Fabrice Bellard
This article is an alpha version. Please send any comments to Fabrice.Bellard@enst.fr
Simon Plouffe explained in [1] a new algorithm to compute the n'th digit of and some other mathematical constants in any base with very little memory. Its running time is . We present here an improvement of this algorithm whose running time is while its memory requirements stay , which makes it practical to compute the millionth digit of for example.
We want to compute the n'th digit in base B of s, where
Let , and . We have
with the Chinese remainder theorem. Hence
The 'th digits of s, where , are the digits in base B of
With the last formula we have
As shown in [1], this equality is interesting because each term can be computed separatly, with a total memory of if we already know b and .
We want now to compute the n'th digit in base B of , where
with
With the first result we have
where
with
and
The key observation is that we can use for all the same modulo, hence
This can be rewritten
To have the digits after the n'th one, in base B, we compute
The running time is and the memory requirements stay if we suppose that:
Given
we can use the result of section 2 because if p is a prime number, we notice that
where is the multiplicity of p in n. It comes from the relation
Hence, if we want the n'th digit of in base B, we may use the following algorithm:
The running time is because there are prime numbers between 2 and 2n . The memory requirements are, as expected, in .
We have presented an algorithm to compute the n'th digit in any base B of whose running time is . It has the same running time as other classical methods for computing (e.g. arctangent formulas), but it uses little memory, it is very simple and does not need high precision computations. It is still slower than the BBP algorithm [2], but it works in any base. As described in [1], the same algorithm may be used to compute other numbers such as , , , and .