A381129 A version of the Josephus problem: a(n) is the surviving integer under the spelling version of the elimination process, called Down-SpellUnder.

1, 2, 2, 3, 3, 2, 7, 8, 4, 6, 4, 6, 7, 11, 10, 3, 14, 4, 17, 11, 3, 16, 7, 16, 7, 22, 2, 8, 24, 27, 7, 21, 13, 28, 30, 8, 3, 37, 12, 7, 8, 33, 7, 33, 44, 11, 32, 8, 6, 43, 2, 18, 49, 8, 32, 54, 26, 43, 44, 30, 40, 52, 26, 44, 8, 27, 60, 16, 11, 61, 70, 14, 58, 55

Offset: 1

Author

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

Keywords

• nonn
• word

Comments

Arrange n people numbered 1, 2, 3, ..., n in a circle, increasing clockwise. Execute the person numbered 1, then spell the letters of O-N-E, moving one person clockwise for each letter. Once you are done, eliminate the next person. Then, spell the letters of T-W-O; in other words, skip three people and eliminate the next person. Following this, spell the letters of T-H-R-E-E; in other words, skip five people and eliminate the next person. Continue until one person remains. The number of this person is a(n).

Examples

			Consider n = 4 people. The first person eliminated is number 1. This leaves the remaining people in the order 2, 3, 4. The second person eliminated is number 2; the people left are in the order 3, 4. The next person eliminated is numbered 4, leaving only the person numbered 3. Thus, a(4) = 3.
		

Crossrefs

Cf. A005589, A006257, A225381, A321298, A378635, A380201, A380202, A380204, A380246, A380247, A380248, A381114, A381127, A381128.

Programs

Python

from num2words import num2words as n2w
def f(n): return sum(1 for c in n2w(n).replace(" and", "") if c.isalpha())
def a(n):
    c, i, J = 1, 0, list(range(1, n+1))
    while len(J) > 1:
        q = J.pop(i)
        i = (i + f(c))%len(J)
        c = c+1
    return J[0]
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Feb 16 2025

Extensions

a(22) and beyond from Michael S. Branicky, Feb 16 2025