1,1

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

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics,Addison-Wesley Publishing Company, 1994. P.9-10.

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 ).

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.

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]];

Cf. A113648, A006257.

Sequence in context: A016646 A160552 A006257 this_sequence A050820 A133179 A146908

Adjacent sequences: A114141 A114142 A114143 this_sequence A114145 A114146 A114147

easy,nonn

Satoshi Hashiba, Daisuke Minematsu and Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp), Feb 03 2006