(m) + ω2(m))qm

m

where ω2(n) is the sum of the odd divisors of n. The series H(q) is the generating function for the number

p(n) of overpartitions of n. These are partitions where the first appearance of a part of each size may or may

not be overlined.

Example 6.8 (Squares, overpartitions). For n > 1,

en =

{

2 if n = m2;

0 if n 6= m2,hn = p(n), pn = σ(n) + ω2(n). (6.25)

We may generalize Example 6.8 by taking the rth power. Write

ϕ(q)r =

∞∑

n=0

Nr(n)qn,

∞∏

j=1

(

1 + qj

1− qj

)r

=

∞∑

n=0

pr(n)qn (6.26)

for any r. When r is a positive integer, Nr(n) is the number of ways to express n as a sum of squares

of r integers (including the order of the integers and their signs), while pr(n) naturally counts r-colored

overpartitions. These are r-colored partitions where the first appearance of a part of each size and color may

be overlined.

Example 6.9 (Representations as sums of squares).

en = Nr(n), hn = pr(n), pn = r(σ(n) + ω2(n)). (6.27)

Discussion. This is Example 1.4 again. See also [Mac95, Ex. 22], [Sun05, Ex. 10]. By the transition

formulas, we obtain (1.12) and

Nr(n) =

n∑

k=0

(−1)n−kAn,k(pr(1), pr(2), pr(3), . . . ). (6.28)

A nice simplification of (6.28) is possible, which may also be inverted:

Proposition 6.10. For n ∈ Z>0 and arbitrary r,

Nr(n) =

n∑

j=0

(−1)n−j

(

n+ 1

j + 1

)

prj(n), pr(n) =

n∑

j=0

(−1)n−j

(

n+ 1

j + 1

)

Nrj(n). (6.29)

Proof. We have

An,k(1, pr(1), pr(2), pr(3), · · · ) = [qn]qk∞∏

j=1

(

1 + qj

1− qj

)rk

= prk(n− k).

24

Hence, by (2.16),

An,k(pr(1), pr(2), . . . ) =

k∑

j=0

(−1)k−j

(

k

j

)

An+j,j(1, pr(1), pr(2), . . . )

=

k∑

j=0

(−1)k−j

(

k

j

)

prk(n− k).

Using this in (6.28) and simplifying with an elementary binomial identity ([GKP94, p. 171]) gives the left

formula in (6.29). The right one is proved the same way.

The r = 1 case of the right formula of Proposition 6.10 has recently appeared in [Jha21] giving a simple

link between overpartitions and representations as sums of squares:

p(n) =

n∑

j=0

(−1)n−j

(

n+ 1

j + 1

)

Nj(n). (6.30)

There are formulas for σ(n) + ω2(n) of the same kind as (6.29), (see (6.35)). An example of a convolution

identity here is

r(σ(n) + ω2(n)) =

n∑

j=1

(−1)j−1j ·Nr(j) · pr(n − j), (6.31)

from (5.7).

Next, take ψ(q) :=∑∞

n=0 qn(n+1)/2 with

1

ψ(−q) =∞∏

j=1

1 + q2j−1

1− q2j=

∞∏

j=1

1

(1− q4j)(1− q2j−1=

∏

j>1, j 6≡2 mod 4

1

1− qj.

Let p(n) be the number of partitions of nwhere no parts have size ≡ 2 mod 4, and recall ωr(n) defined after

(6.20). Then similarly to Example 6.8 we find the symmetric triple:

Example 6.11 (Triangular numbers).

en =

{

1 if n = m(m+1)2 ;

0 otherwise,hn = p(n), pn = σ(n) + ω2(n)− ω4(n). (6.32)

Raising this to the power r, write

ψ(q)r =

∞∑

n=0

∆r(n)qn,

∏

j>1, j 6≡2 mod 4

1

(1− qj)r=

∞∑

n=0

pr(n)qn. (6.33)

When r is a positive integer, ∆r(n) is the number of ways to express n as a sum of triangular numbers

(including the order of the numbers), while pr(n) gives an r-colored version of p(n).

Example 6.12 (Representations as sums of triangular numbers).

en = ∆r(n), hn = pr(n), pn = r(σ(n) + ω2(n)− ω4(n)). (6.34)

Discussion. Analogs of all the identities we saw for Example 6.9 are available here too. For example

r

n(σ(n) + ω2(n)− ω4(n)) =

n∑

j=1

(−1)n−j

j

(

n

j

)

∆rj(n), (6.35)

and this one, for r = 1, is equivalent to the main result in [Jha]. Several of Jha’s recent papers can be

organized and understood by referring to Examples 6.5, 6.9 and 6.12. It is also interesting to consider

generalizations of the last examples, where the squares and triangular numbers are replaced by other sets of

integers S.

25

6.4 Further identities

We may give a De Moivre polynomial version of the q-binomial theorem. Utilizing (6.8) shows that

log(az; q)∞(z; q)∞

=

∞∑

m=1

1− am

m(1− qm)zm,

and so Theorem 6.1 is equivalent to

m∑

k=0

1

k!Am,k

(

1− a

1(1− q),

1− a2

2(1− q2),

1− a3

3(1− q3), . . .

)

=(a; q)m(q; q)m

. (6.36)

Special cases of this are, for m > 0,

m∑

k=0

1

k!Am,k

(

1

1,1

2,1

3, . . .

)

= 1, (6.37)

m∑

k=0

1

k!Am,k

(

1− a

1,1− a2

2,1− a3

3, . . .

)

= 1− a (m > 1), (6.38)

m∑

k=0

1

k!Am,k

(

1

1(1 − q),

1

2(1− q2),

1

3(1 − q3), . . .

)

=1

(q; q)m, (6.39)

where (6.37) is a result of Cauchy and (6.38) is [Mor71, Eq. (11)]. The right side of (6.39) is the generating

function for partitions into at most m parts and MacMahon used (6.39) to simplify expressions for them

[Mac60, Vol. 2, p. 62]. Similar identities to (6.37) arise from the following symmetric triple, as in [Mac95,

Ex. 13].

Example 6.13 (Schur’s identity and generalizations). For r a positive integer,

(

(1 + t)r

1− (−t)r ,1− tr

(1− t)r,

r

1− t− rtr−1

1− tr

)

with pn =

{

0 if r | n;r if r ∤ n.

Discussion. The binomial theorem implies for n > 1

en = (−1)r⌊n/r⌋(

r

n− r⌊n/r⌋

)

+

{

(−1)n−r if r | n;0 if r ∤ n,

hn =

(

n+ r − 1

r − 1

)

−(

n− 1

r − 1

)

. (6.40)

By the transition formula hk ↔ pk,

n∑

k=0

rk

k!An,k

(

1

1,1

2, . . . ,

1

r − 1, 0,

1

r + 1, . . .

)

=

(

n+ r − 1

r − 1

)

−(

n− 1

r − 1

)

, (6.41)

where the entries in An,k at a multiple of r are set to zero. Formula (6.41) is due to Morris, [Mor71, Eq.

(13)], generalizing the r = 2 case which is an identity of Schur:

n∑

k=0

2k

k!An,k

(

1

1, 0,

1

3, 0,

1

5, 0, . . .

)

= 2, (n > 1).

For another interesting case with rational functions, take E(t) to be a quadratic polynomial.

Example 6.14 (Lucas sequences).

(

1 + αt+ βt2,1

1− αt+ βt2,

α− 2βt

1− αt+ βt2

)

.

26

Discussion. Then hn and pn give the usual basis for Lucas sequences: any sequence ℓn satisfying the recur-

sion ℓn+1 = αℓn − βℓn−1 must equal A · hn + B · pn+1 for some fixed A and B. Initial conditions for hnand pn can be taken to be h−1 = 0, h0 = 1, p0 = 2 and p1 = α. The general theory is described in [Cha02,

Sect. 7.5].

By the hn ↔ en transition formula,

hn =

n∑

k=0

(−1)n−kAn,k(α, β, 0, 0, . . . ) =

n∑

k=0

(

k

n− k

)

α2k−n(−β)n−k,

with a similar formula for pn. See [Sun05, Ex. 15] for more on the divisibility of such sequences.

Chebyshev polynomials of the first and second kind, Tn(x) and Un(x) respectively, appear in the special

case α = 2x, β = 1 of Example 6.14. See also [O’S, Sect. 5] and [MR15, Exer. 3.16].

Example 6.15 (Chebyshev polynomials Tn(x), Un(x)).

E(t) = 1 + 2xt+ t2, hn = Un(x), pn = 2Tn(x).

7 Compositional inverses

Let f(t) be a formal power series with a constant term c that is invertible in our ring of coefficients R.

Though f(t) may not have a compositional inverse, F (t) := tf(t) must have one: G(x) = F (x)〈−1〉 so that

G(x) = t ⇐⇒ F (t) = x. See [O’S, Prop. 3.8], for example. Then f∗(t) := G(t)/t is a power series with

constant term 1/c. Following [Mac95, Ex. 24], this defines the ∗ operation on power series with invertible

constant terms:

f∗(t) := (tf(t))〈−1〉/t. (7.1)

If (E(t),H(t), P (t)) is a symmetric triple, then H∗(t) is a series with constant term 1 and we obtain the

’starred’ symmetric triple

(E,H,P )∗ := (E◦(t),H∗(t), P ◦(t)) for E◦(t) :=1

H∗(−t) , P ◦(t) :=d

dtlogH∗(t). (7.2)

Writing the coefficients of the series in (7.2) as e◦n, h∗n and p◦n, we find, as the special case α = β = −1 of

Corollary 8.4, the remarkably similar formulas for n > 1:

(−1)n−1e◦n = [tn]H(t)−(n−1)

n− 1, h∗n = [tn]

H(t)−(n+1)

n+ 1,

p◦nn

= [tn]H(t)−n

n. (7.3)

Applying the ∗ operation again also shows that

(−1)n−1en = [tn]H∗(t)−(n−1)

n− 1, hn = [tn]

H∗(t)−(n+1)

n+ 1,

pnn

= [tn]H∗(t)−n

n. (7.4)

The formulas for e◦1 and e1 above should be interpreted using (8.2). More simply, with (4.4) they can be

replaced by h∗1 and h1 respectively.

Then (7.3) can be used to relate e◦n, h∗n and p◦n back to the original en, hn and pn. To do this, express Hin terms of E or P as in the proof of Theorem 3.2 and then expand using Proposition 2.1. This yields the

following, as a De Moivre polynomial version of [Mac95, Ex. 24].

Proposition 7.1. For n > 1,

e◦n =−1

n− 1

n∑

k=0

(

n− 1

k

)

An,k(e1, e2, . . . ), h∗n =1

n+ 1

n∑

k=0

(−n− 1

k

)

An,k(h1, h2, . . . ),

p◦n =

n∑

k=0

(−n)kk!

An,k

(p11,p22, . . .

)

.

27

Example 7.2 (The logarithm).

(

t

log(1 + t),log(1− t)

−t , −1

t− 1

(1− t) log(1− t)

)

,

with

en =

∫ 1

0

(

x

n

)

dx, hn =1

n+ 1, pn = (−1)n

∫ 1

0

(−xn

)

dx. (7.5)

Discussion. This example comes from applying (4.5) to the Bernoulli Example 5.11, so that H(t) becomes

(e−t − 1)/(−t), and then applying the ∗ operation to find H∗(t) = − log(1 − t)/t directly. The formulas

for en and pn are stated in [Com74, p. 293] and a nice exercise.

We may set C−n := n!en and C+

n := n!pn where the signs indicate the falling and rising factorials in

(7.5). Among other names, C−n and C+

n are known as Cauchy numbers; [Bla16] has more information about

them and their history. The transition formulas give the simple identities

C−n = n!

n∑

k=0

(−1)n−kAn,k

(

1

2,1

3, . . .

)

, C+n = n!

n∑

k=1

(−1)k−1

kAn,k

(

1

2,1

3, . . .

)

, (7.6)

which may also be written with Stirling cycle numbers using (2.20). We will later see the Norlund polynomial

expressions

C−n = − 1

n− 1B(n−1)

n , C+n = (−1)nB(n)

n . (7.7)

Example 7.3 (Tangent, inverse).

(

t

arctan t,arctan t

t, −1

t+

1

(1 + t2) arctan t

)

,

after applying ∗ to Example 5.16, with h2n = (−1)n/(2n + 1) and h2n−1 = 0.

Example 7.4 (Hermite polynomial, inverse). For n > 1,

en = − Hn

(

ix√n− 1

)

(n− 1)1−n/2inn!, hn =

Hn

(

ix√n+ 1

)

(n+ 1)1−n/2(−i)nn! , pn =Hn(ix

√n)

n−n/2(−i)nn! . (7.8)

Discussion. (To find e1 use h1 instead, as e1 = h1.) This is the result of applying ∗ to Example 5.10. Letting

t→ i√αt and x→ i

√αx for α > 0 means

e2xt−t2 =

∞∑

n=0

Hn(x)

n!tn =⇒ [tn]e−α(2xt−t2) = (i

√α)n

Hn(ix√α)

n!,

and we can explicitly compute (7.3) for H(t) = e2xt−t2 to give (7.8).

8 Combining powers and inverses of series

8.1 General series

The results in this section were inspired by Example 25 of [Mac95] as well as the generalized series in

[GKP94, Sect. 5.4].

Definition 8.1. Let ρ(t) = 1 + ρ1t+ ρ2t2 + · · · be a power series and write Dn(α) := [tn]ρ(t)α. For all α,

define the general series associated to ρ as

Qρ,α(t) = Qα(t) := 1 +∞∑

n=1

Dn(αn+ 1)

αn+ 1tn. (8.1)

28

Clearly Q0(t) = ρ(t). Note that ρ(t) and Qα(t) are formal power series for now and we may not have

convergence for any t. The denominator αn + 1 can be zero in (8.1) without causing an issue: for n > 1,

Dn(β) =

n∑

k=1

(

β

k

)

An,k(ρ1, ρ2, . . . ) = β

n∑

k=1

1

k

(

β − 1

k − 1

)

An,k(ρ1, ρ2, . . . ) (8.2)

so that Dn(β)/β is well-defined for all β when n > 1.

Some remarkable properties of these general series are given next. Recall the ∗ operation from (7.1),

related to the compositional inverse.

Theorem 8.2. Let Qα(t) = Qρ,α(t) be the general series associated to a power series ρ with constant term

1. Then it satisfies

Qα(t) = ρ(tQα(t)α). (8.3)

For all α and β we have

Qα(t)β = 1 +

∞∑

n=1

β

αn + βDn(αn + β) · tn, (8.4)

Qα(t)β

(

1 + αtQ′

α(t)

Qα(t)

)

= 1 +

∞∑

n=1

Dn(αn+ β) · tn, (8.5)

(

Qα(t)β)∗

= Qα−β(t)−β . (8.6)

Corollary 8.3. Let ρ(t) be a power series with constant term 1. The two-parameter family Qρ,α(t)β contains

all possible series obtained from ρ(t) by repeatedly taking powers and applying the ∗ operator.

For example, with Qα = Qρ,α,

ρ = Q0, ρβ = Qβ0 , ρ∗ = Q−1

−1, (ρβ)∗ = Q−β−β, (((ρβ)∗)α)∗ = Qβα

β(α−1). (8.7)

Corollary 8.4. Let ρ(t) be a power series with associated general series Qα(t). Then(

Qα(−t)−β , Qα(t)β , βQ′

α(t)/Qα(t))

is a symmetric triple, and for n > 1,

en =(−1)n(−β)αn− β

Dn(αn − β), hn =β

αn + βDn(αn+ β), pn =

β

αDn(αn). (8.8)

Proof. The formulas in (8.8) come from (8.4) and (8.5).

8.2 Lagrange inversion and the proof of Theorem 8.2

Theorem 8.5 (Lagrange inversion). Let F be a formal power series with compositional inverse G so that

G(x) = t ⇐⇒ F (t) = x. Then for all integers n and all formal Laurent series φ, we have

[xn]φ(G(x)) = [t−1]φ(t)F ′(t)

F (t)n+1. (8.9)

This is proved in Theorem 2.1.1 and (2.1.7) of [Ges16]. As explained there, we find another useful form

of (8.9) by setting ψ(t) := φ(t)tF ′(t)/F (t). Then

ψ(G(x)) = φ(G(x))G(x)F ′(G(x))

F (G(x))= φ(G(x))

G(x)

xG′(x)

and by (8.9),

[xn]ψ(G(x))G′(x)

G(x)= [xn+1]φ(G(x)) = [t−1]

φ(t)F ′(t)

F (t)n+2= [t0]

ψ(t)

F (t)n+1.

Writing F (t) = tf(t), where f necessarily has a nonzero constant term, gives the versions we will need:

29

Corollary 8.6. Let F be a formal power series with compositional inverse G. Write F (t) = tf(t). Then for

all integers n and all formal Laurent series φ and ψ we have

[xn]φ(G(x)) = [tn]

(

1 + tf ′(t)

f(t)

)

φ(t)

f(t)n, [xn]ψ(G(x))

G′(x)

G(x)= [tn+1]

ψ(t)

f(t)n+1. (8.10)

Proof of Theorem 8.2. We will use Corollary 8.6, and in its notation write f(t) := ρ(t)−α with G the com-

positional inverse of F (t) = tf(t). Therefore F (G(t)) = t implies

G(t) · ρ(G(t))−α = t. (8.11)

Set Qα(t) to be ρ(G(t)) and we will see that this agrees with Definition 8.1. First note that (8.11) implies

(8.3). Applying Corollary 8.6 with φ(t) = ρ(t)β and f(x) = ρ(x)−α gives

[tn]Qα(t)β = [tn]

(

1 + tf ′(t)

f(t)

)

φ(t)

f(t)n= [tn]

(

ρ(t)αn+β − αtρ′(t)ρ(t)αn+β−1)

.

This equals

[tn]

((

1 +α

αn + β

)

ρ(t)αn+β − α

αn + β

d

dt

(

tρ(t)αn+β)

)

=

(

1 +α

αn+ β

)

Dn(αn+ β)− α(n+ 1)

αn+ βDn(αn + β) =

β

αn + βDn(αn + β).

Therefore (8.4) is true and for β = 1 we see that Qα(t) matches our original definition in (8.1).

From (8.11),

G(t) = tQα(t)α and hence t

G′(t)

G(t)= 1 + αt

Q′α(t)

Qα(t).

Therefore

[tn]Qα(t)β

(

1 + αtQ′

α(t)

Qα(t)

)

= [tn]Qα(t)βtG′(t)

G(t)= [tn−1]ρ(G(t))β

G′(t)

G(t)

and applying Corollary 8.6 with ψ(t) = ρ(t)β and f(x) = ρ(x)−α gives (8.5) directly.

To prove (8.6), write Gu(t) for the compositional inverse of Fu(t) := tρ(t)u. As in (8.11),

Gu(t) · ρ(Gu(t))u = t. (8.12)

From the definition of Qα(t) and (8.12),

tQα(t)β = tρ(G−α(t))

β = G−α(t) · ρ(G−α(t))−α · ρ(G−α(t))

β = Fβ−α(G−α(t)). (8.13)

For (8.6) we need the compositional inverse of (8.13):

t(

Qα(t)β)∗

= F−α(Gβ−α(t)) = Gβ−α(t)ρ(Gβ−α(t))−α

= tρ(Gβ−α(t))α−βρ(Gβ−α(t))

−α = tρ(Gβ−α(t))−β .

Comparing this with the leftmost equality in (8.13) completes the proof of (8.6).

Note that by (8.11) and the definition of G, the general series associated to ρ satisfies

Qα(t) = ρ(

(ρ(t)−α)∗)

=(

(ρ(t)−α)∗)1/α

.

30

9 Further examples of symmetric triples

9.1 Binomial, exponential and Bernoulli generalizations

The simplest example of Theorem 8.2 (besides ρ(t) = 1) is ρ(t) = 1 + t. In that case Dn(α) =(αn

)

. The

series Bα(t) := Qρ,α(t) is called a generalized binomial series in [GKP94, Eq. (5.60)]. Its properties were

first developed by Lambert in the 1750s and Theorem 8.2 tells us it satisfies

Bα(t) = 1 + tBα(t)α, (9.1)

Bα(t)β = 1 +

∞∑

n=1

β

αn+ β

(

αn + β

n

)

tn. (9.2)

Then Corollary 8.4 provides the corresponding symmetric triple, generalizing Example 4.3:

Example 9.1 (General binomial).

en =β

(1− α)n + β

(

(1− α)n + β

n

)

, hn =β

αn + β

(

αn+ β

n

)

, pn =β

α

(

αn

n

)

. (9.3)

This is [Mac95, Ex. 25(a)]. The α and β in (9.3) are arbitrary, and specializing them to α = 2, β = 1,

for example, gives a symmetric triple for the Catalan numbers Cn :=(

2nn

)

/(n + 1).

Example 9.2 (Catalan numbers). For n > 1,

en = (−1)n−1Cn−1, hn = Cn, pn =n+ 1

2Cn. (9.4)

Discussion. The hn ↔ pn transition formula implies the recursion

Cn =

n∑

k=0

1

2kk!An,k

(

2C1

1,3C2

2,4C3

3,5C4

4, . . .

)

. (9.5)

Solving (9.1) in this case gives the familiar generating function

B2(t) =1−

√1− 4t

2t=

∞∑

n=0

Cntn. (9.6)

See [GKP94, p. 203], [Ges16, §§2.3, 3.3, 3.4] for many closely related series.

The next simplest case has ρ(t) = et and Dn(α) = αn/n!. Following [GKP94, Eq. (5.60)] again,

Eα(t) := Qρ,α(t) is called a generalized exponential series. Theorem 8.2 shows

Eα(t) = exp(tEα(t)α), (9.7)

Eα(t)β = 1 +

∞∑

n=1

β(αn + β)n−1

n!tn. (9.8)

The Lambert W function W (t) and the tree function T (t) (see [Ges16, Sect. 3.2]) can be recognized here

from (9.7):

W (t)eW (t) = t, T (t)e−T (t) = t =⇒ W (t) = tE1(−t), T (t) = tE1(t). (9.9)

Corollary 8.4 gives the following symmetric triple, generalizing Example 5.9.

Example 9.3 (General exponential).

en = β(−αn+ β)n−1

n!, hn = β

(αn + β)n−1

n!, pn = β

(αn)n−1

(n − 1)!. (9.10)

31

Discussion. This is [Mac95, Exs. 14, 25(b)]. Special cases of (9.10) are, (see (8.7)),

(et, et, 1) when α = 0, β = 1,

(ext, ext, x) when α = 0, β = x,

(ext, ext, x)∗ when α = −x, β = −x.

The convolution identity (5.7) gives

(

α(n + 1))n

=n∑

j=0

(

n

j

)

β(

β + αj)j−1(

α(n+ 1)− β − αj)n−j

(9.11)

here, which is a special case of Abel’s 1826 generalization of the binomial theorem [Com74, p. 128].

Taking ρ(t) = t/(et − 1) in Theorem 8.2 has Dn(α) = B(α)n /n!, recalling the Norlund polynomials

B(z)n with generating function

(

t

et − 1

)z

=

∞∑

n=0

B(z)n

tn

n!. (9.12)

We may describe Uα(t) := Qρ,α(t) as a generalized Bernoulli series, and Theorem 8.2 implies

et·Uα(t)α = 1 + t · Uα(t)α−1, (9.13)

Uα(t)β = 1 +

∞∑

n=1

βB

(αn+β)n

αn+ β

tn

n!. (9.14)

Then Uα(t) is related to the series Sα(t) in [GKP94, p. 272] by Uα(t) = Sα(−t). The associated symmetric

triple from Corollary 8.4 generalizes Examples 5.11, 7.2 involving Bernoulli numbers and logarithms:

Example 9.4 (General Bernoulli).

en = (−1)n(−β)n!

B(αn−β)n

αn − β, hn =

β

n!

B(αn+β)n

αn + β, pn =

β

n!

B(αn)n

α. (9.15)

The general series Bα(t), Eα(t) and Uα(t) have been treated formally so far. We show next that they

converge for small enough t.

Proposition 9.5. Let α, β and t be any complex numbers. The series (9.2) and (9.14) for Bα(t)β and Uα(t)

β

converge absolutely for |t| < 2−|α|−1. The series (9.8) for Eα(t)β converges absolutely for |t| < 1/(e|α|).

Proof. Take r and ε to be real numbers with r > 0 and |ε| < 1. Then

(1− ε)−r =∞∑

k=0

(−rk

)

(−ε)k =∞∑

k=0

(

k + r − 1

k

)

εk

and(k+r−1

k

)

> 0. Setting ε = 1/2 shows

n∑

k=0

(

k + r − 1

k

)

1

2k6 2r and

(

n+ r − 1

n

)

6 2r+n. (9.16)

Since we easily have |(zk

)

| 6(k+|z|−1

k

)

for all z ∈ C, it follows from the right bound in (9.16) that

∣

∣

∣

∣

β

αn+ β

(

αn+ β

n

)∣

∣

∣

∣

≪ 2n+|αn+β|

and this gives the desired domain of convergence for Bα(t)β .

32

Looking ahead to (9.18), we may apply the bound |An,k(1/2!, 1/3!, . . . )| 6 2n−k from Lemma 2.2 of

[O’S], along with the left bound in (9.16), to see that

∣

∣

∣

∣

∣

B(αn+β)n

αn + β

β

n!

∣

∣

∣

∣

∣

≪n∑

k=0

(

k + |αn+ β| − 1

k

)

2n−k6 2n+|αn+β|.

Lastly, a routine application of Stirling’s formula gives the convergence for Eα(t)β .

The radius of convergence we gave for Eα(t)β is exact by the ratio test (with convergence in all of Cwhen α = 0). It would be interesting to find the exact radii of convergence in the other two cases, and if the

functions can be continued past that.

General series associated to the partition and representation series in (6.10), (6.26) and (6.33) may also

be constructed, along with their symmetric triples.

9.2 Norlund polynomials

To get a better understanding of Example 9.4 we next look at the Norlund polynomials in detail. Expanding

ρ(t) and its reciprocal in (9.12) and then applying (2.6) finds

B(z)n

n!=

n∑

k=0

(

z

k

)

An,k

(

B1

1!,B2

2!,B3

3!, . . .

)

(9.17)

=

n∑

k=0

(−zk

)

An,k

(

1

2!,1

3!,1

4!, . . .

)

, (9.18)

making B(z)n a polynomial in z of degree n with rational coefficients. For small n:

n 0 1 2 3 4

(−2)nB(z)n 1 z z2 − z

3 z3 − z2 z4 − 2z3 + z2

3 + 2z15

They are closely related to the Stirling polynomials σn(z) of [GKP94, Eq. (6.45)] with (−1)nB(z)n =

n!zσn(z). The recursion

zB(z+1)n = (z − n)B(z)

n − znB(z)n−1, (9.19)

as in [Cha02, p. 329], comes from differentiating (9.12).

The next formula for B(z)n is similar to (9.17) and (9.18) but relates to the logarithm. For this, note that

U0(t) = ρ(t) =t

et − 1=⇒ U0(t)

−1 =et − 1

t

=⇒ (U0(t)−1)∗ = U1(t) =

log(1 + t)

t(9.20)

=⇒ U1(t)β =

(

log(1 + t)

t

)β

.

Hence (9.14) provides(

log(1 + t)

t

)z

=

∞∑

n=0

z

n+ zB(n+z)

n

tn

n!. (9.21)

and then, expanding the left side with (2.6),

B(n+z)n

n!= (−1)n

n+ z

z

n∑

k=0

(

z

k

)

An,k

(

1

2,1

3,1

4, . . .

)

. (9.22)

We also see from (9.20) that the logarithmic triple in Example 7.2 is just the case α = β = 1 of Example

9.15 with t→ −t. The formulas (7.7) for the Cauchy numbers follow.

33

The Norlund polynomials are also related to the Stirling numbers. For integers m, k > 0

{

m+ k

k

}

=

(

m+ k

m

)

B(−k)m ,

[

m+ k

k

]

=

(−km

)

B(m+k)m , (9.23)

where the left identity comes from comparing (2.14) and (9.12), while the right one comes from comparing

(2.15) and (9.21). The right sides of both identities in (9.23) are degree 2m polynomials in k, allowing us to

extend the Stirling number definitions. Inserting our formulas (9.17), (9.18) and (9.22) into (9.23) recovers

Kramp’s identities (5.8) and (5.9) in two cases.

Now B(k)m may be given explicitly in different ranges of k ∈ Z. For k close to m we have

B(m−1)m = −(m− 1)C−

m, B(m)m = (−1)mC+

m, (9.24)

B(m+1)m

m!= (−1)m,

B(m+2)m

m!= (−1)mHm+1,

B(m+3)m

m!= H2

m+2 −H(2)m+2, (9.25)

by (7.7), (9.23) and the Stirling cycle number formulas (5.21), (5.26) and (5.27). For k close to 0 we have

B(1)m = Bm, B(2)

m = (1−m)Bm −mBm−1, (9.26)

B(0)m = δm,0,

B(−1)m

m!=

1

(m+ 1)!,

B(−2)m

m!=

2m+2 − 2

(m+ 2)!, (9.27)

with (9.27) following from (9.23) and the Stirling subset number formula (2.21). The first identity in (9.26)

is clear from (9.12); apply the recursion (9.19) for the second.

The following two examples are cases of Example 9.4. The first has α = 0, β = 2 corresponding to the

square of Example 5.11. The second has α = 1, β = 2 corresponding to the square of Example 7.2.

Example 9.6 (Bernoulli numbers, squared).

en = (−1)n2n+2 − 2

(n+ 2)!, hn =

(1− n)Bn − nBn−1

n!, pk = (−1)n−1 2Bn

n!. (9.28)

Example 9.7 (Logarithms, squared).

en = (−1)n−1 2

n− 2

B(n−2)n

n!, hn = (−1)n

2

n+ 2Hn+1, pn =

2C+n

n!.

9.3 Special values of De Moivre polynomials

It is useful to be able to simplify De Moivre polynomials and, in particular, know when they can be evaluated

explicitly. Clearly

Am+k,k(a0, a1, a2, . . . ) = bm ⇐⇒ [xm](

a0 + a1x+ a2x2 + · · ·

)k= bm, (9.29)

for m,k > 0, and so we are looking for power series whose powers are known. Simple examples are

((1 + x)β)k = (1 + x)βk, (eβx)k = eβkx

for β any complex number or indeterminate, giving

Am+k,k

((

β

0

)

,

(

β

1

)

,

(

β

2

)

, . . .

)

=

(

βk

m

)

, Am+k,k

(

β0

0!,β1

1!,β2

2!, . . .

)

=(βk)m

m!. (9.30)

Our work in sections 9.1, 9.2 shows more possibilities. From the generalized exponential series Eα(t) in

(9.8),

Am+k,k

(

1,β

1!, (2α + β)

β

2!, (3α + β)2

β

3!, (4α + β)3

β

4!, . . .

)

= (mα+ kβ)m−1 kβ

m!, (9.31)

34

for arbitrary α and β. The case α = β = 1 is

Am+k,k

(

10

1!,21

2!,32

3!,43

4!, . . .

)

= (m+ k)m−1 k

m!, (9.32)

valid for m, k > 0 and not both 0. This is equivalent to an identity on p. 89 of [Chu19].

From the generalized binomial series Bα(t) in (9.2),

Am+k,k

(

1,β

α+ β

(

α+ β

1

)

,β

2α+ β

(

2α+ β

2

)

,β

3α+ β

(

3α+ β

3

)

, . . .

)

=βk

αm+ βk

(

αm+ βk

m

)

. (9.33)

As we saw in Example 9.2, the case α = 2, β = 1 of this gives the Catalan numbers, so that

Am+k,k(C0, C1, C2, . . . ) =k

2m+ k

(

2m+ k

m

)

. (9.34)

This identity (9.34) is in fact equivalent to Theorem 1.2 of [QSLK17] where they also look at variations and

applications.

The generalized Bernoulli series Uα(t) in (9.14) provides

Am+k,k

(

1,B

(α+β)1

α+ β

β

1!,B

(2α+β)2

2α+ β

β

2!,B

(3α+β)3

3α+ β

β

3!, , . . .

)

=B

(αm+βk)m

αm+ βk

βk

m!. (9.35)

The evaluations in section 9.2 and (9.24) – (9.27) give many explicit cases of this. For example, with α = 0and β = 1,−1,−2,

Am+k,k

(

1,B1

1!,B2

2!,B3

3!, . . .

)

=B

(k)m

m!, (9.36)

Am+k,k

(

1

1!,1

2!,1

3!, . . .

)

=k!

(m+ k)!

{

m+ k

k

}

, (9.37)

Am+k,k

(

22 − 2

2!,23 − 2

3!,24 − 2

4!, , . . .

)

=(2k)!

(m+ 2k)!

{

m+ 2k

2k

}

. (9.38)

For α = 1 and β = 1, 2,

Am+k,k

(

1

1,1

2,1

3, . . .

)

=k!

(m+ k)!

[

m+ k

k

]

, (9.39)

Am+k,k

(

H1

2,H2

3,H3

4, · · ·

)

=(2k)!

2k(m+ 2k)!

[

m+ 2k

2k

]

. (9.40)

The series (6.10), (6.26) and (6.33) may also be used to find similar identities.

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DEPT. OF MATH, THE CUNY GRADUATE CENTER, 365 FIFTH AVENUE, NEW YORK, NY 10016-4309, U.S.A.

E-mail address: cosullivan@gc.cuny.edu

37