“…we have w(5) = c − k + 5 and (13)w(1) = c + 1 if k = 0, c − k + 7 if k = 0.Also, w(4) ≤ w(2) + 2 and w(2) = w(6 − 4) ≤ w(6 − 2) + (6 − 2) = w(4) + 4 by Lemma 15. Combining these two inequalities we get(14) w(4) − 2 ≤ w(2) ≤ w(4) + 4.Let us also note that w(i) ≤ c + 5 for all i ∈ {1, 2, 3, 4, 5}. (i) If c ≡ 0 (mod 6), then w(1) = c + 1 and w(5) = c + 5 by(13).…”