Bernoulli number

In mathematics, the Bernoulli numbers are a sequence of rational numbers with deep connections to number theory. They are closely related to the values of the Riemann zeta function at negative integers.

In Europe, they were first studied by Jakob Bernoulli, after whom they were named by Abraham de Moivre. In Japan, perhaps earlier, they were independently discovered by Seki Kōwa. They appear in the Taylor series expansions of the tangent and hyperbolic tangent functions, in the Euler–Maclaurin formula, and in expressions for certain values of the Riemann zeta function.

In note G of Ada Lovelace's notes on the Analytical Engine from 1842, Lovelace describes an algorithm for generating Bernoulli numbers with Babbage's machine . As a result, the Bernoulli numbers have the distinction of being the subject of the first computer program.

Introduction

The Bernoulli numbers B_n, sometimes denoted b_n. were first discovered in connection with the closed forms of the sums

$\sum _{k=0}^{m-1}k^{n}=0^{n}+1^{n}+2^{n}+\cdots +{(m-1)}^{n}$

for various fixed values of n. The closed forms are always polynomials in m of degree n + 1. The coefficients of these polynomials are closely related to the Bernoulli numbers, in connection with Faulhaber's formula:

$\sum _{k=0}^{m-1}k^{n}={1 \over {n+1}}\sum _{k=0}^{n}{n+1 \choose {k}}B_{k}m^{n+1-k}.$

For example, taking n to be 1,

$0+1+2+\cdots +(m-1)={\frac {1}{2}}\left(B_{0}m^{2}+2B_{1}m^{1}\right)={\frac {1}{2}}\left(m^{2}-m\right).$

Generating functions

The Bernoulli numbers may also be defined using generating functions. Their exponential generating function is x/(e^x − 1), so that:

${\frac {x}{e^{x}-1}}=\sum _{n=0}^{\infty }B_{n}{\frac {x^{n}}{n!}}$

for all values of x of absolute value less than 2π (the radius of convergence of this power series).

These definitions can be shown to be equivalent using mathematical induction. The initial condition $B_{0}=1$ is immediate from L'Hôpital's rule. To obtain the recurrence, multiply both sides of the equation by $e^{x}-1$. Then, using the Taylor series for the exponential function,

$x=\left(\sum _{j=1}^{\infty }{\frac {x^{j}}{j!}}\right)\left(\sum _{k=0}^{\infty }{\frac {B_{k}x^{k}}{k!}}\right).$

By expanding this as a Cauchy product and rearranging slightly, one obtains

$x=\sum _{m=0}^{\infty }\left(\sum _{j=0}^{m}{m+1 \choose j}B_{j}\right){\frac {x^{m+1}}{(m+1)!}}.$

It is clear from this last equality that the coefficients in this power series satisfy the same recurrence as the Bernoulli numbers.

Other forms and conventions

One may also write

$\sum _{k=0}^{m-1}k^{n}={\frac {B_{n+1}(m)-B_{n+1}(0)}{n+1}},$

where B_n + 1(m) is the (n + 1)th-degree Bernoulli polynomial.

Bernoulli numbers may be calculated by using the following recursive formula:

$\sum _{j=0}^{m}{m+1 \choose {j}}B_{j}=0$

for m > 0, and B₀ = 1.

An alternative convention for the Bernoulli numbers is to set B₁ = 1/2 rather than −1/2. If this convention is used, then all the Bernoulli numbers may be calculated by a different recursive formula without further qualification:

$\sum _{j=0}^{m}{m+1 \choose {j}}B_{j}=m+1.$

Dividing both sides by m + 1 then gives a form suggestive of the connection with the Riemann zeta function if the j=0 case is understood as a limit to deal with the pole at ζ(1):

$\sum _{j=0}^{m}{m \choose {j-1}}{\frac {B_{j}}{j}}=1$

The terms of this sum are the coefficients given by Faulhaber's formula for the closed form of $\sum _{x=1}^{n}x^{m}$ and so this recursive definition is merely reflecting the fact that these sums evaluate to 1 when n=1 for any m. In the alternative convention, the generating function is

${\frac {xe^{x}}{e^{x}-1}}.$

Values of the Bernoulli numbers

B_n = 0 for all odd n other than 1. B₁ = 1/2 or −1/2 depending on the convention adopted (see below). The first few non-zero Bernoulli numbers (sequences OEIS: A027641 and OEIS: A027642 in OEIS) and some larger ones are listed below.

n	N	D	Bn = N / D
0	1	1	+1.00000000000
1	-1	2	-0.50000000000
2	1	6	+0.16666666667
4	-1	30	-0.03333333333
6	1	42	+0.02380952381
8	-1	30	-0.03333333333
10	5	66	+0.07575757576
12	-691	2730	-0.25311355311
14	7	6	+1.16666666667
16	-3617	510	-7.09215686275
18	43867	798	+54.9711779448

Efficient computation of Bernoulli numbers

In some applications it is useful to be able to compute the Bernoulli numbers B₀ through B_p − 3 modulo p, where p is a prime; for example to test whether Vandiver's conjecture holds for p, or even just to determine whether p is an irregular prime. It is not feasible to carry out such a computation using the above recursive formulae, since at least (a constant multiple of) p² arithmetic operations would be required. Fortunately, faster methods have been developed (Buhler et al. 2001) which require only O(p (log p)²) operations (see big-O notation).

David Harvey (Harvey 2008) describes an algorithm for computing Bernoulli numbers by computing B_n modulo p for many small primes p, and then reconstructing B_n via the Chinese Remainder Theorem. Harvey writes that the asymptotic time complexity of this algorithm is O(n² log(n)^2+eps) and claims that this implementation is significantly faster than implementations based on other methods. Harvey's implementation is included in Sage since version 3.1. Using this implementation Harvey computed B_n for n = 10⁸, which is a new record (October 2008). Prior to that Bernd Kellner (Kellner 2002) computed B_n to full precision for n = 10⁶ on December 2002 and Oleksandr Pavlyk (Pavlyk 2008) for n = 10⁷ with 'Mathematica' on April 2008.

History of the computation of Bernoulli numbers

Computer	Year	n	Digits
J. Bernoulli	~1689	10	1
L. Euler	1748	30	8
J.C. Adams	1878	62	36
D.E. Knuth, T.J. Buckholtz	1967	360	478
G. Fee, S. Plouffe	1996	10000	27677
G. Fee, S. Plouffe	1996	100000	376755
B.C. Kellner	2002	1000000	4767529
O. Pavlyk	2008	10000000	57675260
D. Harvey	2008	100000000	676752569

Digits is to be understood as the exponent of 10 when B(n) is written as a real in normalized scientific notation.

In 2009, a sixteen year-old Iraqi immigrant living in Sweden submitted a proof to Uppsala University, correctly simplifying the computation of Bernoulli numbers. Mathematicians claim this proof has been discovered before and is well known in the field.

Reconstruction of 'Summae Potestatum'

Jakob Bernoulli's Summae Potestatum, 1713

The Bernoulli numbers were introduced by Jakob Bernoulli in the book 'Ars Conjectandi' published posthumously in 1713. The main formula can be seen in the second half of the facsimile given above. The constant coefficients denoted A, B, C and D by Bernoulli are mapped to the notation which is now prevalent as A = B₂, B = B₄, C = B₆, D = B₈. In the expression c·c−1·c−2·c−3 the small dots are used as grouping symbols, not as signs for multiplication. Using today's terminology these expressions are falling factorial powers $c^{\underline {k}}$. The factorial notation k! as a shortcut for 1×2×...×k was not introduced until 100 years later. The integral symbol on the left hand side goes back to Gottfried Wilhelm Leibniz in 1675 who used it as a long letter S for "summa" (sum). (The Mathematics Genealogy Project shows Leibniz as the doctoral adviser of Jakob Bernoulli. See also the Earliest Uses of Symbols of Calculus .) The letter n on the left hand side is not an index of summation but gives the upper limit of the range of summation which is to be understood as 1,2,...,n. Putting things together, for positive c, today a mathematician is likely to write Bernoulli's formula as

$\sum _{0<k\leq n}k^{c}={\frac {n^{c+1}}{c+1}}+{\frac {1}{2}}n^{c}+\sum _{k\geq 2}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}\ .$

In fact this formula imperatively suggests to set B₁ = 1/2 when switching from the so called 'archaic' enumeration which uses only the even indices 2,4,.. to the modern form (more on different conventions in the next paragraph). Most striking in this context is the fact that the falling factorial $c^{\underline {k-1}}$ has for k = 0 the value 1/(c + 1). (See the explanation in .) Thus Bernoulli's formula can and has to be written

$\sum _{0<k\leq n}k^{c}=\sum _{k\geq 0}{\frac {B_{k}}{k!}}c^{\underline {k-1}}n^{c-k+1}\ .$

if B₁ stands for the value Bernoulli himself has given to the coefficient at that position.

Different viewpoints and conventions

The Bernoulli numbers can be regarded from four main viewpoints:

• as standalone arithmetical objects,
• as combinatorial objects,
• as values of a sequence of certain polynomials,
• as values of the Riemann zeta function.

Each of these viewpoints leads to a set of more or less different conventions.

• Bernoulli numbers as standalone arithmetical objects.
  Associated sequence: 1/6, −1/30, 1/42, −1/30,...
  This is the viewpoint of Jakob Bernoulli. (See the cutout from his Ars Conjectandi, first edition, 1713). The Bernoulli numbers are understood as numbers, recursive in nature, invented to solve a certain arithmetical problem, the summation of powers, which is the paradigmatic application of the Bernoulli numbers. It is misleading to call this viewpoint 'archaic'. For example Jean-Pierre Serre uses it in his highly acclaimed book A Course in Arithmetic which is a standard textbook used at many universities today.
• Bernoulli numbers as combinatorial objects.
  Associated sequence: 1, +1/2, 1/6, 0,....
  This view focuses on the connection between Stirling numbers and Bernoulli numbers and arises naturally in the calculus of finite differences. In its most general and compact form this connection is summarized by the definition of the Stirling polynomials σ_n(x), formula (6.52) in Concrete Mathematics by Graham, Knuth and Patashnik.

  $\left({\frac {ze^{z}}{e^{z}-1}}\right)^{x}=x\sum _{n\geq 0}\sigma _{n}(x)z^{n}$

  In consequence B_n = n! σ_n(1) for n ≥ 0.
• Bernoulli numbers as values of a sequence of certain polynomials.
  Assuming the Bernoulli polynomials as already introduced the Bernoulli numbers can be defined in two different ways:
  B_n = B_n(0). Associated sequence: 1, −1/2, 1/6, 0,....
  B_n = B_n(1). Associated sequence: 1, +1/2, 1/6, 0,....
  The two definitions differ only in the sign of B₁. The choice B_n = B_n(0) is the convention used in the Handbook of Mathematical Functions.
• Bernoulli numbers as values of the Riemann zeta function.
  Associated sequence: 1, +1/2, 1/6, 0,....
  

  

  The Bernoulli numbers as given by the Riemann zeta function.

  This convention agrees with the convention B_n = B_n(1) (for example J. Neukirch and M. Kaneko). The sign '+' for B₁ matches the representation of the Bernoulli numbers by the Riemann zeta function. In fact the identity nζ(1−n) = (−1)ⁿ⁺¹B_n valid for all n > 0 is then replaced by the simpler nζ(1−n) = -B_n. (See the paper of S. C. Woon.)
  (Note that in the foregoing equation for n = 0 and n = 1 the expression −nζ(1 − n) is to be understood as lim_x → n −xζ(1 − x).)

$\ B_{n}=n!\sigma _{n}(1)=B_{n}(1)=-n\zeta (1-n)\quad (n\geq 0)\$

Application of the Bernoulli number

Arguably the most important application of the Bernoulli number in mathematics is their use in the Euler-MacLaurin formula. Assuming that f is a sufficiently often differentiable function the Euler-MacLaurin formula can be written as

$\sum \limits _{a\leq k<b}f(k)=\int _{a}^{b}f(x)dx\ +\sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

This formulation assumes the convention B₁ = -1/2. However, if one sets B₁ = 1/2 then this formula can also be written as

$\sum \limits _{a<k\leq b}f(k)=\int _{a}^{b}f(x)dx\ \ +\ \sum \limits _{k=1}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

Here f⁽⁰⁾ = f which is a commonly used notation identifying the zero-th derivative of f with f. Moreover, let f^(-1) denote an antiderivative of f. By the fundamental theorem of calculus $\int _{a}^{b}f(x)dx\ =f^{(-1)}(b)-f^{(-1)}(a)$. Thus the last formula can be further simplified to the following succinct form of the Euler-Maclaurin formula

$\sum \limits _{a<k\leq b}f(k)=\sum \limits _{k=0}^{m}{\frac {B_{k}}{k!}}\left(f^{(k-1)}(b)-f^{(k-1)}(a)\right)+R(f,m)\ .$

This form is for example the source for the important Euler-MacLaurin expansion of the zeta function

${\begin{aligned}\zeta (s)&=\sum _{k=0}^{m}{\frac {{B}_{k}}{k!}}s^{\overline {k-1}}+R(s,m)\\&={\frac {{B}_{0}}{0!}}s^{\overline {-1}}+{\frac {{B}_{1}}{1!}}s^{\overline {0}}+{\frac {{B}_{2}}{2!}}s^{\overline {1}}+\cdots +R(s,m)\\&={\frac {1}{s-1}}+{\frac {1}{2}}+{\frac {1}{12}}s+\cdots +R(s,m)\ .\\\end{aligned}}$

Here $s^{\overline {k}}$ denotes the rising factorial power.

Combinatorial definitions

The connection of the Bernoulli number to various kinds of combinatorial numbers is based on the classical theory of finite differences and on the combinatorial interpretation of the Bernoulli numbers as an instance of a fundamental combinatorial principle, the inclusion-exclusion principle.

Connection with the Worpitzky number

The definition to proceed with was developed by Julius Worpitzky in 1883. Besides elementary arithmetic only the factorial function n! and the power function k^m is employed. The signless Worpitzky numbers are defined as

$W_{n,k}=\sum _{v=0}^{k}(-1)^{v+k}\left(v+1\right)^{n}{\frac {k!}{v!(k-v)!}}\ .$

They can also be expressed through the Stirling set number

$W_{n,k}=k!\left\{{\begin{matrix}n+1\\k+1\end{matrix}}\right\}.$

A Bernoulli number is then introduced as an inclusion-exclusion sum of Worpitzky numbers weighted by the sequence 1, 1/2, 1/3,...

$B_{n}=\sum _{k=0}^{n}(-1)^{k}{\frac {W_{n,k}}{k+1}}\ =\ \sum _{k=0}^{n}{\frac {1}{k+1}}\sum _{v=0}^{k}(-1)^{v}{k \choose v}\left(v+1\right)^{n}\ .$

This representation has B₁ = 1/2.

Worpitzky's representation of the Bernoulli number
B0	=	1/1
B1	=	1/1 − 1/2
B2	=	1/1 − 3/2 + 2/3
B3	=	1/1 − 7/2 + 12/3 − 6/4
B4	=	1/1 − 15/2 + 50/3 − 60/4 + 24/5
B5	=	1/1 − 31/2 + 180/3 − 390/4 + 360/5 − 120/6
B6	=	1/1 − 63/2 + 602/3 − 2100/4 + 3360/5 − 2520/6 + 720/7

Connection with the Stirling set number

A similar combinatorial representation derives from

$B_{n}=\sum _{k=0}^{n}(-1)^{k+n}{\frac {k!}{k+1}}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}\ .$

Here the Bernoulli numbers are an inclusion-exclusion over the set of length-n words, where the sum is taken over all words of length n with k distinct letters, and normalized by k + 1. The combinatorics of this representation can be seen from:

$B_{0}=+\left\vert \ \left\{\varnothing \right\}\ \right\vert$

$B_{1}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{a\right\}\right\vert$

$B_{2}=+{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert -{\frac {1}{2}}\left\vert \left\{aa\right\}\right\vert +{\frac {1}{3}}\left\vert \left\{ab,ba\right\}\right\vert$

$B_{3}=-{\frac {1}{1}}\left\vert \ \varnothing \ \right\vert +{\frac {1}{2}}\left\vert \left\{aaa\right\}\right\vert -{\frac {1}{3}}\left\vert \left\{aab,aba,baa,abb,bab,bba\right\}\right\vert +{\frac {1}{4}}\left\vert \left\{abc,acb,bac,bca,cab,cba\right\}\right\vert$

Connection with the Stirling cycle number

Let $\left[{\begin{matrix}n\\m\end{matrix}}\right]$ denote the signless Stirling cycle number. The two main formulas relating these number to the Bernoulli number (B₁ = 1/2) are

${\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{\begin{matrix}m+1\\k+1\end{matrix}}\right]B_{k}={\frac {1}{m+1}}\ ,$

and the inversion of this sum (for n ≥ 0, m ≥ 0)

${\frac {1}{m!}}\sum _{k=0}^{m}(-1)^{k}\left[{\begin{matrix}m+1\\k+1\end{matrix}}\right]B_{n+k}=A_{n,m}\ .$

Here the number A_n,m are the rational Akiyama-Tanigawa number, the first few of which are displayed in the following table.

Akiyama-Tanigawa number
n \ m	0	1	2	3	4
0	1	1/2	1/3	1/4	1/5
1	1/2	1/3	1/4	1/5	...
2	1/6	1/6	3/20	...	...
3	0	1/30	...	...	...
4	-1/30	...	...	...	...

These relations lead to a simple algorithm to compute the Bernoulli number. The input is the first row, A_0,m = 1/(m + 1) and the output are the Bernoulli number in the first column A_n,0 = B_n . This transformation is shown in pseudo-code below.

Akiyama-Tanigawa algorithm for Bn

Enter integer n. For m from 0 by 1 to n do     A[m] ← 1/(m + 1)     For j from m by −1 to 1 do         A[j − 1] ← j × (A[j − 1] − A[j]) Return A[0]  (which is Bn).

Connection with the Eulerian number

The Eulerian number are the number of permutations of {1,2,...,n} with m ascents. They were introduced by Leonhard Euler in 1755 and in the notation of D. E. Knuth written as $\left\langle {n \atop m}\right\rangle .$ The two main formulas connecting the Eulerian number to the Bernoulli number are:

$\sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }=2^{n+1}(2^{n+1}-1){\frac {B_{n+1}}{n+1}}\ ,$

$\sum _{m=0}^{n}(-1)^{m}{\left\langle {n \atop m}\right\rangle }{\binom {n}{m}}^{-1}=(n+1)B_{n}\ .$

Both formulas are valid for n ≥ 0 if B₁ is set to 1/2. If B₁ is set to −1/2 they are valid only for n ≥ 1 and n ≥ 2 respectively.

Asymptotic approximation

Leonhard Euler expressed the Bernoulli numbers in terms of the Riemann zeta function as

$B_{2n}=(-1)^{n+1}{\frac {2(2n)!}{(2\pi )^{2n}}}\left[1+{\frac {1}{2^{2n}}}+{\frac {1}{3^{2n}}}+{\frac {1}{4^{2n}}}+\cdots \;\right].$

It then follows from the Stirling formula that, as n goes to infinity,

$|B_{2n}|\sim 4{\sqrt {\pi n}}\left({\frac {n}{\pi e}}\right)^{2n}.$

Integral representation and continuation

The integral

$b(s)=2e^{si\pi /2}\int _{0}^{\infty }{\frac {st^{s}}{1-e^{2\pi t}}}{\frac {dt}{t}}$

has as special values b(2n) = B_2n for n > 0. The integral might be considered as a continuation of the Bernoulli numbers to the complex plane and this was indeed suggested by Peter Luschny in 2004.

For example b(3) = (3/2)ζ(3)Π⁻³Ι and b(5) = −(15/2) ζ(5) Π ⁻⁵Ι. Here ζ(n) denotes the Riemann zeta function and Ι the imaginary unit. It is remarkable that already Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

$p={\frac {3}{2\pi ^{3}}}\left(1+{\frac {1}{2^{3}}}+{\frac {1}{3^{3}}}+{\text{etc.}}\ \right)=0.0581522\ldots \ ,$

$q={\frac {15}{2\pi ^{5}}}\left(1+{\frac {1}{2^{5}}}+{\frac {1}{3^{5}}}+{\text{etc.}}\ \right)=0.0254132\ldots \ .$

Euler's values are unsigned and real, but obviously his aim was to find a meaningful way to define the Bernoulli numbers at the odd integers n > 1.

The relation to the Euler numbers and π

The Euler numbers are a sequence of integers intimately connected with the Bernoulli numbers. Comparing the asymptotic expansions of the Bernoulli and the Euler numbers shows that the Euler numbers E_2n are in magnitude approximately (2/π)(4²ⁿ − 2²ⁿ) times larger than the Bernoulli numbers B_2n. In consequence:

$\pi \ \sim \ 2\left(2^{2n}-4^{2n}\right){\frac {B_{2n}}{E_{2n}}}\ .$

This asymptotic equation reveals that π lies in the common root of both the Bernoulli and the Euler numbers. In fact π could be computed from these rational approximations.

Bernoulli numbers can be expressed through the Euler numbers and vice versa. Since for n odd B_n = E_n = 0 (with the exception B₁), it suffices to regard the case when n is even.

$B_{n}=\sum _{k=0}^{n-1}{\binom {n-1}{k}}{\frac {n}{4^{n}-2^{n}}}E_{k}\quad (n=2,4,6,\ldots )$

$E_{n}=\sum _{k=1}^{n}{\binom {n}{k-1}}{\frac {2^{k}-4^{k}}{k}}B_{k}\quad (n=2,4,6,\ldots ).$

These conversion formulas express an inverse relation between the Bernoulli and the Euler numbers. But more important, there is a deep arithmetic root common to both kinds of numbers, which can be expressed through a more fundamental sequence of numbers, also closely tied to π. These numbers are defined for n > 1 as

$S_{n}=2\left({\frac {2}{\pi }}\right)^{n}\sum _{k=-\infty }^{\infty }\left(4k+1\right)^{-n}\quad (k=0,-1,1,-2,2,\ldots )$

and S₁ = 1 by convention. The magic of these numbers lies in the fact that they turn out to be rational numbers. This was first proved by Leonhard Euler 1734 in a landmark paper `De summis serierum reciprocarum' (On the sums of series of reciprocals) and fascinated mathematicians ever since. The first few of these numbers are

$1,1,{\frac {1}{2}},{\frac {1}{3}},{\frac {5}{24}},{\frac {2}{15}},{\frac {61}{720}},{\frac {17}{315}},{\frac {277}{8064}},{\frac {62}{2835}},\ldots$

The Bernoulli numbers and Euler numbers are best understood as special views of these numbers, selected from the sequence S_n and scaled for use in special applications.

$B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n\ \operatorname {even} \right]{\frac {n!}{2^{n}-4^{n}}}\,S_{n}\ ,\quad (n=2,3,\ldots )\ ,$

$E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n\ \operatorname {even} \right]n!\,S_{n+1}\quad \qquad (n=0,1,\ldots )\ .$

The expression [n even] has the value 1 if n is even and 0 otherwise (Iverson bracket).

These identities show that the quotient of Bernoulli and Euler numbers at the beginning of this section is just the special case of R_n = 2 S_n / S_n+1 when n is even. The R_n are rational approximations to π and two successive terms always enclose the true value of π. Beginning with n = 1 the sequence starts

$2,4,3,{\frac {16}{5}},{\frac {25}{8}},{\frac {192}{61}},{\frac {427}{136}},{\frac {4352}{1385}},\ldots \quad \longrightarrow \pi \ .$

These rational numbers also appear in the last paragraph of Euler's paper cited above. But it was only in September 2007 that this classical sequence found its way into the Encyclopedia of Integer Sequences (A132049).

An algorithmic view: the Seidel triangle

The sequence S_n has another unexpected yet important property: The denominators of S_n divide the factorial (n − 1)!. In other words: the numbers T_n = S_n(n − 1)! are integers.

$T_{n}=1,1,1,2,5,16,61,272,1385,7936,50521,353792,\ldots \quad (n=1,2,\ldots )$

Thus the above representations of the Bernoulli and Euler numbers can be rewritten in terms of this sequence as

$B_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n{\text{ even}}\right]{\frac {n}{2^{n}-4^{n}}}\,T_{n}\ ,\quad (n=2,3,\ldots )\ ,$

$E_{n}=(-1)^{\left\lfloor n/2\right\rfloor }\left[n{\text{ even}}\right]T_{n+1}\quad \quad \qquad (n=0,1,\ldots )\ .$

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers E_n are given immediately by T_2n + 1 and the Bernoulli numbers B_2n are obtained from T_2n by some easy shifting, avoiding rational arithmetic.

What remains is to find a convenient way to compute the numbers T_n. However, already in 1877 Philipp Ludwig von Seidel published an ingenious algorithm which makes it extremely simple to calculate T_n.

Seidel's algorithm for Tn

[begin] Start by putting 1 in row 0 and let k denote the number of the row currently being filled. If k is odd, then put the number on the left end of the row k − 1 in the first position of the row k, and fill the row from the left to the right, with every entry being the sum of the number to the left and the number to the upper. At the end of the row duplicate the last number. If k is even, proceed similar in the other direction. [end]

Seidel's algorithm is in fact much more general (see the exposition of Dominique Dumont (1981)) and was rediscovered several times thereafter.

Similar to Seidel's approach D. E. Knuth and T. J. Buckholtz (1967) gave a recurrence equation for the numbers T_2n and recommended this method for computing B_2n and E_2n ‘on electronic computers using only simple operations on integers’.

V. I. Arnold rediscovered Seidel's algorithm in 1991 and later Millar, Sloane and Young popularized Seidel's algorithm under the name boustrophedon transform.

A combinatorial view: alternating permutations

Main article: Alternating permutations

Around 1880, three years after the publication of Seidel's algorithm, Désiré André proved a now classic result of combinatorial analysis. Looking at the first terms of the Taylor expansion of the trigonometric functions tan x and sec x André made a startling discovery.

$\tan x=1{\frac {x}{1!}}+2{\frac {x^{3}}{3!}}+16{\frac {x^{5}}{5!}}+272{\frac {x^{7}}{7!}}+7936{\frac {x^{9}}{9!}}+\cdots$

$\sec x=1+1{\frac {x^{2}}{2!}}+5{\frac {x^{4}}{4!}}+61{\frac {x^{6}}{6!}}+1385{\frac {x^{8}}{8!}}+50521{\frac {x^{10}}{10!}}+\cdots$

The coefficients are the Euler numbers of odd and even index, respectively. In consequence the ordinary expansion of tan x + sec x has as coefficients the rational numbers S_n.

$\tan x+\sec x=1+1x+{\frac {1}{2}}x^{2}+{\frac {1}{3}}x^{3}+{\frac {5}{24}}x^{4}+{\frac {2}{15}}x^{5}+{\frac {61}{720}}x^{6}+\cdots$

André then succeeded by means of a recurrence argument to show that the alternating permutations of odd size are enumerated by the Euler numbers of odd index (also called tangent numbers) and the alternating permutations of even size by the Euler numbers of even index (also called secant numbers).

Generalizations by polynomials

The Bernoulli polynomials can be regarded as generalizations of the Bernoulli numbers the same as the Euler polynomials are generalizations of the Euler numbers.

Arithmetical properties of the Bernoulli numbers

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as B_n = − nζ(1 − n) for integers n ≥ 0 provided for n = 0 and n = 1 the expression − nζ(1 − n) is understood as the limiting value and the convention B₁ = 1/2 is used. This intimately relates them to the values of the zeta function at negative integers. As such, they could be expected to have and do have deep arithmetical properties, a fact discovered by Kummer in his work on Fermat's last theorem.

Divisibility properties of the Bernoulli numbers are related to the ideal class groups of cyclotomic fields by a theorem of Kummer and its strengthening in the Herbrand-Ribet theorem, and to class numbers of real quadratic fields by Ankeny-Artin-Chowla. We also have a relationship to algebraic K-theory; if c_n is the numerator of B_n/2n, then the order of $K_{4n-2}({\mathbb {Z} })$ is −c_2n if n is even, and 2c_2n if n is odd.

The Agoh-Giuga conjecture postulates that p is a prime number if and only if pB_p−1 is congruent to −1 mod p.

Von Staudt–Clausen theorem

The von Staudt–Clausen theorem was given by Karl Georg Christian von Staudt and Thomas Clausen independently in 1840. It describes the arithmetical structure of the Bernoulli numbers.

The von Staudt–Clausen theorem has two parts. The first one describes how the denominators of the Bernoulli numbers can be computed. Paraphrasing the words of Clausen it can be stated as:

“The denominator of the 2nth Bernoulli number can be found as follows: Add to all divisors of 2n, 1, 2, a, a', ..., 2n the unity, which gives the sequence 2, 3, a + 1, a' + 1, ..., 2n + 1. Select from this sequence only the prime numbers 2, 3, p, p', etc. and build their product.”

Clausen's algorithm translates almost verbatim to a modern computer algebra program, which looks similar to the pseudocode on the left hand site of the following table. On the right hand side the computation is traced for the input n = 88. It shows that the denominator of B₈₈ is 61410. 

Clausen's algorithm for the denominator of Bn
Clausen: function(integer n)	|	n = 88
S = divisors(n);	|	{1, 2, 4, 8, 11, 22, 44, 88}
S = map(k → k + 1, S);	|	{2, 3, 5, 9, 12, 23, 45, 89}
S = select(isprime, S);	|	{2, 3, 5, 23, 89}
return product(S);	|	61410

The second part of the von Staudt–Clausen theorem is a very remarkable representation of the Bernoulli numbers. This representation is given for the first few nonzero Bernoulli numbers in the next table.

Von Staudt–Clausen representation of Bn
B0	=	1
B1	=	1 − 1/2
B2	=	1 − 1/2 − 1/3
B4	=	1 − 1/2 − 1/3 − 1/5
B6	=	1 − 1/2 − 1/3 − 1/7
B8	=	1 − 1/2 − 1/3 − 1/5
B10	=	1 − 1/2 − 1/3 − 1/11

The theorem affirms the existence of an integer I_n such that

$B_{n}={\frac {I_{n}}{2}}-\sum _{(p-1)|n}{\frac {1}{p}}\quad (n\geq 0)\ .$

The sum is over the primes p for which p − 1 divides n. These are the same primes which are employed in the Clausen algorithm. The proposition holds true for all n ≥ 0, not only for even n. I₁ = 2 and for odd n > 1, I_n = 1.

Consequences of the von Staudt–Clausen theorem are: the denominators of the Bernoulli numbers are square-free and for n ≥ 2 divisible by 6.

Why do the odd Bernoulli numbers vanish?

The sum

$\varphi _{k}(n)=\sum _{i=0}^{n}i^{k}-{\frac {n^{k}}{2}}$

can be evaluated for negative values of the index n. Doing so will show that it is an odd function for even values of k, which implies that the sum has only terms of odd index. This and the formula for the Bernoulli sum imply that B_2k+1−m is 0 for m odd and greater than 1; and that the term for B₁ is cancelled by the subtraction. The von Staudt-Clausen theorem combined with Worpitzky's representation also gives a combinatorial answer to this question (valid for n > 1).

From the von Staudt-Clausen theorem it is known that for odd n > 1 the number 2B_n is an integer. This seems trivial if one knows beforehand that in this case B_n = 0. However, by applying Worpitzky's representation one gets

$2B_{n}=\sum _{m=0}^{n}\left(-1\right)^{m}{\frac {2}{m+1}}m!\left\{{\begin{matrix}n+1\\m+1\end{matrix}}\right\}=0\quad \left(n>1\ {\text{odd}}\right)$

as a sum of integers, which is not trivial. Here a combinatorial fact comes to surface which explains the vanishing of the Bernoulli numbers at odd index. Let S_n,m be the number of surjective maps from {1, 2, ..., n} to {1, 2, ..., m}, then $S_{n,m}=m!\left\{{\begin{matrix}n\\m\end{matrix}}\right\}$. The last equation can only hold if

$\sum _{m=1,3,5,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}=\sum _{m=2,4,6,\ldots \leq n}{\frac {2}{m^{2}}}S_{n,m}\quad \left(n>2{\text{ even}}\right).\$

This equation can be proved by induction. The first two examples of this equation are

n = 4 :  2 + 8 = 7 + 3,  n = 6:  2 + 120 + 144 = 31 + 195 + 40.

Thus the Bernoulli numbers vanish at odd index because some non-obvious combinatorial identities are embodied in the Bernoulli numbers.

p-adic continuity

An especially important congruence property of the Bernoulli numbers can be characterized as a p-adic continuity property. If b, m and n are positive integers such that m and n are not divisible by p − 1 and $m\equiv n\,{\bmod {\,}}p^{b-1}(p-1)$, then

$(1-p^{m-1}){B_{m} \over m}\equiv (1-p^{n-1}){B_{n} \over n}\,{\bmod {\,}}p^{b}.$

Since $B_{n}=-n\zeta (1-n)$, this can also be written

$(1-p^{-u})\zeta (u)\equiv (1-p^{-v})\zeta (v)\,{\bmod {\,}}p^{b}\,,$

where u = 1 − m and v = 1 − n, so that u and v are nonpositive and not congruent to 1 mod p − 1. This tells us that the Riemann zeta function, with 1 − p^−s taken out of the Euler product formula, is continuous in the p-adic numbers on odd negative integers congruent mod p − 1 to a particular $a\not \equiv 1\,{\bmod {\,}}p-1$, and so can be extended to a continuous function $\zeta _{p}(s)$ for all p-adic integers ${\mathbb {Z}}_{p},\,$ the p-adic Zeta function.

Bernoulli numbers and the Riemann hypothesis

The connection between the Bernoulli numbers and the Riemann zeta function is strong enough to provide an alternate formulation of the Riemann hypothesis (RH) which uses only the Bernoulli number. In fact in 1916 M. Riesz proved that the RH is equivalent to the following assertion:

For every ε > 1/4 there exists a constant C_ε > 0 (depending on ε) such that |R(x)| < C_ε x^ε as x → ∞. Here R(x) is the Riesz function

$R(x)=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\left(B_{2k}/(2k)\right)}}=2\sum _{k=1}^{\infty }{\frac {k^{\overline {k}}x^{k}}{(2\pi )^{2k}\beta _{2k}}}.\$

$n^{\overline {k}}$ denotes the rising factorial power in the notation of D. E. Knuth. The number β_n = B_n/n occur frequently in the study of the zeta function and are significant because β_n is a p-integer for primes p where p − 1 does not divide n. The β_n are called divided Bernoulli number.

Ramanujan's congruences

The following relations, due to Ramanujan, provide a more efficient method for calculating Bernoulli numbers:

$m\equiv 0\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{m/6}{m+3 \choose {m-6j}}B_{m-6j}$

$m\equiv 2\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}={{m+3} \over 3}-\sum _{j=1}^{(m-2)/6}{m+3 \choose {m-6j}}B_{m-6j}$

$m\equiv 4\,{\bmod {\,}}6\qquad {{m+3} \choose {m}}B_{m}=-{{m+3} \over 6}-\sum _{j=1}^{(m-4)/6}{m+3 \choose {m-6j}}B_{m-6j}.$

Use of Bernoulli numbers in topology

The Kervaire-Milnor formula for the order of the cyclic group of diffeomorphism classes of exotic (4n − 1)-spheres which bound parallelizable manifolds for $n\geq 2$ involves Bernoulli numbers; if B_(n) is the numerator of B_4n/n, then

$2^{2n-2}(1-2^{2n-1})B_{(n)}$

is the number of such exotic spheres. (The formula in the topological literature differs because topologists use a different convention for naming Bernoulli numbers; this article uses the number theorists' convention.)

Assorted identities

• Let n be non-negative and even

  $\zeta (n)={\frac {\left(-1\right)^{{\frac {n}{2}}-1}B_{n}\left(2\pi \right)^{n}}{2n!}}$

• The nth cumulant of the uniform probability distribution on the interval [−1, 0] is B_n/n.
• Let n¡ = 1/n! and n ≥ 1. Then B_n is n! times the determinant of the following matrix.

  Bn =

  1 ─── n¡

  2¡ 1¡ 0 0 0 ... 0 3¡ 2¡ 1¡ 0 0 ... 0 4¡ 3¡ 2¡ 1¡ 0 ... 0 ... ... ... ... ... ... ... (n − 2)¡ ... ... 3¡ 2¡ 1¡ 0 (n − 1)¡ (n − 2)¡ ... ... 3¡ 2¡ 1¡ n¡ (n − 1)¡ (n − 2)¡ ... ... 3¡ 2¡

  Thus the determinant is σ_n(1), the Stirling polynomial at x = 1.

• Let n ≥ 1.

  ${\frac {1}{n}}\sum _{k=1}^{n}{\binom {n}{k}}B_{k}B_{n-k}+B_{n-1}=-B_{n}\quad {\text{(L. Euler)}}$

• Let n ≥ 1. Then (von Ettingshausen 1827)

  $\sum _{k=0}^{n}{\binom {n+1}{k}}(n+k+1)B_{n+k}=0$

• Let n ≥ 1 and m ≥ 1. Then (Carlitz 1968)

  $(-1)^{m}\sum _{r=0}^{m}{\binom {m}{r}}B_{n+r}=(-1)^{n}\sum _{s=0}^{n}{\binom {n}{s}}B_{m+s}$

• Let n ≥ 4 and

  $H_{n}=\sum _{1\leq k\leq n}k^{-1}$

  the harmonic number. Then

  ${\frac {n}{2}}\sum _{k=2}^{n-2}{\frac {B_{n-k}}{n-k}}{\frac {B_{k}}{k}}-\sum _{k=2}^{n-2}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}=H_{n}B_{n}\qquad {\text{(H. Miki, 1978)}}$

• Let n ≥ 4. Yuri Matiyasevich found (1997)

  $(n+2)\sum _{k=2}^{n-2}B_{k}B_{n-k}-2\sum _{l=2}^{n-2}{\binom {n+2}{l}}B_{l}B_{n-l}=n(n+1)B_{n}$

• Let n ≥ 1

  ${\frac {n}{2}}\left(B_{n-1}(x)+\sum _{k=1}^{n-1}{\frac {B_{k}(x)}{k}}{\frac {B_{n-k}(x)}{n-k}}\right)-\sum _{k=0}^{n-1}{\binom {n}{k}}{\frac {B_{n-k}}{n-k}}B_{k}(x)=H_{n-1}B_{n}(x)$

  This is an identity by Faber-Pandharipande-Zagier-Gessel. Choose x = 0 or x = 1 to get a Bernoulli number identity according to your favourite convention.
• The next formula is true for n ≥ 0 if B₁ = B₁(1) = 1/2, but only for n ≥ 1 if B₁ = B₁(0) = -1/2.

  $\sum _{k=0}^{n}{\binom {n}{k}}{\frac {B_{k}}{n-k+2}}={\frac {B_{n+1}}{n+1}}$

• Let n ≥ 0 and [b] = 1 if b is true, 0 otherwise.

  $-1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(1)=2^{n}$

  and

  $-1+\sum _{k=0}^{n}{\binom {n}{k}}{\frac {2^{n-k+1}}{n-k+1}}B_{k}(0)=[n=0]$

See also

• poly-Bernoulli number
• q-Bernoulli number
• Bernoulli polynomials
• Riemann zeta function
• Hurwitz zeta function
• Euler number
• Euler summation

References

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External links

• The first 498 Bernoulli Numbers from Project Gutenberg
• The first 10,000 Bernoulli numbers
• A multimodular algorithm for computing Bernoulli numbers
• The Bernoulli Number Page
• Bernoulli number programs at LiteratePrograms
• Dilcher, K.; Skula, L.; Slavutskii, I. Sh. (1991), "Bernoulli numbers. Bibliography (1713-1990)", Queen's Papers in Pure and Applied Mathematics (87), Kingston, Ontario
• Weisstein, Eric W. "Bernoulli Number". MathWorld.
• The Computation of Irregular Primes
• Online Bernoulli Numbers Generator
• Some properties,sums of Bernoulli-and related numbers