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Part of 1996 Czech And Slovak Olympiad IIIA

Problems(1)

G(n) = n-G(G(n)), G(0) = 0 , G(k) \ge G(k -1) , G(k -1) = G(k) = G(k +1)

Source: Czech And Slovak Mathematical Olympiad, Round III, Category A 1996 p1

2/20/2020
A sequence (Gn)n=0∞(G_n)_{n=0}^{\infty} satisfies G(0)=0G(0) = 0 and G(n)=n−G(G(n−1))G(n) = n-G(G(n-1)) for each n∈Nn \in N. Show that (a) G(k)≥G(k−1)G(k) \ge G(k -1) for every k∈Nk \in N; (b) there is no integer kk for which G(k−1)=G(k)=G(k+1)G(k -1) = G(k) = G(k +1).
Sequencealgebrainequalities