Define polynomials F_n(x) such that for all nonnegative integers n:

F₀(x)=1, F_n+1(0)=0 and

F'_n+1(x) = (n+1)F_n(x+1)

Find the prime factorization of F₃₀₀(7).

Starting with F_0 and working one's way up step by step, one finds F_1(x)=x, F_2(x)=x(x+2), F_3(x)=x(x+3)^2. So one conjectures that F_n(x)=x(x+n)^(n-1), and this is readily seen to satsify the given differential-difference equations.

Hence F_300(7)=7*(307)^299 and this is already the prime factorization since 307 is a prime.

Posted by Richard on 2006-08-07 13:45:55