A376996 Number of odd terms in the Collatz trajectory of n which are > n and are a new record high among its odd terms.

0, 0, 1, 0, 0, 0, 2, 0, 2, 0, 1, 0, 0, 1, 3, 0, 0, 0, 1, 0, 0, 0, 2, 0, 1, 0, 14, 0, 0, 2, 13, 0, 0, 0, 1, 0, 0, 0, 3, 0, 13, 0, 1, 0, 0, 1, 12, 0, 0, 0, 1, 0, 0, 12, 12, 0, 1, 0, 2, 0, 0, 12, 11, 0, 0, 0, 1, 0, 0, 0, 11, 0, 12, 0, 1, 0, 0, 2, 3, 0, 0, 11, 11, 0, 0, 0, 2, 0, 1, 0, 11, 0, 0, 11, 10, 0, 11, 0, 1, 0

Offset: 1

Author

Chia-Ching Chen, Oct 12 2024

Keywords

• nonn

Comments

Similar to A334040, but only count when it comes to a new larger odd number.

Examples

			For n=15, a 4k-1 term, its trajectory is 15, 46, (23), 70, (35), 106, (53), 160, 80, 40, 20, 10, 5, 1. The numbers in the parentheses are numbers which make larger odd numbers. There are 3 of them, so a(15) = 3.
For n=9, a 4k+1 term, its trajectory is 9, 28, 14, 7, 22, (11), 34, (17), 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. There are 2 new largest odd terms, so a(9) = 2. It is noticeable that before the first new largest odd term 11, it reaches 7, which is a 4k-1 term smaller than 9, so a(9) = a(7).
For n=10, a 2k term, its trajectory is 10, 5, 16, 8, 4, 2, 1. There are no odd terms > 10, so a(10) = 0.
		

Crossrefs

Cf. A006370, A139391, A334040.

Programs

Python

def a(num:int) -> int:
    count = 0
    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