A337835 -id:A337835 - cp's OEIS frontend

A337014 a(1) = 0 and for n > 1, a(n+1) = (k(n) - a(n))*(-1)^(n+1) where k(n) is the number of terms equal to a(n) among the first n terms.

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

Offset: 1

Bence Bernáth, Nov 21 2020

The graph of the sequence shows interesting, "Christmas tree"-like shapes.

			For n = 4, a(4) = 2, which appeared only once before, so a(5)=(1-2)*(-1)^5 = 1.

Bence Bernáth, Table of n, a(n) for n = 1..10000

Cf. A337835, A329985.

length=10000;
sequence(1)=0;
for n=2:1:length
     sequence(n)=((nnz(sequence==sequence(end)))-(sequence(n-1)))*(-1)^n;
end

a = {0}; Do[AppendTo[a, (-1)^k*(Count[a, a[[-1]]] - a[[-1]])], {k, 0, 81}]; a (* Amiram Eldar, Nov 21 2020 *)

{ for (n=1, #a=vector(83), print1(a[n]=if (n==1, 0, k=sum(k=1, n-1, a[k]==a[n-1]); (k-a[n-1])*(-1)^n) ", ")) } \\ Rémy Sigrist, Nov 22 2020

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an, k, sign = 0, Counter(), -1
    while True:
        yield an
        k[an] += 1
        sign *= -1
        an = (k[an] - an)*sign
print(list(islice(agen(), 83))) # Michael S. Branicky, Nov 12 2022

A358338 a(n) = abs(a(n-1) - count(a(n-1))) where count(a(n-1)) is the number of times a(n-1) has appeared so far in the sequence, a(1)=0.

Original entry on oeis.org

0, 1, 0, 2, 1, 1, 2, 0, 3, 2, 1, 3, 1, 4, 3, 0, 4, 2, 2, 3, 1, 5, 4, 1, 6, 5, 3, 2, 4, 0, 5, 2, 5, 1, 7, 6, 4, 1, 8, 7, 5, 0, 6, 3, 3, 4, 2, 6, 2, 7, 4, 3, 5, 1, 9, 8, 6, 1, 10, 9, 7, 3, 6, 0, 7, 2, 8, 5, 2, 9, 6, 1, 11, 10, 8, 4, 4, 5, 3, 7, 1, 12, 11, 9, 5, 4

Offset: 1

Clément Vovard, Nov 10 2022

This sequence is related to the inventory sequence (A342585) as it uses the number of times a number has occurred so far in the sequence.
The following comments are only empirical observations:
ceiling(sqrt(2n)) is an excellent envelope of a(n) with no exceptions found in the first 50000 terms.
When x > 3 appears for the first time, it seems to always be preceded by a 1 and followed by x-1. Also, x-1 will already have occurred earlier in the sequence (new highest terms grow by 1).
The number of times x > 0 appears in the first k terms seems to approximately equal sqrt(2k)-x-1. Therefore, 1 appears approximately sqrt(2k) times. The highest term that has appeared in k terms is then approximately sqrt(2k), which also makes sense considering the number of times 1 appears and the fact that a new number is preceded by 1. The only exception is 0, which appears approximately sqrt(2k)/2 times.

			For n=2, a(2-1)=0 and 0 has occurred 1 time so far so a(2)=abs(0-1)=1.
For n=12, a(12-1)=1 and 1 has occurred 4 times so far so a(12)=abs(1-4)=3.

Clément Vovard, Table of n, a(n) for n = 1..10000

Cf. A337835, A340488, A342585.

from collections import Counter
def aupton(terms):
    alst, inventory = [0], Counter([0])
    for n in range(2, terms+1):
        c = abs(alst[-1] - inventory[alst[-1]])
        alst.append(c); inventory[c] += 1
    return alst
print(aupton(85)) # Michael S. Branicky, Nov 10 2022

cp's OEIS Frontend

A337014 a(1) = 0 and for n > 1, a(n+1) = (k(n) - a(n))*(-1)^(n+1) where k(n) is the number of terms equal to a(n) among the first n terms.

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