Meowth 寫到:
Q2: 3x+4y+6z=2003 有多少正整數解?
Main Estimation= (2003-(3+4+6)/2)^2/2/3/4/6=27680.64063
f(x)=1/(1-x^3)(1-x^4)(1-x^6)
=1/(1-x)^3/(1+x)^2/(1-wx)^2/(1-wwx)^2/(1-x+xx)/(1+xx)
=a/(1-x)+b/(1-x)^2+c/(1-x)^2
+d/(1+x)+e/(1+x)^2
+g/(1-wx)+g'/(1-wwx))+h/(1-wx)^2+h'/(1-wwx)^2)
+...
e=1/(1-(-1)^3)/4/6=1/48
h'=1/3/(1-w^4)/6=1/72+i/72sqrt3
Main variation =(2003-3-4-6+1)*( (-1)^(2003-3-4-6)/48+2cos(2*(2003-3-4-6)pi/3)/72+2sin(2*(2003-3-4-6)pi/3)/72sqrt3) =41.47916667
N~27680.64063+41.47916667=27722.11979~ 27722
error=0
