A354971 -id:A354971 - cp's OEIS frontend

A363193 a(1)=1, and thereafter a(n) = number of occurrences of a(k) among terms a(1..k), where k = n-a(n-1).

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

Offset: 1

Neal Gersh Tolunsky, May 20 2023

			For n=5, a(5) is the number of times a(5 - a(5-1)) = a(5 - a(4)) = a(3) = 2 has occurred in the sequence among a(1..3). It has occurred 1 time up to that point, so a(5)=1.

Neal Gersh Tolunsky, Table of n, a(n) for n = 1..10000
Neal Gersh Tolunsky, Scatterplot of first 30000 terms
Neal Gersh Tolunsky, Scatterplot of first 100000 terms

Cf. A354971.

from bisect import bisect
from itertools import count, islice
def agen(): # generator of terms
    an, a, locs = 1, [None, 1], {1: [1]}
    yield 1
    for n in count(2):
        k = n-an
        an = bisect(locs[a[k]], k) # sum(1 for i in locs[a[k]] if i <= k)
        a.append(an)
        if an not in locs: locs[an] = []
        locs[an].append(n)
        yield an
print(list(islice(agen(), 86))) # Michael S. Branicky, May 23 2023

A366548 a(0) = 0; for n > 0, a(n) is the number of terms prior to the term a(n-1-a(n-1)) that equal a(n-1-a(n-1)).

Original entry on oeis.org

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

Offset: 0

Scott R. Shannon, Oct 13 2023

In the first 10 million terms the value 4 appears the most often, 11838 times, although the count of neighboring values is less than 2% different. It is unknown if this stays the most common term as n increases. In the same range on thirty-eight occasions there are three consecutive equal terms, the first time being a(105) = a(106) = a(107) = 8. It is unknown if four or more consecutive terms eventually appear.

			a(2) = 1 as a(2-1-a(2-1)) = a(1-0) = a(1) = 0, and there is one term prior to a(1) that equals 0, namely a(0).
a(6) = 1 as a(6-1-a(6-1)) = a(5-2) = a(3) = 1, and there is one term prior to a(3) that equals 1, namely a(2).

Scott R. Shannon, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 10 million terms.

Cf. A366549, A363193, A354971, A181391, A342585.

A366549 a(0) = 0; for n > 0, a(n) is the number of terms prior to and including the term a(n-1-a(n-1)) that equal a(n-1-a(n-1)).

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 3, 3, 1, 2, 2, 4, 2, 4, 3, 1, 3, 2, 5, 2, 6, 3, 1, 5, 1, 2, 2, 7, 1, 1, 8, 6, 8, 7, 9, 8, 1, 3, 1, 6, 2, 11, 1, 1, 12, 2, 13, 2, 11, 6, 13, 6, 11, 1, 3, 5, 2, 7, 2, 13, 1, 3, 14, 2, 8, 13, 3, 15, 3, 4, 4, 9, 1, 2, 2, 16, 3, 16, 8, 4, 1, 5, 11, 2, 4, 17, 10, 11, 11, 2, 5, 1, 5

Offset: 0

Scott R. Shannon, Oct 13 2023

In the first 10 million terms the value 1 appears the most often, 13584 times, although the count of neighboring values is less than 2% different. It is unknown if this stays the most common term as n increases. In the same range on fifty occasions there are three consecutive equal terms, the second time, after the three 1's at the start of the sequence, being a(406) = a(407) = a(408) = 11. It is unknown if four or more consecutive terms eventually appear.

			a(2) = 1 as a(2-1-a(2-1)) = a(1-1) = a(0) = 0, and there is one term prior to or equal to a(0) that equals 0, namely a(0).
a(6) = 3 as a(6-1-a(6-1)) = a(5-2) = a(3) = 1, and there are three terms prior to or equal to a(3) that equal 1, namely a(1), a(2) and a(3).

Scott R. Shannon, Table of n, a(n) for n = 0..10000
Scott R. Shannon, Image of the first 10 million terms.

Cf. A366548, A363193, A354971, A181391, A342585.

cp's OEIS Frontend

A363193 a(1)=1, and thereafter a(n) = number of occurrences of a(k) among terms a(1..k), where k = n-a(n-1).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Python

A366548 a(0) = 0; for n > 0, a(n) is the number of terms prior to the term a(n-1-a(n-1)) that equal a(n-1-a(n-1)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

A366549 a(0) = 0; for n > 0, a(n) is the number of terms prior to and including the term a(n-1-a(n-1)) that equal a(n-1-a(n-1)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs