# [seqfan] A000041, the partitions : the classical example of p(200), macmahon, hardy and ramanujan

Simon Plouffe simon.plouffe at gmail.com
Mon Mar 14 17:59:11 CET 2011

```
Hello,

I have been musing around the expansion of certain numbers
in the exp(Pi) base.

Some results are I think interesting.

If I use F(x) = product(1/(1-x^n),n=1..infinity), then
when x -> exp(-Pi*x) the value for F(10) is easily found to
be

1/2 1/2
GAMMA(3/4) (5 + 5 5   )
-------------------------- = F(10).
1/4     5 Pi
Pi    exp(----)
12

and lprinted : F(10) =

GAMMA(3/4)*(5+5*5^(1/2))^(1/2)/(Pi^(1/4)*exp(5/12*Pi))

If I expand this number in base exp(Pi*10) then the first
205 coefficients can be obtained. In other words, I can
compute p(n) for up to n=205 using 1 mathematical constant
and enough precision. note : all the coefficients from
n=1 to n=205 are computed at once.

The <enough precision> is 2800 digits, unfortunately.

I can only think at the Major Percy MacMahon that
took 6 months to compute p(200) by hand and later Hardy
and Ramanujan, using a very clever method to get
that number!

The method I use is not very clever, I only use
a numerical trick, but it does work.

With the base exp(Pi*16) I could obtain the
algebraic value also and get :

F(16) =
2*2^(3/4)*(560+396*2^(1/2)+3*(69708+49291*2^(1/2))^(1/2))^(1/8)*GAMMA(3/4)/Pi^(1/4)/exp(2/3*Pi)

well, by using 11000 digits, I can get the first 500 values,

NOTE 2) the computation of F(x) when x is small is
more suitable when I use the infinite product, it converges
quite fast, it is nice to have
the exact value of F(x) but faster to use the
classical formula.

Best regards,

Simon Plouffe

```