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

1,7

Comments

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

Links

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

Example

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.

Prog

(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

Crossrefs

Cf. A006370, A139391, A334040.

Sequence in context: A324735 A219495 A242041 * A224937 A035193 A004556

Adjacent sequences: A376993 A376994 A376995 * A376997 A376998 A376999

Keyword

nonn

Author

Chia-Ching Chen, Oct 12 2024

Status

approved