Please prove, give more symbolic or numeric support (counterexample!?), simplify or drop me a reference (or some vague hunch).


We have 
$$B\_n = n!\sum\_{\lambda} \frac{\lambda^{2n}}{p'\_n(\lambda)}$$ 
where $\lambda$ ranges over the roots of the polynomial $p\_n$ and $p\_n'$ is the derivative of $p\_n$. $B\_n$ is the $n$-th Bernoulli number.


The polynomial $p\_n$ is defined as follows. 
$$p\_n(x):=x^{n+2}t\_n(1/x)$$
where
$$t\_n(x) = \sum\_{k=0}^{n+2} \frac{x^k}{k!} -1$$
is the truncated Taylor polynomial of $\exp-1$ to power $n+2$.


Then $p\_n$ is the [reciprocal polynomial](http://en.wikipedia.org/wiki/Reciprocal_polynomial) of $t\_n$ just without conjugation (though it probably makes no difference if there is conjugation or not - since $p$ is a polynomial with real coefficients and the roots come in conjugate pairs).


Examples (already tested for $n=0..18$ with a symbolic solver and for $n=0..62$ numerically):


$n=0$


$$t\_0(x)=\frac{x^2}{2}+x$$


$$p\_0(x)=x^2t(1/x)=x^2(\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{2}+x$$


root of $p\_0$ is $-1/2$.


$$p\_0'(x)=1$$


$$B\_0=0!\cdot \frac{(-1/2)^{2\cdot 0}}{1}=1$$


$n=1$


$$t\_1(x)=\frac{x^3}{3!}+\frac{x^2}{2!}+x=\frac{x^3}{6}+\frac{x^2}{2}+x$$
$$p\_1(x)=x^3t\_1(1/x)=x^3(\frac{1}{6x^3}+\frac{1}{2x^2}+\frac{1}{x})=\frac{1}{6}+\frac{x}{2}+x^2$$


Roots of $p\_1$ are $${\lambda}\_{1,2}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt {15}i}{12}$$


We have $$\lambda\_{1,2}^2=-\frac{1}{24}\frac{-}{+}\frac{\sqrt{15}i}{24}$$


$$p\_1'(x)=2x+\frac{1}{2}$$ so
$$\frac{1}{p\_1'(\lambda\_{1,2})}=\frac{-}{+}\frac{2 \sqrt{15}i}{5}$$
$$\frac{\lambda\_{1,2}^2}{p\_{1}'(\lambda\_{1,2})}=-\frac{1}{4}\frac{+}{-}\frac{\sqrt{15}i}{60}$$


$$B\_1=1!(\frac{\lambda\_{1}^2}{p'\_{1}(\lambda\_1)}+\frac{\lambda\_{2}^2}{p'\_{2}(\lambda\_2)})=-\frac{1}{4}-\frac{1}{4}=-\frac{1}{2}$$



Bernoulli number formula involving roots of Taylor polynomial of $\exp-1$
Start by the definition of the Bernoulli numbers via their generating function: we have
$$\big(e^x-1\big)\sum\_{k=0}^\infty B \_ k\frac{x^k}{k!}=x\, .$$
Therefore, writing with your notation $ t \_ n(x)= \sum\_{k=1}^{n+2}\frac{x^k}{k!}$,


$$ t \_ n (x) \sum \_ {k=0}^{n } B \_ k \frac {x ^ k}{k!} =x + x^{n+1}S \_ n(x) + \frac {B \_ n}{n!(n+2)!}x ^ {2n+2}\, ,$$
for some polynomial $S \_ n (x)$ of degree not larger than $n$. Passing to the reciprocal polynomials we obtain an equation in form of a division by $p\_n$ with remainder:


$$x^{2n+1}=p \_ n (x) q \_ n(x) + r \_ n(x)$$


for a polynomial $q \_ n(x)$ of degree $n$ ( precisely, it's $q \_ n(x):= B\_{n }(x)/n!$) and a polynomial $r \_ n(x)$ with $\mathrm{deg\, } r \le n+1 < \mathrm{deg\,} p\_n$ and $r \_ n (0) = - \frac{B \_ n}{n!(n+2)!} $ .


Finally, let's write the partial fraction decomposition of the (proper) fraction $\frac{r \_n(x)}{p\_n(x)}$. We have, summing over the $n+2$ (simple) roots of $p \_ n(x)$, by a general elementary identity


$$\frac{r\_n(x)}{p \_ n (x)}=\sum \_ \lambda \frac{r \_ n(\lambda )}{p' \_ n (\lambda ) }\frac{1}{x-\lambda }\, . $$


Computing the identity at $x=0$ we finally find your identity:


$$ B \_n =n!\sum \_ \lambda \frac{ \lambda^ {2n} }{p' \_ n (\lambda ) } . $$


