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\def\wB{\widetilde B} \def\O{\text O} \def\e{\varepsilon} \def\h{\frac12} \def\G{\Gamma} \def\l{\ell}
\topmatter\title A modified Bernoulli number\endtitle\author D\. Zagier\endauthor\endtopmatter\document   

 The classical Bernoulli numbers $B_n$, defined by the generating function 
 $$\frac x{e^x-1}=\sum_{n=0}^\infty B_n\,\frac{x^n}{n!}\,,\tag1$$
have many famous and beautiful properties, including the following three: \smallskip

\noindent(i) $\;B_n\,=\, 0$ {\it for odd} $n>1$.

\noindent(ii) {\it The fractional part of $B_n$ is given by}
$$ B_n\equiv-\sum\Sb(p-1)|n\\\text{$p$ prime}\endSb\frac1p\pmod1\qquad\quad(\text{$n>0,\;n$ even})\,.$$   

\noindent(iii) $\;B_n\,$ {\it is given asymptotically for large even $n$ by the formula}
$$ B_n\,\sim\,(-1)^{n/2-1}\,\frac{n!}{2^{n-1}\pi^n}\qquad\quad\text{($n\to\infty$, $n$ even).}$$
\medskip
 In this note we show that the rational numbers defined by
  $$ B^*_n=\sum_{r=0}^n\binom{n+r}{2r}\,\frac{B_r}{n+r}\qquad\qquad(n>0)\tag2$$ 
satisfy the following amusing variants of the above three properties: \medskip

\noindent(I) {\it The value of $B^*_n$ for $n$ odd is periodic; more precisely, it is given by} 
 $$\vbox{\offinterlineskip\halign{\hfil#\hfil&\ \vrule#\ &&$\hfil\quad#\quad\hfil$\cr  
  $n$ (mod 12)&&1&3&5&7&9&11\cr\omit&height2pt&\omit&\cr\noalign{\hrule}\omit&height2pt&\omit&\cr
   $B^*_n$&&3/4&-1/4&-1/4&1/4&1/4&-3/4\cr}}$$  

\noindent(II) {\it The fractional part of the number $\wB_n:=2nB^*_n\!-\!B_n$ is given by} 
  $$\wB_n\equiv\sum\Sb(p+1)|n\\\text{$p$ prime}\endSb\frac1p\pmod1\qquad\quad(\text{$n>0,\;n$ even})\,.$$ 

\noindent(III) $B^*_n$ {\it is asymptotically equal to $(-1)^{n/2-1}\dfrac{(n-1)!}{(2\pi)^n}$ for 
$n$ large and even, and is given much more precisely by the approximation}
  $$ B^*_n\,\approx\, (-1)^{n/2}\,\pi\,Y_n(4\pi)\qquad\quad\text{($n\to\infty$, $n$ even)},$$  
{\it where $Y_n(x)$ denotes the $n$th Bessel function of the second kind.}  

\newpage
The proofs of (I)--(III) will be given in the next three sections, after which we will give
the statement and proof of a fourth property, an exact formula for $B^*_n$ refining the asymptotic
formula (III). The proofs, especially those of (I) and (II), are quite fun and the reader is
invited to try to find them him/herself before proceeding.  

We end the introduction with a small table of the numbers $B^*_n$ and $\wB_n$.
 $$\vbox{\offinterlineskip\halign{\hfil#\hfil&\ \vrule#\ &&$\hfil\,#\,\hfil$\cr  
  $n$ &&2&4&6&8&10&12&14&16&18&20\cr\omit&height2pt&\omit&\cr\noalign{\hrule}\omit&height2pt&\omit&\cr
   $B^*_n$&&\frac1{24}&-\frac{27}{80}&-\frac{29}{1260}&\frac{451}{1120}&-\frac{65}{264}&-\frac{6571}{12012}
    &\frac{571}{312}&-\frac{181613}{38080}&\frac{23663513}{1220940}&-\frac{10188203}{83600}\cr  \omit&height2pt&\omit&\cr
   $\wB_n$&&0&-\frac83&-\frac3{10}&\frac{136}{21}&-5&-\frac{4249}{330}&\frac{651}{13}&-\frac{3056}{21}
  &\frac{109269}{170}&-\frac{247700}{57}\cr}}$$  

\bigskip\bigskip\noindent{\bf Proof of (I).} 
Instead of using the familiar generating function (1), we represent 
the Bernoulli numbers by the generating function
  $$ F(x)=\sum_{r=1}^\infty\frac{B_r}r\,x^r\,\in\,\Q[[x]]\,.$$
This formal power series does not converge anywhere, but occurs in the asymptotic formula
  $$ \frac{\G'(X)}{\G(X)} \;\sim\; \log X\,-\,\frac1X\,-\,F\bigl(\frac1X\bigr)\qquad\quad(X\to\infty)$$
for the logarithmic derivative of the gamma function, and the functional equation
$\G(X+1)=X\G(X)$ of the gamma function implies the functional equation 
 $$ F\bigl(\frac x{1-x}\bigr)=F(x)+x+\log(1-x)\quad\in\;\Q[[x]]\tag3$$
of the power series $F$. An elementary proof of (3), or of the equivalent but simpler functional
equation $G\bigl(\dfrac x{1-x}\bigr)=G(x)-x^2$ for the simpler power series 
$G(x)=\sum_{r=0}^\infty B_rx^{r+1}=x+x^2F'(x)$, can be obtained by noting that either one of these
functional equations is equivalent to the standard recursion formula
  $$ \sum_{k=0}^{n-1}\binom nkB_k=0\qquad(n>1) \tag4$$
for the Bernoulli numbers, which is in turn an easy restatement of the definition~(1). 

Now introduce a new power series $F_\lambda(x)$, depending on a parameter $\lambda$, by
  $$  F_\lambda(x)\;=\;F\bigl(\frac x{1-\lambda x+x^2}\bigr)\,-\log(1-\lambda x+x^2)\quad\in\;\Q[[x]]\,.  $$
For $\lambda=2$ this specializes to
  $$ F_2(x)\,=\,\sum_{r=1}^\infty\frac{B_r}r\,\frac {x^r}{(1-x)^{2r}}-2\log(1-x)
      =2\,\sum_{n=1}^\infty B^*_n x^n\,.\tag5$$
On the other hand, the functional equation (3) together with the symmetry property $F(-x)=F(x)+x$, which
is a restatement of (i), give the functional equations
  $$ F_{\lambda+1}(x) \,=\, F_\lambda(x)+\frac x{1-\lambda x+x^2}\,=\,F_{-\lambda}(-x) $$
for the power series $F_\lambda$.  We deduce
  $$\align F_2(&x)-F_2(-x)=\bigl(F_2(x)-F_1(x)\bigr)\,+\,\bigl(F_1(x)-F_0(x)\bigr)\,+\,\bigl(F_0(x)-F_{-1}(x)\bigr)\\
     &=\,\frac x{1-x+x^2}+\frac x{1+x^2}+\frac x{1+x+x^2}\,
      =\,\frac{3x-x^3-x^5+x^7+x^9-3x^{11}}{1-x^{12}}\,.\endalign$$
Statement (I) follows.

\bigskip\bigskip\noindent{\bf Proof of (II).}  
Rather surprisingly, this property is a consequence of the analogous property (ii) for the usual
Bernoulli numbers, the divisibility by $p-1$ being metamorphosed into the divisibility by $p+1$ by
the magic of generating functions.  We  begin by rewriting the definition of $\wB_n$ as
  $$ \wB_n=\sum_{r=0}^{n-1}\frac{(n+r)+(n-r)}{n+r}\,\binom{n+r}{2r}\, B_r=\beta_n+\beta_{n-1}-B_n\,,\tag 6$$ 
where $$\beta_n=\sum_{r=0}^n\binom{n+r}{2r}\, B_r\qquad\qquad(n\ge0)\,.$$
Fix a prime $p$.  We want to show that $p\wB_n$ is $p$-integral for all $n$ and is congruent to~1 mod $p$
if $p+1$ divides $n$ and to 0 otherwise. We suppose that $p>2$.   (The case $p=2$ is similar but easier.)
By (ii) we know that $pB_r$ is $p$-integral and is congruent to $-1$ mod $p$ if $p-1$ divides $r>0$ and to 0 
otherwise. Equation (6) then immediately gives the $p$-integrality of $p\wB_n$ and the congruence
  $$ p\wB_n \,\equiv \,-\gamma_n-\gamma_{n-1}+\cases 1&\text{if $(p-1)|n$}\\ 0&\text{otherwise}\endcases\tag7$$
(here and from now on all congruences are modulo $p$), where 
  $$\gamma_n=\sum_{r>0,\;(p-1)|r} \binom{n+r}{2r} \,.$$
As usual, we use generating functions.  From the definition of $\gamma_n$ we have
  $$ \sum_{n=0}^\infty\gamma_n\,x^n=\sum_{r>0,\;(p-1)|r} \frac{x^r}{(1-x)^{2r+1}}
     =\frac{x^{p-1}}{(1-x)^{2p-1}-x^{p-1}(1-x)}\,.$$
Hence (7) gives
  $$\align\sum_{n>0} p\wB_n\,x^n &\equiv-(1+x)\sum_{n=0}^\infty\gamma_n\,x^n+\frac{x^{p-1}}{1-x^{p-1}}\\
     &\equiv -(1+x)\,\frac{x^{p-1}(1-x)}{(1-x^p)^2-x^{p-1}(1-x)^2}+\frac{x^{p-1}}{1-x^{p-1}}\,.\endalign$$
The expression on the right simplifies to $\dfrac{x^{p+1}}{1-x^{p+1}}$, completing the proof.

\bigskip\bigskip\noindent{\bf Proof of (III).} 
As in the proofs of (I) and (II), we deduce property (III) from its classical Bernoulli analogue.
Suppose $n$ is even and large.  Then for $r=n-k$ with $k$ fixed (and even) we have
  $$ \binom{2n+r}{2r}\frac{B_r}{2n+r}  = \frac{(-1)^{n/2-1}(n-1)!}{(2\pi)^n}\,
       \biggl(\frac{(-1)^{k/2}(4\pi)^k}{k!}+\O\bigl(\frac1n\bigr)\biggr),$$
and since $\sum\dfrac{(-1)^{k/2}(4\pi)^k}{k!}=\cos(4\pi)=1$ this gives the asymptotic formula in (III).
The same argument in conjunction with the binomial coefficient identity
  $$ \binom{2n+r}{2r}\frac1{2n+r}  = \frac1{2r!}\sum_{0\le h\le k/2}\frac{(-1)^h2^{k-2h}(n-h-1)!}{h!(k-2h)!}\qquad(r=n-k)$$
(which  holds because both sides express the coefficient of $x^k$ in $(1-2x+x^2)^{-r}/2r$) lets us replace 
the asymptotic formula for $B_n^*$ by the full asymptotic expansion
 $$ (-1)^{n/2-1}B_n^*\,\sim\,\,\,\sum_{h\ge0}\frac{(n-h-1)!}{h!}\,(2\pi)^{-n+2h}\;.$$
Consulting standard reference works,  we discover that the expression on the right is also the asymptotic development
of $-\pi Y_n(4\pi)$, since $Y_n$ is defined by
$$ -\pi\,Y_n(2x)= \sum_{h=0}^{n-1}\frac{(n-h-1)!}{h!}x^{-n+2h} 
  +\sum_{h=0}^\infty(c_h+c_{n+h}-2\log x)\,\frac{(-1)^hx^{2h+n}}{h!(n+h)!}$$
with $c_h=1+\h+\cdots+\frac1h-\gamma=\O(\log h)$. (Cf.~[1], 7.2(32).)
This proves the second assertion in (III), though without any estimate of the error. 

Here are a few numerical values to illustrate the accuracy of the two approximations for $B_n^*$ in (III) :
 $$\vbox{\offinterlineskip\halign{$\hfil#\hfil$&\ \vrule#\ && \hfil\quad$#$\quad\hfil\cr  
    n && 20 & 30 & 50 \cr\omit&height2pt&\omit&\cr\noalign{\hrule}\omit&height3pt&\omit&\cr
   (n-1)!\,(2\pi)^{-n}&& \phantom013.228 & 10026347.89 & 0.75008667460769577\times10^{23}\cr \omit&height3pt&\omit&\cr
     -\pi\,Y_n(4\pi)&&  122.319 & 40532569.11 & 1.69052138468088825\times10^{23}\cr\omit&height3pt&\omit&\cr
   (-1)^{n/2-1}B^*_n&&  121.868 & 40532573.81 & 1.69052138468090709\times10^{23}\cr}}$$  
The poorness of the approximations in the first row is explained by the above asymptotic expansion, which shows
that the ratio of $|B_n^*|$ to $(n-1)!(2\pi)^{-n}$ is $1+Cn^{-1}+\O(n^{-2})$ with $C=4\pi^2\approx40$.

\newpage\noindent{\bf An exact formula.}  
In the case of the usual Bernoulli numbers, the rough asymptotic formula (iii) can be replaced by the exact formula
  $$  B_n\;=\;\frac{(-1)^{n/2-1}\,n!}{2^{n-1}\,\pi^n}\,\biggl(1+\frac1{2^n}+\frac1{3^n}+\cdots\biggr)
          \qquad\quad\text{($n>0$, $n$ even)}\tag8$$
due to Euler.  It is reasonable to look for a corresponding exact formula for $B_n^*$.

We start with numerical data.  From the table above we have $B_{50}^*+\pi Y_{50}(4\pi)\approx 1,884\times 10^9$,
the exact value being $1884415006.56\ldots$.  Guessing that this difference might be related to the value $-\pi Y_{50}(8\pi)$,
we compute this latter number, which turns out to be 1884414704.76\dots, suggesting that we are on the right track.
Going one step further, we find that the difference $B_{50}^*+\pi Y_{50}(4\pi)+\pi Y_{50}(8\pi)$ equals $301.79\ldots$,
rather close to the value $-\pi Y_{50}(12\pi)=300.89\ldots$.  This suggests that $B_n^*$ might be very well approximated by, 
or even equal to, the number $U_n=\sum_{\l=1}^\infty(-1)^{n/2}\pi Y_n(4\pi\l)$.  However, this sum diverges, since 
$(-1)^{n/2}\pi Y_n(4\pi\l)$ behaves like $-\h\l^{-1/2}$ for $\l$ large, so we must renormalize it, setting
  $$  U_n:=\sum_{\l=1}^\infty\biggl((-1)^{n/2}\pi Y_n(4\pi\l)\,+\,\frac1{2\sqrt\l}\biggr)
            -\h\,\zeta\bigl(\h\bigr).\tag9$$
The series converges only like $\sum\l^{-3/2}$, but can be replaced by the expression
$$  \align U_n&=\sum_{\l=1}^\infty\biggl((-1)^{n/2}\pi Y_n(4\pi\l)\,+\,\sum_{k=0}^K\frac{c_{n,k}}{\l^{k+1/2}}\biggr)
            -\,\sum_{k=0}^K c_{n,k}\,\zeta\bigl(k+\h\bigr)\,,\\
&\qquad c_{n,k}=\frac{(-1)^{[\frac{k+1}2]}}{2(8\pi)^k\,k!}\,(n+k-\tfrac12)(n+k-\tfrac32)\cdots(n-k+\tfrac12)\endalign  $$
for any $K\ge0$, where the $\l$th term is $\,\O(\l^{-K-3/2})$, so the numerical value of $U_n$ can be computed easily.  
Comparing it with $B_n^*$, we find the following table:
  $$\vbox{\offinterlineskip\halign{$\hfil#\hfil$&\ \vrule#\ && \hfil\quad$#$\quad\hfil\cr  
    n &&   B_n^*-U_n \cr\omit&height2pt&\omit&\cr\noalign{\hrule}\omit&height3pt&\omit&\cr
   2 && -0.886968175\ldots\cr \omit&height4pt&\omit&\cr 4 && -1.988273972\ldots\cr \omit&height4pt&\omit&\cr
  10 && -4.999969846\ldots\cr \omit&height2pt&\omit&\cr
  30 && -15+1.292\ldots\times10^{-13}\cr \omit&height2pt&\omit&\cr  50 && -25+5.646\ldots\times10^{-22}\cr}}$$
suggesting a formula of the form $$ B_n^*+\frac n2=U_n+\e_n\qquad\quad(n>0,\;\text{$n$ even}),\tag10 $$ 
where $\e_n$ is positive and goes to 0 rapidly as $n$ tends to infinity. (Notice that the quantity $B_n^*+n/2$
occurring on the left is given by the same expression as in (2), but with $B_r$ replaced by $(-1)^rB_r$.)
We now prove this and give an exact formula for the error term $\e_n$.
\bigskip\noindent(IV) {\it Define $B_n^*$ by $(2)$ and $U_n$ by $(9)$. Then equation $(10)$ holds with} 
 $$ \e_n\;=\;\sum_{k=1}^\infty\frac1{\sqrt{k(k+4)}}\,\biggl(\frac{\sqrt k+\sqrt{k+4}}2\biggr)^{-2n}\;\,\sim\;\,
\frac1{\sqrt5}\,\biggl(\frac{3+\sqrt5}2\biggr)^{-n}\,.\tag11$$ 


\newpage\noindent{\bf Proof of (IV).} 
Using the relation (8) between Bernoulli numbers and zeta values, we find
 $$ B_n^*=\frac1n-\frac n4-\sum\Sb2\le r\le n\\r\text{ even}\endSb\binom{n+r-1}{2r-1}\,\frac{(r-1)!}{(2\pi i)^r}\,\zeta(r)$$ 
and hence
 $$ B_n^*-U_n=\frac1n-\frac n4+\h\zeta\bigl(\h\bigr)+\sum_{\l=1}^\infty b_{n,\l}\,,\tag12$$
where 
 $$ b_{n,\l}:=(-1)^{n/2-1}\,\pi\,Y_n(4\pi\l) - \sum\Sb2\le r\le n\\r\text{ even}\endSb\binom{n+r-1}{2r-1}\,
        \frac{(r-1)!}{(2\pi i\l)^r} - \frac1{2\sqrt\l}\,.$$    
On the other hand, standard formulas for Bessel functions (cf.~[1], 7.2(15) and 7.3(16)) give
 $$ \align &(-1)^{n/2-1}\pi\,Y_n(x)=2\,\Re\bigl(K_n(ix)\bigr)\\&\quad=\frac{\sqrt{\pi/2}}{\G(n+\h)}\,\int_0^\infty
    \frac{e^{-ix}(t/2+ix)^{n-\h}+e^{ix}(t/2-ix)^{n-\h}}{(ix)^n}\,e^{-t}\,t^{n-\h}\,dt\endalign$$
for $n$ even and positive, and hence, after some simple manipulations, the formula
 $$ b_{n,\l}=\frac{2\,\G(\h)}{\G(n+\h)}\,\int_0^\infty t^{n-1}\,e^{-t}\,
   \biggl(f_n\bigl(\frac{8\pi\l}t\bigr)-\frac14\,\sqrt{\frac t{\pi\l}}\,\biggr)\,dt\tag13$$
where $f_n(x)$ is defined for $x>0$ by
  $$ f_n(x)=\frac{(1+ix)^{n-\h}+(1-ix)^{n-\h}}{2\,(ix)^n}-\sum\Sb2\le r\le n\\r\text{ even}\endSb\binom{n-\h}{n-r}(ix)^{-r}\,.$$
Note that $f(x)=(2x)^{-1/2}+\text O(x^{-3/2})$ as $x\to\infty$, so that $b_{n,\l}=\text O(\l^{-3/2})$, as we already know.
Clearly $f_n(x)$ extends to $\C^0:=\C\smallsetminus(-i\infty,-i]\smallsetminus[i,i\infty)$ as an even holomorphic function, 
and the binomial theorem gives the Taylor expansion
$$  f_n(x) = \sum_{r\ge0}(-1)^r\,\binom{n-\h}{n+2r}\,x^{2r}\qquad(|x|<1)\,.$$
The beta integral identity
 $$\binom{n-\h}{n+2r}=\frac1\pi\,\frac{\G(n+\h)\G(2r+\h)}{\G(n+2r+1)}=\frac1\pi\,\int_0^1 u^{2r-\h}\,(1-u)^{n-\h}\,du$$
now gives the integral representation
$$ f_n(x)=\frac1\pi\,\int_0^1 \frac{u^{-\h}\,(1-u)^{n-\h}}{1+x^2u^2}\,du $$
for $|x|<1$, and by analytic continuation this holds for all $x\in\C^0$.  Consider the more general integral
$$ f_n(x,s)=\frac1\pi\,\int_0^1 \frac{u^{s-1}\,(1-u)^{n-\h}}{1+x^2u^2}\,du\qquad(x\in\C^0,\;s\in\C,\;0<\Re(s)<2)\,, $$
so that $f_n(x)=f_n(x,\h)$.  It can be estimated for large $x$ by
$$ f_n(x,s)= \frac1\pi\int_0^\infty\frac{u^{s-1}\,du}{1+x^2u^2}\,+\,\text O(x^{-s-1})
  =\frac{x^{-s}}{2\sin(\pi s/2)}\,+\,\text O(x^{-s-1})\;. $$
In particular, $\sum_\l f_n(\l x,s)$ converges for $\Re(s)>1$ and all $x>0$, and using the Poisson summation identity
$$ \sum_{\l=-\infty}^\infty\frac1{1+\l^2\,t^2}=\frac{\pi/t}{\tanh\pi/t}
  =\frac\pi t\,\biggl(1+2\,\sum_{k=1}^\infty e^{-2\pi k/t}\biggr)\qquad(t>0)$$
we find
$$\sum_{\l=-\infty}^\infty f_n(\l x,s)=\frac1x\,\biggl(\frac{\G(s-1)\G(n+\h)}{\G(s+n-\h)}\,
    +\,2\,\sum_{k=1}^\infty\,\int_0^1 u^{s-2}\,(1-u)^{n-\h}\,e^{-2\pi k/xu}\,du\biggr)$$
for $\Re(s)>1$. Writing the left-hand side of this identity as 
 $$f_n(0,s)+\sum_{\l=1}^\infty\biggl(2f_n(\l x,s)-\frac{(\l x)^{-s}}{\sin(\pi s/2)}\biggr)
  +\frac{\zeta(s)\,x^{-s}}{\sin(\pi s/2)}$$
gives its analytic continuation to $\Re(s)>0$, and setting $s=\dfrac12$, $x=\dfrac{8\pi}t$ we obtain
$$\align \sum_{\l=1}^\infty\biggl(f_n\bigl(\frac{8\pi\l}t\bigr)-\frac14\sqrt{\frac t{\pi\l}}\biggr)
  =&-\frac{\G(n+\h)}{8\G(\h)\G(n)}\,t\,-\frac{\zeta(\h)}{4\sqrt\pi}\,t^\h\,-\frac{\G(n+\h)}{2\G(\h)\G(n+1)}\\
  &\;+\,\frac t{8\pi}\,\sum_{k=1}^\infty\,\int_0^1 u^{-\frac32}\,(1-u)^{n-\h}\,e^{-kt/4u}\,du\,.\endalign$$
Combining this with (13) and performing the integrations over $t$ we get
  $$ \sum_{\l=1}^\infty b_{n,\l} = -\frac n4-\frac{\zeta(\h)}2-\frac1n
  +\frac{\G(n+1)}{4\G(\h)\G(n+\h)}\,\sum_{k=1}^\infty\int_0^1\frac{ u^{-\frac32}\,(1-u)^{n-\h}}{(1+k/4u)^{n+1}}\,du\;. $$
The desired result now follows by inserting this into (12) and using the identity
 $$ \frac{\G(n+1)}{4\G(\h)\G(n+\h)}\,\int_0^1\frac{ u^{-\frac32}\,(1-u)^{n-\h}}{(1+k/4u)^{n+1}}\,du  
  =  \frac1{\sqrt{k(k+4)}}\,\biggl(\frac{\sqrt k+\sqrt{k+4}}2\biggr)^{-2n}\,,$$
which can be proved either from standard hypergeometric formulas or by expanding
both sides as power series in $1/k$ for $k>4$ and then using analytic continuation.  


\newpage\noindent{\bf Remarks.}  We end this paper with a number of remarks.

\subheading 1 The formula given in (i), which can be rewritten in the form
 $$ B_n^*=\h\,\biggl(\frac{-3}n\biggr)\,+\,\frac14\,\biggl(\frac{-4}n\biggr)\qquad\quad\text{($n$ odd)}\,,$$
occurred originally in [4] in the context of the Eichler-Selberg trace formula for the traces of the Hecke operator
$T_\l$ acting on modular forms on $SL_2(\Z)$.  The method of proof there gave a formula for these traces which had
a somewhat different form from the classical one, and in particular involved Bernoulli numbers. The specialization
of the formula to the case $\l=1$ gave the dimension of $M_k(SL_2(\Z))$ in terms of $B^*_{k-1}$, and the 
equality of this expression with the standard dimension formula required the periodicity property (I).

\subheading 2  The same idea as was used to prove (I) leads to another simple expression for the modified Bernoulli 
numbers $B^*_n$.  Specifically, from 
  $$ F_2(x) =  F_0(x)+\frac x{1-x+x^2}+\frac x{1+x^2}   $$
and the binomial formula we get the identity
  $$  2B^*_{2n}=\biggl(\frac{-3}n\biggr)+\sum_{r=0}^n(-1)^{n+r}\binom{n+r}{2r}\frac{B_{2r}}{n+r}\,,$$
whose second term has a pleasing similarity to the original formula (2) defining $B^*_n$.


\subheading 3  Next, we mention that the definition (2) can also be inverted to express the ordinary Bernoulli numbers
in terms of the modified ones, should we for any reason wish to do so.  Indeed, from (5) we have
  $$ \sum_{n=1}^\infty\frac{B_n}n\,x^n= 2\,\log\bigl(\frac{\sqrt{1+4x}-1}{2x}\bigr)
+ 2\sum_{r=1}^\infty B_r^*\,\biggl(\frac{1+2x-\sqrt{1+4x}}{2x}\biggr)^r\,,$$
and comparing coefficients of $B_n$ we get
  $$ B_n=(-1)^n\binom{2n}n + 2\,\sum_{r=1}^n(-1)^{n-r}\,r\binom{2n}{n-r}\,\,B_r^*\,.$$

\subheading 4  Finally, we mention that there are (at least) two ways of calculating Bernoulli
numbers which are faster than the standard recursion (4). The first is due to M.~Kaneko [3].
The second I noticed myself, but Kaneko has informed me that it is in fact a classical
identity going back to Kronecker. (See [2] for a historical survey.)   Nevertheless, these 
formulas are both pretty and useful, so for the sake of popularization we reproduce them here. 

\noindent{\bf a.} One can replace (4) by a recursion of the same type, but with only half as many terms,
namely, setting $b_n=(n+1)B_n$,
  $$ b_{2n}=-\frac1{n+1}\sum_{i=0}^{n-1}\binom{n+1}i\,b_{n+i}\tag 14$$
together with the conditions $b_1=-1$ and $b_{2n+1}=0$ for $n>0$.
 Equation (14) can be seen as a special case of the following fact:  Define an 
involution~$^*$ on the set of sequences  $\{a_0,\,a_1,\,a_2,\ldots\}$  by  $A^*(x)\,=\,e^{-x}\,A(-x)$, 
where $A(x):=\sum_{n=0}^\infty a_n x^n/(n+1)!$, or more explicitly by
$a_n^* = (-1)^n \sum_{i=0}^n \binom{n+1}{i+1} a_i$. Then the expression $\sum_{i=0}^n \binom{n}{i} a_{n+i-1}$ 
 $(n\ge 1)$  is anti-invariant under $^*$ and hence vanishes if  $A^* = A$.  
(Note that $A(x)=x/(e^x-1)=A^*(x)$ for $a_n\!=\!b_n$.)

\medskip\noindent{\bf b.} The Bernoulli numbers can be calculated directly, rather than recursively, by the closed formula
  $$  B_k= \sum_{n=0}^k(-1)^n\binom{k+1}{n+1}\frac{0^k+\cdots+n^k}{n+1} \,.\tag 15 $$
To prove this, we apply Bernoulli's famous formula for $1^k+\cdots+n^k$ to get
  $$\frac{0^k+\cdots+n^k}{n+1} =\frac{B_{k+1}(n+1)-B_{k+1}}{(k+1)(n+1)}
=\text{(polynomial of degree $k$ in $n$)}+\frac{B_k}{n+1}\,,$$
where $B_r(x)$ denotes the $r$th Bernoulli polynomial; taking the $(k+1)$st difference of both sides kills the
polynomial on the right, leaving only an easily computed multiple of $B_k$.   Formula (15) is  much more convenient 
for numerical computations than the recursion formula, at least if one wants to compute individual Bernoulli 
numbers rather than a table up to some limit, since the number of steps needed to compute $B_k$ is
$\,\text O(k)\,$ rather than $\,\text O(k^2)$ (each term in the sum can be computed from its
predecessor in $\,\text O(1)\,$ steps). Indeed, even for computing a table, (15) is sometimes more useful
than the recursion (4), since the time required is about the same but the storage requirements are reduced from 
$\,\text O(k)\,$ to $\,\text O(1)$.  Here is a one-line PARI program implementing the formula (15) (for $k>0$):
\bigskip {\font\big=cmr10 scaled \magstep1\parindent=6pt\tt
  B(k)=h=0;s=1;c=k+1;for(n=2,k+1,c=c*(n-k-2)/n;h=h+c*s/n;s=s+n\^{\kern.2pt}k);h \par\ \leftskip=0pt }
\bigskip \medskip


{\bf References.} 
\smallskip\noindent [1]  A.~Erd\'elyi, W.~Magnus, F.~Oberhettinger and F.~Tricomi, {\it Higher Transcendental
Functions} (Bateman Manuscript Project), Vol.~II, McGraw-Hill, New York (1953)
\smallskip\noindent [2] H.W.~Gould, Explicit formulas for Bernoulli numbers, 
   Amer.~Math.~Monthly {\bf 79} (1972) 44--51
\smallskip\noindent [3] M.~Kaneko, A recurrence formula for the Bernoulli numbers, 
  Proc\. J\. Acad\. {\bf 71} (1995) 192--193 
\smallskip\noindent [4] D.~Zagier, {\it Hecke operators and periods of modular forms},
  Israel Math. Conf. Proc. {\bf 3} (1990) 321--336

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