Editing Bernoulli number (section)

=== Assorted identities ===
{{unordered list
|1 = [[Umbral calculus]] gives a compact form of Bernoulli's formula by using an abstract symbol {{math|'''B'''}}:

: <math>S_m(n) = \frac 1 {m+1} ((\mathbf{B} + n)^{m+1} - B_{m+1}) </math>

where the symbol {{math|'''B'''<sup>''k''</sup>}} that appears during binomial expansion of the parenthesized term is to be replaced by the Bernoulli number {{math|''B<sub>k</sub>''}} (and {{math|''B''<sub>1</sub> {{=}} +{{sfrac|1|2}}}}). More suggestively and mnemonically, this may be written as a definite integral:

:<math>S_m(n) = \int_0^n (\mathbf{B}+x)^m\,dx </math>

Many other Bernoulli identities can be written compactly with this symbol, e.g.

:<math> (1-2\mathbf{B})^m  = (2-2^m) B_m  </math>

|2 = Let {{math|''n''}} be non-negative and even

:<math> \zeta(n) = \frac{(-1)^{\frac{n}{2} - 1} B_n (2\pi)^n}{2(n!)}</math>

|3 = The {{math|''n''}}th [[cumulant]] of the [[uniform distribution (continuous)|uniform]] [[probability distribution]] on the interval [−1,&nbsp;0] is {{math|{{sfrac|''B''<sub>''n''</sub>|''n''}}}}.

|4 = Let {{math|''n''? {{=}} {{sfrac|1|''n''!}}}} and {{math|''n'' ≥ 1}}. Then {{math|''B''<sub>''n''</sub>}} is the following {{math|(''n'' + 1) × (''n'' + 1)}} determinant:{{r|Malenfant2011}}

: <math>
\begin{align}
B_n & = n! \begin{vmatrix}
1 & 0 & \cdots & 0 & 1 \\
2? & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & & \vdots & \vdots \\
n? & (n-1)? & \cdots & 1 & 0 \\
(n+1)? & n? & \cdots & 2? & 0
\end{vmatrix} \\[8pt]
& = n! \begin{vmatrix}
1 & 0 & \cdots & 0 & 1 \\
\frac{1}{2!} & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & & \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}
\end{align}
</math>

Thus the determinant is {{math|''σ''<sub>''n''</sub>(1)}}, the [[Stirling polynomial]] at {{math|''x'' {{=}} 1}}.

|5 = For even-numbered Bernoulli numbers, {{math|''B''<sub>2''p''</sub>}} is given by the {{math|(''p'' + 1) × (''p'' + 1)}} determinant::{{r|Malenfant2011}}

:<math> B_{2p} = -\frac{(2p)!}{2^{2p} - 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 & & \vdots& \vdots \\
\frac{1}{(2p+1)!} & \frac{1}{(2p-1)!} & \frac{1}{(2p-3)!} &\cdots & \frac{1}{3!} & 0
\end{vmatrix}</math>

|6 = Let {{math|''n'' ≥ 1}}. Then ([[Leonhard Euler]])<ref>Euler, E41, Inventio summae cuiusque seriei ex dato termino generali </ref>

: <math> \frac{1}{n} \sum_{k=1}^n \binom{n}{k}B_k B_{n-k}+B_{n-1}=-B_n </math>

|7 = Let {{math|''n'' ≥ 1}}.  Then{{r|vonEttingshausen1827}}

: <math> \sum_{k=0}^n \binom{n+1}k (n+k+1)B_{n+k}=0 </math>

|8 = Let {{math|''n'' ≥ 0}}.  Then ([[Leopold Kronecker]] 1883)

: <math> B_n = - \sum_{k=1}^{n+1} \frac{(-1)^k}{k} \binom{n+1}{k} \sum_{j=1}^k j^n </math>

|9 = Let {{math|''n'' ≥ 1}} and {{math|''m'' ≥ 1}}.  Then{{r|Carlitz1968}}

: <math> (-1)^m \sum_{r=0}^m \binom{m}{r} B_{n+r}=(-1)^n \sum_{s=0}^n \binom{n}{s} B_{m+s} </math>

|10 = Let {{math|''n'' ≥ 4}} and

: <math> H_n=\sum_{k=1}^n k^{-1} </math>

the [[harmonic number]]. Then (H. Miki 1978)

: <math> \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</math>

|11 = Let {{math|''n'' ≥ 4}}. [[Yuri Matiyasevich]] found (1997)

: <math> (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 </math>

|12 = ''Faber–[[Rahul Pandharipande|Pandharipande]]–[[Zagier]]–Gessel identity'': for {{math|''n'' ≥ 1}},

: <math> \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).</math>

Choosing {{math|''x'' {{=}} 0}} or {{math|''x'' {{=}} 1}} results in the Bernoulli number identity in one or another convention.

|13 = The next formula is true for {{math|''n'' ≥ 0}} if {{math|''B''<sub>1</sub> {{=}} ''B''<sub>1</sub>(1) {{=}} {{sfrac|1|2}}}}, but only for {{math|''n'' ≥ 1}} if {{math|''B''<sub>1</sub> {{=}} ''B''<sub>1</sub>(0) {{=}} −{{sfrac|1|2}}}}.

:<math> \sum_{k=0}^n \binom{n}{k} \frac{B_k}{n-k+2} = \frac{B_{n+1}}{n+1} </math>

|14 = Let {{math|''n'' ≥ 0}}. Then

:<math> -1 + \sum_{k=0}^n \binom{n}{k} \frac{2^{n-k+1}}{n-k+1}B_k(1) = 2^n </math>

and

:<math> -1 + \sum_{k=0}^n \binom{n}{k} \frac{2^{n-k+1}}{n-k+1}B_{k}(0) = \delta_{n,0} </math>

|15 = A reciprocity relation of M.&nbsp;B.&nbsp;Gelfand:{{r|AgohDilcher2008}}

: <math> (-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)!} </math>

}}