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A165556
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A symmetric version of the Josephus problem read modulo 2.
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3
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1, 1, 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, 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, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1
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OFFSET
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1,1
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COMMENTS
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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 numbers, ... are to be eliminated.
We suppose that the first process comes first and the second process second at every stage.
We denote the position of the last survivor by JI(n). If we use this sequence under mod 2, then we get the above sequence with 1 and 0.
Old name was "{1,1}, {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, 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, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0}, ... In this way the two patterns {1,1} and {0,1} take turns in subsequences with the length of 2, 4, 8, 16, 64,...".
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LINKS
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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.
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FORMULA
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(1) JI(8*n) = 4*JI(2*n) - 1 - [JI(2*n)/(n+1)].
(2) JI(8*n+1) = 8*n + 5 - 4*JI(2n).
(3) JI(8*n+2) = 4*JI(2*n) - 3 - [JI(2*n)/(n + 2)] .
(4) JI(8*n+3) = 8*n + 7 - 4JI(2*n).
(5) JI(8*n+4) = 8*n + 8 - 4*JI(2*n+1) + [JI(2*n+1)/(n + 2)].
(6) JI(8*n+5) = 4*JI(2*n+1) - 1.
(7) JI(8*n+6) = 8*n + 10 - 4*JI(2*n+1) + [(JI(2*n+1)/(n + 2)].
(8) JI(8*n+7) = 4*JI(2*n+1) - 3,
where [ ] is the floor function.
Conjecture: a(n) = (1 - (-1)^(n + (n + 1)*floor(log_2(n + 1))))/2. - Velin Yanev, Nov 23 2016
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EXAMPLE
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Suppose that there are n = 14 numbers.
Then the 2nd, 4th, and 6th numbers will be eliminated by the first process. Similarly the 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, 12 and the second process will eliminate 5, 1.
After this the first process will eliminate 3, 14, and the second process will eliminate 10. The number that remains is 7. Therefore JI(14) = 7 and JI(14) = 1 (mod 2).
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MATHEMATICA
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initialvalue = {1, 1, 3, 4, 3, 6, 1, 3}; Table[JI[n] = initialvalue[[n]], {n, 1, 8}]; JI[m_] := JI[m] = Block[{n, h}, h = Mod[m, 8]; n = (m - h)/8; Which[h == 0, 4 JI[2 n] - 1 - Floor[JI[2 n]/(n + 1)], h == 1, 8 n + 5 - 4 JI[2 n], h == 2, 4 JI[2 n] -3 -Floor[JI[2 n]/(n + 2)], h == 3, 8 n + 7 - 4 JI[2 n], h == 4, 8 n + 8 - 4 JI[2 n + 1] + Floor[JI[2 n + 1]/(n + 2)], h == 5, 4 JI[2 n + 1] - 1, h == 6, 8 n + 10 - 4 JI[2 n + 1] + Floor[JI[2 n + 1]/(n + 2)], h == 7, 4 JI[2 n + 1] - 3]]; Table[Mod[JI[n], 2], {n, 1, 62}]
Flatten[Table[{PadRight[{}, 2^n, {1}], PadRight[{}, 2^(n+1), {1, 0}]}, {n, 1, 5, 2}], 1] (* Harvey P. Dale, Mar 24 2013 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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