A049067 LaBar's conjecture (steps to return n to 1 after division by 3 and, if needed, multiplication by 2, addition of 1 or 2).

18, 44, 4, 216, 34, 50, 181, 68, 13, 126, 125, 228, 278, 256, 49, 364, 82, 68, 180, 575, 202, 1033, 245, 92, 403, 140, 40, 520, 499, 156, 872, 214, 158, 1400, 221, 264, 399, 368, 317, 1157, 390, 298, 648, 376, 94, 594, 1155, 412, 1983, 500, 133, 808, 226, 122

Offset: 1

Author

Enoch Haga

Keywords

• easy
• nonn
• changed

Comments

Begin with n and repeat the following procedure until 1 is reached: If n is divisible by 3 then apply n -> n/3. Otherwise, alternately apply n -> 2*n+2 and n -> 2*n+1. Then a(n) is the sum of the numbers produced by this process, including n and 1. - David Radcliffe, Aug 28 2025

Examples

			Beginning at n=1, algorithm produces s+t+a=18.
a(2) = 44 because the trajectory of n=2 is (2, 6, 2, 5, 12, 4, 9, 3, 1) and these numbers sum to 44. - _David Radcliffe_, Aug 28 2025
		

Links

• Enoch Haga, Problem, School Science and Mathematics, Nov 1983, vol. 83, no 7, page 628.
• Sean A. Irvine, Java program (github)
• LaBar, Problem #3929, School Science and Mathematics, Dec 1982, vol. 82 no 8, page 715.

Crossrefs

Cf. A049074.

Programs

Python

def A049067(n):
    s = n
    c = 2
    while n > 1 or s == n:
        if n % 3 == 0:
            n //= 3
        else:
            n = 2*n + c
            c = 3 - c
        s += n
    return s # David Radcliffe, Aug 28 2025