“…For 3 < k ≤ ⌊n/2⌋ − 1 we have that k ≤ ⌊ℓm 1 /2⌋ + ⌊ℓm 2 /2⌋, so there are values r, s ≥ 2, r ≤ ⌊ℓm 1 /2⌋ and s ≤ ⌊ℓm 2 /2⌋, such that k = r + s. Finally, when k = ⌊n/2⌋, by the parity conditions we have that at least one of ℓm 1 , ℓm 2 is even, so we can choose (r, s) = (ℓm 1 /2, ⌊ℓm 2 /2⌋) or (⌊ℓm 1 /2⌋, ℓm 2 /2). {(5, 6), (5,10), (5,14), (6,6), (6,7), (6,9), (6,11), (6,13), (7, 10)}. Now, case ℓ = 2 is straightforward, since p 2i (2, m) = p 2i+1 (2, m) for all i < n/4.…”