A329985 -id:A329985 - cp's OEIS frontend

A330334 Number of times A329985(n) has appeared in A329985 (including the present occurrence).

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

Offset: 1

N. J. A. Sloane, Dec 16 2019

nonn

Rémy Sigrist, Table of n, a(n) for n = 1..25000

Cf. A329985.

A358921 a(1) = 1; a(n) is the smallest positive number not among the terms a(n-c .. n-1) where c = the number of times a(n-1) has occurred.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 3, 1, 4, 1, 5, 1, 2, 3, 4, 1, 6, 1, 7, 1, 5, 2, 3, 4, 1, 8, 1, 9, 1, 6, 2, 3, 4, 1, 5, 2, 6, 1, 7, 2, 3, 4, 5, 1, 8, 2, 6, 3, 7, 1, 9, 2, 4, 5, 3, 6, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 8, 2, 3, 4, 5, 6, 1, 7, 2, 9, 1, 15, 1, 16, 1, 17, 1

Offset: 1

Samuel Harkness, Dec 06 2022

A new number other than 1 is always followed by a 1, so a(n) < n/2 for n > 4.

			For a(6), a(5) = 1 has occurred 3 times, so the smallest positive integer not in {a(5), a(4), a(3)} = {1, 3, 1} is 2, thus a(6) = 2.
Next, for a(7), a(6) = 2 has occurred 2 times, so the smallest positive integer not in {a(6), a(5)} = {2, 1} is 3, thus a(7) = 3.
Then, for a(8), a(7) = 3 has occurred 2 times, so the smallest positive integer not in {a(7), a(6)} = {3, 2} is 1, thus a(8) = 1.
Now, for a(9), a(8) = 1 has occurred 4 times, so the smallest positive integer not in {a(8), a(7), a(6), a(5)} = {1, 3, 2, 1} is 4, thus a(9) = 4.
The first terms, alongside the number of times they have occurred o(n), are:
  n  a(n)  o(n)
  -  ----  ----
  1     1     1
  2     2     1
  3     1     2
  4     3     1
  5     1     3
  6     2     2
  7     3     2
  8     1     4
  9     4     1
  10    1     5

Samuel Harkness, Table of n, a(n) for n = 1..10000
Samuel Harkness, Scatterplot of the first 2000000 terms

Cf. A133622, A268696, A329985.

V = {1} While[Length[V] < 84, b = 1; While[MemberQ[Take[V, -Count[V, Last[V]]], b], b++ ]; AppendTo[V, b]]; Print[V]

{ a = o = vector(84); v = 1; for (n=1, #a, print1 (a[n]=v", "); v=setminus([1..n+1], Set(a[n-o[a[n]]+++1..n]))[1]) } \\ Rémy Sigrist, Jan 09 2023

A329981 a(1) = 0, and for n > 0, a(n+1) = floor(k / 3) where k is the number of terms equal to a(n) among the first n terms.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1, 2, 0, 2, 0, 2, 1, 2, 1, 2, 1, 3, 0, 2, 2, 2, 2, 3, 0, 3, 1, 3, 1, 3, 1, 4, 0, 3, 2, 3, 2, 3, 2, 4, 0, 3, 3, 3, 3, 4, 1, 4, 1, 4, 1, 5, 0, 4, 2, 4, 2, 4, 2, 5, 0, 4, 3, 4, 3, 4, 3, 5, 1, 5, 1, 5, 1, 6, 0, 4, 4, 4, 4, 5, 2, 5, 2

Offset: 1

Rémy Sigrist, Nov 26 2019

nonn

In other words, for n > 0, a(n+1) = floor(o(n) / 3) where o is the ordinal transform of the sequence.
Every nonnegative number appears infinitely many times in the sequence.
The second difference of the positions of the zeros in the sequence appears to be eventually 6-periodic.

			The first terms, alongside their ordinal transform, are:
  a   a(n)  o(n)
  --  ----  ----
   1     0     1
   2     0     2
   3     0     3
   4     1     1
   5     0     4
   6     1     2
   7     0     5
   8     1     3
   9     1     4
  10     1     5
  11     1     6
  12     2     1

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Colored scatterplot of the first 10000 terms (where the color denotes the parity of n)

See A329982, A329984, A329985 and A330004 for similar sequences.

o = vector(7); v=0; for (n=1, 87, print1 (v", "); v=o[1+v]++\3)

A354971 a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence.

Original entry on oeis.org

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

Offset: 1

Neal Gersh Tolunsky, Jun 14 2022

Conjecture: every positive integer eventually appears in the sequence.
Conjecture: The longest run of same, consecutive a(n) is 4 terms.
A003056 can be generated using the same rules but starting with 0.

			For n=9, a(9) is the number of times a(9 - 1 - a(9-1)) = a(8 - a(8)) = a(3) = 1 has appeared in the sequence, so a(9)=5.

Michael De Vlieger, Table of n, a(n) for n = 1..10000
Michael De Vlieger, Log-log scatterplot of a(n), n = 1..2^16, showing records in red, first entries of other m in green, and the latest entries of m in blue.
Neal Gersh Tolunsky, Illustration of first 100 terms

Cf. A003056, A329985

function a = A354971( max_n )
    a = [1 0]; c = zeros(1,max_n); c(1:2) = [1 1];
    for n = 3:max_n
        m = a(n-1-a(n-1))+1;
        a = [a c(m)];
        c(c(m)+1) = c(c(m)+1)+1;
    end
end % Thomas Scheuerle, Jun 15 2022

nn = 2^20; c[] = 0; a[1] = c[0] = c[1] = 1; a[2] = j = 0; Do[k = c[a[n - j - 1]]; a[n] = j = k; c[k]++, {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger, Jun 25 2022 *)

{ nb = [0]; for (n=1, #a=vector(81), print1 (a[n] = if (n==1, 1, n==2, 0, nb[1+a[n-1-a[n-1]]])", "); if (#nb < 1+a[n], nb = concat(nb, vector(#nb))); nb[1+a[n]]++) } \\ Rémy Sigrist, Jun 18 2022

from collections import Counter
from itertools import count, islice
def agen():
    a = [None, 1, 0]; inventory = Counter(a); yield from a[1:]
    for n in count(3):
        c = inventory[a[n-1-a[n-1]]]
        a.append(c); inventory.update([c]); yield c
print(list(islice(agen(), 81))) # Michael S. Branicky, Jun 18 2022

A337835 a(1) = 0 and for n > 1, a(n+1) = k - a(n) where k is the number of terms equal to a(n) among the first n terms.

Original entry on oeis.org

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

Offset: 1

Bence Bernáth, Nov 17 2020

Empirical observation: -A000194(n) <= a(n) <= A000194(n).

			For n = 4, a(4) = 2, which appeared only once in the sequence so a(5)=1-2=-1.

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

Cf. A000194, A329985.

length=1000;
sequence(1)=0;
for n=2:1:length
        sequence(n)=nnz(sequence==sequence(end))-sequence(n-1);
end

a = {0}; Do[AppendTo[a, Count[a, a[[-1]]] - a[[-1]]], {100}]; a (* Amiram Eldar, Nov 18 2020 *)
a[1] = 0; a[n_Integer?Positive] := a[n] = Count[Array[a, n - 1], a[n - 1]] - a[n - 1]; Array[a, 101] (* Jan Mangaldan, Nov 27 2020 *)

lista(nn) = {my(va = vector(nn)); for (n=2, nn, va[n] = #select(x->(x==va[n-1]), vector(n-1, k, va[k])) - va[n-1];); va;} \\ Michel Marcus, Nov 18 2020

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    an, k = 0, Counter()
    while True:
        yield an
        k[an] += 1
        an = k[an] - an
print(list(islice(agen(), 86))) # Michael S. Branicky, Nov 12 2022

A363083 a(0)=a(1)=1. For n>1, if the number of occurrences of a(n-1) is less than abs(a(n-1)), then a(n)=a(n-1)-a(n-2). Otherwise, a(n)=a(n-1)+a(n-2).

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 3, -2, -5, -3, 2, -1, 1, 0, 1, 1, 2, 3, 5, 2, 7, 5, -2, 3, 1, 4, 3, 7, 4, -3, -7, -4, 3, -1, 2, 1, 3, 4, 1, 5, 4, 9, 5, 14, 9, -5, -14, -9, 5, -4, -9, -5, 4, -1, 3, 2, 5, 7, 2, 9, 7, -2, 5, 3, 8, 5, 13, 8, -5, -13, -8, 5, -3, 2, -1, 1, 0

Offset: 0

Gavin Lupo, May 18 2023

			a(0) = 1
a(1) = 1
a(2) = 2. Two 1's in the list so far.    2 > abs(1).   1 + 1 = 2.
a(3) = 1. One 2 in the list so far.      1 < abs(2).   2 - 1 = 1.
a(4) = 3. Three 1's in the list so far.  3 > abs(1).   1 + 2 = 3.

Gavin Lupo, Table of n, a(n) for n = 0..50000
Gavin Lupo, Image of the first 100000 terms.
Gavin Lupo, Image of the first 1000000 terms.
Gavin Lupo, Image of the first 10000000 terms.
Gavin Lupo, Image of the first 25000000 terms.

Cf. A329985, A362746, A362890, A363086.

from itertools import islice
from collections import Counter
def agen(): # generator of terms
    anprev, an, c = 1, 1, Counter([1])
    yield 1
    while True:
        yield an
        c[an] += 1
        anprev, an = an, an-anprev if c[an] < abs(an) else an+anprev
print(list(islice(agen(), 80))) # Michael S. Branicky, May 18 2023

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.

Original entry on oeis.org

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

cp's OEIS Frontend

A330334 Number of times A329985(n) has appeared in A329985 (including the present occurrence).

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A358921 a(1) = 1; a(n) is the smallest positive number not among the terms a(n-c .. n-1) where c = the number of times a(n-1) has occurred.

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A329981 a(1) = 0, and for n > 0, a(n+1) = floor(k / 3) where k is the number of terms equal to a(n) among the first n terms.

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A354971 a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence.

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A337835 a(1) = 0 and for n > 1, a(n+1) = k - a(n) where k is the number of terms equal to a(n) among the first n terms.

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A363083 a(0)=a(1)=1. For n>1, if the number of occurrences of a(n-1) is less than abs(a(n-1)), then a(n)=a(n-1)-a(n-2). Otherwise, a(n)=a(n-1)+a(n-2).

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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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