A381667 Triangle read by row: T(n,k) is the number of the k-th eliminated person in the variation of the Josephus elimination process for n people where two people are skipped each step.

1, 1, 2, 3, 1, 2, 3, 2, 4, 1, 3, 1, 5, 2, 4, 3, 6, 4, 2, 5, 1, 3, 6, 2, 7, 5, 1, 4, 3, 6, 1, 5, 2, 8, 4, 7, 3, 6, 9, 4, 8, 5, 2, 7, 1, 3, 6, 9, 2, 7, 1, 8, 5, 10, 4, 3, 6, 9, 1, 5, 10, 4, 11, 8, 2, 7, 3, 6, 9, 12, 4, 8, 1, 7, 2, 11, 5, 10, 3, 6, 9, 12, 2, 7, 11, 4, 10, 5, 1, 8, 13

Offset: 1

Author

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 03 2025

Keywords

• nonn
• tabl

Comments

In this variation of the Josephus elimination process, the numbers 1 through n are arranged in a circle. A pointer starts at position 1. With each turn, the pointer skips two numbers and the next number is eliminated. The process repeats until no numbers remain. This sequence represents the triangle T(n, k), where n is the number of people in the circle, and T(n, k) is the elimination order of the k-th number in the circle.

This variation of the Josephus problem is related to under-under-down card dealing, where the cards of a deck are dealt by alternately cycling two cards from the top "under", and then dealing the next card "down". In particular, T(n,k) is the k-th card dealt in under-under-down dealing if the deck begins in order 1,2,3,...,n.

Examples

			Consider 4 people in a circle. Initially, people numbered 1 and 2 are skipped, and person 3 is eliminated. The remaining people are now in order 4, 1, 2. Then 4 and 1 are skipped, and person 2 is eliminated. The remaining people are in order 4, 1. Now, 4 and 1 are skipped, and 4 is eliminated. Person 1 is eliminated last. Thus, the fourth row of the triangle is 3, 2, 4, 1.
Triangle begins;
  1;
  1, 2;
  3, 1, 2;
  3, 2, 4, 1;
  3, 1, 5, 2, 4;
  3, 6, 4, 2, 5, 1;
  3, 6, 2, 7, 5, 1, 4;
  3, 6, 1, 5, 2, 8, 4, 7;
		

Crossrefs

Cf. A006257, A054995, A225381, A321298, A378635, A380195, A008585, A381591 A381667.

Programs

Python

def UUDES(n):
    l=[]
    for i in range(n):
        l.append(i+1)
    index = 0
    P=[]
    for i in range(n):
        index+=2
        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
sequence = []
for i in range(1,20):
    sequence += [str(v) for v in UUDES(i)]
print(", ".join(sequence))

Python

def row(n):
    i, J, out = 0, list(range(1, n+1)), []
    while len(J) > 0:
        i = (i + 2)%len(J)
        out.append(J.pop(i))
    return out
print([e for n in range(1, 14) for e in row(n)]) # Michael S. Branicky, Mar 24 2025