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
.