391)
f(x)=1/(1-x)
a(0)=x and a(n+1)=f(a(n))
n=0 gives us a(1)=f(x)
so a(2)=f(a(1))=f(f(x))
and a(3)=f(f(f(x)))
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a(n)=f(f(f(.......f(x))))))))))) (n times f occurs)
a(1)=f(x)=1/1-x
a(2)=f(f(x))=f(1/(1-x))=(x-1)/x
a(3)=f(f(f(x)))=f((x-a)/x)=x
a(4) will similarly be 1/(1-x)
and series follows
1/(1-x) , (x-1)/x , x
these keep on occuring one after another
so a(n) willd epend on what remainder n leaves when divided by 3
if its 1 , then a(n)=1/(1-x)
if its 2, then a(n)=(x-1)/x
if its 0, then a(n)=x
