“…This also proves that a j+1,j+1 = a j−1,j−1 = −ζ, if and only if a j,j = 2ζ, for 1 ≤ j ≤ m. On the other hand, if m is odd, it follows from ( 7) that, a j,j = −ζ for all j, which implies ζ = 0. Thus, let us assume that m = 2ℓ, with ℓ ∈ N. From (7), we know that (m − 2j)a j,j = −(m − 2j)ζ, for all j. Whence, a j,j = −ζ for all j = ℓ, and thus…”