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 a(n) = J(3*n).
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley Publishing Company, 1994, pp. 9-10.
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 a(5) = J(15) = 8.
MATHEMATICA
(*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,more
AUTHOR
Satoshi Hashiba, Daisuke Minematsu and Ryohei Miyadera, Feb 03 2006
STATUS
approved