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# 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 Takakazu. 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 Byron's notes on the analytical engine from 1842, there is an algorithm for computer-generated Bernoulli numbers. This distinguishes the Bernoulli numbers as being the subject of the first ever published computer program.

## Introduction

The Bernoulli numbers Bn (often denoted bn to distinguish them from the Bell numbers) were first discovered in connection with the closed forms of the sums

$sum_\left\{k=0\right\}^\left\{m-1\right\} k^n = 0^n + 1^n + 2^n + cdots + \left\{\left(m-1\right)\right\}^n$

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:

$sum_\left\{k=0\right\}^\left\{m-1\right\} k^n = \left\{1over\left\{n+1\right\}\right\}sum_\left\{k=0\right\}^n\left\{n+1choose\left\{k\right\}\right\} B_k m^\left\{n+1-k\right\}.$

For example, taking n to be 1,

$0 + 1 + 2 + ... + \left(m-1\right) = frac\left\{1\right\}\left\{2\right\}left\left(B_0 m^2+2B_1 m^1right\right) = frac\left\{1\right\}\left\{2\right\}left\left(m^2-mright\right).$

### Generating functions

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


frac{x}{e^x-1} = sum_{n=0}^{infin} 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,

$x = left\left(sum_\left\{j=1\right\}^\left\{infty\right\} frac\left\{x^j\right\}\left\{j!\right\} right\right) left\left(sum_\left\{k=0\right\}^\left\{infty\right\} frac\left\{B_k x^k\right\}\left\{k!\right\} right\right).$

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

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

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

$sum_\left\{k=0\right\}^\left\{m-1\right\} k^n = frac\left\{B_\left\{n+1\right\}\left(m\right)-B_\left\{n+1\right\}\left(0\right)\right\}\left\{n+1\right\},$

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

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

$sum_\left\{j=0\right\}^m\left\{m+1choose\left\{j\right\}\right\}B_j = 0$

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:

$sum_\left\{j=0\right\}^m\left\{m+1choose\left\{j\right\}\right\}B_j = m+1.$

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):

$sum_\left\{j=0\right\}^m\left\{mchoose\left\{j-1\right\}\right\}frac\left\{B_j\right\}\left\{j\right\} = 1$

The terms of this sum are the coefficients given by Faulhaber's formula for the closed form of $sum_\left\{x=1\right\}^n x^m$ 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

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

## Values of the Bernoulli numbers

Bn = 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 A027641 and A027642 in OEIS) and some larger ones are listed below.

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 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

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 n = 10k N D Bn = N / D 2 -9.4598037819... × 1082 33330 -2.8382249570... × 1078 3 -1.8243104738... × 101778 342999030 -5.3187044694... × 101769 4 -2.1159583804... × 1027690 2338224387510 -9.0494239636... × 1027677 5 -5.4468936061... × 10376771 9355235774427510 -5.8222943146... × 10376755 6 -2.0950366959... x 104767553 936123257411127577818510 -2.2379923576... × 104767529 7 -4.7845869850... × 1057675291 9601480183016524970884020224910 -4.9831764414... × 1057675260 8 -1.8637053533... × 10676752608 394815332706046542049668428841497001870 -4.7204482677... × 10676752569

Bn for n = 106 was computed to full precision by Bernd Kellner in December 2002 in 13.3 hours.

On April 29, 2008 on the 'Wolfram Blog' Oleksandr Pavlyk announced to have computed Bn for n = 107 with 'Mathematica'. He waited almost 6 days for the result. Some days later the computation was redone with PARI/GP and Bn for n = 107 + 4. The computation run for about 2 days and 12 hours (announced at the sage-devel newsgroup on May 5).

The displayed values for n = 107 and n = 108 were computed in less than one second with the von Staudt-Clausen formula and the asymptotic formula given below.

## Different viewpoints and conventions

 Jakob Bernoulli's Summae Potestatum, 1713

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 σn(x),`
`  formula (6.52) in Concrete Mathematics by Graham, Knuth and Patashnik.`

$left\left(frac\left\{ze^\left\{z\right\}\right\}\left\{e^\left\{z\right\}-1\right\} right\right)^\left\{x\right\} = xsum_\left\{ngeq0\right\}sigma_\left\{n\right\}\left(x\right)z^\left\{n\right\}$

In consequence Bn = n! σn(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: `
`  Bn = Bn(0). Associated sequence: 1, −1/2, 1/6, 0,.... `
`  Bn = Bn(1). Associated sequence: 1, +1/2, 1/6, 0,....`
`  The two definitions differ only in the sign of B1.`
`  The choice Bn = Bn(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,.... `
`  This convention agrees with the convention Bn = Bn(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+1Bn`
valid for all n > 0 is then replaced by the simpler
`  nζ(1−n) = −Bn. (See the paper of S. C. Woon.)`

 $B_n = n!sigma_\left\{n\right\}\left(1\right) = B_n\left(1\right) = -nzeta\left(1-n\right) quad \left(n geq 0\right)$
(Note that in the foregoing equation for n = 0 and n = 1 the expression −nζ(1 − n) is to be understood as limx → n −xζ(1 − x).)

## Combinatorial definitions

The definition to proceed with was developed by Julius Worpitzky in 1883. It 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. Besides elementary arithmetic only the factorial function n! and the power function km is employed.

The signless Worpitzky numbers are defined as

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

They can also be expressed through the Stirling set number

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

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

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

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 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

This representation has B1 = 1/2. A similar combinatorial representation derives from

$B_\left\{n\right\}=sum_\left\{k=0\right\}^\left\{n\right\}\left(-1\right)^\left\{k+n\right\}frac\left\{k!\right\}\left\{k+1\right\}$
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:

$B_\left\{0\right\}= +leftvert left\left\{ varnothing right\right\} rightvert$
$B_\left\{1\right\}= -frac\left\{1\right\}\left\{1\right\} leftvert varnothing rightvert + frac\left\{1\right\}\left\{2\right\}leftvert left\left\{aright\right\}rightvert$

$B_\left\{2\right\}= +frac\left\{1\right\}\left\{1\right\} leftvert varnothing rightvert - frac\left\{1\right\}\left\{2\right\}leftvert left\left\{aaright\right\}rightvert +frac\left\{1\right\}\left\{3\right\}leftvert left\left\{ab,baright\right\} rightvert$

$B_\left\{3\right\}= -frac\left\{1\right\}\left\{1\right\} leftvert varnothing rightvert +frac\left\{1\right\}\left\{2\right\}leftvert left\left\{aaaright\right\}rightvert -frac\left\{1\right\}\left\{3\right\}leftvert left\left\{aab,aba,baa,abb,bab,bbaright\right\}rightvert +frac\left\{1\right\}\left\{4\right\}leftvert left\left\{abc,acb,bac,bca,cab,cbaright\right\} rightvert$

## Asymptotic approximation

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

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

The first few Bernoulli numbers might lead one to assume that they are all small. Later values belie this assumption, however. In fact, since the factor in the squared brackets is greater than 1 from this representation follows

$|B_\left\{2n\right\}| > frac\left\{2 \left(2n\right)!\right\}\left\{\left(2 pi\right)^\left\{2 n\right\}\right\}$

so that the sequence of Bernoulli numbers diverges quite rapidly for large indices. Substituting an asymptotic approximation for the factorial function in this formula gives an asymptotic approximation for the Bernoulli numbers. For example

$|B_\left\{2 n\right\}| sim 4 sqrt\left\{pi n\right\} left\left(frac\left\{n\right\}\left\{ pi e\right\} cdot frac\left\{480 n^2 + 9\right\}\left\{480 n^2 -1\right\}right\right)^\left\{2n\right\}.$

This formula (Peter Luschny, 2007) is based on the connection of the Bernoulli numbers with the Riemann zeta function and on an approximation of the factorial function given by Gergő Nemes in 2007. For example this approximation gives

$|B\left(1000\right)| approx 5.318704469415522033ldotstimes 10^\left\{1769\right\} ,$

which is off only by three units in the least significant digit displayed.

## Inequalities

The following two inequalities (Peter Luschny, 2007) hold for n > 8 and the arithmetic mean of the two bounds is an approximation of order n−3 to the Bernoulli numbers B2n.

$4 sqrt\left\{ pi n\right\} left\left(frac\left\{n\right\}\left\{pi e\right\} right\right)^\left\{2n\right\}$
left[1 + frac{1}{24n}right] < |B_{2 n}| < 4 sqrt{ pi n} left(frac{n}{ pi e} right)^{2n} left[1+frac{1}{24n}left(1+frac{1}{24n}right)right]

Deleting the squared brackets on both sides and replacing on the right hand side the factor 4 by 5 gives simple inequalities valid for n > 1. These inequalities can be compared to related inequalities for the Euler numbers.

For example the low bound for 2n = 1000 is 5.31870445... × 101769, the high bound is 5.31870448... × 101769 and the mean is 5.31870446942... × 101769.

## Integral representation and continuation.

The integral

$b\left(s\right) = 2e^\left\{s i pi/2\right\}int_\left\{0\right\}^\left\{infty\right\} frac\left\{st^\left\{s\right\}\right\}\left\{1-e^\left\{2pi t\right\}\right\} frac\left\{dt\right\}\left\{t\right\}$

has as special values b(2n) = B2n 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

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

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

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 E2n are in magnitude approximately (2/π)(42n − 22n) times larger than the Bernoulli numbers B2n. In consequence

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

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 Bn = En = 0 (with the exception B1), it suffices to regard the case when n is even.

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

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

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

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

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

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

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

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

$E_\left\{n\right\} =\left(-1\right)^\left\{leftlfloor n/2rightrfloor \right\}left\left[n operatorname\left\{even\right\}right\right] n! , S_\left\{n+1\right\} quadqquad \left(n=0,1,ldots\right) .$

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 Rn = 2 Sn / Sn+1 when n is even. The Rn are rational approximations to π and two successive terms always enclose the true value of π. Beginning with n = 1 the sequence starts

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

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

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

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

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

$E_\left\{n\right\} =\left(-1\right)^\left\{leftlfloor n/2rightrfloor \right\}left\left[ntext\left\{ even\right\}right\right] T_\left\{n+1\right\} quadquadqquad\left(n=0,1,ldots\right) .$

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers En are given immediately by T2n + 1 and the Bernoulli numbers B2n are obtained from T2n by some easy shifting, avoiding rational arithmetic.

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

 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 T2n and recommended this method for computing B2n and E2n ‘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

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.

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

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

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 Sn.

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

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. However, the most beautiful generalization of this kind is the sequence of the Euler–Worpitzky–Chen polynomials Wn(x), which have only integer coefficients, in contrast to the rational coefficients of the Bernoulli and Euler polynomials. These polynomials are closely related to whole family of numbers under consideration here.

The sequence Wn(0) gives the signed tangent numbers and the sequence Wn(1) the signed secant numbers, the coefficients of the hyperbolic functions tanh(−x) and sech(−x) in exponential expansion, respectively. And the sequence Wn − 1(0) n / (2n − 4n)  gives the Bernoulli numbers for n > 1.

## Arithmetical properties of the Bernoulli numbers

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as Bn = − 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 cn is the numerator of Bn/2n, then the order of $K_\left\{4n-2\right\}\left(Bbb\left\{Z\right\}\right)$ is −c2n if n is even, and 2c2n if n is odd.

The Agoh-Giuga conjecture postulates that p is a prime number if and only if pBp−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 2n-th 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.

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 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.

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 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 In such that

$B_\left\{n\right\}=frac\left\{I_n\right\}\left\{2\right\} - sum_\left\{\left(p-1\right)|n\right\} frac1p quad \left(ngeq0\right) .$

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 In = 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
$phi_k\left(n\right) = sum_\left\{i=0\right\}^n i^k - frac\left\{n^k\right\}\left\{2\right\}$
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 B2k+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 2Bn is an integer. This seems trivial if one knows beforehand that in this case Bn = 0. However, by applying Worpitzky's representation one gets

$2B_\left\{n\right\}=sum_\left\{m=0\right\}^\left\{n\right\}left\left(-1right\right)^\left\{m\right\}frac\left\{2\right\}\left\{m+1\right\}m!$
left{begin{matrix} n+1 m+1 end{matrix}right} =0quadleft(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 Sn,m be the number of surjective maps from {1,2,...,n} to {1,2,...,m}, then $S_\left\{n,m\right\}=m! left\left\{begin\left\{matrix\right\} n m end\left\{matrix\right\}right\right\}$. The last equation can only hold if

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

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.

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 $m equiv n, bmod,p^\left\{b-1\right\}\left(p-1\right)$, then

$\left(1-p^\left\{m-1\right\}\right)\left\{B_m over m\right\} equiv \left(1-p^\left\{n-1\right\}\right)\left\{B_n over n\right\} ,bmod, p^b.$

Since $B_n = -nzeta\left(1-n\right)$, this can also be written

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

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 − ps 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 $a notequiv 1, bmod, p-1$, and so can be extended to a continuous function $zeta_p\left(s\right)$ for all p-adic integers $Bbb\left\{Z\right\}_p,,$ the p-adic Zeta function.

### Ramanujan's congruences

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

$mequiv 0,bmod,6qquad \left\{\left\{m+3\right\}choose\left\{m\right\}\right\}B_m=\left\{\left\{m+3\right\}over3\right\}-sum_\left\{j=1\right\}^\left\{m/6\right\}\left\{m+3choose\left\{m-6j\right\}\right\}B_\left\{m-6j\right\}$

$mequiv 2,bmod,6qquad \left\{\left\{m+3\right\}choose\left\{m\right\}\right\}B_m=\left\{\left\{m+3\right\}over3\right\}-sum_\left\{j=1\right\}^\left\{\left(m-2\right)/6\right\}\left\{m+3choose\left\{m-6j\right\}\right\}B_\left\{m-6j\right\}$

$mequiv 4,bmod, 6qquad\left\{\left\{m+3\right\}choose\left\{m\right\}\right\}B_m=-\left\{\left\{m+3\right\}over6\right\}-sum_\left\{j=1\right\}^\left\{\left(m-4\right)/6\right\}\left\{m+3choose\left\{m-6j\right\}\right\}B_\left\{m-6j\right\}.$

### Efficient computation of Bernoulli numbers mod p

In some applications it is useful to be able to compute the Bernoulli numbers B0 through Bp − 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 (see Buhler et al) which require only O(p (log p)2) operations (see big-O notation).

## 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 $n ge 2$ involves Bernoulli numbers; if B(n) is the numerator of B4n/n, then

$2^\left\{2n-2\right\}\left(1-2^\left\{2n-1\right\}\right)B_\left\{\left(n\right)\right\}$

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

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

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

• Let n ≥ 1.

$frac\left\{1\right\}\left\{n\right\} sum_\left\{k=1\right\}^\left\{n\right\}B_\left\{k\right\}B_\left\{n-k\right\}+B_\left\{n-1\right\}=-B_\left\{n\right\} quad text\left\{\left(L. Euler\right)\right\}$
• Let n ≥ 1.

$sum_\left\{k=0\right\}^\left\{n\right\}binom\left\{n+1\right\}\left\{k\right\}\left(n+k+1\right)B_\left\{n+k\right\}=0 qquadtext\left\{\left(von Ettingshausen, 1827\right)\right\}$

• Let n ≥ 1 and m ≥ 1.

$\left(-1\right)^\left\{m\right\}sum_\left\{r=0\right\}^\left\{m\right\}binom\left\{m\right\}\left\{r\right\}B_\left\{n+r\right\}=\left(-1\right)^\left\{n\right\}sum_\left\{s=0\right\}^\left\{n\right\}binom\left\{n\right\}\left\{s\right\}B_\left\{m+s\right\}qquadtext\left\{\left(L. Carlitz\right)\right\}$

• Let n ≥ 4 and

$H_\left\{n\right\}=sum_\left\{1leq kleq n\right\}k^\left\{-1\right\}$

the harmonic number. Then

$frac\left\{n\right\}\left\{2\right\}sum_\left\{k=2\right\}^\left\{n-2\right\}frac\left\{B_\left\{n-k\right\}\right\}\left\{n-k\right\}frac\left\{B_\left\{k\right\}\right\}\left\{k\right\}-sum_\left\{k=2\right\}$

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

$\left(n+2\right)sum_\left\{k=2\right\}^\left\{n-2\right\}B_\left\{k\right\}B_\left\{n-k\right\}-2sum_\left\{l=2\right\}^\left\{n-2\right\}binom\left\{n+2\right\}\left\{l\right\}B_\left\{l\right\} B_\left\{n-l\right\}=n\left(n+1\right)B_\left\{n\right\}$

• Let n ≥ 1

$frac\left\{n\right\}\left\{2\right\}left\left(B_\left\{n-1\right\}\left(x\right)+sum_\left\{k=1\right\}^\left\{n-1\right\}frac\left\{B_\left\{k\right\}\left(x\right)\right\}\left\{k\right\}$
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.

## References

• L. Euler, "De summis serierum reciprocarum", Opera Omnia I.14, E 41, 73-86.
• Thomas Clausen, "Lehrsatz aus einer Abhandlung über die Bernoullischen Zahlen", Astr. Nachr., 17 (1840), 351--352.
• K. G. Ch. von Staudt, "Beweis eines Lehrsatzes, die Bernoullischen Zahlen betreffend"', Journal für die reine und angewandte Mathematik, 21 (1840), 372--374.
• K. G. Ch. von Staudt, "De numeris Bernoullianis, commentationem alteram", Erlangen, 1845.
• L. Seidel, "Über eine einfache Entstehungsweise der Bernoullischen Zahlen und einiger verwandten Reihen", Sitzungsber. Münch. Akad., 4 (1877), 157--187.
• D. André, "Développements de sec x et tan x.", Comptes Rendus Acad. Sci., 88 (1879), 965--967.
• D. André, "Mémoire sur les permutations alternées", J. Math., 7 (1881), 167--184.
• J. Worpitzky, "Studien über die Bernoullischen und Eulerschen Zahlen", Journal für die reine und angewandte Mathematik, 94 (1883), 203--232.
• Charles Jordan, "Calculus of Finite Differences", Chelsea Publ. Co., New York, 1950.
• Entringer R. C., "A combinatorial interpretation of the Euler and Bernoulli numbers", Nieuw. Arch. V. Wiskunde, 14 (1966), 241--6.
• D. E. Knuth and T. J. Buckholtz, "Computation of Tangent, Euler, and Bernoulli Numbers", Mathematics of Computation, 21 (1967), 663--688.
• Abramowitz, M. and Stegun, C. A. (Eds.). "Bernoulli and Euler Polynomials and the Euler-Maclaurin Formula'' §23.1 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 804--806, 1972.
• John W. Milnor and James D. Stasheff, "Appendix B: Bernoulli Numbers", p. 281-287, in: Characteristic Classes (Annals of Mathematics Studies 76), Princeton University Press and University of Tokyo Press, 1974.
• D. Dumont and G. Viennot, "A combinatorial interpretation of Seidel generation of Genocchi numbers", Ann. Discrete Math., 6 (1980), 77--87.
• D. Dumont, "Matrices d'Euler-Seidel", Séminaire Lotharingien de Combinatoire, 1981.

• http://emis.u-strasbg.fr/journals/SLC/opapers/s05dumont.html

• A. Ayoub, "Euler and the Zeta Function", Amer. Math. Monthly, 74 (1981), 1067--1086.
• R. L. Graham, D. E. Knuth, and O. Patashnik, "Concrete Mathematics", Addison-Wesley, 1989.
• V. I. Arnold, "Bernoulli-Euler updown numbers associated with function singularities, their combinatorics and arithmetics", Duke Math. J., 63 (1991), 537--555.
• Ilya Sh. Slavutskii, "Staudt and arithmetical properties of Bernoulli numbers", Historia Scientiarum, 2 (1995), 69--74.
• John Conway and Richard Guy, "The Book of Numbers", Springer-Verlag, 1996.
• S. C. Woon, "Generalization of a relation between the Riemann zeta function and Bernoulli numbers", Dec 1998,

• http://arXiv.org/abs/math.NT/9812143

• D. Arlettaz, "Die Bernoulli-Zahlen: eine Beziehung zwischen Topologie und Gruppentheorie", Math. Semesterber. 45, 1998, 61-75.
• M. Kaneko, "The Akiyama-Tanigawa algorithm for Bernoulli numbers", Journal of Integer Sequences 12 (2000).

• http://www.cs.uwaterloo.ca/journals/JIS/vol3.html

• Buhler, J., Crandall, R., Ernvall, R., Metsankyla, T., and Shokrollahi, M. "Irregular Primes and Cyclotomic Invariants to 12 Million." Journal of Symbolic Computation, Volume 31, Issues 1–2, January 2001, pages 89–96.
• Zhi-Wei Sun, "Some curious results on Bernoulli and Euler polynomials", 2005/2006,

• http://pweb.nju.edu.cn/zwsun

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