A372196 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)).

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, 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, 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

Offset

1,3

Comments

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).

All positive integers appear in the sequence.

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).

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

Links

Table of n, a(n) for n=1..100.

Thomas Scheuerle, Scatter plot of a(1 to 2000). Yellow: a(n) = a(a(n-1)), blue: a(n) indicates how often the value of a(n-1) appeared previously.

Example

a(1) = 1 by definition.
For n = 2, a(1) = 1 has appeared only 1 time, so a(2) = a(a(1)) = a(1) = 1.
For n = 3, a(2) = 1 has appeared 2 times, so a(3) = 2.
For n = 4, a(3) = 2 has appeared only 1 time, so a(4) = a(a(3)) = a(2) = 1.
For n = 5, a(4) = 1 has appeared 3 times, so a(5) = 3.
For n = 6, a(5) = 3 has appeared only 1 time, so a(6) = a(a(5)) = a(3) = 2.
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).
.
   n  a(n-1)  m  a(n)  comments
  --  ------  -  ----  -----------------------------------
   1     -    -     1  (by definition)
   2     1    1     1  m = 1, so a(2) = a(a(1)) = a(1) = 1
   3     1    2     2  m > 1, so a(3) = m
   4     2    1     1  m = 1, so a(4) = a(a(3)) = a(2) = 1
   5     1    3     3  m > 1, so a(5) = m
   6     3    1     2  m = 1, so a(6) = a(a(5)) = a(3) = 2
   7     2    2     2  m > 1, so a(7) = m
   8     2    3     3  m > 1, so a(8) = m
   9     3    2     2  m > 1, so a(9) = m
  10     2    4     4  m > 1, so a(10) = m

Prog

(Python)
def generate_sequence(n):
    sequence = [1]
    for i in range(1, n):
        last_number = sequence[-1]
        if sequence.count(last_number) == 1:
            next_number = sequence[last_number - 1]
        else:
            next_number = sequence.count(last_number)
        sequence.append(next_number)
    return sequence
n = 30
print(generate_sequence(n))
(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

Crossrefs

Inspired by A181391.

Sequence in context: A345058 A191654 A327983 * A205784 A066272 A237130

Adjacent sequences: A372193 A372194 A372195 * A372197 A372198 A372199

Keyword

nonn

Author

Mario Cortés, Jul 03 2024

Status

approved