Simon Plouffe>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. (rwg)>Clever! SP> 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) (rwg)>F(16) = GAMMA(3/4)*2^(57/32)*%E^-(Pi*2/3)/(Pi^(1/4)*(-3*2^(1/2)-4*2^(1/4)+9)^(1/8)) Also, F(24) = 2^(25/48)*3^(3/8)*((3^(1/4)+SQRT(2))*SQRT(3)+3^(1/4)+2)*Pi^(3/4)/ (exp(Pi)*(SQRT(2)-1)^(3/4)*(SQRT(3)-1)^(1/6)*GAMMA(1/4)) isn't too bad. SP>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, As fast as F(x) = 1/SUM((-1)^n*(1-exp(-(2*n+1)*Pi*x))*exp(-n*(3*n+1)*Pi*x/2),n,0,INF) ? it is nice to have the exact value of F(x) but faster to use the classical formula. Best regards, Simon Plouffe (rwg)>If you use a series or product formula, you can use F(arbitrarily large).