In mathematics, the Bernoulli numbers B_n are a sequence of rational numbers which occur frequently in analysis. The Bernoulli numbers appear in (and can be defined by) the Taylor series expansions of the tangent and hyperbolic tangent functions, in Faulhaber's formula for the sum of m-th powers of the first n positive integers, in the Euler–Maclaurin formula, and in expressions for certain values of the Riemann zeta function.

The values of the first 20 Bernoulli numbers are given in the adjacent table. Two conventions are used in the literature, denoted here by ${\displaystyle B_{n}^{-{))}$ and ${\displaystyle B_{n}^{+{))}$; they differ only for n = 1, where ${\displaystyle B_{1}^{-{))=-1/2}$ and ${\displaystyle B_{1}^{+{))=+1/2}$. For every odd n > 1, B_n = 0. For every even n > 0, B_n is negative if n is divisible by 4 and positive otherwise. The Bernoulli numbers are special values of the Bernoulli polynomials ${\displaystyle B_{n}(x)}$, with ${\displaystyle B_{n}^{-{))=B_{n}(0)}$ and ${\displaystyle B_{n}^{+}=B_{n}(1)}$.^[1]

The Bernoulli numbers were discovered around the same time by the Swiss mathematician Jacob Bernoulli, after whom they are named, and independently by Japanese mathematician Seki Takakazu. Seki's discovery was posthumously published in 1712^[2][3][4] in his work Katsuyō Sanpō; Bernoulli's, also posthumously, in his Ars Conjectandi of 1713. Ada Lovelace's note G on the Analytical Engine from 1842 describes an algorithm prepared by Babbage for generating Bernoulli numbers with Babbage's machine.^[5] As a result, the Bernoulli numbers have the distinction of being the subject of the first published complex computer program.

Notation

The superscript ± used in this article distinguishes the two sign conventions for Bernoulli numbers. Only the n = 1 term is affected:

• B⁻
  _n with B⁻
  ₁ = −1/2 (OEIS: A027641 / OEIS: A027642) is the sign convention prescribed by NIST and most modern textbooks.^[6]
• B⁺
  _n with B⁺
  ₁ = +1/2 (OEIS: A164555 / OEIS: A027642) was used in the older literature,^[1] and (since 2022) by Donald Knuth^[7] following Peter Luschny's "Bernoulli Manifesto".^[8]

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

Since B_n = 0 for all odd n > 1, and many formulas only involve even-index Bernoulli numbers, only a few authors write "B_n" instead of B_2n . 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), Abu Bakr al-Karaji (d. 1019, Persia) and Abu Ali al-Hasan ibn al-Hasan 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 the sums of the pth powers of the first n positive integers for p = 0, 1, 2, ..., k.

The Swiss mathematician Jakob 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.^[9]

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.^[2] 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^[9] a rigorous proof of Faulhaber's formula was first published by Carl Jacobi in 1834.^[10] Knuth's in-depth study of Faulhaber's formula concludes (the nonstandard notation on the LHS is explained further on):

"Faulhaber never discovered the Bernoulli numbers; i.e., he never realized that a single sequence of constants B₀, B₁, B₂, ... would provide a uniform

${\textstyle \sum n^{m}={\frac {1}{m+1))\left(B_{0}n^{m+1}-{\binom {m+1}{1))B_{1}n^{m}+{\binom {m+1}{2))B_{2}n^{m-1}-\cdots +(-1)^{m}{\binom {m+1}{m))B_{m}n\right)}$

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 n^m from polynomials in N to polynomials in n."^[11]

In the above Knuth meant ${\displaystyle B_{1}^{-))$; instead using ${\displaystyle B_{1}^{+))$ the formula avoids subtraction:

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

Reconstruction of "Summae Potestatum"

The Bernoulli numbers OEIS: A164555(n)/OEIS: A027642(n) were introduced by Jakob Bernoulli in the book Ars Conjectandi published posthumously in 1713 page 97. The main formula can be seen in the second half of the corresponding facsimile. 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. The expression c·c−1·c−2·c−3 means c·(c−1)·(c−2)·(c−3) – the small dots are used as grouping symbols. Using today's terminology these expressions are falling factorial powers c^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).^[b] 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:

${\displaystyle \sum _{k=1}^{n}k^{c}={\frac {n^{c+1)){c+1))+{\frac {1}{2))n^{c}+\sum _{k=2}^{c}{\frac {B_{k)){k!))c^{\underline {k-1))n^{c-k+1}.}$

This formula suggests setting B₁ = 1/2 when switching from the so-called 'archaic' enumeration which uses only the even indices 2, 4, 6... 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^k−1 has for k = 0 the value 1/c + 1.^[12] Thus Bernoulli's formula can be written

${\displaystyle \sum _{k=1}^{n}k^{c}=\sum _{k=0}^{c}{\frac {B_{k)){k!))c^{\underline {k-1))n^{c-k+1))$

if B₁ = 1/2, recapturing the value Bernoulli gave to the coefficient at that position.

The formula for ${\displaystyle \textstyle \sum _{k=1}^{n}k^{9))$ in the first half of the quotation by Bernoulli above contains an error at the last term; it should be ${\displaystyle -{\tfrac {3}{20))n^{2))$ instead of ${\displaystyle -{\tfrac {1}{12))n^{2))$.

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.^[13]

Recursive definition

The Bernoulli numbers obey the sum formulas^[1]

${\displaystyle {\begin{aligned}\sum _{k=0}^{m}{\binom {m+1}{k))B_{k}^{-{))&=\delta _{m,0}\\\sum _{k=0}^{m}{\binom {m+1}{k))B_{k}^{+{))&=m+1\end{aligned))}$

where ${\displaystyle m=0,1,2...}$ and δ denotes the Kronecker delta. Solving for ${\displaystyle B_{m}^{\mp {))}$ gives the recursive formulas

${\displaystyle {\begin{aligned}B_{m}^{-{))&=\delta _{m,0}-\sum _{k=0}^{m-1}{\binom {m}{k)){\frac {B_{k}^{-{))}{m-k+1))\\B_{m}^{+}&=1-\sum _{k=0}^{m-1}{\binom {m}{k)){\frac {B_{k}^{+)){m-k+1)).\end{aligned))}$

Explicit definition

In 1893 Louis Saalschütz listed a total of 38 explicit formulas for the Bernoulli numbers,^[14] usually giving some reference in the older literature. One of them is (for ${\displaystyle m\geq 1}$):

${\displaystyle {\begin{aligned}B_{m}^{-{))&=\sum _{k=0}^{m}\sum _{v=0}^{k}(-1)^{v}{\binom {k}{v)){\frac {v^{m)){k+1))\\B_{m}^{+}&=\sum _{k=0}^{m}\sum _{v=0}^{k}(-1)^{v}{\binom {k}{v)){\frac {(v+1)^{m)){k+1)).\end{aligned))}$

Generating function

The exponential generating functions are

${\displaystyle {\begin{alignedat}{3}{\frac {t}{e^{t}-1))&={\frac {t}{2))\left(\operatorname {coth} {\frac {t}{2))-1\right)&&=\sum _{m=0}^{\infty }{\frac {B_{m}^{-{))t^{m)){m!))\\{\frac {t}{1-e^{-t))}&={\frac {t}{2))\left(\operatorname {coth} {\frac {t}{2))+1\right)&&=\sum _{m=0}^{\infty }{\frac {B_{m}^{+}t^{m)){m!)).\end{alignedat))}$

where the substitution is ${\displaystyle t\to -t}$. If we let ${\displaystyle F(t)=\sum _{i=1}^{\infty }f_{i}t^{i))$ and ${\displaystyle G(t)=1/(1+F(t))=\sum _{i=0}^{\infty }g_{i}t^{i))$ then

${\displaystyle G(t)=1-F(t)G(t).}$

Then ${\displaystyle g_{0}=1}$ and for ${\displaystyle m>0}$ the m^th term in the series for ${\displaystyle G(t)}$ is:

${\displaystyle g_{m}t^{i}=-\sum _{j=0}^{m-1}f_{m-j}g_{j}t^{m))$

If

${\displaystyle F(t)={\frac {e^{t}-1}{t))-1=\sum _{i=1}^{\infty }{\frac {t^{i)){(i+1)!))}$

then we find that

${\displaystyle G(t)=t/(e^{t}-1)}$

${\displaystyle {\begin{aligned}m!g_{m}&=-\sum _{j=0}^{m-1}{\frac {m!}{j!)){\frac {j!g_{j)){(m-j+1)!))\\&=-{\frac {1}{m+1))\sum _{j=0}^{m-1}{\binom {m+1}{j))j!g_{j}\\\end{aligned))}$

showing that the values of ${\displaystyle i!g_{i))$ obey the recursive formula for the Bernoulli numbers ${\displaystyle B_{i}^{-))$.

The (ordinary) generating function

${\displaystyle z^{-1}\psi _{1}(z^{-1})=\sum _{m=0}^{\infty }B_{m}^{+}z^{m))$

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:

${\displaystyle B_{2n}=4n(-1)^{n+1}\int _{0}^{\infty }{\frac {t^{2n-1)){e^{2\pi t}-1))\mathrm {d} t}$

Bernoulli numbers and the Riemann zeta function

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

B⁺
_n = −nζ(1 − n)           for n ≥ 1 .

Here the argument of the zeta function is 0 or negative.

By means of the zeta functional equation and the gamma reflection formula the following relation can be obtained:^[15]

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

Now the argument of the zeta function is positive.

It then follows from ζ → 1 (n → ∞) and Stirling's formula that

${\displaystyle |B_{2n}|\sim 4{\sqrt {\pi n))\left({\frac {n}{\pi e))\right)^{2n}\quad }$ 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^[16] which require only O(p (log p)²) operations (see big O notation).

David Harvey^[17] 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 n = 10⁸. Harvey's implementation has been included in SageMath since version 3.1. Prior to that, Bernd Kellner^[18] computed B_n to full precision for n = 10⁶ in December 2002 and Oleksandr Pavlyk^[19] for n = 10⁷ with Mathematica in April 2008.

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	1672	3330
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 number in normalized scientific notation.

A possible algorithm for computing Bernoulli numbers in the Julia programming language is given by^[14]

b    = Array{Float64}(undef, n+1)
b[1] = 1
b[2] = -0.5
for m=2:n 
    for k=0:m
        for v=0:k
            b[m+1] += (-1)^v * binomial(k,v) * v^(m) / (k+1)
        end
    end
end
return b

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^[20]

${\displaystyle \sum _{k=a}^{b-1}f(k)=\int _{a}^{b}f(x)\,dx+\sum _{k=1}^{m}{\frac {B_{k}^{-)){k!))(f^{(k-1)}(b)-f^{(k-1)}(a))+R_{-}(f,m).}$

This formulation assumes the convention B⁻
₁ = −1/2. Using the convention B⁺
₁ = +1/2 the formula becomes

${\displaystyle \sum _{k=a+1}^{b}f(k)=\int _{a}^{b}f(x)\,dx+\sum _{k=1}^{m}{\frac {B_{k}^{+)){k!))(f^{(k-1)}(b)-f^{(k-1)}(a))+R_{+}(f,m).}$

Here ${\displaystyle f^{(0)}=f}$ (i.e. the zeroth-order derivative of ${\displaystyle f}$ is just ${\displaystyle f}$). Moreover, let ${\displaystyle f^{(-1)))$ denote an antiderivative of ${\displaystyle f}$. By the fundamental theorem of calculus,

${\displaystyle \int _{a}^{b}f(x)\,dx=f^{(-1)}(b)-f^{(-1)}(a).}$

Thus the last formula can be further simplified to the following succinct form of the Euler–Maclaurin formula

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

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

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

Here s^k denotes the rising factorial power.^[21]

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.

${\displaystyle \psi (z)\sim \ln z-\sum _{k=1}^{\infty }{\frac {B_{k}^{+)){kz^{k))))$

Sum of powers

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

${\displaystyle S_{m}(n)=\sum _{k=1}^{n}k^{m}=1^{m}+2^{m}+\cdots +n^{m}.}$

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:

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

where (^{m + 1}
_k) denotes the binomial coefficient.

For example, taking m to be 1 gives the triangular numbers 0, 1, 3, 6, ... OEIS: A000217.

${\displaystyle 1+2+\cdots +n={\frac {1}{2))(B_{0}n^{2}+2B_{1}^{+}n^{1})={\tfrac {1}{2))(n^{2}+n).}$

Taking m to be 2 gives the square pyramidal numbers 0, 1, 5, 14, ... OEIS: A000330.

${\displaystyle 1^{2}+2^{2}+\cdots +n^{2}={\frac {1}{3))(B_{0}n^{3}+3B_{1}^{+}n^{2}+3B_{2}n^{1})={\tfrac {1}{3))\left(n^{3}+{\tfrac {3}{2))n^{2}+{\tfrac {1}{2))n\right).}$

Some authors use the alternate convention for Bernoulli numbers and state Bernoulli's formula in this way:

${\displaystyle S_{m}(n)={\frac {1}{m+1))\sum _{k=0}^{m}(-1)^{k}{\binom {m+1}{k))B_{k}^{-{))n^{m+1-k}.}$

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.^[22]

Taylor series

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

$${\displaystyle {\begin{aligned}\tan x&={\hphantom {1 \over x))\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n)){(2n)!))\;x^{2n-1},&&\left|x\right|<{\frac {\pi }{2)).\\\cot x&={1 \over x}\sum _{n=0}^{\infty }{\frac {(-1)^{n}B_{2n}(2x)^{2n)){(2n)!)),&0<&|x|<\pi .\\\tanh x&={\hphantom {1 \over x))\sum _{n=1}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{2n)){(2n)!))\;x^{2n-1},&&|x|<{\frac {\pi }{2)).\\\coth x&={1 \over x}\sum _{n=0}^{\infty }{\frac {B_{2n}(2x)^{2n)){(2n)!)),&0<&|x|<\pi .\end{aligned))}$$

Laurent series

The Bernoulli numbers appear in the following Laurent series:^[23]

Digamma function: ${\displaystyle \psi (z)=\ln z-\sum _{k=1}^{\infty }{\frac {B_{k}^{+{))}{kz^{k))))$

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

${\displaystyle {\textit {ES))_{n}=(2^{2n-2}-2^{4n-3})\operatorname {Numerator} \left({\frac {B_{4n)){4n))\right).}$

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 k^m is employed. The signless Worpitzky numbers are defined as

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

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

${\displaystyle 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, 1/2, 1/3, ...

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

B₀ = 1

B₁ = 1 − 1/2

B₂ = 1 − 3/2 + 2/3

B₃ = 1 − 7/2 + 12/3 − 6/4

B₄ = 1 − 15/2 + 50/3 − 60/4 + 24/5

B₅ = 1 − 31/2 + 180/3 − 390/4 + 360/5 − 120/6

B₆ = 1 − 63/2 + 602/3 − 2100/4 + 3360/5 − 2520/6 + 720/7

This representation has B⁺
₁ = +1/2.

Consider the sequence s_n, n ≥ 0. From Worpitzky's numbers OEIS: A028246, OEIS: A163626 applied to s₀, s₀, s₁, s₀, s₁, s₂, s₀, s₁, s₂, s₃, ... is identical to the Akiyama–Tanigawa transform applied to s_n (see Connection with Stirling numbers of the first kind). This can be seen via the table:

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 OEIS: A198631 (n) / OEIS: A006519 (n + 1):

E₀ = 1

E₁ = 1 − 1/2

E₂ = 1 − 3/2 + 2/4

E₃ = 1 − 7/2 + 12/4 − 6/8

E₄ = 1 − 15/2 + 50/4 − 60/8 + 24/16

E₅ = 1 − 31/2 + 180/4 − 390/8 + 360/16 − 120/32

E₆ = 1 − 63/2 + 602/4 − 2100/8 + 3360/16 − 2520/32 + 720/64

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

${\displaystyle B_{n}={\frac {n}{2^{n+1}-2))\sum _{k=0}^{n-1}(-2)^{-k}\,W_{n-1,k}.}$

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

OEIS: A164555 (n + 1) / OEIS: A027642(n + 1) = n + 1/2^{n + 2} − 2 × OEIS: A198631(n) / OEIS: A006519(n + 1)

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

1/2, 1/6, 0, −1/30, 0, 1/42, ... = (1/2, 1/3, 3/14, 2/15, 5/62, 1/21, ...) × (1, 1/2, 0, −1/4, 0, 1/2, ...)

The numerators of the first parentheses are OEIS: A111701 (see Connection with Stirling numbers of the first kind).

Connection with Stirling numbers of the second kind

If one defines the Bernoulli polynomials B_k(j) as:^[24]

${\displaystyle B_{k}(j)=k\sum _{m=0}^{k-1}{\binom {j}{m+1))S(k-1,m)m!+B_{k))$

where B_k for k = 0, 1, 2,... are the Bernoulli numbers.

One also has the following for Bernoulli polynomials,^[24]

${\displaystyle B_{k}(j)=\sum _{n=0}^{k}{\binom {k}{n))B_{n}j^{k-n}.}$

The coefficient of j in (^j
_{m + 1}) is (−1)^m/m + 1.

Comparing the coefficient of j in the two expressions of Bernoulli polynomials, one has:

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

(resulting in B₁ = +1/2) which is an explicit formula for Bernoulli numbers and can be used to prove Von-Staudt Clausen theorem.^[25][26][27]

Connection with Stirling numbers of the first kind

The two main formulas relating the unsigned Stirling numbers of the first kind [ⁿ
_m] to the Bernoulli numbers (with B₁ = +1/2) are

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

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

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

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

Akiyama–Tanigawa number

m n	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	...	...	...	...

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 OEIS: A051714/OEIS: A051715.

An autosequence is a sequence which has its inverse binomial transform equal to the signed sequence. If the main diagonal is zeroes = OEIS: A000004, the autosequence is of the first kind. Example: OEIS: A000045, the Fibonacci numbers. If the main diagonal is the first upper diagonal multiplied by 2, it is of the second kind. Example: OEIS: A164555/OEIS: A027642, the second Bernoulli numbers (see OEIS: A190339). The Akiyama–Tanigawa transform applied to 2⁻ⁿ = 1/OEIS: A000079 leads to OEIS: A198631 (n) / OEIS: A06519 (n + 1). Hence:

Akiyama–Tanigawa transform for the second Euler numbers

m n	0	1	2	3	4
0	1	1/2	1/4	1/8	1/16
1	1/2	1/2	3/8	1/4	...
2	0	1/4	3/8	...	...
3	−1/4	−1/4	...	...	...
4	0	...	...	...	...

See OEIS: A209308 and OEIS: A227577. OEIS: A198631 (n) / OEIS: A006519 (n + 1) are the second (fractional) Euler numbers and an autosequence of the second kind.

(OEIS: A164555 (n + 2)/OEIS: A027642 (n + 2) = 1/6, 0, −1/30, 0, 1/42, ...) × ( 2^{n + 3} − 2/n + 2 = 3, 14/3, 15/2, 62/5, 21, ...) = OEIS: A198631 (n + 1)/OEIS: A006519 (n + 2) = 1/2, 0, −1/4, 0, 1/2, ....

Also valuable for OEIS: A027641 / OEIS: A027642 (see Connection with Worpitzky numbers).

Connection with Pascal's triangle

There are formulas connecting Pascal's triangle to Bernoulli numbers^[c]

${\displaystyle B_{n}^{+}={\frac {|A_{n}|}{(n+1)!))~~~}$

where ${\displaystyle |A_{n}|}$ is the determinant of a n-by-n Hessenberg matrix part of Pascal's triangle whose elements are: ${\displaystyle a_{i,k}={\begin{cases}0&{\text{if ))k>1+i\\{i+1 \choose k-1}&{\text{otherwise))\end{cases))}$

Example:

${\displaystyle B_{6}^{+}={\frac {\det {\begin{pmatrix}1&2&0&0&0&0\\1&3&3&0&0&0\\1&4&6&4&0&0\\1&5&10&10&5&0\\1&6&15&20&15&6\\1&7&21&35&35&21\end{pmatrix))}{7!))={\frac {120}{5040))={\frac {1}{42))}$

Connection with Eulerian numbers

There are formulas connecting Eulerian numbers ⟨ⁿ
_m⟩ to Bernoulli numbers:

${\displaystyle {\begin{aligned}\sum _{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle &=2^{n+1}(2^{n+1}-1){\frac {B_{n+1)){n+1)),\\\sum _{m=0}^{n}(-1)^{m}\left\langle {n \atop m}\right\rangle {\binom {n}{m))^{-1}&=(n+1)B_{n}.\end{aligned))}$

Both formulae 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.

A binary tree representation

The Stirling polynomials σ_n(x) are related to the Bernoulli numbers by B_n = n!σ_n(1). S. C. Woon described an algorithm to compute σ_n(1) as a binary tree:^[28]

Woon's recursive algorithm (for n ≥ 1) starts by assigning to the root node N = [1,2]. Given a node N = [a₁, a₂, ..., a_k] of the tree, the left child of the node is L(N) = [−a₁, a₂ + 1, a₃, ..., a_k] and the right child R(N) = [a₁, 2, a₂, ..., a_k]. A node N = [a₁, a₂, ..., a_k] 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

${\displaystyle N!=a_{1}\prod _{k=2}^{\operatorname {length} (N)}a_{k}!.}$

Restricted to the nodes N of a fixed tree-level n the sum of 1/N! is σ_n(1), thus

${\displaystyle B_{n}=\sum _{\stackrel {N{\text{ node of))}((\text{ tree-level ))n)){\frac {n!}{N!)).}$

For example:

B₁ = 1!(1/2!)

B₂ = 2!(−1/3! + 1/2!2!)

B₃ = 3!(1/4! − 1/2!3! − 1/3!2! + 1/2!2!2!)

Integral representation and continuation

The integral

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

has as special values b(2n) = B_2n for n > 0.

For example, b(3) = 3/2ζ(3)π⁻³i and b(5) = −15/2ζ(5)π⁻⁵i. 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

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

Another similar integral representation is

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

The relation to the Euler numbers and 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:

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

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 odd n, B_n = E_n = 0 (with the exception B₁), it suffices to consider the case when n is even.

${\displaystyle {\begin{aligned}B_{n}&=\sum _{k=0}^{n-1}{\binom {n-1}{k)){\frac {n}{4^{n}-2^{n))}E_{k}&n&=2,4,6,\ldots \\[6pt]E_{n}&=\sum _{k=1}^{n}{\binom {n}{k-1)){\frac {2^{k}-4^{k)){k))B_{k}&n&=2,4,6,\ldots \end{aligned))}$

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 π. These numbers are defined for n ≥ 1 as^[29][30]

${\displaystyle S_{n}=2\left({\frac {2}{\pi ))\right)^{n}\sum _{k=0}^{\infty }{\frac {(-1)^{kn)){(2k+1)^{n))}=2\left({\frac {2}{\pi ))\right)^{n}\lim _{K\to \infty }\sum _{k=-K}^{K}(4k+1)^{-n}.}$

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.^[31] The first few of these numbers are

${\displaystyle S_{n}=1,1,{\frac {1}{2)),{\frac {1}{3)),{\frac {5}{24)),{\frac {2}{15)),{\frac {61}{720)),{\frac {17}{315)),{\frac {277}{8064)),{\frac {62}{2835)),\ldots }$ (OEIS: A099612 / OEIS: A099617)

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.

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

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 = 2S_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 (OEIS: A132049 / OEIS: A132050):

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

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

Consider the Akiyama–Tanigawa transform for the sequence OEIS: A046978 (n + 2) / OEIS: A016116 (n + 1):

0	1	1/2	0	−1/4	−1/4	−1/8	0
1	1/2	1	3/4	0	−5/8	−3/4
2	−1/2	1/2	9/4	5/2	5/8
3	−1	−7/2	−3/4	15/2
4	5/2	−11/2	−99/4
5	8	77/2
6	−61/2

From the second, the numerators of the first column are the denominators of Euler's formula. The first column is −1/2 × OEIS: A163982.

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 T_n = S_{n + 1} n!, sometimes called Euler zigzag numbers, are integers.

${\displaystyle T_{n}=1,\,1,\,1,\,2,\,5,\,16,\,61,\,272,\,1385,\,7936,\,50521,\,353792,\ldots \quad n=0,1,2,3,\ldots }$ (OEIS: A000111). See (OEIS: A253671).

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

${\displaystyle \sum _{n=0}^{\infty }T_{n}{\frac {x^{n)){n!))=\tan \left({\frac {\pi }{4))+{\frac {x}{2))\right)=\sec x+\tan x}$.

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

${\displaystyle {\begin{aligned}B_{n}&=(-1)^{\left\lfloor {\frac {n}{2))\right\rfloor }[n{\text{ even))]{\frac {n}{2^{n}-4^{n))}\,T_{n-1}\ &n&\geq 2\\E_{n}&=(-1)^{\left\lfloor {\frac {n}{2))\right\rfloor }[n{\text{ even))]T_{n}&n&\geq 0\end{aligned))}$

These identities make it easy to compute the Bernoulli and Euler numbers: the Euler numbers E_2n are given immediately by T_2n and the Bernoulli numbers B_2n are fractions obtained from T_{2n - 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.^[32]

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

Seidel's algorithm for T_n

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 ^[33]) 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 T_2n and recommended this method for computing B_2n and E_2n 'on electronic computers using only simple operations on integers'.^[34]

V. I. Arnold^[35] rediscovered Seidel's algorithm and later Millar, Sloane and Young popularized Seidel's algorithm under the name boustrophedon transform.