A383846 A version of the Josephus problem: a(n) is the surviving integer under the eliminate-eliminate-skip version of the elimination process.

1, 2, 3, 3, 3, 6, 6, 3, 9, 6, 3, 9, 6, 12, 9, 15, 12, 18, 15, 3, 18, 6, 21, 9, 24, 12, 27, 15, 3, 18, 6, 21, 9, 24, 12, 27, 15, 30, 18, 33, 21, 36, 24, 39, 27, 42, 30, 45, 33, 48, 36, 51, 39, 54, 42, 3, 45, 6, 48, 9, 51, 12, 54, 15, 57, 18, 60, 21, 63, 24, 66, 27

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, May 12 2025

Keywords

• nonn

Comments

This variation of the Josephus problem is related to down-down-under card dealing.

Crossrefs

Cf. A006257, A383845, A383846, A001651, A383847, A337191, A381051.

Programs

Maple

Consider 4 people in a circle in order 1,2,3,4. In the first round, person 1 is eliminated, then person 2 is eliminated, then person 3 is skipped. Now people are in order 4,3. In the second round, person 4 is eliminated. The last person, person 3, is freed. Thus, a(4) = 3.

Python

def a(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        J.pop(i)
        i = i%len(J)
        if len(J) > 1:
            J.pop(i)
        i = i%len(J)
        i = (i + 1)%len(J)
    return J[0]
print([a(n) for n in range(1, 73)])