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A090569
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The survivor w(n,2) in a modified Josephus problem, with a step of 2.
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2
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1, 1, 3, 3, 1, 1, 3, 3, 9, 9, 11, 11, 9, 9, 11, 11, 1, 1, 3, 3, 1, 1, 3, 3, 9, 9, 11, 11, 9, 9, 11, 11, 33, 33, 35, 35, 33, 33, 35, 35, 41, 41, 43, 43, 41, 41, 43, 43, 33, 33, 35, 35, 33, 33, 35, 35, 41, 41, 43, 43, 41, 41, 43, 43, 1, 1, 3, 3, 1, 1, 3, 3, 9, 9, 11, 11, 9, 9, 11, 11, 1, 1
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Arrange n persons {1,2,...,n} consecutively on a line rather than around in a circle. Beginning at the left end of the line, we remove every qth person until we reach the end of the line. At this point we immediately reverse directions, taking care not to "double count" the person at the end of the line and continue to eliminate every qth person, but now moving right to left. We continue removing people in this back and forth manner until there remains a lone survivor w(n,q).
Or a(n) is in A145812 such that 2n+1-2a(n) is in A145812 as well. Note also that 2a(n)+A088442(n-1)=2n+1. [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Oct 20 2008]
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REFERENCES
| Chris Groer, The Mathematics of Survival: From Antiquity to the Playground, Am. Math. Monthly 110(2003)812-825.
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FORMULA
| w(n, 2)=1+Sum[b(j)*(2^j), j=1...k, j odd], where Sum[b(j)*(2^j), j=0...k] is the binary expansion of N, with N either n or n-1, whichever is odd.
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EXAMPLE
| W(12,2)=11, since people are eliminated in the order 2, 4, 6, 8, 10, 12, 9, 5, 1, 7, 3, leaving 11 as the survivor.
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CROSSREFS
| Cf. A088443, A088452.
A063695(n-1) + 1.
Sequence in context: A096433 A084101 A053386 * A160324 A197928 A109439
Adjacent sequences: A090566 A090567 A090568 * A090570 A090571 A090572
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KEYWORD
| nonn
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Dec 02 2003
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