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

Author

Chia-Ching Chen, Oct 12 2024

Keywords

• nonn

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).

Examples

			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.
		

Links

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

Crossrefs

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

Programs

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

Formula

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

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