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)

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