A181281 A version of 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 4 places clockwise from i. Repeat, counting 4 places from the next undeleted integer, until only one integer remains.

1, 2, 1, 2, 2, 1, 6, 3, 8, 3, 8, 1, 6, 11, 1, 6, 11, 16, 2, 7, 12, 17, 22, 3, 8, 13, 18, 23, 28, 3, 8, 13, 18, 23, 28, 33, 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 1, 6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 3, 8, 13, 18, 23, 28, 33, 38

Offset: 1

Author

Paul Weisenhorn, Oct 10 2010

Keywords

• nonn

Examples

			a(7) = 6: (^1,2,3,4,5,6,7) -> (1,2,3,4,^6,7) -> (1,2,^4,6,7) -> (1,^4,6,7) -> (1,^6,7) -> (^1,6) -> (^6).
a(14) = 11 => a(15) = (a(14)+4) mod 15 + 1 = 1.
		

References

• Paul Weisenhorn, Josephus und seine Folgen, MNU Journal (Der mathematische und naturwissenschaftliche Unterricht), 59 (2006), 18-19.

Links

• Index entries for sequences related to the Josephus Problem

Crossrefs

Cf. A006257, A054995, A088333.

Programs

Maple

a:= proc(n) option remember;
      `if` (n=1, 1, (a(n-1)+4) mod n +1)
    end:
seq (a(n), n=1..100);

Mathematica

a[1] = 1; a[n_] := a[n] = Mod[a[n-1]+4, n]+1; Table[a[n], {n, 1, 80}] (* Jean-François Alcover, Oct 18 2013 *)

Formula

a(n) = (a(n-1) + 4) mod n + 1 if n>1, a(1) = 1.

Extensions

Edited by Alois P. Heinz, Sep 06 2011