12/07/10 20:51:50.83
(1)
f(n)は1≦a≦b≦c…(i)、a+b＞c…(ii)、a+b+c=nの整数解(a,b,c)の個数に等しい
(i),(ii)より、b≦c≦a+b-1]
Σ[k=0,∞]f(k)x^k
=Σ[1≦a≦b≦c , a+b＞c]x^(a+b+c)
=Σ[b≦c≦a+b-1 , 1≦a≦b]x^c*x^(a+b)
=Σ[1≦a≦b](x^(a+b)(Σ[c=b,a+b-1]x^c))
=Σ[1≦a≦b]x^(a+b)(1-x^a)x^b/(1-x)
=1/(1-x)Σ[1≦a≦b]x^(2b)(x^a-x^(2b))
=1/(1-x)Σ[1≦b](x^(2b)(Σ[a=1,b]x^a-x^(2a)))
=1/(1-x)Σ[1≦b]x^(2b)((1-x^b)x/(1-x)-(1-x^(2b)x^2/(1-x^2))
=x/((1-x)(1-x^2))Σ[1≦b](x^(2b)-x^(3b)-x^(3b+1)+x^(4b+1))
=…
=x^3/((1-x^2)(1-x^3)(1-x^4))
=x^3/((1-x)^3*(1+x)^2*(1+x^2)*(1+x+x^2))
(2)
0≦m≦2n , 0≦n≦2m
⇔m/2≦n≦2m , 0≦m
mが偶数のとき、m=2k(k≧0)とおけて、これを満たすnはk～4k
mが奇数のとき、m=2k+1(k≧0)とおけて、これを満たすnはk+1～4k+2
S(x,y)
=Σ[0≦m≦2n , 0≦n≦2m]x^m*y^n
=Σ[k=0,∞](x^(2k)Σ[n=k,4k]y^n)+Σ[k=0,∞](x^(2k+1)Σ[n=k+1,4k+2]y^n)
=…
=(x^2*y^2+xy+1)/((1-x^2*y)(1-x*y^2))