A165728 If we divide the sequence into these subsequences, the pattern is obvious. {{1,1}, {0,1}, {1,1}}, {{0,1,0,1}, {1,1,1,1}, {0,1,0,1}}, {{1,1,1,1,1,1,1,1}, {0,1,0,1,0,1,0,1}, {1,1,1,1,1,1,1,1}}, {{0,1,0,1,0,1,0,1,0,1,0,1,0,1,0}, {1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1}, {0,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1}}, ...

1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1

Offset: 1

Author

Ryohei Miyadera and Masakazu Naito, Sep 25 2009

Keywords

• nonn

Comments

This is a sequence made by a variant of the Josephus Problem mod 2, but we use the last number to eliminate instead the last number that remains.

We put n numbers in a circle, and in this variant two numbers are to be eliminated at the same time.

These two processes of elimination go in different directions. Suppose that there are n numbers. Then the first process of elimination starts with the first number and the 2nd, 4th, 6th numbers, ... are to be eliminated.

The second process starts with the n-th number, and the (n-1)-st, (n-3)-rd, (n-5)-th number, ... are to be eliminated.

We suppose that the first process comes first and the second process second at every stage.

We denote by JI2(n) the position of the last number to be eliminated when we have n numbers.

If we use this sequence JI2(n) for n = 6,7,8,... under mod 2, then we get the above sequence with 1 and 0.

Note that we have to omit the first 5 terms to get this sequence with its beautiful pattern.

Examples

			Suppose that there are n = 14 numbers. Then the 2nd, 4th, and 6th numbers will be eliminated by the first process. Similarly 13th, 11th, and 9th numbers will be eliminated by the second process. Now two directions are going to overlap. The first process will eliminate the 8 and 12 and the second process will eliminate 5 and 1. After this the first process will eliminate 3 and 14, and the second process will eliminate 10. The number that remains is 7, and hence the last number to be eliminated is 14. Therefore JI2(14) = 14. JI2(14) = 0 (mod 2).
		

Links

• Hiroshi Matsui, Toshiyuki Yamauchi, Soh Tatsumi, Takahumi Inoue, Masakazu Naito and Ryohei Miyadera, Interesting Variants of the Josephus Problem, Computer Algebra - Design of Algorithms, Implementations and Applications, Kokyuroku, The Research Institute of Mathematical Science, No. 1652, (2009), 44-54.
• Masakazu Naito and Ryohei Miyadera, The Josephus Problem in Both Directions, The Wolfram Demonstrations Project.
• Masakazu Naito, Daisuke Minematsu and Ryohei Miyadera, The Self-Similarity of the Josephus Problem and its Variants, Visual Mathematics, 11(2) (2009).
• Index entries for sequences related to the Josephus Problem

Crossrefs

Cf. A113648, A114144, A165556.

Programs

Mathematica

last2 = {1, 2, 1, 1, 1, 3, 5, 6, 5, 5}; Table[JI2[n] = last2[[n]], {n, 1, 10}]; JI2[m_] := JI2[m] = Block[{n, h}, h = Mod[m, 8]; n = (m - h)/8; Which[h == 0, 4 JI2[2 n] - 1 - Floor[JI2[2 n]/(n + 1)], h == 1, 8 n + 5 - 4 JI2[2 n], h == 2, 4 JI2[2 n] -3 -Floor[JI2[2 n]/(n + 2)], h == 3, 8 n + 7 - 4 JI2[2 n], h == 4, 8 n + 8 - 4 JI2[2 n + 1] + Floor[JI2[2 n + 1]/(n + 2)], h == 5, 4 JI2[2 n + 1] - 1, h == 6, 8 n + 10 - 4 JI2[2 n + 1] + Floor[JI2[2 n + 1]/(n + 2)], h == 7, 4 JI2[2 n + 1] - 3]]; Table[Mod[JI2[n], 2], {n, 6, 95}]

Formula

{JI2(n): n = 1,2,3,4,5,6,7,8} = {1, 2, 1, 1, 1, 3, 5}.

(1) JI2(8*n) = 4*JI2(2*n) - 1 - [JI2(2*n)/(n+1) ].

(2) JI2(8*n+1) = 8*n + 5 - 4*JI2(2*n).

(3) JI2(8*n+2) = 4*JI2(2*n) - 3 - [JI2(2*n)/(n + 2)] .

(4) JI2(8*n+3) = 8*n + 7 - 4*JI2(2*n).

(5) JI2(8*n+4) = 8*n + 8 - 4*JI2(2*n+1) + [JI2(2*n+1)/(n+2)].

(6) JI2(8*n+5) = 4*JI2(2*n+1) - 1.

(7) JI2(8*n+6) = 8*n + 10 - 4*JI2(2*n+1) + [JI2(2*n+1)/(n+2)].

(8) JI2(8*n+7) = 4*JI2(2*n+1) - 3,

Note that recurrence relations are the same as those of A165556, but initial values are different.