Bernoulli number

Notation

The superscript ± used in this article distinguishes the two sign conventions for Bernoulli numbers. Only the term is affected:

• with ( / ) is the sign convention prescribed by NIST and many modern textbooks.
• with ( / ) was used in the older literature, and (since 2022) by Donald KnuthDonald Knuth (2022), Recent News (2022): Concrete Mathematics and Bernoulli. {{blockquote|But last year I took a close look at Peter Luschny's Bernoulli manifesto, where he gives more than a dozen good reasons why the value of $B_1$ should really be plus one-half. He explains that some mathematicians of the early 20th century had unilaterally changed the conventions, because some of their formulas came out a bit nicer when the negative value was used. It was their well-intentioned but ultimately poor choice that had led to what I'd been taught in the 1950s. […] By now, hundreds of books that use the “minus-one-half” convention have unfortunately been written. Even worse, all the major software systems for symbolic mathematics have that 20th-century aberration deeply embedded. Yet Luschny convinced me that we have all been wrong, and that it's high time to change back to the correct definition before the situation gets even worse.

In the formulas below, one can switch from one sign convention to the other with the relation B_n^{+}=(-1)^n B_n^{-}, or for integer n = 2 or greater, simply ignore it.

Since for all odd n 1, and many formulas only involve even-index Bernoulli numbers, a few authors write "B" instead of B. This article does not follow that notation.

History

Early history

The Bernoulli numbers are rooted in the early history of the computation of sums of integer powers, which have been of interest to mathematicians since antiquity.

Methods to calculate the sum of the first n positive integers, the sum of the squares and of the cubes of the first n positive integers were known, but there were no real 'formulas', only descriptions given entirely in words. Among the great mathematicians of antiquity to consider this problem were Pythagoras (c. 572–497 BCE, Greece), Archimedes (287–212 BCE, Italy), Aryabhata (b. 476, India), Al-Karaji (d. 1019, Persia) and Ibn al-Haytham (965–1039, Iraq).

During the late sixteenth and early seventeenth centuries mathematicians made significant progress. In the West Thomas Harriot (1560–1621) of England, Johann Faulhaber (1580–1635) of Germany, Pierre de Fermat (1601–1665) and fellow French mathematician Blaise Pascal (1623–1662) all played important roles.

Thomas Harriot seems to have been the first to derive and write formulas for sums of powers using symbolic notation, but even he calculated only up to the sum of the fourth powers. Johann Faulhaber gave formulas for sums of powers up to the 17th power in his 1631 Academia Algebrae, far higher than anyone before him, but he did not give a general formula.

Blaise Pascal in 1654 proved Pascal's identity relating (n+1)k+1 to the sums of the pth powers of the first n positive integers for .

The Swiss mathematician Jacob Bernoulli (1654–1705) was the first to realize the existence of a single sequence of constants B0, B1, B2,... which provide a uniform formula for all sums of powers.

The joy Bernoulli experienced when he hit upon the pattern needed to compute quickly and easily the coefficients of his formula for the sum of the cth powers for any positive integer c can be seen from his comment. He wrote:

:"With the help of this table, it took me less than half of a quarter of an hour to find that the tenth powers of the first 1000 numbers being added together will yield the sum 91,409,924,241,424,243,424,241,924,242,500."

Bernoulli's result was published posthumously in Ars Conjectandi in 1713. Seki Takakazu independently discovered the Bernoulli numbers and his result was published a year earlier, also posthumously, in 1712. However, Seki did not present his method as a formula based on a sequence of constants.

Bernoulli's formula for sums of powers is the most useful and generalizable formulation to date. The coefficients in Bernoulli's formula are now called Bernoulli numbers, following a suggestion of Abraham de Moivre.

Bernoulli's formula is sometimes called Faulhaber's formula after Johann Faulhaber who found remarkable ways to calculate sum of powers but never stated Bernoulli's formula. According to Knuth a rigorous proof of Faulhaber's formula was first published by Carl Jacobi in 1834. Knuth's in-depth study of Faulhaber's formula concludes:

:"Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a single sequence of constants B0, B1, B2, ... would provide a uniform ::\sum n^m = \frac 1{m+1}\left( B_0n^{m+1}-\binom{m+1} 1 B_1 n^m+\binom{m+1} 2B_2n^{m-1}-\cdots +(-1)^m\binom{m+1}mB_mn\right) :for all sums of powers. He never mentioned, for example, the fact that almost half of the coefficients turned out to be zero after he had converted his formulas for Σ nm from polynomials in N to polynomials in n."

In the above Knuth meant B_1^-; instead using B_1^+ the formula avoids subtraction: : \sum n^m = \frac 1{m+1}\left( B_0n^{m+1}+\binom{m+1} 1 B^+_1 n^m+\binom{m+1} 2B_2n^{m-1}+\cdots+\binom{m+1}mB_mn\right).

Reconstruction of "Summæ Potestatum"

|chapter-url=https://archive.org/details/sourcebookinmath00smit/page/85

The Bernoulli numbers (n)/(n) were introduced by Jacob Bernoulli in the book Ars Conjectandi published posthumously in 1713.{{cite book

: \sum_{k=1}^n k^c = \frac{n^{c+1}}{c+1}+\frac 1 2 n^c+\sum_{k=2}^c \frac{B_k}{k!} c^{\underline{k-1}}n^{c-k+1}.

This formula suggests setting when switching from the so-called 'archaic' enumeration which uses only the even indices 2, 4, 6... to the modern form. Most striking in this context is the fact that the falling factorial c has for the value . Thus Bernoulli's formula can be written : \sum_{k=1}^n k^c = \sum_{k=0}^c \frac{B_k}{k!}c^{\underline{k-1}} n^{c-k+1}

if , recapturing the value Bernoulli gave to the coefficient at that position.

The formula for \textstyle \sum_{k=1}^n k^9 on page 97 of Bernoulli's Ars Conjectandi contains an error at the last term; it should be -\tfrac {3}{20}n^2 instead of -\tfrac {1}{12}n^2.

Definitions

Many characterizations of the Bernoulli numbers have been found in the last 300 years, and each could be used to introduce these numbers. Here only four of the most useful ones are mentioned:

• a recursive equation,
• an explicit formula,
• a generating function,
• an integral expression.

For the proof of the equivalence of the four approaches, see or .

Recursive definition

The Bernoulli numbers obey the sum formulas : \begin{align} \sum_{k=0}^{m}\binom {m+1} k B^{-{}}k &= \delta{m, 0} \ \sum_{k=0}^{m}\binom {m+1} k B^{+{}}_k &= m+1 \end{align} where m=0,1,2... and δ denotes the Kronecker delta.

The first of these is sometimes written as the formula (for m 1) (B+1)^m-B_m=0, where the power is expanded formally using the binomial theorem and B^k is replaced by B_k.

Solving for B^{\mp{}}m gives the recursive formulas : \begin{align} B_m^{-{}} &= \delta{m, 0} - \sum_{k=0}^{m-1} \binom{m}{k} \frac{B^{-{}}k}{m - k + 1} \ B_m^+ &= 1 - \sum{k=0}^{m-1} \binom{m}{k} \frac{B^+_k}{m - k + 1}. \end{align}

Explicit definition

In 1893 Louis Saalschütz listed a total of 38 explicit formulas for the Bernoulli numbers, usually giving some reference in the older literature. One of them is (for m\geq 1): :\begin{align} B^-m &= \sum{k=0}^m \frac1{k+1} \sum_{j=0}^k \binom{k}{j} (-1)^j j^m \ B^+m &= \sum{k=0}^m \frac1{k+1} \sum_{j=0}^k \binom{k}{j} (-1)^j (j + 1)^m. \end{align}

Generating function

The exponential generating functions are :\begin{alignat}{3} \frac{t}{e^t - 1} &= \frac{t}{2} \left( \operatorname{coth} \frac{t}{2} -1 \right) &&= \sum_{m=0}^\infty \frac{B^{-{}}m t^m}{m!}\ \frac{te^t}{e^t - 1} = \frac{t}{1 - e^{-t}} &= \frac{t}{2} \left( \operatorname{coth} \frac{t}{2} +1 \right) &&= \sum{m=0}^\infty \frac{B^+_m t^m}{m!}. \end{alignat} where the substitution is t \to - t. The arithmetic difference between the generating functions for B^+m and B^{-}m is t. If we let F(t)=\sum{i=1}^\infty f_it^i and G(t)=1/(1+F(t))=\sum{i=0}^\infty g_it^i then

:G(t)=1-F(t)G(t).

Then g_0=1 and for m0 the m term in the series for G(t) is:

:g_mt^m=-\sum_{j=0}^{m-1}f_{m-j}g_jt^m

If

:F(t)=\frac{e^t-1}t-1=\sum_{i=1}^\infty \frac{t^i}{(i+1)!}

then we find that

:G(t)=t/(e^t-1)

:\begin{align} m!g_m&=-\sum_{j=0}^{m-1}\frac{m!}{j!}\frac{j!g_j}{(m-j+1)!}\ &=-\frac 1{m+1}\sum_{j=0}^{m-1}\binom{m+1}jj!g_j\ \end{align}

showing that the values of i!g_i obey the recursive formula for the Bernoulli numbers B^-_i.

The (ordinary) generating function : z^{-1} \psi_1(z^{-1}) = \sum_{m=0}^{\infty} B^+_m z^m

is an asymptotic series. It contains the trigamma function ψ1.

Integral Expression

From the generating functions above, one can obtain the following integral formula for the even Bernoulli numbers:

:B_{2n} = 4n (-1)^{n+1} \int_0^{\infty} \frac{t^{2n-1}}{e^{2 \pi t} -1 } \mathrm{d} t

Bernoulli numbers and the Riemann zeta function

The Bernoulli numbers can be expressed in terms of the Riemann zeta function:

: B_n^+ = -n, \zeta(1-n) \quad for n ≥ 1 .

Here the argument of the zeta function is *0 *or negative. As \zeta(k) is zero for negative even integers (the trivial zeroes), if n1 is odd, \zeta(1-n) is zero.

By means of the zeta functional equation and the gamma reflection formula the following relation can be obtained:

: B_{2n} = \frac {(-1)^{n+1}2(2n)!} {(2\pi)^{2n}} \zeta(2n) \quad for integers n ≥ 1 .

Now the argument of the zeta function is positive.

It then follows from ζ → 1 (n → ∞) and Stirling's formula that : |B_{2 n}| \sim 4 \sqrt{\pi n} \left(\frac{n}{ \pi e} \right)^{2n} \quad for n → ∞ .

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 which require only O(p (log p)2) operations (see big O notation).

David Harvey 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 + ε) and claims that this implementation is significantly faster than implementations based on other methods. Using this implementation Harvey computed B**n for . Harvey's implementation has been included in SageMath since version 3.1. Prior to that, Bernd Kellner computed B**n to full precision for in December 2002 and Oleksandr Pavlyk for with Mathematica in April 2008.

:{| class="wikitable defaultright col1left" ! 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 || || |- | G. Fee, S. Plouffe || 1996 || || |- | G. Fee, S. Plouffe || 1996 || || |- | B. C. Kellner || 2002 || || |- | O. Pavlyk || 2008 || || |- | D. Harvey || 2008 || || |} ::* Digits is to be understood as the exponent of 10 when B**n is written as a real number in normalized scientific notation.

Applications of the Bernoulli numbers

Asymptotic analysis

Arguably the most important application of the Bernoulli numbers 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_{k=a}^{b-1} f(k) = \int_a^b f(x),dx + \sum_{k=1}^m \frac{B^-k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R-(f,m).

This formulation assumes the convention . Using the convention the formula becomes

: \sum_{k=a+1}^{b} f(k) = \int_a^b f(x),dx + \sum_{k=1}^m \frac{B^+k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R+(f,m).

Here f^{(0)}=f (i.e. the zeroth-order derivative of f is just 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_{k=a+1}^{b} f(k)= \sum_{k=0}^m \frac{B_k}{k!} (f^{(k-1)}(b)-f^{(k-1)}(a))+R(f,m).

This form is for example the source for the important Euler–Maclaurin expansion of the zeta function

: \begin{align} \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{align}

Here s denotes the rising factorial power.

Bernoulli numbers are also frequently used in other kinds of asymptotic expansions. The following example is the classical Poincaré-type asymptotic expansion of the digamma function ψ.

:\psi(z) \sim \ln z - \sum_{k=1}^\infty \frac{B^+_k}{k z^k}

Sum of powers

Main article: Faulhaber's formula

Bernoulli numbers feature prominently in the closed form expression of the sum of the mth powers of the first n positive integers. For m, n ≥ 0 define

:S_m(n) = \sum_{k=1}^n k^m = 1^m + 2^m + \cdots + n^m.

This expression can always be rewritten as a polynomial in n of degree m + 1. The coefficients of these polynomials are related to the Bernoulli numbers by Bernoulli's formula: : S_m(n) = \frac{1}{m + 1} \sum_{k=0}^m \binom{m + 1}{k} B^+k n^{m + 1 - k} = m! \sum{k=0}^m \frac{B^+_k n^{m + 1 - k}}{k! (m+1-k)!} , where denotes the binomial coefficient.

For example, taking m to be 1 gives the triangular numbers 0, 1, 3, 6, ... .

: 1 + 2 + \cdots + n = \frac{1}{2} (B_0 n^2 + 2 B^+_1 n^1) = \tfrac12 (n^2 + n).

Taking m to be 2 gives the square pyramidal numbers 0, 1, 5, 14, ... .

: 1^2 + 2^2 + \cdots + n^2 = \frac{1}{3} (B_0 n^3 + 3 B^+_1 n^2 + 3 B_2 n^1) = \tfrac13 \left(n^3 + \tfrac32 n^2 + \tfrac12 n\right).

Some authors use the alternate convention for Bernoulli numbers and state Bernoulli's formula in this way: : S_m(n) = \frac{1}{m + 1} \sum_{k=0}^m (-1)^k \binom{m + 1}{k} B^{-{}}_k n^{m + 1 - k}.

Bernoulli's formula is sometimes called Faulhaber's formula after Johann Faulhaber who also found remarkable ways to calculate sums of powers.

Faulhaber's formula was generalized by V. Guo and J. Zeng to a q-analog.

Taylor series

The Bernoulli numbers appear in the Taylor series expansion of many trigonometric functions and hyperbolic functions.

\begin{align} \tan x &= \hphantom \sum_{n=1}^\infty \frac{(-1)^{n-1} 2^{2n} (2^{2n}-1) B_{2n} }{(2n)!}; x^{2n-1}, && \left|x \right| \cot x &= {1\over x} \sum_{n=0}^\infty \frac{(-1)^n B_{2n} (2x)^{2n}}{(2n)!}, & 0 \tanh x &= \hphantom \sum_{n=1}^\infty \frac{2^{2n}(2^{2n}-1)B_{2n}}{(2n)!};x^{2n-1}, && |x| \coth x &= {1\over x} \sum_{n=0}^\infty \frac{B_{2n} (2x)^{2n}}{(2n)!}, & 0 \end{align}

Laurent series

The Bernoulli numbers appear in the following Laurent series:

Digamma function: \psi(z)= \ln z- \sum_{k=1}^\infty \frac {B_k^{+{}}} {k z^k}

Use 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 involves Bernoulli numbers. Let ES**n be the number of such exotic spheres for n ≥ 2, then

:\textit{ES}n = (2^{2n-2}-2^{4n-3}) \operatorname{Numerator}\left(\frac{B{4n}}{4n} \right) .

The Hirzebruch signature theorem for the L genus of a smooth oriented closed manifold of dimension 4n also involves Bernoulli numbers.

Connections with combinatorial numbers

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 Worpitzky numbers

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

: W_{n,k}=\sum_{v=0}^k (-1)^{v+k} (v+1)^n {k \choose v}\ .

They can also be expressed through the Stirling numbers of the second kind

: W_{n,k}=k! \left{ {n+1\atop k+1} \right}.

A Bernoulli number is then introduced as an inclusion–exclusion sum of Worpitzky numbers weighted by the harmonic sequence 1, , , ...

: 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 (v+1)^n {k \choose v}\ .

: : : : : : :

This representation has .

Consider the sequence sn, n ≥ 0. From Worpitzky's numbers , applied to s0, s0, s1, s0, s1, s2, s0, s1, s2, s3, ... is identical to the Akiyama–Tanigawa transform applied to sn (see Connection with Stirling numbers of the first kind). This can be seen via the table:

:{| style="text-align:center" |+ Identity of Worpitzky's representation and Akiyama–Tanigawa transform |- |1|| || || || || ||0||1|| || || || ||0||0||1|| || || ||0||0||0||1|| || ||0||0||0||0||1|| |- |1||−1|| || || || ||0||2||−2|| || || ||0||0||3||−3|| || ||0||0||0||4||−4|| || || || || || || |- |1||−3||2|| || || ||0||4||−10||6|| || ||0||0||9||−21||12|| || || || || || || || || || || || || |- |1||−7||12||−6|| || ||0||8||−38||54||−24|| || || || || || || || || || || || || || || || || || || |- |1||−15||50||−60||24|| || || || || || || || || || || || || || || || || || || || || || || || || |- |}

The first row represents s0, s1, s2, s3, s4.

Hence for the second fractional Euler numbers (n) / (n + 1):

: : : : : : :

A second formula representing the Bernoulli numbers by the Worpitzky numbers is for n ≥ 1

: B_n=\frac n {2^{n+1}-2}\sum_{k=0}^{n-1} (-2)^{-k}, W_{n-1,k} .

The simplified second Worpitzky's representation of the second Bernoulli numbers is:

(n + 1) / (n + 1) = × (n) / (n + 1)

which links the second Bernoulli numbers to the second fractional Euler numbers. The beginning is:

:

The numerators of the first parentheses are (see Connection with Stirling numbers of the first kind).

Connection with Stirling numbers of the second kind

Stirling numbers of the second kind, S(k,m), have the property that:

: x^k=\sum_{m=0}^k {x^{\underline{m}}} S(k,m)

where x denotes the falling factorial function.

The Bernoulli polynomials Bk(x) can be written as:

: B_k(x)=k\sum_{m=0}^{k-1}\binom{x}{m+1}S(k-1,m)m!+B_k

where Bk for are the Bernoulli numbers.

The following property of the binomial coefficient:

: \binom{x}{m}=\binom{x+1}{m+1}-\binom{x}{m+1}

thus implies that

: x^k=\frac{B_{k+1}(x+1)-B_{k+1}(x)}{k+1}.

One also has the following for Bernoulli polynomials,

: B_k(x)=\sum_{n=0}^k \binom{k}{n} B_n x^{k-n}.

The coefficient of x in is .

The coefficient of x in the first expression for B_k(x) is k\sum_{m=0}^{k-1}\frac{(-1)^m}{m+1}S(k-1,m)m! whereas in the second expression it is kB_{k-1}. Replacing k-1 by k this yields:

: B_k=\sum_{m=0}^k (-1)^m \frac{m!}{m+1} S(k,m)

(resulting in ) which is an explicit formula for Bernoulli numbers and can be used to prove Von-Staudt Clausen theorem.

Connection with Stirling numbers of the first kind

The two main formulas relating the unsigned Stirling numbers of the first kind to the Bernoulli numbers (with ) are

: \frac{1}{m!}\sum_{k=0}^m (-1)^{k} \left[{m+1\atop k+1}\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[{m+1\atop k+1}\right] B_{n+k} = A_{n,m}.

Here the number A**n,m are the rational Akiyama–Tanigawa numbers, the first few of which are displayed in the following table.

:{| class="wikitable" style="text-align:center" |+ Akiyama–Tanigawa number ! !!0!!1!!2!!3!!4 |- ! 0 | 1 || || || || |- ! 1 | || || || || ... |- ! 2 | || || || ... || ... |- ! 3 | 0 || || ... || ... || ... |- ! 4 | − || ... || ... || ... || ... |}

The Akiyama–Tanigawa numbers satisfy a simple recurrence relation which can be exploited to iteratively compute the Bernoulli numbers. This leads to the algorithm shown in the section 'algorithmic description' above. See /.

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. If the main diagonal is zeroes = , the autosequence is of the first kind. Example: , the Fibonacci numbers. If the main diagonal is the first upper diagonal multiplied by 2, it is of the second kind. Example: /, the second Bernoulli numbers (see ). The Akiyama–Tanigawa transform applied to 2−n = 1/ leads to (n) / (n + 1). Hence:

:{| class="wikitable" style="text-align:center"

See and . (n) / (n + 1) are the second (fractional) Euler numbers and an autosequence of the second kind.

:( = , 0, −, 0, , ...) × ( = 3, , , , 21, ...) = = , 0, −, 0, , ....

Also valuable for / (see Connection with Worpitzky numbers).

Connection with Pascal's triangle

There are formulas connecting Pascal's triangle to Bernoulli numbers : B^{+}n=\frac{|A_n|}{(n+1)!}~~~ where |A_n| is the determinant of a n-by-n Hessenberg matrix part of Pascal's triangle whose elements are: a{i, k} = \begin{cases} 0 & \text{if } k1+i \ {i+1 \choose k-1} & \text{otherwise} \end{cases}

Example:

: B^{+}_6 =\frac{\det\begin{pmatrix} 1& 2& 0& 0& 0& 0\ 1& 3& 3& 0& 0& 0\ 1& 4& 6& 4& 0& 0\ 1& 5& 10& 10& 5& 0\ 1& 6& 15& 20& 15& 6\ 1& 7& 21& 35& 35& 21 \end{pmatrix}}{7!}=\frac{120}{5040}=\frac 1 {42} .

Connection with Eulerian numbers

There are formulas connecting Eulerian numbers to Bernoulli numbers:

:\begin{align} \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. \end{align}

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

A binary tree representation

The Stirling polynomials σ**n(x) are related to the Bernoulli numbers by . S. C. Woon described an algorithm to compute σ**n(1) as a binary tree:

:[[File:SCWoonTree.png]]

Woon's recursive algorithm (for n ≥ 1) starts by assigning to the root node . Given a node of the tree, the left child of the node is and the right child . A node is written as ±[a2, ..., a**k] in the initial part of the tree represented above with ± denoting the sign of a1.

Given a node N the factorial of N is defined as

: N! = a_1 \prod_{k=2}^{\operatorname{length}(N)} a_k!.

Restricted to the nodes N of a fixed tree-level n the sum of is σ**n(1), thus

: B_n = \sum_\stackrel{N \text{ node of}}{\text{ tree-level } n} \frac{n!}{N!}.

For example: : : :

Integral representation and continuation

The integral : b(s) = 2e^{s i \pi/2}\int_0^\infty \frac{st^s}{1-e^{2\pi t}} \frac{dt}{t} = \frac{s!}{2^{s-1}}\frac{\zeta(s)}(-i)^s= \frac{2s!\zeta(s)}{(2\pi i)^s} has as special values for n 0.

For example, and . Here, ζ is the Riemann zeta function, and i is the imaginary unit. Leonhard Euler (Opera Omnia, Ser. 1, Vol. 10, p. 351) considered these numbers and calculated

: \begin{align} p &= \frac{3}{2\pi^3}\left(1+\frac{1}{2^3}+\frac{1}{3^3}+\cdots \right) = 0.0581522\ldots \ q &= \frac{15}{2\pi^5}\left(1+\frac{1}{2^5}+\frac{1}{3^5}+\cdots \right) = 0.0254132\ldots \end{align}

Another similar integral representation is : b(s) = -\frac{e^{s i \pi/2}}{2^{s}-1}\int_0^\infty \frac{st^{s}}{\sinh\pi t} \frac{dt}{t}= \frac{2e^{s i \pi/2}}{2^{s}-1}\int_0^\infty \frac{e^{\pi t}st^s}{1-e^{2\pi t}} \frac{dt}{t}.

The relation to the Euler numbers and {{pi}}

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 (42n − 22n) times larger than the Bernoulli numbers B2n. In consequence:

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

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

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

:\begin{align} B_n &= \sum_{k=0}^{n-1}\binom{n-1}{k} \frac{n}{4^n-2^n}E_k & n&=2, 4, 6, \ldots \[6pt] E_n &= \sum_{k=1}^n \binom{n}{k-1} \frac{2^k-4^k}{k} B_k & n&=2,4,6,\ldots \end{align}

These conversion formulas express a connection 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 . These numbers are defined for n ≥ 1 as{{citation

: S_n = 2 \left(\frac{2}{\pi}\right)^n \sum_{k = 0}^\infty \frac{ (-1)^{kn} }{(2k+1)^n} = 2 \left(\frac{2}{\pi}\right)^n \lim_{K\to \infty} \sum_{k = -K}^K (4k+1)^{-n}.

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

: S_n = 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 ( / )

These are the coefficients in the expansion of sec x + tan x.

The Bernoulli numbers and Euler numbers can be understood as special views of these numbers, selected from the sequence S**n and scaled for use in special applications.

: \begin{align} B_{n} &= (-1)^{\left\lfloor \frac{n}{2}\right\rfloor} [ n \text{ even}] \frac{n! }{2^n - 4^n}, S_{n}\ , & n&= 2, 3, \ldots \ E_n &= (-1)^{\left\lfloor \frac{n}{2}\right\rfloor} [ n \text{ even}] n! , S_{n+1} & n &= 0, 1, \ldots \end{align}

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 when n is even. The R**n are rational approximations to and two successive terms always enclose the true value of . Beginning with the sequence starts ( / ):

: 2, 4, 3, \frac{16}{5}, \frac{25}{8}, \frac{192}{61}, \frac{427}{136}, \frac{4352}{1385}, \frac{12465}{3968}, \frac{158720}{50521},\ldots \quad \longrightarrow \pi.

These rational numbers also appear in the last paragraph of Euler's paper cited above.

Consider the Akiyama–Tanigawa transform for the sequence (n + 2) / (n + 1):

:{| class="wikitable" style="text-align:right;" ! 0 |1||||0||−||−||−||0 |- ! 1 | || 1|| || 0|| −|| −|| |- ! 2 | −|| || || || || || |- ! 3 | −1|| −|| −|| || || || |- ! 4 | || −|| −|| || || || |- ! 5 | 8|| || || || || || |- ! 6 | −|| || || || || || |}

From the second, the numerators of the first column are the denominators of Euler's formula. The first column is − × .

An algorithmic view: the Seidel triangle

The sequence S**n has another unexpected yet important property: The denominators of S**n+1 divide the factorial n!. In other words: the numbers , sometimes called Euler zigzag numbers, are integers.

: T_n = 1,,1,,1,,2,,5,,16,,61,,272,,1385,,7936,,50521,,353792,\ldots \quad n=0, 1, 2, 3, \ldots (). See ().

Their exponential generating function is the sum of the secant and tangent functions.

: \sum_{n=0}^\infty T_n \frac{x^n}{n!} = \tan \left(\frac\pi4 + \frac x2\right) = \sec x + \tan x.

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

:\begin{align} B_n &= (-1)^{\left\lfloor \frac{n}{2}\right\rfloor} [n\text{ even}] \frac{n }{2^n-4^n}, T_{n-1}\ & n &\geq 2 \ E_n &= (-1)^{\left\lfloor \frac{n}{2}\right\rfloor} [n\text{ even}] T_{n} & n &\geq 0 \end{align}

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers E2n are given immediately by T2n and the Bernoulli numbers B2n are fractions obtained from T2n - 1 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 simple to calculate T**n.

\begin{array}{crrrcc} { } & { } & {\color{red}1} & { } & { } & { } \ { } & {\rightarrow} & {\color{blue}1} & {\color{red}1} & { } \ { } & {\color{red}2} & {\color{blue}2} & {\color{blue}1} & {\leftarrow} \ {\rightarrow} & {\color{blue}2} & {\color{blue}4} & {\color{blue}5} & {\color{red}5} \ {\color{red}16} & {\color{blue}16} & {\color{blue}14} & {\color{blue}10} & {\color{blue}5} & {\leftarrow} \end{array}

1. Start by putting 1 in row 0 and let k denote the number of the row currently being filled
2. 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
3. At the end of the row duplicate the last number.
4. If k is even, proceed similar in the other direction.

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

Similar to Seidel's approach D. E. Knuth and T. J. Buckholtz 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 and later Millar, Sloane and Young popularized Seidel's algorithm under the name boustrophedon transform.

Triangular form:

:{| style="text-align:right" | || || || || || || 1|| || || || || || |- | || || || || || 1|| || 1|| || || || || |- | || || || || 2|| || 2|| || 1|| || || || |- | || || || 2|| || 4|| || 5|| || 5|| || || |- | || || 16|| || 16|| || 14|| || 10|| || 5|| || |- | || 16|| || 32|| || 46|| || 56|| || 61|| || 61|| |- |272|| ||272|| ||256|| ||224|| ||178|| ||122|| || 61 |}

Only , with one 1, and , with two 1s, are in the OEIS.

Distribution with a supplementary 1 and one 0 in the following rows:

:{| style="text-align:right" | || || || || || || 1|| || || || || || |- | || || || || || 0|| || 1|| || || || || |- | || || || || −1|| || −1|| || 0|| || || || |- | || || || 0|| || −1|| || −2|| || −2|| || || |- | || || 5|| || 5|| || 4|| || 2|| || 0|| || |- | || 0|| || 5|| || 10|| || 14|| || 16|| || 16|| |- |−61|| ||−61|| ||−56|| ||−46|| ||−32|| ||−16|| || 0 |}

This is , a signed version of . The main andiagonal is . The main diagonal is . The central column is . Row sums: 1, 1, −2, −5, 16, 61.... See . See the array beginning with 1, 1, 0, −2, 0, 16, 0 below.

The Akiyama–Tanigawa algorithm applied to (n + 1) / (n) yields:

:{| style="text-align:right" | 1|| 1|| || 0|| −|| −|| − |- | 0|| 1|| || 1|| 0|| − |- | −1|| −1|| || 4|| |- | 0|| −5|| −|| 1 |- | 5|| 5|| − |- | 0|| 61 |- | −61 |}

1. The first column is . Its binomial transform leads to:

:{| style="text-align:right" |- | 1|| 1|| 0|| −2|| 0|| 16|| 0 |- |0||−1||−2||2||16||−16 |- |−1||−1||4||14||−32 |- |0||5||10||−46 |- |5||5||−56 |- |0||−61 |- |−61 |}

The first row of this array is . The absolute values of the increasing antidiagonals are . The sum of the antidiagonals is

2. The second column is 1 1 −1 −5 5 61 −61 −1385 1385.... Its binomial transform yields:

:{| style="text-align:right" |- | 1|| 2|| 2|| −4|| −16|| 32|| 272 |- |1||0||−6||−12||48||240 |- |−1||−6||−6||60||192 |- |−5||0||66||32 |- |5||66||66 |- |61||0 |- |−61 |}

The first row of this array is 1 2 2 −4 −16 32 272 544 −7936 15872 353792 −707584.... The absolute values of the second bisection are the double of the absolute values of the first bisection.

Consider the Akiyama-Tanigawa algorithm applied to (n) / ( (n + 1) = abs( (n)) + 1 = 1, 2, 2, , 1, , , , 1, , , ....

:{| style="text-align:right" |1||2||2||||1|||| |- |−1||0||||2||||0 |- |−1||−3||−||3|| |- |2||−3||−||−13 |- |5||21||− |- |−16||45 |- |−61 |}

The first column whose the absolute values are could be the numerator of a trigonometric function.

is an autosequence of the first kind (the main diagonal is ). The corresponding array is:

:{| style="text-align:right" |0||−1||−1||2||5||−16||−61 |- |−1||0||3||3||−21||−45 |- |1||3||0||−24||−24 |- |2||−3||−24||0 |- |−5||−21||24 |- |−16||45 |- |−61 |}

The first two upper diagonals are −1 3 −24 402... = (−1)n + 1 × . The sum of the antidiagonals is 0 −2 0 10... = 2 × (n + 1).

− is an autosequence of the second kind, like for instance / . Hence the array:

:{| style="text-align:right" |- |2||1||−1||−2||5||16||−61 |- |−1||−2||−1||7||11||−77 |- |−1||1||8||4||−88 |- |2||7||−4||−92 |- |5||−11||−88 |- |−16||−77 |- |−61 |}

The main diagonal, here 2 −2 8 −92..., is the double of the first upper one, here . The sum of the antidiagonals is 2 0 −4 0... = 2 × (n +1). − = 2 × .

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.

:\begin{align} \tan x &= x + \frac{2x^3}{3!} + \frac{16x^5}{5!} + \frac{272x^7}{7!} + \frac{7936x^9}{9!} + \cdots\[6pt] \sec x &= 1 + \frac{x^2}{2!} + \frac{5x^4}{4!} + \frac{61x^6}{6!} + \frac{1385x^8}{8!} + \frac{50521x^{10}}{10!} + \cdots \end{align}

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 + x + \tfrac{1}{2}x^2 + \tfrac{1}{3}x^3 + \tfrac{5}{24}x^4 + \tfrac{2}{15}x^5 + \tfrac{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).

Related sequences

The arithmetic mean of the first and the second Bernoulli numbers are the associate Bernoulli numbers: , , , , , / . Via the second row of its inverse Akiyama–Tanigawa transform , they lead to Balmer series / .

The Akiyama–Tanigawa algorithm applied to (n + 4) / (n) leads to the Bernoulli numbers / , / , or without B1, named intrinsic Bernoulli numbers B**i(n).

:{| style="text-align:center; padding-left; padding-right: 2em;" |- |1|||||||| |- ||||||||| |- |0|||||||| |- |−||−||−||−||0 |- |0||−||−||−||− |}

Hence another link between the intrinsic Bernoulli numbers and the Balmer series via (n).

(n − 2) = 0, 2, 1, 6,... is a permutation of the non-negative numbers.

The terms of the first row are f(n) = + . 2, f(n) is an autosequence of the second kind. 3/2, f(n) leads by its inverse binomial transform to 3/2 −1/2 1/3 −1/4 1/5 ... = 1/2 + log 2.

Consider g(n) = 1/2 – 1 / (n+2) = 0, 1/6, 1/4, 3/10, 1/3. The Akiyama-Tanagiwa transforms gives:

:{| style="text-align:center; padding-left; padding-right:2em;" |- |0||||||||||||... |- |−||−||−||−||−||−||... |- |0||−||−||−||−||−||... |- |||||||||0||−||... |}

0, g(n), is an autosequence of the second kind.

Euler (n) / (n + 1) without the second term () are the fractional intrinsic Euler numbers The corresponding Akiyama transform is:

:{| style="text-align:center; padding-left; padding-right: 2em;" |- |1||1|||||| |- |0|||||||| |- |−||−||0|||| |- |0||−||−||−||− |- |||||−||−||− |}

The first line is Eu(n). Eu(n) preceded by a zero is an autosequence of the first kind. It is linked to the Oresme numbers. The numerators of the second line are preceded by 0. The difference table is:

:{| style="text-align:center; padding-left; padding-right: 2em;" |- |0||1||1|||||||| |- |1||0||−||−||−||−||− |- |−1||−||0|||||||| |}

Arithmetical properties of the Bernoulli numbers

The Bernoulli numbers can be expressed in terms of the Riemann zeta function as for integers n ≥ 0 provided for the expression −nζ(1 − n) is understood as the limiting value and the convention 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. For example, the Agoh–Giuga conjecture postulates that p is a prime number if and only if pB**p − 1 is congruent to −1 modulo p. 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.

The Kummer theorems

The Bernoulli numbers are related to Fermat's Last Theorem (FLT) by Kummer's theorem, which says:

:If the odd prime p does not divide any of the numerators of the Bernoulli numbers B2, B4, ..., B**p − 3 then has no solutions in nonzero integers.

Prime numbers with this property are called regular primes. Another classical result of Kummer are the following congruences.

Main article: Kummer's congruence

:Let p be an odd prime and b an even number such that p − 1 does not divide b. Then for any non-negative integer k :: \frac{B_{k(p-1)+b}}{k(p-1)+b} \equiv \frac{B_{b}}{b} \pmod{p}.

A generalization of these congruences goes by the name of p-adic continuity.

{{math|''p''}}-adic continuity

If b, m and n are positive integers such that m and n are not divisible by p − 1 and m ≡ n (mod p**b − 1 (p − 1)), then

:(1-p^{m-1})\frac{B_m}{m} \equiv (1-p^{n-1})\frac{B_n} n \pmod{p^b}.

Since , this can also be written

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

where and , so that u and v are nonpositive and not congruent to 1 modulo 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 modulo p − 1 to a particular a ≢ 1 mod (p − 1), and so can be extended to a continuous function ζ**p(s) for all p-adic integers \mathbb{Z}_p, the p-adic zeta function.

Ramanujan's congruences

The following relations, due to Ramanujan, provide a method for calculating Bernoulli numbers that is more efficient than the one given by their original recursive definition:

:\binom{m+3}{m} B_m=\begin{cases} \frac{m+3}{3}-\sum\limits_{j=1}^\frac{m}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 0\pmod 6;\ \frac{m+3}{3}-\sum\limits_{j=1}^\frac{m-2}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 2\pmod 6;\ -\frac{m+3}{6}-\sum\limits_{j=1}^\frac{m-4}{6}\binom{m+3}{m-6j}B_{m-6j}, & \text{if } m\equiv 4\pmod 6.\end{cases}

Von Staudt–Clausen theorem

Main article: Von Staudt–Clausen theorem

The von Staudt–Clausen theorem was given by Karl Georg Christian von Staudt and Thomas Clausen independently in 1840. The theorem states that for every n 0, : B_{2n} + \sum_{(p-1),\mid,2n} \frac1p is an integer. The sum extends over all primes p for which p − 1 divides 2n.

A consequence of this is that the denominator of B2n is given by the product of all primes p for which p − 1 divides 2n. In particular, these denominators are square-free and 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 B2k + 1 − m is 0 for m even and 2k + 1 − m 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 the integer in question is zero. However, by applying Worpitzky's representation one gets

: 2B_n =\sum_{m=0}^n (-1)^m \frac{2}{m+1}m! \left{{n+1\atop m+1} \right} = 0\quad(n1 \text{ is odd})

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 } to }, then }. The last equation can only hold if

: \sum_{\text{odd }m=1}^{n-1} \frac 2 {m^2}S_{n,m}=\sum_{\text{even } m=2}^n \frac{2}{m^2} S_{n,m} \quad (n2 \text{ is even}).

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

:, :.

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

A restatement of 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 numbers. In fact Marcel Riesz proved that the RH is equivalent to the following assertion:

:For every ε there exists a constant C**ε 0 (depending on ε) such that εxε 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(\frac{B_{2k}}{2k}\right)} = 2\sum_{k=1}^\infty \frac{k^{\overline{k}}x^k}{(2\pi)^{2k}\beta_{2k}}.

n denotes the rising factorial power in the notation of D. E. Knuth. The numbers 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 numbers.

Generalized Bernoulli numbers

The generalized Bernoulli numbers are certain algebraic numbers, defined similarly to the Bernoulli numbers, that are related to special values of Dirichlet L-functions in the same way that Bernoulli numbers are related to special values of the Riemann zeta function.

Let χ be a Dirichlet character modulo f. The generalized Bernoulli numbers attached to χ are defined by

: \sum_{a=1}^f \chi(a) \frac{te^{at}}{e^{ft}-1} = \sum_{k=0}^\infty B_{k,\chi}\frac{t^k}{k!}.

Apart from the exceptional , we have, for any Dirichlet character χ, that if χ(−1) ≠ (−1)k.

Generalizing the relation between Bernoulli numbers and values of the Riemann zeta function at non-positive integers, one has the for all integers k ≥ 1:

: L(1-k,\chi)=-\frac{B_{k,\chi}}k,

where L(s,χ) is the Dirichlet L-function of χ.

Eisenstein–Kronecker number

Main article: Eisenstein–Kronecker number

Eisenstein–Kronecker numbers are an analogue of the generalized Bernoulli numbers for imaginary quadratic fields. They are related to critical L-values of Hecke characters.

Appendix

Assorted identities

: S_m(n) = \frac 1 {m+1} ((\mathbf{B} + n)^{m+1} - B_{m+1}),

where the symbol Bk that appears during binomial expansion of the parenthesized term is to be replaced by the Bernoulli number Bk (and ). More suggestively and mnemonically, this may be written as

:S_m(n) = \int_0^n (\mathbf{B}+x)^m,dx.

Other Bernoulli identities can be written compactly with this symbol. For example,

: (1-2\mathbf{B})^m = (2-2^m) B_m.

: \zeta(n) = \frac{(-1)^{\frac{n}{2} - 1} B_n (2\pi)^n}{2(n!)}.

: B_n = n! \begin{vmatrix} 1 & 0 & \cdots & 0 & 1 \ \frac{1}{2!} & 1 & \cdots & 0 & 0 \ \vdots & \vdots & \ddots & \vdots & \vdots \ \frac{1}{n!} & \frac{1}{(n-1)!} & \cdots & 1 & 0 \ \frac{1}{(n+1)!} & \frac{1}{n!} & \cdots & \frac{1}{2!} & 0 \end{vmatrix}.

So, the determinant is σ**n(1), the Stirling polynomial at .

: B_{2n} = -\frac{(2n)!}{2^{2n} - 2} \begin{vmatrix} 1 & 0 & 0 & \cdots & 0 & 1 \ \frac{1}{3!} & 1 & 0 & \cdots & 0 & 0 \ \frac{1}{5!} & \frac{1}{3!} & 1 & \cdots & 0 & 0 \ \vdots & \vdots & \vdots & \ddots & \vdots& \vdots \ \frac{1}{(2n+1)!} & \frac{1}{(2n-1)!} & \frac{1}{(2n-3)!} &\cdots & \frac{1}{3!} & 0 \end{vmatrix}.

: \frac{1}{n} \sum_{k=1}^n \binom{n}{k}B_k B_{n-k}+B_{n-1}=-B_n.

: \sum_{k=0}^n \binom{n+1}k (n+k+1)B_{n+k}=0.

: B_n = - \sum_{k=1}^{n+1} \frac{(-1)^k}{k} \binom{n+1}{k} \sum_{j=1}^k j^n.

: (-1)^m \sum_{r=0}^m \binom{m}{r} B_{n+r}=(-1)^n \sum_{s=0}^n \binom{n}{s} B_{m+s}.

: H_n=\sum_{k=1}^n k^{-1}.

Then (H. Miki 1978)

: \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.

: (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.

: \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).

Choosing or results in the Bernoulli number identity.

: \sum_{k=0}^n \binom{n}{k} \frac{B_k}{n-k+2} = \frac{B_{n+1}}{n+1}.

: -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) = \delta_{n,0}.

: (-1)^{m+1} \sum_{j=0}^k \binom{k}{j} \frac{B_{m+1+j}}{m+1+j} + (-1)^{k+1} \sum_{j=0}^m \binom{m}{j}\frac{B_{k+1+j}}{k+1+j} = \frac{k!m!}{(k+m+1)!}.

Notes

References

Bibliography

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