第二題

I admit that the following method is the first one coming to my mind.

   Let a[n] be the number of partitions of a positive integer n into distinct terms, and let a[0] = 1,  then the generating function  A(x) = sum a[n]x^n = (1+x)(1+x^2)(1+x^3)...... (1+x^m).... for all positive integer m.

  Let b[n] be the number of partitions of a positive integer n into odd terms,  and let b[0]=1, then the generating function B(x)=sum b[n] x^n = 1/(1-x) * 1/(1-x^3) * 1/(1-x^5).... *1/(1-x^k).... for all positive odd integer k.

   And we can prove that A(x)=B(x) by proving that    (1+ y)(1+y^2)(1+y^4)... = 1/(1-y) and then let y=x, x^3, x^5......   done.


Then if I want to cheat, I can give an "elementary" proof here, which is an interpretation of the above proof. Basically, for a partition of n into distinct terms, n=c_1+c_2+... Suppose C_i=2^{ki}*ti, where ki is a non-negative integer and ti is odd, then we have a partition of n into odd terms, n=(t1+t1+...) + (t2+t2....)+... We need to prove this mapping is 1-1, which is not very complicated.

       To illustrate,   one mapping is, 8=2+6  => 8=(1+1) + (3+3)