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, not bad. 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