In Euler–Maclaurin formula Bernoulli numbers express a finite sum through the integral. In my generalization a finite sum is expressed through another finite sum with a different step. All that is needed for that is to change the generating function: instead of $\frac{x}{e^x-1}$ we do the substitution $e = (1+1/n)^n$. And the generating function for generalized Bernoulli numbers is $\frac{x}{(1+x/n)^{n}-1}$

If $n\longrightarrow \infty$ then my numbers become Bernoulli

But I couldn't find anything where my Euler–Maclaurin generalization be useful. Do you have ideas where expressing a finite sum through another finite sum with another step could be useful?

  • 4
    $\begingroup$ you might want to write down your formula? $\endgroup$ Commented Jul 7 at 20:35
  • $\begingroup$ It was long ago but it's easy to derive. $\endgroup$ Commented Jul 7 at 20:38

1 Answer 1


These numbers and corresponding polynomials were introduced by Korobov, see second edition of his book "Number theory methods in numerical analysis". He found some examples, where these "discrete" polynomials are more useful than "continuous" ones. They work in some boundary cases where infinite constructions do not converge.

Similar construction is also known as degenerate Bernoulli polynomials, which were introduced earlier by Carlitz.

Some related links are

On summation and interpolation formulas,

A Discrete Analog of Euler's Summation Formula

On one generalization of Stirling numbers

Korobov polynomials and umbral calculus.

For a fixed $p$, Korobov numbers $K_n$ and Korobov polynomials $K_n(x)$ can be defined via generating functions \begin{gather} \label{Pr_Fu_P}F_K(t)=\sum\limits_{n=0}^{\infty} K_n\,\dfrac{t^n}{n!} = \dfrac{pt}{(t+1)^p -1},\\ \label{Pr_Fu_Px}F_K(x,t)=\sum\limits_{n=0}^{\infty} K_n(x)\,\dfrac{t^n}{n!}=\dfrac{pt(t+1)^x}{(t+1)^p-1}. \end{gather} In particular $$K_0=1,\quad K_1=-\dfrac{p-1}{2},\quad K_2=\dfrac{p^2-1}{6},\quad K_3=-\dfrac{p^2-1}{4},\\ \label{ExK} K_0(x)=1,\quad K_1(x)=x-\dfrac{p-1}{2},\quad K_2(x)=x^2-px+\dfrac{p^2-1}{6},\\ \nonumber K_3(x)=x^3-\dfrac{3(p+1)}{2}x^2+\dfrac{p(p+3)}{2}x-\dfrac{p^2-1}{4}. $$ Probably the main feature of Korobov numbers and polynomials is that they connect real and $p$-adic worlds. In the real world they become close to usual Bernoulli numbers and polynomials as $p\to\infty$: $$\lim\limits_{p\to\infty}p^{-n}K_n=B_n,\qquad\lim\limits_{p\to\infty}p^{-n}K_n(px)=B_n(x)\qquad(n\ge0).$$ If $p=0$ then Korobov numbers and polynomials coinside with Bernoulli numbers and polynomials of the second kind \begin{gather*} b_0=1,\quad b_1=\dfrac{1}{2},\quad b_2=\dfrac{1}{6},\quad b_3=\dfrac{1}{4},\\ b_0(x)=1,\quad b_1(x)=x+\dfrac{1}{2},\quad b_2(x)=x^2-\dfrac{1}{6},\\ b_3(x)=x^3-\dfrac{3}{2}x^2+\dfrac{1}{4}. \end{gather*} They can be defined by generating functions \begin{gather} \label{Def_b} F_b(t)=\sum_{n=0}^{\infty} b_n \frac{t^n}{n!}=\dfrac{t}{\ln(1+t)},\\ \label{Def_b(x)}F_b(x,t)=\sum_{n=0}^{\infty} b_n(x) \frac{t^n}{n!}=\dfrac{t(1+t)^x}{\ln(1+t)}. \end{gather} It means that $K_n(x)$ and $b_n(x)$ are close to each other in $p$-adic sense.

The same is true for corresponding Stirling numbers $\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{\!p}$ and $\left[\displaystyle{\vphantom{M}n\atop k}\right]_{\!p}$ are defined by \begin{gather*}\label{St9} x^{\underline{n}}=\sum\limits_{k=0}^{n}\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{\!p}p^k\left(\dfrac{x}{p}\right)^{\underline{k}},\\ \label{St10}p^n\left(\dfrac{x}{p}\right)^{\overline{n}} =\sum\limits_{k=0}^{n}\left[\displaystyle{\vphantom{M}n\atop k}\right]_{\!p}x^{\overline{k}}, \end{gather*} where $$\begin{array}{c} x^{\underline{n}}=x(x-1)\ldots(x-n+1),\\ x^{\overline{n}}=x(x+1)\ldots(x+n-1). \end{array}$$ These numbers are dual to each other $$\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{\!\frac{1}{p}}=\left(-\dfrac{1}{p}\right)^{n-k}\left[\displaystyle{\vphantom{M}n\atop k}\right]_{\!p},\qquad \left[\displaystyle{\vphantom{M}n\atop k}\right]_{\!\frac{1}{p}}=\left(-\dfrac{1}{p}\right)^{n-k}\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{\!p}\qquad(p\ne0),$$ and lie somewhere between usual Stirling numbers \begin{align*} \label{St16}&\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{0}=(-1)^{n-k}\left[\displaystyle{\vphantom{M}n\atop k}\right],&&\left[\displaystyle{\vphantom{M}n\atop k}\right]_{0}=(-1)^{n-k}\left\{\displaystyle{\vphantom{M}n\atop k}\right\},\\ \label{St17}&\lim\limits_{p\to\infty}\dfrac{1}{p^{n-k}}\left\{\displaystyle{\vphantom{M}n\atop k}\right\}_{\!p}=\left\{\displaystyle{\vphantom{M}n\atop k}\right\},&&\lim\limits_{p\to\infty}\dfrac{1}{p^{n-k}}\left[\displaystyle{\vphantom{M}n\atop k}\right]_{\!p}=\left[\displaystyle{\vphantom{M}n\atop k}\right].\end{align*}

  • $\begingroup$ Thanks, once I tried to find something about this polinomials, but it's hard to google a polinomial. OEIS is very good but there's nothing if you wish to know about a polinomial. $\endgroup$ Commented Jul 8 at 15:04
  • $\begingroup$ @МаратРамазанов Yes, they are not included in OEIS. I've added some information in my answer. $\endgroup$ Commented Jul 10 at 12:15

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