A198791 Least count k that deletes the alternate (odd) numbers in the Josephus problem for a circle of numbers 1, 2, 3, ... 2*n, leaving the even numbers undeleted.

1, 5, 19, 27, 87, 989, 3119, 5399, 8189, 99663, 57455, 222397, 2603047, 8476649, 117917347, 290190179, 360064247, 1344262919, 3181391639, 39179386959, 204692414215, 165424388219, 2254874520599, 2922139183443, 594630799853

Offset: 1

Author

William Rex Marshall, Nov 21 2011

Keywords

• nonn
• more

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.
• Index entries for sequences related to the Josephus Problem

Crossrefs

Cf. A198788, A198789.