Is there any general explicit formula for Bernoulli numbers ? Something like:
$$f(x)=B_x$$
Where $B_x$ is the $x$-th Bernoulli number ?
Searching the internet I only found the so-called "generating formula" or recursive relations but can there be an explicit formula ? And if it can't exist why ?
On
Hmm, something like the explicite formula for $e^x$ perhaps, an infinite series...?
This is possible (as in the comment of Raymond Manzoni), however, the infinite series is divergent (but alternating, so Cesaro or Eulersummable).
Here is such an explicite formula:
$$B_{x+1} = f(x) = \left(1^x -2^x +3^x-4^x + ... - ...\right) \cdot{ 1+x\over 1-2 \cdot 2^x} $$
On
In the paper [2], the Bernoulli polynomials $B_n(x)$ were determinantally expressed as \begin{equation} B_n(x)=\frac{(-1)^n}{(n-1)!} \begin{vmatrix} 1 & x & x^2 & x^3 & \dotsm & x^{n-1} & x^n\\ 1 & \frac12 & \frac13 & \frac14 & \dotsm & \frac1n & \frac1{n+1}\\ 0 & 1 & 1 & 1 & \dotsm & 1 & 1\\ 0 & 0 & 2 & 3 & \dotsm & n-1 & n\\ 0 & 0 & 0 & \binom32 & \dotsm & \binom{n-1}2 & \binom{n}2\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \dotsm & \binom{n-1}{n-2} & \binom{n}{n-2} \end{vmatrix}, \quad n\in\mathbb{N}. \end{equation} In the paper [1] below, the Bernoulli polynomials $B_n(x)$ were represented as \begin{align*} B_n(x)&=(-1)^nn! \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x}{1!} & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ \frac{x^2}{2!} & \frac1{3!} & \frac1{2!} & 1 & 0 & 0 & \dotsm & 0\\ \frac{x^3}{3!} & \frac1{4!} & \frac1{3!} & \frac1{2!} & 1 & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ \frac{x^n}{n!} & \frac1{(n+1)!} & \frac1{n!} & \frac1{(n-1)!} & \frac1{(n-2)!}&\frac1{(n-3)!}&\dotsm& 1 \end{vmatrix}\\ &=(-1)^n\frac{n!}{\prod_{k=1}^nk!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & 3! & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!}&\frac{n!}{(n-3)!}&\dotsm& n! \end{vmatrix}\\ &=(-1)^n\prod_{k=1}^{n-1}\frac{(k-1)!}{k!} \begin{vmatrix} 1 & 1 & 0 & 0 & 0 & 0 & \dotsm & 0\\ x & \frac1{2!} & 1 & 0 & 0 & 0 & \dotsm & 0\\ x^2 & \frac{2!}{3!} & 1 & 2! & 0 & 0 & \dotsm & 0\\ x^3 & \frac{3!}{4!} & 1 & \frac{3!}{2!} & \frac{3!}{2!} & 0 & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm & \dotsm\\ x^n & \frac{n!}{(n+1)!} & 1 & \frac{n!}{(n-1)!} & \frac{n!}{(n-2)!2!}&\frac{n!}{(n-3)!3!}&\dotsm& \frac{n!}{2!(n-2)!} \end{vmatrix}. \end{align*} In the paper [4], the Bernoulli polynomials $B_n(x)$ was explicitly expressed as \begin{multline}\label{BP-Stirl-form} B_n(x)=\sum_{k=1}^nk!\sum_{r+s=k}\sum_{\ell+m=n}(-1)^m\binom{n}{\ell} \frac{\ell!}{(\ell+r)!}\frac{m!}{(m+s)!}\\ \times\Biggl[\sum_{i=0}^r\sum_{j=0}^s(-1)^{i+j}\binom{\ell+r}{r-i}\binom{m+s}{s-j}S(\ell+i,i)S(m+j,j)\Biggr]x^{m+s}(1-x)^{\ell+r} \end{multline} and was determinantally represented as \begin{equation}\label{Bern-Polyn-determ} B_n(x)=(-1)^n\biggl|\frac1{\ell+1}\binom{\ell+1}{m} \bigl[(1-x)^{\ell-m+1}-(-x)^{\ell-m+1}\bigr]\biggr|_{1\le \ell\le n,0\le m\le n-1} \end{equation} for $n\in\mathbb{N}$, where $S(n,k)$ denotes the Stirling numbers of the second kind and $|\cdot|_{1\le \ell\le n,0\le m\le n-1}$ denotes a $n\times n$ determinant.
In the paper [5], an alternative determinantal expression \begin{equation}\label{Bernoulli-Polyn-Det-Erew} B_n(x)=\frac{(-1)^n}{(n+1)!} \begin{vmatrix} 1&1&0&0&\dotsm&0&0&0\\ \frac{x}2&\binom{2}0&\frac12&0&\dotsm&0&0&0\\ \frac{x^2}3&\binom{3}0&\binom{3}1&\frac13&\dotsm&0&0&0\\ \frac{x^3}4&\binom{4}0&\binom{4}1&\binom{4}2&\dotsm&0&0&0\\ \vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots&\vdots\\ \frac{x^{n-2}}{n-1}&\binom{n-1}0&\binom{n-1}1&\binom{n-1}2&\dotsm &\binom{n-1}{n-3}&\frac1{n-1}&0\\ \frac{x^{n-1}}{n}&\binom{n}0&\binom{n}1&\binom{n}2&\dotsm &\binom{n}{n-3}&\binom{n}{n-2}&\frac1{n}\\ \frac{x^n}{n+1}&\binom{n+1}0&\binom{n+1}1&\binom{n+1}2&\dotsm &\binom{n+1}{n-3}&\binom{n+1}{n-2}&\binom{n+1}{n-1} \end{vmatrix}, \quad n\ge0 \end{equation} for the Bernoulli numbers $B_n(x)$ was derived.
Theorem 1.1 in [3] reads that, for all integers $n,r\ge0$, the Bernoulli polynomials $B_n(r)$ can be computed in terms of the $r$-Stirling numbers of the second kind $S_r(n,k)$ by \begin{equation}\label{Bernoulli-Poly-r-Stirling-eq} B_n(r)=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S_r(n,k). \end{equation}
When taking $x=0$ in the above determinatal expressions and closed-form expressions, we can arrive at closed-form formulas for the Bernoulli numbers $B_n=B_n(0)$.
References
The explicit formula \begin{equation}\label{Higgins-Gould-B}\tag{1} B_n=\sum_{k=0}^n\frac1{k+1}\sum_{j=0}^k(-1)^j\binom{k}{j}j^n,\quad n\ge0, \end{equation} has a long history, it appeared in the paper
It is a special case of the formula (2.5) in the paper
Its equivalent form is \begin{equation}\label{Bernoulli-Stirling-eq}\tag{2} B_n=\sum_{k=0}^n(-1)^k\frac{k!}{k+1}S(n,k), \quad n\ge0, \end{equation} where $S(n,k)$ is the Stirling numbers of the second kind. See page 29, Remark 2 in the paper
There existed at least seven alternative proofs of the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} in the paper [1, 2] above and in the following monographs and papers:
To the best of my knowledge, except the formulas \eqref{Higgins-Gould-B} and \eqref{Bernoulli-Stirling-eq} above, there are also the following explicit formulas for the Bernoulli numbers $B_n$: \begin{align}\label{Higgins-Gould-B(11)}\tag{3} B_n&=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{n!}{(n+j)!}\sum_{k=0}^j(-1)^{j-k}\binom{j}{k}k^{n+j}, \quad n\ge0;\\ B_n&=\sum_{i=0}^n(-1)^{i}\frac{\binom{n+1}{i+1}}{\binom{n+i}{i}}S(n+i,i), \quad n\ge0;\label{Bernoulli-Stirling-formula}\tag{4}\\ B_{2k}&=1+\sum_{m=1}^{2k-1}\frac{S(2k+1,m+1) S(2k,2k-m)}{\binom{2k}{m}}\\ &\quad-\frac{2k}{2k+1}\sum_{m=1}^{2k}\frac{S(2k,m)S(2k+1,2k-m+1)}{\binom{2k}{m-1}}, \quad k\in\mathbb{N};\tag{5}\\ B_{2k}&=\frac{(-1)^{k-1}k}{2^{2(k-1)}(2^{2k}-1)}\sum_{i=0}^{k-1}\sum_{\ell=0}^{k-i-1} (-1)^{i+\ell}\binom{2k}{\ell}(k-i-\ell)^{2k-1}, \quad k\in\mathbb{N};\tag{6}\\ B_{2m}&=(-1)^{m-1}\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\Biggl[\sum_{k=0}^{m-1} (-1)^k\binom{2m}{k}(m-k)^{2m-1}\\ &\quad+2\sum_{k=1}^{m-1}(-1)^k\sum_{\ell=0}^{m-k-1} (-1)^{\ell}\binom{2m}{\ell}(m-k-\ell)^{2m-1}\Biggr],\quad m\in\mathbb{N};\\ B_{2m}&=\frac{m} {2^{2m-1}\bigl(2^{2m}-1\bigr)}\sum_{\ell=1}^{2m}\frac{(-1)^{\ell-1}}{2^\ell} \biggl(\frac1\ell-\frac1{m+1}\biggr) \binom{2m+1}{\ell} \sum_{q=0}^\ell\binom{\ell}{q}(2q-\ell)^{2m}, \quad m\in\mathbb{N};\\ B_{2k}&= \frac12 - \frac1{2k+1} - 2k \sum_{i=1}^{k-1} \frac{A_{2(k-i)}}{2(k - i) + 1},\quad k\in\mathbb{N};\tag{7} \end{align} where $A_m$ is defined by \begin{equation*} \sum_{m=1}^nm^k=\sum_{m=0}^{k+1}A_mn^{m}. \end{equation*} The formulas \eqref{Higgins-Gould-B(11)} and \eqref{Bernoulli-Stirling-formula} are also equivalent to eah other.
By the way, I would like to mention two intereting double inequalities related to the Bernoulli numbers $B_{2n}$ as follows.
More related references