A382355 A version of the Josephus problem: a(n) is the surviving integer under the skip-eliminate-skip version of the elimination process.

1, 1, 1, 4, 3, 6, 3, 6, 9, 3, 6, 9, 12, 1, 4, 7, 10, 13, 16, 19, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 49, 52, 55, 58, 61, 64, 67, 70, 3, 6

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 22 2025

Keywords

• nonn

Comments

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

Examples

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

Crossrefs

Cf. A006257, A225381, A321298, A378635, A382354, A382356, A382358.

Programs

Python

def J(n,A):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=A[i]
        index=index%len(l)
        P.append(l[index])
        l.pop(index)
    return P
def invPerm(p):
    inv = []
    for i in range(len(p)):
        inv.append(None)
    for i in range(len(p)):
        inv[p[i]-1]=i+1
    return inv
def survivor(n, A):
    return J(n, A)[n-1]
def UDU(n):
    return [1] + [2 for i in range(n)]
seq = []
for i in range(1,20):
    seq += [survivor(i, UDU(i))]
print(", ".join([str(v) for v in seq]))

Python

def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        i = (i + 1)%len(J)
        q = J.pop(i)
        i = (i + 1)%len(J) # skip the third
    return J[0]
print([a(n) for n in range(1, 73)]) # Michael S. Branicky, Mar 24 2025