%I
%S 3,1,3,5,8,11,14,17,21,25,29,33,37,41,45,1,3,5,7,9
%N A variant of the Josephus Problem in which three persons are to be eliminated at the same time.
%C This is a variant of the Josephus Problem. When there are 3m persons, the first process of elimination starts with the first person, the second with the (m+1)st person and the third with the (2m+1)st person. We suppose that the first process comes first, the second process secondly and the third process thirdly. J(n) is the position of the survivor when there are n persons. Our sequence is {J(3), J(6), J(9), J(12), .....} = {3, 1, 3, 5, 8, 11, 14, 17, 21, 25, 29, 33
%D R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, AddisonWesley Publishing Company, 1994, pp. 910.
%H <a href="/index/J#Josephus">Index entries for sequences related to the Josephus Problem</a>
%F The function J(n) is defined only for integers n that have 3 as a factor. J(6m+3) = 2J(3m)+2m+2 (if J(3m) <= m), J(6m+3) = 2J(3m)+2m+3 (if m+1 <= J(3m) <= 2m) and J(6m+3) = 2J(3m)4m+1 (if 2m+1 <= J(3m)). J(6m) = 2J(3m)+2m1 (if J(3m) <= 2m) and J(6m) = 2J(3m)4m1 (if J(3m) > 2m).
%e If there are 15 persons, then 2, 7, 12, 4, 9, 14, 6, 11, 1, 10, 15, 5, 3, 13 are to be eliminated and the survivor is 8. Therefore J(15) = 8.
%t Clear[jose];(*This function is defined only for numbers that are multiples of 3.*)jose[3] = 3; jose[n_?(IntegerQ[ #/3] &)] := If[Mod[n, 6] == 0, If[jose[n/2] < n/3 + 1, 2jose[n/2] + n/3  1,2jose[n/2]  2n/3  1], Which[jose[(n  3)/2] < (n  3)/6 +1, 2jose[(n  3)/2] + (n  3)/3 + 2, (n  3)/6 < jose[(n  3)/2] < (n  3)/3 + 1, 2jose[(n  3)/2] + (n  3)/3 + 3, (n  3)/3 < jose[(n  3)/2], 2jose[(n  3)/2]  2(n  3)/3 + 1]];
%Y Cf. A006257, A113648.
%K easy,nonn
%O 1,1
%A Satoshi Hashiba, Daisuke Minematsu and _Ryohei Miyadera_, Feb 03 2006
