A360268 A version of the Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer 5 places clockwise from i. Repeat, counting 5 places from the next undeleted integer, until only one integer remains.

1, 1, 1, 3, 4, 4, 3, 1, 7, 3, 9, 3, 9, 1, 7, 13, 2, 8, 14, 20, 5, 11, 17, 23, 4, 10, 16, 22, 28, 4, 10, 16, 22, 28, 34, 4, 10, 16, 22, 28, 34, 40, 3, 9, 15, 21, 27, 33, 39, 45, 51, 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 1, 7, 13, 19, 25

Offset: 1

Author

Benjamin Lilley, Jan 31 2023

Keywords

• nonn

Examples

			a(7) = 3 because the elimination process gives (^1,2,3,4,5,6,7) -> (1,2,3,4,5,^7) -> (1,2,3,4,^7) -> (^1,2,3,4) -> (1,^3,4) -> (^3,4) -> (3), where ^ denotes the counting reference position.
a(13) = 9 => a(14) = (a(13) + 5) mod 14 + 1 = 1.
		

Links

• Wikipedia, Josephus problem.
• Index entries for sequences related to the Josephus Problem.

Crossrefs

6th column of A198789.

Cf. A006257, A054995, A088333, A181281.

Programs

Mathematica

a[1] = 1; a[n_] := a[n] = Mod[a[n - 1] + 5, n] + 1; Array[a, 100] (* Amiram Eldar, Feb 03 2023 *)

Python

def A360268_up_to_n(n):
  val = 1
  return [val := (val + 5) % i + 1 for i in range(1,n+1)]

Formula

a(n) = (a(n-1) + 5) mod n + 1 if n > 1, a(1) = 1.