# Is there an explicit formula for the Bernoulli numbers that doesn't implicitly recapitulate Faulhaber's formula?

So, I'm familiar with the "standard" explicit formula for the Bernoulli numbers:

$$B_m (n) = \sum^m_{k=0}\sum^k_{v=0}(-1)^v {k \choose v} {(n+v)^m \over k+1}$$

where choosing $n=0$ gives the Bernoulli numbers of the first kind and $n=1$ gives the Bernoulli numbers of the second kind. The term $(n+v)^m$ used in this formula basically means calculating Faulhaber's formula in a slight disguise, which of course depends on the Bernoulli numbers again. Wikipedia claims that the paper by Louis Saalschütz found at http://digital.library.cornell.edu/cgi/t/text/text-idx?c=math;idno=00450002 has some 38 other explicit formulae for the Bernoulli numbers, but unfortunately I can't read German and would not trust my understanding of the given formulae as a result. Is there any known explicit formula that doesn't rely on Faulhaber's formula the way the given formula does? Alternately, is there an English translation of that paper accessible anywhere?

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In your mind does $e=\sum_{k\ge0}1/k!$ not 'implicitly recapitulate' the other definition $e=\lim(1+1/n)^n$ as $n\to\infty$? – oldrinb Sep 18 '13 at 15:47
Not necessarily, although of course they are related. The fact is that you can calculate those expressions separately from one another, whereas using the expression I gave to calculate the coefficients for Faulhaber's formula requires you to employ Faulhaber's formula, which is the sort of dependence I would like to avoid. – David Sep 18 '13 at 15:51
These aren't Bernoulli numbers, they are Bernoulli polynomials. Which would you like an expression for, the former or the latter? In either case, do some of the formulas on the wiki page entice you? en.wikipedia.org/wiki/Bernoulli_polynomials – Alex R. Sep 18 '13 at 19:14