On Thu, 2007-07-26 at 13:10 -0700, Bill Hart wrote:
>
> Perhaps if one had a fast way of evaluating the Dedekind sum, one
> might have a chance.
>
> Bill.
>
I think this gives a faster way to compute it:
Write the sum as
s(h,k) = sum_{j=1}^{k-1} j/k [hj/k - floor(hj/k) - 1/2]
(This isn't strictly correct in general, but in our case hj/k will never
be an integer, so we are ok.)
Then if we separate this into three different sums and use some simple
summation formulas, we get
s(h,k) = h(k - 1)(2k - 1)/(6k) - (k-1)/4 -
(1/k) sum_{j=1}^{k-1} j*floor(hj/k).
(To compute the floor in the inner sum, you can just use integer
division.)
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