OFFSET

1,2

COMMENTS

Arrange 1, 2, 3, ... 2*n clockwise in a circle. Starting the count at 1, delete every k-th integer clockwise until exactly half of the numbers have been deleted. a(n) is the least positive integer k for which the deleted numbers are the odd numbers.

Deleting the alternate (even) numbers from a circle of 2*n numbers leaving the odd numbers is trivially achieved with k = 2 for all n >= 1.

LINKS

P. Schumer, The Josephus Problem: Once More Around, Mathematics Magazine, Vol. 75:1 (2002), 12-17.

CROSSREFS

KEYWORD

nonn,more

AUTHOR

William Rex Marshall, Nov 21 2011

STATUS

approved