A377009 Number of odd terms in the Collatz trajectory of k = 4n-1 which are a new record high among its odd terms.

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

Offset

1,2

Comments

For these k = 4n-1, the first odd number is A139391(k) > k so it is the first record high.

The trajectory of k starts with A001511(n) successive initial records, so that a(n) >= A001511(n) (and reaches A351974(n) at that point).

Links

Table of n, a(n) for n=1..100.

Index entries for sequences related to 3x+1 (or Collatz) problem

Formula

a(n) = A376996(4n-1).

a(n) = A001511(n) + A376996(A351974(n)).

Example

For n = 7, which corresponds to the Collatz trajectory started from 27, the trajectory reaches larger maximum odd numbers at the following points: 41, 47, 71, 107, 233, 263, 395, 593, 719, 1079, 1619, 2429, 3077. Since there are 14 instances where a new maximum odd number is reached, we have a(7)=14.

Prog

(Python)
def a(num:int) -> int:
    count = 0
    num = num * 4 - 1
    maxnum = num
    while num > 1:
        if num%2 == 1:
            num = num*3 + 1
        while num%2 == 0:
            num //= 2
        if num > maxnum:
            count += 1
            maxnum = num
    return count

Crossrefs

Cf. A139391, A334040, A376996, A001511, A351974.

Sequence in context: A368759 A240694 A258643 * A287561 A046924 A015710

Adjacent sequences: A377006 A377007 A377008 * A377010 A377011 A377012

Keyword

nonn

Author

Chia-Ching Chen, Oct 12 2024

Status

approved