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For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = (n-m)^a(n); otherwise a(n+1) = 0. Start with a(1)=1, a(2)=2.
0

%I #22 Aug 16 2019 00:51:29

%S 1,2,0,0,1,4,0,1,3,0,1,3,27,0,1,4,10000,0,1,4,256,0,1,4,256

%N For n >= 1, if there exists an m < n such that a(m) = a(n), take the largest such m and set a(n+1) = (n-m)^a(n); otherwise a(n+1) = 0. Start with a(1)=1, a(2)=2.

%C This sequence quickly generates terms of immense size.

%C Changing what a(n+1) is set to when a new term arises changes the sequence nontrivially.

%t Nest[Function[{a, n}, Append[a, If[Length@ # == 0, 0, (n - #[[-1, 1]])^Last@ a] &@ Position[Most@ a, _?(# == Last@ a &)] ]] @@ {#, Length@ #} &, {1, 2}, 24] (* _Michael De Vlieger_, Jul 08 2019 *)

%o (Python)

%o def Prog(length):

%o L = 2

%o seq = [1,2]

%o while L < length:

%o x = len(seq)-1

%o while x > 0:

%o if seq[-1] == seq[x-1]:

%o m_minus_n = len(seq)-x

%o a_n = seq[L-1]

%o seq.append((m_minus_n)**a_n)

%o x = -1

%o else:

%o x -= 1

%o if x == 0:

%o seq.append(0)

%o L += 1

%o return seq

%Y Cf. A181391.

%K nonn

%O 1,2

%A _Philip Kalisman_, Jul 07 2019