

A054995


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


16



1, 2, 2, 1, 4, 1, 4, 7, 1, 4, 7, 10, 13, 2, 5, 8, 11, 14, 17, 20, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34, 37, 40, 43, 46, 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 1, 4, 7, 10, 13, 16, 19, 22, 25
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

If one counts only one place (rather than two) at each stage to determine the element to be deleted, the Josephus survivors (A006257) are obtained.


LINKS

Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
Ph. Dumas, Algebraic aspects of Bregular series. [Broken link]
Philippe Dumas, Algebraic aspects of Bregular series, Research Report, RR1931, INRIA, 1993.
Ph. Dumas, Algebraic aspects of Bregular series, in: International Colloquium on Automata, Languages and Programming, ICALP 1993 (A. Lingas, R. Karlsson, S. Carlsson, eds.), pp. 457468, Lecture Notes in Computer Science, vol. 700, Springer, Berlin, 1993.
L. Halbeisen and N. Hungerbühler, The Josephus Problem, J. Théor. Nombres Bordeaux 9 (1997), no. 2, 303318.
Alasdair MacFhraing, Aireamh Muinntir Fhinn Is Dhubhain, Agus Sgeul Josephuis Is An Da Fhichead Iudhaich, [Gaelic with English summary], Proc. Royal Irish Acad., Vol. LII, Sect. A., No. 7, 1948, 8793.
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33, 235240, 1991.
Index entries for sequences related to the Josephus Problem


FORMULA

a(n) = 3*n + 1  floor(K(3)*(3/2)^(ceiling(log((2*n+1)/K(3))/log(3/2)))) where K(3) = (3/2)*K = 1.622270502884767... (K is the constant described in A061419); a(n) = 3n + 1  A061419(k+1) where A061419(k+1) is the least integer such that A061419(k+1) > 2n.
a(1) = 1 and, for n > 1, a(n) = (a(n1) + 3) mod n, if this value is nonzero, n otherwise.
a(n) = (a(n1) + 2) mod n + 1.  Paul Weisenhorn, Oct 10 2010


EXAMPLE

a(5) = 4 because the elimination process gives (1^,2,3,4,5) > (1,2,4^,5) > (2^,4,5) > (2^,4) > (4), where ^ denotes the counting reference position.
a(13) = 13 => a(14) = (a(13) + 2) mod 14 + 1 = 2.  Paul Weisenhorn, Oct 10 2010


MATHEMATICA

(* First do *) Needs["Combinatorica`"] (* then *) f[n_] := Last@ InversePermutation@ Josephus[n, 3]; Array[f, 70] (* Robert G. Wilson v, Jul 31 2010 *)
Table[Nest[Rest@RotateLeft[#, 2] &, Range[n], n  1], {n, 72}] // Flatten (* Arkadiusz Wesolowski, Jan 14 2013 *)


CROSSREFS

Cf. A032434, A005427, A005428, A006257, A007495, A000960, A056530.
Cf. A181281 (with s=5).  Paul Weisenhorn, Oct 10 2010
Sequence in context: A268193 A238606 A325612 * A018219 A174714 A116633
Adjacent sequences: A054992 A054993 A054994 * A054996 A054997 A054998


KEYWORD

nonn


AUTHOR

John W. Layman, May 30 2000


STATUS

approved



