|
| |
|
|
A114144
|
|
A variant of the Josephus Problem in which three persons are to be eliminated at the same time.
|
|
3
|
|
|
|
3, 1, 3, 5, 8, 11, 14, 17, 21, 25, 29, 33, 37, 41, 45, 1, 3, 5, 7, 9
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
COMMENTS
|
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
|
|
|
REFERENCES
|
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley Publishing Company, 1994, pp. 9-10.
|
|
|
LINKS
|
|
|
|
FORMULA
|
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).
|
|
|
EXAMPLE
|
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.
|
|
|
MATHEMATICA
|
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]];
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
easy,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
