(In reply to
Solution by Praneeth)
You have, for example
S(2) = C(1,0) S(0)+ C(1,1) S(1) = e1+e=2e1
However, the first four terms of S(2) are:
1^2/1! + 2^2 / 2! + 3^2 / 3! + 4^2 / 4!
= 1 + 2 + 3/2 + 16/24
= 1 + 2 + 3/2 + 2/3
which totals 5 + 1/6, which already exceeds 2e  1. None of the terms are negative, so the infinite sum must be larger, and I have computed as 2e. Likewise the remaining ones are all integral multiples of e without any integer subtracted.

Posted by Charlie
on 20080220 11:04:57 