%I A114144
%S A114144 3,1,3,5,8,11,14,17,21,25,29,33,37,41,45,1,3,5,7,9
%N A114144 A variant of Josephus Problem in which three persons are to be eliminated
at the same time.
%C A114144 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 A114144 R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics,Addison-Wesley
Publishing Company, 1994. P.9-10.
%F A114144 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)+2m-1 (if J(3m) =< 2m) and J(6m) = 2J(3m)-4m-1 (if
J(3m) > 2m ).
%e A114144 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 A114144 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 A114144 Cf. A113648, A006257.
%Y A114144 Sequence in context: A016646 A160552 A006257 this_sequence A050820 A133179
A146908
%Y A114144 Adjacent sequences: A114141 A114142 A114143 this_sequence A114145 A114146
A114147
%K A114144 easy,nonn
%O A114144 1,1
%A A114144 Satoshi Hashiba, Daisuke Minematsu and Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp),
Feb 03 2006
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