The problem D in the latest Open Cup involves function f(n) which is defined as the minimum sum of sequence b₁,...,b_k such that any sequence a₁,...,a_l 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.