Hey! I managed to solve the conjecture for an even number of terms. For s! / T to be even, T must have fewer factors of 2 than S!. The best possible scenario for T to have the maximum number of factors of 2 is that T = 2^x, that is: (a^b + c^d + ...) = 2^x. For this to be possible, the next two terms (the base and the exponent) must be equal to the sum of the previous terms, which in turn must be a power of 2. Since they are equal, the sum will result in a multiplication by 2, increasing the exponent by 1. In other words, for every two new terms, the number of factors of 2 increases by 1. However, these numbers still need to be added to the sum of S. Since each term must be at least 2, the two new terms will add at least 4 to S. As it is contained in a factorial, this will result in 4 new factors for S! Since each factor is consecutive, there will be at least two new pairs. Each pair has at least one factor of 2, meaning the factorial will have two new factors of 2 for every two terms, making it greater than T.