A309023 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.

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

Offset

1,2

Comments

This sequence quickly generates terms of immense size.

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

Links

Table of n, a(n) for n=1..25.

Mathematica

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 *)

Prog

(Python)
def Prog(length):
    L = 2
    seq = [1, 2]
    while L < length:
        x = len(seq)-1
        while x > 0:
            if seq[-1] == seq[x-1]:
                m_minus_n = len(seq)-x
                a_n = seq[L-1]
                seq.append((m_minus_n)**a_n)
                x = -1
            else:
                x -= 1
            if x == 0:
                seq.append(0)
        L += 1
    return seq

Crossrefs

Cf. A181391.

Sequence in context: A109077 A378178 A355916 * A353429 A137585 A301344

Adjacent sequences: A309020 A309021 A309022 * A309024 A309025 A309026

Keyword

nonn

Author

Philip Kalisman, Jul 07 2019

Status

approved