A334040 -id:A334040 - cp's OEIS frontend

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

Chia-Ching Chen, Oct 12 2024

nonn

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

			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.

Cf. A006370, A139391, A334040.

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

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

Original entry on oeis.org

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

Chia-Ching Chen, Oct 12 2024

nonn

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).

			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.

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

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

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

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

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

cp's OEIS Frontend

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

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