A239281 a(1) = 0; for n>1, a(n) is one more than the value of the sequence at index the number of times a(n-1) has previously appeared in the sequence.

0, 1, 1, 2, 1, 2, 2, 2, 3, 1, 3, 2, 2, 3, 2, 3, 3, 2, 3, 3, 3, 3, 4, 1, 2, 4, 2, 2, 4, 2, 3, 2, 3, 4, 3, 3, 3, 4, 2, 4, 3, 3, 4, 3, 4, 3, 3, 4, 4, 2, 3, 4, 4, 3, 4, 3, 4, 4, 3, 5, 1, 3, 2, 4, 4, 4, 3, 3, 5, 2, 4, 4, 4, 4, 4, 5, 2, 3, 3, 3, 5, 3, 3, 4, 2, 4, 3, 3, 4, 5, 2, 4, 3, 5, 3, 4, 3, 4, 5, 3

Offset

1,4

Comments

The same sequence is obtained by looking at the values following each occurrence of each positive integer; and that sequence is this sequence plus one.

Every positive integer occurs infinitely many times. - Ivan Neretin, May 02 2016

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Maple

a:= proc() local a, c, t;
      c:= proc() 0 end; c(0):=1;
      a:= proc(n) option remember;
            if n=1 then 0
          else t:= 1+a(c(a(n-1)));
               c(t):= c(t)+1; t
            fi
          end
    end():
seq(a(n), n=1..120);  # Alois P. Heinz, Mar 15 2014

Mathematica

a[_] = 0; nmax = 120;
Do[a[i+1] = 1 + a[Sum[Boole[a[j] == a[i]], {j, 1, i}]], {i, 1, nmax-1}];
Table[a[n], {n, 1, nmax}] (* Jean-François Alcover, Apr 28 2022 *)

Prog

(PARI) al(n)={local(r=vector(n), i, j);
for(i=1, n-1, r[i+1]=1+r[sum(j=1, i, if(r[j]==r[i], 1, 0))]);
r}
(Python)
from collections import Counter
from itertools import count, islice
def agen(): # generator of terms
    an, a, c = 0, dict(), Counter()
    for n in count(1):
        yield an
        a[n] = an
        c[an] += 1
        an = a[c[an]] + 1
print(list(islice(agen(), 100))) # Michael S. Branicky, Jul 12 2024

Crossrefs

Sequence in context: A065531 A366574 A256558 * A024936 A144590 A057368

Adjacent sequences: A239278 A239279 A239280 * A239282 A239283 A239284

Keyword

nonn

Author

Franklin T. Adams-Watters, Mar 14 2014

Status

approved