A330004 -id:A330004 - cp's OEIS frontend

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.

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)

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

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI