A381051 A version of the Josephus problem: a(n) is the surviving integer under the eliminate-skip-eliminate version of the elimination process.

1, 2, 2, 2, 5, 5, 2, 8, 5, 2, 8, 5, 11, 8, 14, 11, 17, 14, 2, 17, 5, 20, 8, 23, 11, 26, 14, 2, 17, 5, 20, 8, 23, 11, 26, 14, 29, 17, 32, 20, 35, 23, 38, 26, 41, 29, 44, 32, 47, 35, 50, 38, 53, 41, 2, 44, 5, 47, 8, 50, 11, 53, 14, 56, 17, 59, 20, 62, 23, 65, 26, 68

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Apr 14 2025

Keywords

• nonn

Comments

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

Examples

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

Crossrefs

Cf. A006257, A032766, A225381, A321298, A378635, A381050, A383076.

Programs

Python

def a(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 1:
        i = i%len(J)
        J.pop(i)
        i = (i + 1)%len(J)
        i = i%len(J)
        if len(J) > 1:
            J.pop(i)
    return J[0]
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Apr 28 2025