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From Yifan Xie, Jul 09 2024: (Start)

Theorem 1: a(n+1) - a(n) = 0 or 1.

Proof: We use induction. For n = 1 and 2, the formula holds. Suppose that it holds for all n < k (k >= 3). Then a(k+1) - a(k) = (a(a(k-1))-a(a(k-2))) + (a(k+1 - a(k-1)) - a(k - a(k-2))). We know that a(k-1) - a(k-2) = 0 or 1. If a(k-1) = a(k-2), Then a(a(k-1))-a(a(k-2)), (k+1 - a(k-1)) - (k - a(k-2)) = 1, thus a(k+1 - a(k-1)) - a(k - a(k-2)) = 0 or 1 (note that k - a(k-2) < k), hence a(k+1) - a(k) = 0 or 1. Similarly, if a(k-1) = a(k-2) + 1, a(k+1) - a(k) = 0 or 1. Therefore the formula holds for all n's.

(End)

Theorem 2: If a(n) = a(n+1), a(n-1) = a(n-2) + 1.

Proof: Since 0 = a(n+1) - a(n) = (a(a(n-1))-a(a(n-2))) + (a(n+1 - a(n-1)) - a(n - a(n-2))) and a(a(n-1)) - a(a(n-2)) >= 0, a(n+1 - a(n-1)) <= a(n - a(n-2)). Therefore, n+1 - a(n-1) <= n - a(n-2), hence a(n-1) - a(n-2) >= 1. Applying theorem 1, we conclude that a(n-1) = a(n-2) + 1.

Theorem 3: There cannot be four consecutive equal terms.

Proof: Applying theorem 2 we conclude that a(n) = a(n+1) and a(n+2) = a(n+3) is a contradiction.

From Yifan Xie, Jul 09 2024: (Start)

Theorem 1: a(n+1) - a(n) = 0 or 1.

Proof: We use induction. For n = 1 and 2, the formula holds. Suppose that it holds for all n < k (k >= 3). Then a(k+1) - a(k) = (a(a(k-1))-a(a(k-2))) + (a(k+1 - a(k-1)) - a(k - a(k-2))). We know that a(k-1) - a(k-2) = 0 or 1. If a(k-1) = a(k-2), Then a(a(k-1))-a(a(k-2)), (k+1 - a(k-1)) - (k - a(k-2)) = 1, thus a(k+1 - a(k-1)) - a(k - a(k-2)) = 0 or 1 (note that k - a(k-2) < k), hence a(k+1) - a(k) = 0 or 1. Similarly, if a(k-1) = a(k-2) + 1, a(k+1) - a(k) = 0 or 1. Therefore the formula holds for all n's.

Theorem 2: If a(n) = a(n+1), a(n-1) = a(n-2) + 1.

Proof: Since 0 = a(n+1) - a(n) = (a(a(n-1))-a(a(n-2))) + (a(n+1 - a(n-1)) - a(n - a(n-2))) and a(a(n-1)) - a(a(n-2)) >= 0, a(n+1 - a(n-1)) <= a(n - a(n-2)). Therefore, n+1 - a(n-1) <= n - a(n-2), hence a(n-1) - a(n-2) >= 1. Applying theorem 1, we conclude that a(n-1) = a(n-2) + 1.

Theorem 3: There cannot be four consecutive equal terms.

Proof: Applying theorem 2 we conclude that a(n) = a(n+1) and a(n+2) = a(n+3) is a contradiction.

(End)

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A005229:= proc(n) option remember;

A005229 := proc(n) option remember; if n<=2 then 1 else A005229(A005229(n-2)) +A005229(n-A005229(n-2)); fi; end;

fi; end;

seq(A005229(n), n=1..70)

(PARI) a(n)=an[ n ]; an=vector(100, n, 1); for(n=3, 100, an[ n ]=a(a(n-2))+a(n-a(n-2)))

(Sage)

@CachedFunction

def a(n): # A005229

if (n<3): return 1

else: return a(a(n-2)) + a(n-a(n-2))

[a(n) for n in (1..100)] # G. C. Greubel, Mar 27 2022

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