A372196 - OEIS

%I #57 Jul 25 2024 14:16:38

%S 1,1,2,1,3,2,2,3,2,4,1,4,2,5,3,3,4,3,5,2,6,2,7,2,8,3,6,2,9,2,10,4,4,5,

%T 3,7,2,11,1,5,4,6,3,8,2,12,4,7,3,9,2,13,2,14,5,5,6,4,8,3,10,2,15,3,11,

%U 2,16,3,12,2,17,4,9,3,13,2,18,3,14,2,19,5,7,4,10,3,15,2,20,2,21,6,5,8,4,11,3,16,2,22

%N a(1) = 1. For n > 1, let m be the number of times that the value a(n-1) has appeared (including its appearance at a(n-1)); if m > 1 then a(n) = m, otherwise a(n) = a(a(n-1)).

%C The sequence refers to itself. If k = a(n-1) is a novel term, then a(n) equals the term whose index is k; otherwise a(n) is the number of times k has appeared in a(1)..a(n-1).

%C All positive integers appear in the sequence.

%C For any n, there exist numbers m > n such that a(m) = a(n), and there exist numbers m > n such that a(m) > a(n).

%C The graph of the largest number a(n) in the interval n = 1..j grows approximately as j^0.66.

%H Thomas Scheuerle, <a href="/A372196/a372196.png">Scatter plot of a(1 to 2000)</a>. Yellow: a(n) = a(a(n-1)), blue: a(n) indicates how often the value of a(n-1) appeared previously.

%e a(1) = 1 by definition.

%e For n = 2, a(1) = 1 has appeared only 1 time, so a(2) = a(a(1)) = a(1) = 1.

%e For n = 3, a(2) = 1 has appeared 2 times, so a(3) = 2.

%e For n = 4, a(3) = 2 has appeared only 1 time, so a(4) = a(a(3)) = a(2) = 1.

%e For n = 5, a(4) = 1 has appeared 3 times, so a(5) = 3.

%e For n = 6, a(5) = 3 has appeared only 1 time, so a(6) = a(a(5)) = a(3) = 2.

%e In the table below, m is the number of times that the value a(n-1) has appeared among the terms a(1)..a(n-1).

%e .

%e    n  a(n-1)  m  a(n)  comments

%e   --  ------  -  ----  -----------------------------------

%e    1     -    -     1  (by definition)

%e    2     1    1     1  m = 1, so a(2) = a(a(1)) = a(1) = 1

%e    3     1    2     2  m > 1, so a(3) = m

%e    4     2    1     1  m = 1, so a(4) = a(a(3)) = a(2) = 1

%e    5     1    3     3  m > 1, so a(5) = m

%e    6     3    1     2  m = 1, so a(6) = a(a(5)) = a(3) = 2

%e    7     2    2     2  m > 1, so a(7) = m

%e    8     2    3     3  m > 1, so a(8) = m

%e    9     3    2     2  m > 1, so a(9) = m

%e   10     2    4     4  m > 1, so a(10) = m

%o (Python)

%o def generate_sequence(n):

%o     sequence = [1]

%o     for i in range(1, n):

%o         last_number = sequence[-1]

%o         if sequence.count(last_number) == 1:

%o             next_number = sequence[last_number - 1]

%o         else:

%o             next_number = sequence.count(last_number)

%o         sequence.append(next_number)

%o     return sequence

%o n = 30

%o print(generate_sequence(n))

%o (PARI) lista(nn) = my(va= vector(nn)); va[1] = 1; for (n=2, nn, my(nb = #select(x->(x==va[n-1]), va)); if (nb<=1, va[n] = va[va[n-1]], va[n] = nb);); va; \\ _Michel Marcus_, Jul 04 2024

%Y Inspired by A181391.

%K nonn

%O 1,3

%A _Mario Cortés_, Jul 03 2024