INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A31

AN INFINITE FAMILY OF DUAL SEQUENCE IDENTITIES

Robin Chapman

Department of Mathematics, University of Exeter, Exeter, EX4 4QE, UK

                            rjc@maths.ex.ac.uk

Received: 7/27/05, Accepted: 11/13/05, Published: 12/9/05

Abstract

We subsume three identities of Sun [2] into an infinite family of identities, and consider some special cases.

1. Introduction

Let (a_{n})_{n=0}^{∞} be a sequence of complex numbers. We call the sequence (a_{n}^{∗})_{n=0}^{∞} defined by

a_{n}^{∗} = ∑_{i=0}^{n} \binom{n}{i}(−1)^{i}a_{i}

the dual sequence of (a_{n}). Repeating this operation gives rise to the original sequence: a_{n}^{∗∗} = a_{n} for all n [1, 192–193].

A sequence (a_{n}) satisfying a_{n}^{∗} = a_{n} is called self-dual. An important example of a self-dual sequence is ((−1)^{n}B_{n}) where the Bernoulli numbers B_{n} satisfy

∑_{n=0}^{∞} B_{n} \frac{t^{n}}{n!} = \frac{t}{exp(t) − 1}.

Sun [2] considered the polynomials

A_{n}(x) = ∑_{i=0}^{n} \binom{n}{i}(−1)^{i}a_{i}x^{n−i} and A_{n}^{∗}(x) = ∑_{i=0}^{n} \binom{n}{i}(−1)^{i}a_{i}^{∗}x^{n−i}.  (1)

When a_{k} = (−1)^{k}B_{k} then A_{n}(x) = B_{n}(x), the classical Bernoulli polynomial. Sun’s main theorem [2, Theorem 1.1] gives three identities involving the polynomials A_{n} and A_{n}^{∗}. Using these identities he unifies and generalizes various identities concerning Bernoulli numbers and polynomials in the recent literature. The main theorem here gives an infinite family of identities subsuming Sun’s identities.

2. The main theorem

We begin with preliminaries concerning exponential generating functions. Throughout we let (a_{n})_{n=0}^{∞} be a sequence and (a_{n}^{∗})_{n=0}^{∞} be its dual. Also we let A_{n} and A_{n}^{∗} be the polynomials defined by (1).

We define

α(t) = ∑_{n=0}^{∞} a_{n} \frac{t^{n}}{n!} and α^{∗}(t) = ∑_{n=0}^{∞} a_{n}^{∗} \frac{t^{n}}{n!}

as the exponential generating functions of the sequences (a_{n}) and (a_{n}^{∗}). Then

α^{∗}(t) = ∑_{n=0}^{∞} ∑_{i=0}^{n} \binom{n}{i}(−1)^{i}a_{i} \frac{t^{n}}{n!} = ∑_{i=0}^{∞} a_{i} \frac{(−t)^{i}}{i!} ∑_{n=i}^{∞} \frac{t^{n−i}}{(n − i)!} = exp(t)α(−t).

As a consequence,

exp(t)α^{∗}(−t) = exp(t) exp(−t)α(t) = α(t)

which recovers the fact that a_{n}^{∗∗} = a_{n}.

It is well-known that the Bernoulli polynomials B_{n}(x) are symmetric or anti-symmetric about 1/2 according to the parity of n, to wit

B_{n}(1 − x) = (−1)^{n}B_{n}(x).

The following lemma generalizes this to arbitrary sequences, relating the polynomials A_{n} and A_{n}^{∗}.

Lemma 1 (Duality principle) For each n, A_{n}^{∗}(x) = (−1)^{n}A_{n}(1 − x).

Proof. First

∑_{n=0}^{∞} A_{n}(x)\frac{t^{n}}{n!} = ∑_{n=0}^{∞} ∑_{i=0}^{n} \binom{n}{i}(−1)^{i}a_{i}x^{n−i} \frac{t^{n}}{n!}
= ∑_{i=0}^{∞} a_{i} \frac{(−t)^{i}}{i!} ∑_{n=i}^{∞} \frac{(xt)^{n−i}}{(n − i)!} = exp(xt)α(−t).

Similarly

∑_{n=0}^{∞} A_{n}^{∗}(x)\frac{t^{n}}{n!} = exp(xt)α^{∗}(−t) = exp((x − 1)t)α(t) = ∑_{n=0}^{∞} A_{n}(1 − x)\frac{(−t)^{n}}{n!} .

Hence A_{n}^{∗}(x) = (−1)^{n}A_{n}(1 − x) for all n. □

Sun’s main theorem uses three variables x, y and z satisfying x + y + z = 1 and expresses three sums featuring polynomials A_{m}(y) and A_{n}^{∗}(z) with coefficients in x as expressions depending only on x. Our main theorem generalizes the first two of these identities. We later see how to generalize the third.

Theorem 2 Let x + y + z = 1. Then for all integers k, l, r ≥ 0,

(−1)^{k} ∑_{j=0}^{k} \binom{k}{j}x^{k−j} \frac{(l + j)!A_{l+j+r}(y)}{(l + j + r)!}  (2)

− (−1)^{j+r} ∑_{j=0}^{l} \binom{l}{j}x^{l−j} \frac{(k + j)!A_{k+j+r}^{∗}(z)}{(k + j + r)!}

= k!l!(−x)^{k+l+1} ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(k + l + r − i)!}

× ∑_{j=0}^{r−1−i} \binom{k + r − i − 1 − j}{k}\binom{l + j}{l}y^{r−i−1−j}(1 − z)^{j}.

Proof. We first recast (2) in a more symmetric form. We replace z by 1 − z and use the duality principle. Then x = 1 − y − z must be replaced by z − y. After some rearrangement we find that (2) is equivalent to

∑_{j=0}^{k} \binom{k}{j}(z − y)^{k−j} \frac{(l + j)!A_{l+j+r}(y)}{(l + j + r)!}  (3)

− ∑_{j=0}^{l} \binom{l}{j}(y − z)^{l−j} \frac{(k + j)!A_{k+j+r}(z)}{(k + j + r)!}

= (−1)^{k}k!l!(y − z)^{k+l+1} ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(k + l + r − i)!}

× ∑_{j=0}^{r−1−i} \binom{k + r − i − 1 − j}{k}\binom{l + j}{l}y^{r−i−1−j}z^{j}.

Define

A_{r}(k, l; y, z) = ∑_{j=0}^{k} \binom{k}{j}(z − y)^{k−j}(l + j)!\frac{A_{l+j+r}(y)}{(l + j + r)!}

so that (3) is concerned with A_{r}(k, l; y, z) − A_{r}(l, k; z, y). We consider the exponential generating function of the A_{r}(k, l; y, z). We calculate

∑_{k,l=0}^{∞} A_{r}(k, l; y, z)\frac{t^{k}u^{l}}{k!l!}

= ∑_{k,l=0}^{∞} ∑_{j=0}^{k} \binom{k}{j}(z − y)^{k−j}(l + j)!\frac{A_{l+j+r}(y)}{(l + j + r)!} \frac{t^{k}u^{l}}{k!l!}
= ∑_{j,l=0}^{∞} \binom{l + j}{j}t^{j}u^{l} \frac{A_{l+j+r}(y)}{(l + j + r)!} ∑_{k=j}^{∞} \frac{(z − y)^{k−j}t^{k−j}}{(k − j)!}
= exp((z − y)t) ∑_{p=0}^{∞}(t + u)^{p} \frac{A_{p+r}(y)}{(p + r)!}

= \frac{exp((z − y)t)}{(t + u)^{r}} [exp(y(t + u))α(−t − u) − ∑_{s=0}^{r−1} A_{s}(y)\frac{(t + u)^{s}}{s!} ]
= \frac{exp(zt + yu)α(−t − u)}{(t + u)^{r}} − \frac{exp((z − y)t)}{(t + u)^{r}} ∑_{s=0}^{r−1} A_{s}(y)\frac{(t + u)^{s}}{s!} .

Hence

∑_{k,l=0}^{∞}[A_{r}(k, l; y, z) − A_{r}(l, k; z, y)]\frac{t^{k}u^{l}}{k!l!}

= \frac{1}{(t + u)^{r}} ∑_{s=0}^{r−1}[exp((y − z)u)A_{s}(z) − exp((z − y)t)A_{s}(y)]\frac{(t + u)^{s}}{s!} .

As

∑_{s=0}^{r−1} A_{s}(x)\frac{v^{s}}{s!} = ∑_{s=0}^{r−1} ∑_{i=0}^{s}(−1)^{i} \frac{a_{i}}{i!} \frac{x^{s−i}}{(s − i)!}v^{s} = ∑_{i=0}^{r−1}(−1)^{i}a_{i} \frac{v^{i}}{i!} ∑_{s=i}^{r−1} \frac{(xv)^{s−i}}{(s − i)!}

then

∑_{k,l=0}^{∞}[A_{r}(k, l; y, z) − A_{r}(l, k; z, y)]\frac{t^{k}u^{l}}{k!l!}

= ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(t + u)^{r−i}} ∑_{s=i}^{r−1} \frac{(t + u)^{s−i}}{(s − i)!} [z^{s−i} exp((y − z)u) − y^{s−i} exp((z − y)t)]

= ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(t + u)^{r−i}} ∑_{s=0}^{r−1−i} \frac{(t + u)^{s}}{s!} [z^{s} exp((y − z)u) − y^{s} exp((z − y)t)]

= ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!} B_{r−i}(t, u, y, z)

where

B_{q}(t, u, y, z) = \frac{1}{(t + u)^{q}} ∑_{s=0}^{q−1} \frac{(t + u)^{s}}{s!} [z^{s} exp((y − z)u) − y^{s} exp((z − y)t)].

We claim that

B_{q}(t, u, y, z)  (4)
= ∑_{k,l=0}^{∞} \frac{(y − z)^{k+l+1}}{(k + l + r)!} (−t)^{k}u^{l} ∑_{j=0}^{q−1} \binom{k + q − 1 − j}{q − 1 − j}\binom{l + j}{j}y^{q−1−j}z^{j}

= ∑_{m=0}^{∞} \frac{(y − z)^{m+1}}{(m + q)!} F_{m,q}(t, u, y, z)

where

F_{m,q}(t, u, y, z) = ∑_{k=0}^{m} ∑_{j=0}^{q−1} \binom{k + q − 1 − j}{q − 1 − j}\binom{m − k + j}{j}(−t)^{k}u^{m−k}y^{q−1−j}z^{j}.

We remark that

F_{m,q}(t, u, y, z) = ∑_{j=0}^{q−1} G_{m,q−1−j,j}(t, u)y^{q−1−j}z^{j}

where

G_{m,i,j}(t, u) = ∑_{k=0}^{m} \binom{k + i}{i}\binom{m − k + j}{j}(−t)^{k}u^{m−k}.

We need to calculate (t + u)G_{m,i,j}(t, u). In doing this we adopt the convention that \binom{x}{i} is defined for arbitrary x by insisting that it be a polynomial of degree i. Adopting this convention ensures that \binom{−1}{i} = \binom{i+1}{i} = 0. We must take particular care with the cases where i = 0 or j = 0. If i > 0 and j > 0 then

(t + u)G_{m,i,j}(t, u)

= ∑_{k=0}^{m+1} \binom{k + i}{i}\binom{m − k + j}{j}(−t)^{k}u^{m+1−k}

− ∑_{k=−1}^{m} \binom{k + i}{i}\binom{m − k + j}{j}(−t)^{k+1}u^{m−k}

= ∑_{k=0}^{m+1} [\binom{k + i}{i}\binom{m − k + j}{j} − \binom{k − 1 + i}{i}\binom{m − k + 1 + j}{j}]

× (−t)^{k}u^{m+1−k}

= ∑_{k=0}^{m+1} [\binom{k + i − 1}{i − 1}\binom{m − k + 1 + j}{j} − \binom{k + i}{i}\binom{m − k + 1 + j − 1}{j − 1}]

×(−t)^{k}u^{m+1−k}

= G_{m+1,i−1,j}(t, u) − G_{m+1,i,j−1}(t, u).

If i > 0 then

(t + u)G_{m,i,0} = ∑_{k=0}^{m} \binom{k + i}{i}(−t)^{k}u^{m+1−k} − ∑_{k=−1}^{m} \binom{k + i}{i}(−t)^{k+1}u^{m−k}
= ∑_{k=0}^{m} \binom{k + i}{i}(−t)^{k}u^{m+1−k} − ∑_{k=0}^{m+1} \binom{k − 1 + i}{i}(−t)^{k}u^{m+1−k}
= ∑_{k=0}^{m+1} \binom{k + i − 1}{i − 1}(−t)^{k}u^{m+1−k} − \binom{m + 1 + i}{i}(−t)^{m+1}

− \binom{m + i}{i}(−t)^{m+1}

= G_{m+1,i−1,0}(t, u) − \binom{m + 1 + i}{i}(−t)^{m+1}.

Similarly, if j > 0 then

(t + u)G_{m,0,j}(t, u) = \binom{m + 1 + j}{j}u^{m+1} − G_{m+1,0,j−1}(t, u).

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A31 6

Finally,

(t + u)G_{m,0,0}(t, u) = (t + u) ∑_{k=0}^{m}(−t)^{k}u^{m+1−k} = u^{m+1} − (−t)^{m+1}.

It follows that

(t + u)F_{m,1}(t, u, y, z) = (t + u)G_{m,0,0}(t, u) = u^{m+1} − (−t)^{m+1},

(t + u)F_{m,2}(t, u, y, z)
= (t + u)(G_{m,0,1}(t, u)z + G_{m,1,0}(t, u)y)
= (m + 2)u^{m+1}z + G_{m,0,0}(t, u)(y − z) − (m + 2)(−t)^{m+1}y,

and for r ≥ 3,

(t + u)F_{m,r}(t, u, y, z)

= \binom{m + r}{r − 1}u^{m+1}z^{r−1} − G_{m+1,0,r−2}(t, u)z^{r−1}

+ G_{m+1,r−2,0}(t, u)y^{r−1} − \binom{m + r}{r − 1}(−t)^{m+1}y^{r−1}

+ ∑_{i=1}^{r−2}[G_{m+1,i−1,r−1−i}(t, u) − G_{m+1i,r−2−i}(t, u)]y^{i}z^{r−1−i}

= (y − z)F_{m+1,r−1}(t, u, y, z) + \binom{m + r}{r − 1}(u^{m+1}z^{r−1} − (−t)^{m+1}y^{r−1}).

Therefore

(t + u)^{r}F_{m,r}(t, u, y, z) = ∑_{l=1}^{r} \binom{m + r}{r − l}(y − z)^{r−l}(u^{m+r}z^{r−l} − (−t)^{m+r}y^{r−l}).

This establishes (4).

Consequently

A_{r}(k, l; y, z) − A_{r}(l, k; z, y)

= (−1)^{k}k!l!(y − z)^{k+l+1} ∑_{i=0}^{r−i−1} \frac{(−1)^{i}a_{i}}{i!(k + l + r − i)!}

× ∑_{j=0}^{r−i−1} \binom{k + r − 1 − i − j}{r − 1 − i − j}\binom{m − k + j}{j}y^{r−i−1−j}z^{j}

as required. □

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 5(1) (2005), #A31 7

3. Examples and applications The identities (1.4) and (1.5) in [1] are respectively the r = 1 and r = 0 cases of Theorem 2.

The next special case of Theorem 2, with r = 2 yields, after some rearrangement,

(−1)^{k} ∑_{j=0}^{k} \binom{k}{j}x^{k−j} \frac{A_{l+j+2}(y)}{(l + j + 1)(l + j + 2)}

− (−1)^{j} ∑_{j=0}^{l} \binom{l}{j}x^{l−j} \frac{A_{k+j+2}^{∗}(z)}{(k + j + 1)(k + j + 2)}

= k!l!(−x)^{k+l+1} (\frac{a_{0}}{(k + l + 2)!} [(k + 1)y + (l + 1)(1 − z)] − \frac{a_{1}}{(k + l + 1)!})
= \frac{(−x)^{k+l+1}}{\binom{k+l}{k}} (\frac{a_{0} [(k + 1)y + (l + 1)(1 − z)]}{(k + l + 1)(k + l + 2)} − \frac{a_{1}}{k + l + 1}) .

Taking l = k and z = y in (2) gives the following corollary. Corollary 3 For all integers k, r ≥ 0,

∑_{j=0}^{k} \binom{k}{j}(1 − 2y)^{k−j} \frac{(k + j)![A_{k+j+r}(y) − (−1)^{r}A_{k+j+r}^{∗}(y)]}{(k + j + r)!}

= (−1)^{k}k!^{2}(2y − 1)^{2k+1} ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(2k + r − i)!}

× ∑_{j=0}^{r−1−i} \binom{k + r − i − 1 − j}{k}\binom{k + j}{k}y^{r−i−1−j}z^{j}.

In addition if r is odd and (a_{n}) is self-dual, or if r is even and a_{n}^{∗} = −a_{n} for all n then

2 ∑_{j=0}^{k} \binom{k}{j}(1 − 2y)^{k−j} \frac{(k + j)!A_{k+j+r}(y)}{(k + j + r)!}

= (−1)^{k}k!^{2}(2y − 1)^{2k+1} ∑_{i=0}^{r−1} \frac{(−1)^{i}a_{i}}{i!(2k + r − i)!}

× ∑_{j=0}^{r−1−i} \binom{k + r − i − 1 − j}{k}\binom{k + j}{k}y^{r−i−1−j}z^{j}.

This corollary naturally gives rise to Bernoulli polynomial identities by taking a_{n} = (−1)^{n}B_{n}. For example, when r is odd we have

2 ∑_{j=0}^{k} \binom{k}{j}(1 − 2y)^{k−j} \frac{(k + j)!B_{k+j+r}(y)}{(k + j + r)!}

= (−1)^{k}k!^{2}(2y − 1)^{2k+1} ∑_{i=0}^{r−1} \frac{B_{i}}{i!(2k + r − i)!}

× ∑_{j=0}^{r−1−i} \binom{k + r − i − 1 − j}{k}\binom{k + j}{k}y^{r−i−1−j}z^{j}.

The following theorem which subsumes the second and third identities [2, (1.5), (1.6)] in Sun’s main theorem follows from Theorem 2.

Theorem 4 Let x + y + z = 1. Then for all integers k, l, r ≥ 0,

(−1)^{k} ∑_{j=0}^{k} \binom{k}{j}\binom{l + j}{r}x^{k−j}A_{l+j−r}(y)  (5)

= (−1)^{l+r} ∑_{j=0}^{l} \binom{l}{j}\binom{k + j}{r}x^{l−j}A_{k+j−r}^{∗}(z).

Proof. In this identity we interpret A_{m}(x)/m! as zero whenever m < 0.

Replacing z by 1 − z and using the duality principle we see that (5) is equivalent to

∑_{j=0}^{k} \binom{k}{j}\binom{l + j}{r}(z − y)^{k−j}A_{l+j−r}(y) = ∑_{j=0}^{l} \binom{l}{j}\binom{k + j}{r}(y − z)^{l−j}A_{k+j−r}(z)  (6)

The r = 0 case of (6) is the same as the r = 0 case of (3). It is immediate from the definition of A_{m}(x) that A′_{m}(x) = mA_{m−1}(x). The case r > 0 of (6) follows from the r = 0 case by applying the partial differential operator

\frac{1}{r!} (\frac{∂}{∂y} + \frac{∂}{∂z} )^{r} ,

which annihilates all powers of (y − z), to both sides. □

References

[1] R.L. Graham, D.E. Knuth & O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989.

[2] Z.-W. Sun, ‘Combinatorial identities in dual sequences’, European J. Combin. 24 (2003) 709–718.