Bernoulli Numbers and a New Binomial Transform Identity

H. W. Gould Department of Mathematics West Virginia University Morgantown, WV 26506 USA

                      gould@math.wvu.edu

Jocelyn Quaintance Department of Mathematics Rutgers University Piscataway, NJ 08854 USA

                   quaintan@math.rutgers.edu

Abstract Let (b_{n})_{n≥0} be the binomial transform of (a_{n})_{n≥0}. We show how a binomial transformation identity of Chen proves a symmetrical Bernoulli number identity attributed to Carlitz. We then modify Chen’s identity to prove a new binomial transformation identity.

Carlitz [1] posed as a problem the remarkable symmetric Bernoulli number identity

(−1)^{m} ∑_{k=0}^{m} \binom{m}{k}B_{n+k} = (−1)^{n} ∑_{k=0}^{n} \binom{n}{k}B_{m+k},  (1)

valid for arbitrary m, n ≥ 0. The published solution by Shannon [2] used mathematical induction on m and n. The identity was rediscovered recently by Vassilev and Vassilev-Missana [10], but stated in the form

(−1)^{m} ∑_{k=0}^{m−1} \binom{m}{k}B_{n+k} = (−1)^{n} ∑_{k=0}^{n−1} \binom{n}{k}B_{m+k},  (2)

valid for arbitrary positive integers m and n. Identity (2) is equivalent to Identity (1) since [(−1)^{m} − (−1)^{n}] B_{m+n} = 0. Their proof used the symmetry of a function f_{k}(x, y) involving Bernoulli numbers introduced in a separate paper [9]. They give no reference to Carlitz’s or to Shannon’s proof.

An alternative proof of Equation (1) is derived through an application of a binomial transformation identity discovered by Chen [3]. Let (a_{n}) be any sequence of numbers, and define the binomial transform of (a_{n}) to be the sequence (b_{n}), where b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}. A corollary of [3, Thm. 2.1] is

∑_{k=0}^{m} \binom{m}{k}a_{n+k} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}b_{m+k}.  (3)

The Bernoulli numbers satisfy the recurrence ∑_{k=0}^{n} \binom{n}{k}B_{k} = (−1)^{n}B_{n} for n ≥ 0. Setting a_{k} = B_{k}, we then have b_{n} = (−1)^{n}B_{n}, so that Equation (3) becomes

∑_{k=0}^{m} \binom{m}{k}B_{n+k} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}(−1)^{m+k}B_{m+k},

which is precisely Identity (1) of Carlitz.

Chen’s proof of Equation (3) relies on certain properties of Seidel matrices. We present a direct proof which relies on the hypergeometric identity

∑_{k=0}^{m}(−1)^{k}\binom{m}{k}\binom{x + k}{r} = (−1)^{m}\binom{x}{r − m};  (4)

see [6, Identity 3.47, p. 27]. In Equation (4) we require that m and r be nonnegative integers and x be a complex number.

Since the binomial transform inverts to give a_{n} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}b_{k} we find that

∑_{k=0}^{m} \binom{m}{k}a_{n+k} = ∑_{k=0}^{m} \binom{m}{k} ∑_{j=0}^{n+k}(−1)^{n+k−j}\binom{n + k}{j}b_{j}

= ∑_{j=0}^{n+m}(−1)^{−j}b_{j} ∑_{k=j−n}^{m}(−1)^{n+k}\binom{m}{k}\binom{n + k}{j}

= ∑_{j=0}^{n+m}(−1)^{−j}b_{j} ∑_{k=0}^{m}(−1)^{n+k}\binom{m}{k}\binom{n + k}{j}

= ∑_{j=0}^{n+m}(−1)^{n+m−j}\binom{n}{j − m}b_{j} = ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}b_{j+m}.

A careful analysis of this preceding proof yields a short proof of [3, Thm. 3.2], where Chen relies on lengthy induction arguments. We will instead use Equation (4).

Theorem 1. [3, Thm. 3.2] Let b_{n} be the binomial transform of a_{n}. Then

∑_{k=0}^{m} \binom{m}{k}\binom{n + k}{s}a_{n+k−s} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}\binom{m + k}{s}b_{m+k−s},  (5)

for arbitrary nonnegative m, n, and s.

Proof. By definition b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}. This implies that a_{n} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}b_{k}. Hence

∑_{k=0}^{m} \binom{m}{k}\binom{n + k}{s}a_{n+k−s} = ∑_{k=0}^{m} \binom{m}{k}\binom{n + k}{s} ∑_{j=0}^{n+k−s}(−1)^{n+k−s−j}\binom{n + k − s}{j}b_{j}

= ∑_{j=0}^{n+m−s}(−1)^{n−s−j}b_{j} ∑_{k=0}^{m}(−1)^{k}\binom{m}{k}\binom{n + k}{s}\binom{n + k − s}{j}

= ∑_{j=0}^{n+m−s}(−1)^{n−s−j}\binom{s + j}{s}b_{j} ∑_{k=0}^{m}(−1)^{k}\binom{m}{k}\binom{n + k}{s + j}

= ∑_{j=m−s}^{n+m−s}(−1)^{m+n−j−s}\binom{s + j}{s}\binom{n}{j + s − m}b_{j}

= ∑_{j=0}^{n}(−1)^{n−j}\binom{n}{j}\binom{m + j}{s}b_{m+j−s},

where the fourth equality follows by Equation (4).

Equation (5) allows us to establish a generalization of the curious formula

∑_{k=0}^{n} \binom{n}{k}\binom{n + k}{n}x^{k} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}\binom{n + k}{n}(1 + x)^{k},  (6)

discovered by Simons [8]. A quick proof of this was given by Gould [7] using elementary properties of Legendre polynomials. Instead, choose a_{n} = x^{n} for all n ≥ 0. Then b_{n} = (1+x)^{n} and Identity (5) tells us that

∑_{k=0}^{m} \binom{m}{k}\binom{n + k}{s}x^{n+k−s} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}\binom{m + k}{s}(1 + x)^{m+k−s}.

Letting m = s = n recovers Identity (6).

Through an induction argument Chen proves

Theorem 2. [3, Thm. 3.1] Let b_{n} be the binomial transform of a_{n}. Then

∑_{k=0}^{m} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}a_{n+k+s} = ∑_{k=0}^{n}(−1)^{n−k} \begin{matrix}( & {n \atop k} & ) \\ ( & {m+k+s \atop s} & ) \end{matrix}b_{m+k+s}

+ ∑_{j=0}^{s−1} ∑_{i=0}^{s−1−j} \binom{s − 1 − j}{i}\binom{s − 1}{j}\frac{(−1)^{n+1+i}sa_{j}}{(m + n + 1 + i)\binom{m+n+i}{n}},  (7)

where m, n, and s are nonnegative integers.

If we use Equation (4) and the following hypergeometric identity attributed to Frisch [4], [5, p. 337],

∑_{k=0}^{n}(−1)^{k}\binom{n}{k}\frac{1}{\binom{b+k}{c}} = \frac{c}{n + c} \frac{1}{\binom{n+b}{b−c}}, b ≥ c > 0, (8)

[6, Identity 4.2, p. 46], we are able to prove the following new binomial transformation identity. Theorem 3. Let b_{n} be the binomial transform of a_{n}. Let m, n, and s be nonnegative integers. Then

∑_{j=0}^{s} \frac{\binom{s}{j}a_{j}}{(m + n + s + 1 − j)\binom{m+n+s−j}{m}} = ∑_{j=0}^{s} \frac{(−1)^{s−j}\binom{s}{j}b_{j}}{(m + n + s + 1 − j)\binom{m+n+s−j}{n}}.  (9)

Proof. By definition b_{n} = ∑_{k=0}^{n} \binom{n}{k}a_{k}. Hence a_{n} = ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k}b_{k} and

∑_{k=0}^{m} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}a_{n+k+s} = ∑_{k=0}^{m} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix} ∑_{j=0}^{n+k+s}(−1)^{n+k+s−j}\binom{n + k + s}{j}b_{j}

= ∑_{j=0}^{m+n+s}(−1)^{n+s−j}b_{j} ∑_{k=0}^{m}(−1)^{k} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}\binom{n + k + s}{j}

= ∑_{j=s}^{m+n+s}(−1)^{n+s−j} \frac{b_{j}}{\binom{j}{s}} ∑_{k=0}^{m}(−1)^{k}\binom{m}{k}\binom{n + k}{j − s} + ∑_{j=0}^{s−1}(−1)^{n+s−j}b_{j} ∑_{k=0}^{m}(−1)^{k} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}\binom{n + k + s}{j}

= ∑_{j=s+m}^{m+n+s}(−1)^{m+n+s−j} \begin{matrix}( & {n \atop j−s−m} & ) \\ ( & {j \atop s} & ) \end{matrix}b_{j} + ∑_{j=0}^{s−1}(−1)^{n+s−j}b_{j} ∑_{k=0}^{m}(−1)^{k} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}\binom{n + k + s}{j}

= ∑_{j=0}^{n}(−1)^{n−j} \begin{matrix}( & {n \atop j} & ) \\ ( & {m+j+s \atop s} & ) \end{matrix}b_{m+j+s} + ∑_{j=0}^{s−1}(−1)^{n+s−j}b_{j} ∑_{k=0}^{m}(−1)^{k} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}\binom{n + k + s}{j}

= ∑_{j=0}^{n}(−1)^{n−j} \begin{matrix}( & {n \atop j} & ) \\ ( & {m+j+s \atop s} & ) \end{matrix}b_{m+j+s} + ∑_{j=0}^{s−1}(−1)^{n+s−j}\binom{s}{j}b_{j} ∑_{k=0}^{m}(−1)^{k} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s−j \atop s−j} & ) \end{matrix}

= ∑_{j=0}^{n}(−1)^{n−j} \begin{matrix}( & {n \atop j} & ) \\ ( & {m+j+s \atop s} & ) \end{matrix}b_{m+j+s} + ∑_{j=0}^{s−1}(−1)^{n+s−j} \frac{(s − j)\binom{s}{j}}{(m + s − j)\binom{m+n+s−j}{n}}b_{j}

= ∑_{j=0}^{n}(−1)^{n−j} \begin{matrix}( & {n \atop j} & ) \\ ( & {m+j+s \atop s} & ) \end{matrix}b_{m+j+s} + ∑_{j=0}^{s−1}(−1)^{n+s−j} \frac{s\binom{s−1}{j}}{(m + n + s − j)\binom{m+n+s−j−1}{n}}b_{j}.

The fourth line follows from Equation (4) while the seventh follows from Equation (8). In summary, we have shown that

∑_{k=0}^{m} \begin{matrix}( & {m \atop k} & ) \\ ( & {n+k+s \atop s} & ) \end{matrix}a_{n+k+s} = ∑_{j=0}^{n} \frac{(−1)^{n−j}\binom{n}{j}}{\binom{m+j+s}{s}}b_{m+j+s} + ∑_{j=0}^{s−1} \frac{(−1)^{n+s−j}s\binom{s−1}{j}}{(m + n + s − j)\binom{m+n+s−j−1}{n}}b_{j}, (10)

If we compare Identity (7) to Identity (10), we conclude that

∑_{j=0}^{s−1} \frac{(−1)^{n+s−j}s\binom{s−1}{j}}{(m + n + s − j)\binom{m+n+s−j−1}{n}}b_{j}

= ∑_{j=0}^{s−1} ∑_{i=0}^{s−1−j} \binom{s − 1 − j}{i}\binom{s − 1}{j}\frac{(−1)^{n+1+i}sa_{j}}{(m + n + 1 + i)\binom{m+n+i}{n}}.  (11)

Equation (11) can be furthered simplified by applying Equation (8). In particular,

∑_{j=0}^{s−1} ∑_{i=0}^{s−1−j} \binom{s − 1 − j}{i}\binom{s − 1}{j}\frac{(−1)^{n+1+i}sa_{j}}{(m + n + 1 + i)\binom{m+n+i}{n}}

= ∑_{j=0}^{s−1}(−1)^{n+1} \frac{s}{n + 1}\binom{s − 1}{j}a_{j} ∑_{i=0}^{s−1−j}(−1)^{i} \begin{matrix}( & {s−1−j \atop i} & ) \\ ( & {m+n+1+i \atop n+1} & ) \end{matrix}

= ∑_{j=0}^{s−1}(−1)^{n+1} \frac{s}{n + 1}\binom{s − 1}{j}a_{j} \frac{n + 1}{n + s − j} \frac{1}{\binom{m+n+s−j}{m}}

= (−1)^{n+1} ∑_{j=0}^{s−1} \frac{s\binom{s−1}{j}}{(n + s − j)\binom{m+n+s−j}{m}}a_{j}.

These calculations show that Equation (11) is equivalent to

− ∑_{j=0}^{s−1} \frac{\binom{s−1}{j}}{(n + s − j)\binom{m+n+s−j}{m}}a_{j} = ∑_{j=0}^{s−1}(−1)^{s−j} \frac{\binom{s−1}{j}}{(m + n + s − j)\binom{m+n+s−j−1}{n}}b_{j}. (12)

Set s → s + 1 to obtain

∑_{j=0}^{s} \frac{\binom{s}{j}a_{j}}{(n + s + 1 − j)\binom{m+n+s+1−j}{m}} = ∑_{j=0}^{s} \frac{(−1)^{s−j}\binom{s}{j}b_{j}}{(m + n + s + 1 − j)\binom{m+n+s−j}{n}}.  (13)

Since (n + s + 1 − j)\binom{m+n+s+1−j}{m} = (m + n + s + 1 − j)\binom{m+n+s−j}{m}, we see that Equation (13) is equivalent to Equation (9).

References

[1] L. Carlitz, Problem 795, Math. Mag. 44 (1971), 107.

[2] A. G. Shannon, Solution of Problem 795, Math. Mag. 45 (1972), 55–56.

[3] K. W. Chen, Identities from the binomial transform, J. Number Theory 124 (2007), 142–150.

[4] R. Frisch, Sur les semi-invariants et moments employés dans l’étude des distributions statistiques, Skrifter utgitt av Det Norske Videnskaps-Akademi i Oslo, II. Historisk-Filosofisk Klasse, 1926, No. 3, 87 pp.

[5] Eugen Netto, Lehrbuch der Combinatorik, 2nd edition, 1927. Reprinted by Chelsea, 1958.

[6] H. W. Gould, Combinatorial Identities, A Standardized Set of Tables Listing 500 Binomial Coefficient Summations, revised edition. Published by the author, Morgantown, WV, 1972.

[7] H. W. Gould, A curious identity which is not so curious. Math. Gaz. 88 (2004), 87.

[8] S. Simons, A curious identity, Math. Gaz. 85 (2001), 296–298.

[9] P. Vassilev and M. Vassilev-Missana, On the sum of equal powers of the first n terms of an arbitrary arithmetic progression, Notes on Number Theory and Discrete Mathematics 11 (2005), 15–21.

[10] P. Vassilev and M. Vassilev-Missana, On one remarkable identity involving Bernoulli numbers, Notes on Number Theory and Discrete Mathematics 11 (2005), 22–24.

2010 Mathematics Subject Classification: Primary 11B68; Secondary 05A10, 11B65. Keywords: Bernoulli number, binomial transform.

(Concerned with sequences \underline{A027641} and \underline{A027642}.)

Received October 2 2013; revised version received January 3 2014. Published in Journal of Integer Sequences, January 3 2014.

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