A329934 -id:A329934 - cp's OEIS frontend

A316774 a(n) = n for n < 2, a(n) = freq(a(n-1),n) + freq(a(n-2),n) for n >= 2, where freq(i,j) is the number of times i appears in [a(0),a(1),...,a(j-1)].

0, 1, 2, 2, 4, 3, 2, 4, 5, 3, 3, 6, 4, 4, 8, 5, 3, 6, 6, 6, 8, 6, 7, 6, 7, 8, 5, 6, 10, 8, 5, 8, 9, 6, 9, 10, 4, 7, 8, 9, 9, 8, 11, 8, 9, 13, 6, 10, 12, 4, 7, 10, 8, 13, 11, 4, 9, 13, 9, 10, 12, 7, 7, 12, 9, 11, 11, 8, 14, 11, 6, 15, 11, 7, 13, 11, 11, 16, 9, 10

Offset: 0

Peter Illig, Jul 12 2018

In other words, a(n) = (number of times a(n-1) has appeared) plus (number of times a(n-2) has appeared). - N. J. A. Sloane, Dec 13 2019
What is the asymptotic behavior of this sequence?
Does it contain every positive integer at least once?
Does it contain every positive integer at most finitely many times?
Additional comments from Peter Illig's "Puzzles" link below (Start):
Sometimes referred to as "The Devil's Sequence" (by me), due to the early presence of three consecutive 6's (and my inability to understand it). The next time a number occurs three times in a row isn't until a(355677).
If each n does appear only finitely many times, approximately how many times does it appear? (It seems to be close to 2n.)
What are the best possible upper/lower bounds on a(n)?
Let r(k) be the smallest n such that {0,1,2,...,k} is contained in {a(0),...,a(n)}. What is the asymptotic behavior of r(k)? (It seems to be close to k^2/2.)
(End)

			For n=4, a(n-1) = a(n-2) = 2, and 2 appears twice in the first 4 terms. So a(4) = 2 + 2 = 4.

Alois P. Heinz, Table of n, a(n) for n = 0..65536
"Horseshoe_Crab" Reddit User, Properties of a Strange, Rather Meta Sequence. [In case this link breaks, the main point of the discussion is to propose the sequence and suggest other initial values. - _N. J. A. Sloane_, Dec 13 2019]
Peter Illig, Problems. [No date, probably 2018]
Samuel B. Reid, Density plot of one billion terms
Rémy Sigrist, Density plot of the first 10000000 terms

Cf. A001462, A316973 (freq(n)), A316905 (when n appears), A316984 (when n last appears), A330439 (total number of times a(n) has appeared so far).
For records see A330330, A330331.
See A306246 and A329934 for similar sequences with different initial conditions.
A330332 considers the frequencies of the three previous terms.

b:= proc() 0 end:
a:= proc(n) option remember; local t;
      t:= `if`(n<2, n, b(a(n-1))+b(a(n-2)));
      b(t):= b(t)+1; t
    end:
seq(a(n), n=0..200);  # Alois P. Heinz, Jul 12 2018

a = prev = {0, 1};
Do[
AppendTo[prev, Count[a, prev[[1]]] + Count[a, prev[[2]]]];
AppendTo[a, prev[[3]]];
prev = prev[[2 ;;]] , {78}]
a (* Peter Illig, Jul 12 2018 *)

from itertools import islice
from collections import Counter
def agen():
    a = [0, 1]; c = Counter(a); yield from a
    while True:
        a = [a[-1], c[a[-1]] + c[a[-2]]]; c[a[-1]] += 1; yield a[-1]
print(list(islice(agen(), 80))) # Michael S. Branicky, Oct 13 2022

Definition clarified by N. J. A. Sloane, Dec 13 2019

A364835 a(1) = 1; a(n+1) = (number of times a(n)+1 has appeared) - (number of times a(n) has appeared).

Original entry on oeis.org

1, -1, -1, -2, 1, -2, 0, 1, -3, 1, -4, 0, 2, -1, -1, -2, 1, -4, -1, -3, 1, -5, 1, -6, 0, 4, -1, -3, 0, 3, 0, 2, -1, -2, 3, -1, -3, 0, 1, -6, -1, -3, -1, -4, 2, -1, -5, 1, -6, -1, -6, -2, 7, -1, -7, 3, -2, 7, -2, 6, 1, -7, 2, -1, -8, 1, -7, 1, -8, 1, -9, 1, -10, 0, 7, -3, 1, -11, 0

Offset: 1

Rok Cestnik, Aug 28 2023

Irregular until n=149695 at which point it follows a simple pattern (see formula).

			a(5) = 2 - 1 = 1 because a(4) + 1 = -1 has appeared twice before and a(4) = -2 has appeared once.
a(7) = 2 - 2 = 0 because a(6) + 1 = -1 and a(6) = -2 have both appeared twice before.

Rok Cestnik, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A330004, A329934.

nmax=78; a={1}; For[n=1, n<=nmax, n++, AppendTo[a,Count[a,Part[a,n]+1]-Count[a,Part[a,n]]]]; a (* Stefano Spezia, Aug 29 2023 *)

a=[1]
for n in range(1000):
    a.append(a.count(a[n]+1)-a.count(a[n]))

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an = 1; c = Counter([1])
    while True: yield an; an = c[an+1] - c[an]; c[an] += 1
print(list(islice(agen(),1001))) # Michael S. Branicky, Aug 29 2023

For n >= 149695: a(n) = 49456 - n/3 if (n mod 3) = 0, otherwise a(n) = (n mod 3) - 1.

cp's OEIS Frontend

A316774 a(n) = n for n < 2, a(n) = freq(a(n-1),n) + freq(a(n-2),n) for n >= 2, where freq(i,j) is the number of times i appears in [a(0),a(1),...,a(j-1)].

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A364835 a(1) = 1; a(n+1) = (number of times a(n)+1 has appeared) - (number of times a(n) has appeared).

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Mathematica

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Python

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