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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.
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%I #14 Jun 23 2020 19:04:21

%S 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,

%T 10,14,18,22,26,30,34,38,3,7,11,15,19,23,27,31,35,39,43,47,51,3,7,11,

%U 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

%N 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.

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

%D See A054995 for references and links.

%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>

%F 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;

%F 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

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

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

%K nonn,easy

%O 1,3

%A _N. J. A. Sloane_, Nov 13 2003

%E More terms from _David Wasserman_, Aug 08 2005