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Search: id:A113648
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| A113648 |
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A variant of Jesephus Problem in which 2 persons are to be eliminated at the same time. |
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+0
5
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| 1, 3, 6, 1, 3, 5, 7, 9, 12, 15, 18, 21, 24, 27, 30, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 1, 3, 5, 7
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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a(n) is defined as follows. Write the numbers 1 through 2n in a circle, start at 1 and n+1. Cross off every other number until only one number is left. The process that starts with 1 should be the first at any stage. For example we cross off 2, n+2, 4, n+4, 6, n+6, ..... The remaining number is a(n). This function is defined only for even arguments.
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REFERENCES
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R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley
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FORMULA
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The sequence a(m) is defined for any even number m as follows: a(2) = 1. a(4n) = 2a(2n)-2n-1 ( if a(2n) > n) and a(4n) = 2a(2n)+2n-1 ( if a(2n) <= n). a(4n+2) = 2a(2n+2)-2n-5 ( if a(2n+2) >= n+3), a(4n+2) = 2a(2n+2)+2n-2 ( if n+3 > a(2n+2) >= 2) and a(4n+2) = 2n+1 ( if a(2n+2) = 1).
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EXAMPLE
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For a(8): we are to cross off 2, 6, 4, 8, 7, 3, 5 and 1 is left. Therefore a(8) = 1.
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MATHEMATICA
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jose2[2] = 1; jose2[n_] := If[Mod[n, 4] == 0, If[jose2[n/2] <= (n/4), 2(n/4) + 2jose2[n/2] - 1, 2jose2[n/2] - 2(n/4) - 1], Which[jose2[(n + 2)/2] == 1, n/2, 1 < jose2[(n + 2)/2] < (n + 10)/4, 2jose2[(n + 2)/2] + (n - 2)/2 - 2, (n + 6)/4 < jose2[(n + 2)/2], 2jose2[(n + 2)/2] - (n + 8)/2]];
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CROSSREFS
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Cf. A006257.
Sequence in context: A032660 A102257 A091425 this_sequence A104612 A088392 A163327
Adjacent sequences: A113645 A113646 A113647 this_sequence A113649 A113650 A113651
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KEYWORD
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easy,nonn
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AUTHOR
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Satoshi Hashiba, Daisuke Minematsu and Ryohei Miyadera (miyadera1272000(AT)yahoo.co.jp), Jan 15 2006
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