A088333 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 3 places clockwise from i. Repeat, counting 3 places from the next undeleted integer, until only one integer remains.

1, 1, 2, 2, 1, 5, 2, 6, 1, 5, 9, 1, 5, 9, 13, 1, 5, 9, 13, 17, 21, 3, 7, 11, 15, 19, 23, 27, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42

Offset: 1

Author

N. J. A. Sloane, Nov 13 2003

Keywords

• nonn
• easy

Comments

If one counts only one place (resp. two places) at each stage to determine the element to be deleted, we get A006257 (resp. A054995).

References

• See A054995 for references and links.

Links

• Index entries for sequences related to the Josephus Problem

Crossrefs

Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.

Formula

It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n-1)+4) mod n. The conjecture is false; counterexample: a(21)=21; a(20)=17; (a(20)+4)mod 21=0; corrected formula: a(n)=(a(n-1)+3) mod n +1;

The conjecture is true. After removing the 4th number, we are reduced to the n-1 case, but starting with 5 instead of 1. - David Wasserman, Aug 08 2005

a(n) = A032434(n,4) if n>=4. - R. J. Mathar, May 04 2007

Extensions

More terms from David Wasserman, Aug 08 2005