The problem D in the latest Open Cup involves function *f*(*n*) which is defined as the minimum sum of sequence *b*_{1}, ..., *b*_{k} such that any sequence *a*_{1}, ..., *a*_{l} with sum less than or equal to *n* can be dominated by some subsequence of *b*. One sequence dominates the other, if they have the same length any every element of the first sequence greater than or equal to corresponding element of the second sequence. It turns out that *f*(*n*) = *n* + *f*(*k*) + *f*(*n* − 1 − *k*) where *k* = [(*n* − 1) / 2]. But why? 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.