A380195 -id:A380195 - cp's OEIS frontend

A381591 Elimination order of the first person in a variation of the Josephus problem, where there are n people total and two people are skipped each time.

1, 1, 2, 4, 2, 6, 6, 3, 9, 6, 4, 7, 11, 5, 11, 15, 6, 13, 11, 7, 12, 16, 8, 23, 18, 9, 22, 16, 10, 17, 31, 11, 27, 30, 12, 35, 21, 13, 22, 37, 14, 30, 35, 15, 32, 26, 16, 27, 35, 17, 47, 37, 18, 53, 31, 19, 32, 47, 20, 57, 56, 21, 51, 36, 22, 37, 65, 23, 49, 70

Offset: 1

Tanya Khovanova, Nathan Sheffield, and the MIT PRIMES STEP junior group, Mar 02 2025

nonn

a(3k-1) = k.

			Consider n = 4 people. The first person eliminated is number 3. This leaves the remaining people in order 4, 1, 2. The second person eliminated is number 2. Thus, the remaining people in order 4, 1. The next person eliminated is number 4. On the fourth step, person number 1 is eliminated, implying that the order of elimination of the first person is 4: a(4) = 4.

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

def UUD(n):
    return invPerm(UUDES(n))
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, 71):
    sequence += [str(UUD(i)[0])]
print(", ".join(sequence))

def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 0:
        i = (i + 2)%len(J)
        q = J.pop(i)
        if q == 1: return c
        c = c+1
print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Mar 24 2025

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.

Original entry on oeis.org

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

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

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.

			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;

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

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))

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

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A381591 Elimination order of the first person in a variation of the Josephus problem, where there are n people total and two people are skipped each time.

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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.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Python

Python