The problem D in the latest Open Cup involves function f(n) which is defined as the minimum sum of sequence b1,...,bk such that any sequence a1,...,al with sum less than or equal to n can be dominated by some subsequence of b. It turns out that f(n) = n + f(k) + f(n − 1 − k) where k = [(n − 1) / 2]. If this equation still keeps you up at night, you can finally sleep well now. I have found a wonderful proof of this statement which fits the bounds of this site.
Two sides of the coin.
Theorem 1. Let B1 be a sequence that covers all sequences with sum less than or equal to k. Let B2 be a sequence that covers all sequences with sum less than or equal to l. Let n = k + l + 1. Then sequence B1, n, B2 covers all sequences with sum less than of equal to n.
Proof.