Bernoulli number

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.

A page from Seki Takakazu's Katsuyō Sanpō (1712), tabulating binomial coefficients and Bernoulli numbers

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:

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

Jakob Bernoulli's "Summae Potestatum", 1713[lower-alpha 1]

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).[lower-alpha 2] 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 ${\displaystyle -{\tfrac {1}{12}}n^{2}}$.

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

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

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 ${\displaystyle m\geq 1}$):

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

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.

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

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

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

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

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.

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

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]

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

Tangent

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

Cotangent

${\displaystyle {\begin{aligned}\cot x&{}={\frac {1}{x}}\sum _{n=0}^{\infty }{\frac {(-1)^{n}B_{2n}(2x)^{2n}}{(2n)!}},&\qquad 0<|x|<\pi .\end{aligned}}}$

Hyperbolic tangent

${\displaystyle {\begin{aligned}\tanh x&=\sum _{n=1}^{\infty }{\frac {2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}}\;x^{2n-1},&|x|&<{\frac {\pi }{2}}.\end{aligned}}}$

Hyperbolic cotangent

${\displaystyle {\begin{aligned}\coth x&{}={\frac {1}{x}}\sum _{n=0}^{\infty }{\frac {B_{2n}(2x)^{2n}}{(2n)!}},&\qquad \qquad 0<|x|<\pi .\end{aligned}}}$

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

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

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.

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).

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]

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).

There are formulas connecting Pascal's triangle to Bernoulli numbers[lower-alpha 3]

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

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:

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

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.