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A165735
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The pattern is obvious. The sequence can be divided into subsequence of {1,1,1,...} and {2,2,2,...}.
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0
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1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 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, 1, 1, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| Let n be a natural number. We put n numbers in a circle, and we are going to remove every third number. Let J3(n) be the last number that remains. This is the traditional Josephus Problem. Let J3 ( mod 3) be the residue of the sequence J3(n) under mod 3. J3 (mod 3) produces the sequence {1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2,...}.
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REFERENCES
| Hiroshi Matsui,Masakazu Naito and Naoyuki Totani, The Period and the Distribution of the Fibonacci-like Sequence Under Various Moduli, Undergraduate Math Journal, Rose-Hulman Institute of Technology, Vol. 10, Issue 1, 2009.
M. Naito, R. Miyadera and D. Minematsu, The Self-Similarity of the Josephus Problem and its Variants, Visual Mathematics, Volume 11, No. 2, 2009.
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LINKS
| Masakazu Naito and Ryohei Miyadera, The Self-Similarity of the Josephus Problem and its Variants, VISUAL MATHEMATICS, Volume 11, No.2, 2009.
Wolfram MathWorld, Josephus Problem
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FORMULA
| (1) J3(1) = 1 and J3(2) = 2.
(2) J3(3m) = J3(2m) + [(J3(2m)-1)/2].
(3a) J3(3m+1) = 3m + 1 (if J3(2m + 1) = 1).
(3b) J3(3m+1) = J3(2m+1) + [J3(2m+1)/2] - 2 (if J3(2m + 1) > 1).
(4) J3(3m+2) = J3(2m+1) + [J3(2m+1)/2] + 1
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EXAMPLE
| If we use n = 10, then we put numbers 1,2,3,4,5,6,7,8,9,10 in a circle. We eliminate 3,6,9,2,7,1,8,5,10, and the last number that remains is 4. Therefore J3(10) = 4 and J3(10) = 1 mod 3.
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MATHEMATICA
| J3[1] = 1; J3[2] = 2; J3[n_] := J3[n] = Block[{m, t}, t = Mod[n, 3]; m = (n - t)/3; Which[t == 0, J3[2 m] + Floor[(J3[2 m] - 1)/2], t == 1, If[J3[2 m + 1] == 1, 3 m + 1, J3[2 m + 1] + Floor[J3[2 m + 1]/2] - 2], t == 2, J3[2 m + 1] + Floor[J3[2 m + 1]/2] + 1]]; Table[Mod[J3[n], 3], {n, 1, 200}]
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CROSSREFS
| Cf. A114144, A113648, A165556
Sequence in context: A030409 A030407 A098357 * A083888 A102681 A172086
Adjacent sequences: A165732 A165733 A165734 * A165736 A165737 A165738
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KEYWORD
| nonn
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AUTHOR
| Ryohei Miyadera and Masakazu Naito (Miyadera127(AT)aol.com), Sep 25 2009
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