A381048 - OEIS

A381048

Elimination order of the first person in a variation of the Josephus problem, where there are n people total. During each round the first person is skipped, and the second and the third person are eliminated. Then the process repeats.

1, 2, 3, 3, 4, 6, 5, 6, 9, 7, 8, 11, 9, 10, 14, 11, 12, 18, 13, 14, 19, 15, 16, 22, 17, 18, 27, 19, 20, 27, 21, 22, 30, 23, 24, 35, 25, 26, 35, 27, 28, 38, 29, 30, 44, 31, 32, 43, 33, 34, 46, 35, 36, 54, 37, 38, 51, 39, 40, 54, 41, 42, 61, 43, 44, 59, 45, 46, 62, 47

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,2

COMMENTS

This elimination process is related to the under-down-down card dealing.

LINKS

Table of n, a(n) for n=1..70.

Eric Huang, Tanya Khovanova, Timur Kilybayev, Ryan Li, Brandon Ni, Leone Seidel, Samarth Sharma, Nathan Sheffield, Vivek Varanasi, Alice Yin, Boya Yun, and William Zelevinsky, Card Dealing Math, arXiv:2509.11395 [math.NT], 2025. See p. 17.

FORMULA

a(3k+1) = 2k+1; a(3k-1) = 2k.

EXAMPLE

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

PROG

(Python)

def a(n):

i, J, out, c = 0, list(range(1, n+1)), [], 0

while len(J) > 1:

i = (i + 1)%len(J)

q = J.pop(i)

c += 1

if q == 1:

return c

i = i%len(J)

if len(J) > 1:

q = J.pop(i)

c += 1

if q == 1:

return c

return c+1

print([a(n) for n in range(1, 71)]) # Michael S. Branicky, Apr 28 2025

CROSSREFS

Cf. A006257, A007494, A225381, A321298, A378635, A337191, A381049, A382528.

Sequence in context: A207100 A281365 A304705 * A131187 A099072 A382079

Adjacent sequences: A381045 A381046 A381047 * A381049 A381050 A381051

KEYWORD

nonn

AUTHOR

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

STATUS

approved