Abstract

In this note, starting with a little-known result of Kuo, I derive a recurrence relation for the Bernoulli numbers B_2n , n being a positive integer. This formula is shown to be advantageous in comparison to other known formulae for the exact symbolic computation of B_2n. Interestingly, it is suitable for large values of n since it allows the computation of both B_4n and B_4n+2 from only B₀, B₂, ..., B_2n.

1. Introduction

The Bernoulli numbers B_n, n being a nonnegative integer, can be defined by the generating function

(1)

which holds for all complex z with |z| < 2. The first few values are B₀ = 1, B₁ = -1/2, and B₂ = 1/6. It is well-known that B_n = 0 for odd values of n, n>1. For even values of n, n>1, the numbers B_n form a subsequence of non-null rational numbers such that (-1)^m+1 B_2m>0 for all positive integer m. In other words, the entire subsequence of non-null Bernoulli numbers up to any B_n, n>1, consists of B₀ and B₁, given above, and the preceding numbers B_2m, m = 1, ..., n/2. The basic properties of these numbers can be found, e.g., in Sec. 9.61 of ¹.

The numbers B_n appear in many instances in pure and applied mathematics, most notably in number theory, finite differences calculations, and asymptotic analysis. They also appear as coefficients in the Euler-Maclaurin formula, thus being important for accurate numerical evaluation of special functions. Therefore, numerical computations of B_n are of great mathematical interest. To this end, recurrence formulae were soon recognized as the most efficient tool ². One of the simplest such formulas is found by multiplying both sides of Eq. (1) by e^x-1, using the Cauchy product with the Maclaurin series for e^x-1, and equating the coefficients, which yields

(2)

This kind of recurrence has the disadvantage of demanding the previous knowledge of every B₀, B₁, ...,B_n-1 for the computation of B_n. On searching for more efficient formulae in literature, one finds shortened recurrence relations of two different types. The first type consists of lacunary recurrences, in which B_n is determined only from every second, or every third, etc., preceding B_n (see, e.g., the lacunary formula by Ramanujan in ³). The second type demands the knowledge of the second-half of Bernoulli numbers up to B_n-1 in order to compute B_n ⁴. For an extensive study of these and other recurrence relations involving the numbers B_n, see ⁵.

Here in this note, I apply the Euler's formula relating the even zeta value (2n) to B_2n to a recurrence formula for (2n) obtained by Kuo in ⁶ in order to convert it into a new recurrence formula for B_2n. This reveals a third type of recurrence for the Bernoulli numbers, in the sense that it allows us to compute both B_4n and B_4n+2 from only the first-half of the preceding numbers, i.e. B₀, B₂, ..., B_2n.

2. Recurrence Relation for Bernoulli Numbers

For complex values of s with Re(s)>1, the Riemann zeta function is defined as In this domain, the convergence of this series is guaranteed by the integral test. For s=1, one has the well-known harmonic serie , which diverges to infinity. For positive even values of s, one has a celebrated formula by Euler ⁷:

(3)

Since (-1)^n-1 B_2n is a positive rational for all integer n>0, then (2n) is a positive rational multiple of ²ⁿ. The function (s), as defined above, can be extended to the entire complex plane (except for the only simple pole at s=1) by analytic continuation, which yields (0) = -1/2 ⁸. Alternatively, we can use a globally convergent series representation due to Hasse (1930) ⁹:

(4)

valid for all complex s 1 +(2m)i/ln(2), m being an integer. At s=0, it reduces to (0) = -1/2, which shows that Eq. (3) is also valid for n=0.

These are the ingredients we need to state our first lemma, which comes from a little-known recurrence formula by Kuo (1949) ⁶, written here in terms of even zeta values.

Lemma 1 (Kuo's recurrence formula). For any positive integer n, one has

This formula is proved in ⁶ by developing successive integrations (from 0 to x) of the Fourier series , which converges to (-x)/2 for all x (0,2), and then applying Parseval's theorem.

We are now in a position to state and prove our main result. In what follows, (λ)_n:= λ (λ+1) (λ+n-1) is the Pochhammer symbol.

Theorem 1 (Recurrence for B_2n). For any integer n>0, the following formula holds:

where

Proof. On dividing both sides of Kuo's formula, our Lemma 1, by (2)²ⁿ, one finds

(5)

From Eq. (3), we know that , which is valid for all integer m 0. By substituting this in Eq. (5), one finds

(6)

Now, by factoring (-1)^n-1 one finds

(7)

On writing the first term in the right-hand side as (2n)!/[4(2n-1)(n-1)!²] and putting (2n)!/(n-1)! in evidence (out of the square brackets), the desired result follows.

For instance, when n=1, since (1)₁ = 1 and B₀=1, one readily finds

(8)

For larger values of n, a more efficient, fast computation is achieved when one avoids the repeated computation of the factorials in our Theorem 1. For this, it is advantageous to make use of the algorithm given below. A short Maple code implementation of this algorithm is freely available in the journal website (file B4mB4mp2.mws).

Of course, more efficient numerical routines for floating-point computation of B_2n can be found in some recent works (see, e.g., ^{10, 11, 12}), but we are interested here only in exact symbolic computations of the rational numbers B_2n, which are important for implementation in modern mathematical software such as Maple and Mathematica. Admittedly, most algorithms in ¹² can, a priori, be implemented in a manner to yield exact symbolic results.

For completion, the algorithm corresponding to an efficient implementation of the recurrence formula given in our Theorem 1 is presented below.

Acknowledgements

The author wishes to thank M. R. Javier for checking computationally all results in this work.

References