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a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.
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%I #14 Jun 28 2019 08:31:31

%S 2,3,5,7,10,7,13,15,22,23,12,6,15,18,13,25,41,37,10,22,17,40,47,40,81,

%T 38,22,53,85,134,51,29,156,215,23,47,46,35,69,98,144,81,108,116,102,

%U 37,47,37,72,75,85,104,217,111,10,15,37,60,40,147,197,51,110

%N a(n) = a(a(n-1) mod n) + a(a(n-2) mod n) with a(0)=2 and a(1)=3.

%C a(0) and a(1) are chosen to be the smallest starting numbers greater than 1 that are believed to result in a sequence that doesn't cycle.

%C Empirical observation of the first 10^8 terms suggests that the sequence doesn't enter a cycle.

%C Conjectures: (i) This sequence doesn't enter a cycle. (ii) There is an integer greater than 1 that can never appear in this sequence.

%e a(2) = a(a(2-1) mod 2) + a(a(2-2) mod 2) = a(a(1) mod 2) + a(a(0) mod 2) = a(3 mod 2) + a(2 mod 2) = a(1) + a(0) = 3 + 2 = 5.

%o (Python)

%o a = [2, 3]

%o for n in range(2, 10**4 + 3):

%o a.append(a[(a[n - 1] % n)] + a[(a[n - 2] % n)])

%o print((n - 2), ",", a[n - 2], sep="")

%Y Cf. A005185, A046698, A003160.

%Y Cf. A000027 (if a(0)=1 and a(1)=2).

%K nonn

%O 0,1

%A _Arran Ireland_, Jun 26 2019