1. The problem statement, all variables and given/known data are an = afloor(n-2) + afloor(2n/3) + n a0 = 1 Prove that for all n ≥ 3, an > 4n 2. Relevant equations 3. The attempt at a solution Since this is induction, I start out with a base case: Base Case (n = 3): a3 = a1 + a2 + 3 = 3 + 8 + 3 = 14 4(n) = 4(3) = 12 14 > 12 Therefore, the base case is true. a3 > 4(3) Inductive Hypothesis: Let some integer m ≥ 3 and assume am > 4m for all integers m. Also assume there is an integer k, where k > 4, so that 3 ≤ m ≤ k Inductive Step: We want to prove that ak+1 > 4(k+1) Starting out with writing the equation of ak+1: ak+1 = afloor( (k+1) / 2 ) + afloor( 2(k+1)/3 ) + k + 1 By the inductive hypothesis: Since floor( (k+1)/2 ) and floor (2(k+1)/3) are both between 3 and k, we can assume that: afloor( (k+1) / 2 ) and afloor( 2(k+1)/3 ) are both greater than 4 floor( (k + 1)/2) and 4floor( 2(k+1)/3) respectively. Therefore, all I have to do is prove that: 4 floor ( (k + 1)/2) + 4 floor ( 2(k+1)/3) + k + 1 > 4(k+1) However, I am stuck on this step, I know I just have to simplify this equation, but up until now, I've been proving these types of equations without the floor in it, so I can just simplify with algebra and prove the inequality. Could anyone clarify how I would continue this solution? Thank you in advance.