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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.

%I A088333 #18 May 19 2025 08:16:49
%S A088333 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 A088333 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 A088333 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 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.
%C A088333 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 A088333 See A054995 for references and links.
%H A088333 <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%F A088333 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 A088333 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 A088333 a(n) = A032434(n,4) if n>=4. - _R. J. Mathar_, May 04 2007
%Y A088333 Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.
%K A088333 nonn,easy
%O A088333 1,3
%A A088333 _N. J. A. Sloane_, Nov 13 2003
%E A088333 More terms from _David Wasserman_, Aug 08 2005