This is clear from Lemma 1:
F(n,m) = F(n-1,m) + F(n-1,m-1) + F(n-1,m-2) + ..... + 1
Put m = m-1 and you get:
F(n,m-1) = F(n-1,m-1) + F(n-1,m-2) + ..... + 1
But this is the same as the series for F(n,m) with the first term missing. Hence
F(n,m) = F(n-1,m) + F(n,m-1)
Back to NumberOfCoefficients.