# How quickly can we estimate the $n$th Bernoulli number to $m$ digits of precision?

Regardless of form, I want to know how I can get some interpretation (closed formula preferred) of $B_n$ that is equivalent to $m$ digits of precision, and I'd like to know the fastest way to do it (in terms of calculations required to get the estimate). Can someone please help?

For your information, you may want to consider Wikipedia's page on Bernoulli numbers. I've been attempting to use the "explicit definition" that they give, but I've been running into problems. I'll keep you informed of the progress that I make...

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What exactly is your question? Does not en.wikipedia.org/wiki/… answer it, with references? –  lhf Nov 2 '11 at 16:23
@lhf: No, they give formulas for how to get $B_n$ - but they don't tell which method is fastest for a given amount of precision. I want to know the fastest way to get $B_n$ to $m$ digits of precision. –  Matt Groff Nov 2 '11 at 16:28
Have you seen Fillebrown's and Harvey's papers? –  Guess who it is. Nov 2 '11 at 16:40
@J.M.: Thanks for the links! I am looking for an estimate, though, which may be easier and quicker to calculate... –  Matt Groff Nov 2 '11 at 16:59

## 1 Answer

I will assume that $m$ is small. Then a good approximation is obtained from the following $$B_{2n} = (-1)^{n-1} 2 \frac{(2n)!}{(2\pi)^{2n}} \zeta(2n) = (-1)^{n-1} 2 \frac{(2n)!}{(2\pi)^{2n}} \prod_{i \ge 1} \left( 1 - \frac{1}{p_i^{2n}} \right)^{-1}$$

To obtain sufficient approximation you truncate the product over primes accordingly. See these slides from W. Stein.

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Can someone give an asymptotic bounds for precision as a function of $i$, or vice versa? I'd like to see how this compares, if possible... –  Matt Groff Nov 2 '11 at 16:41