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A088333
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A version of Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer 3 places clockwise from i. Repeat, counting 3 places from the next undeleted integer, until only one integer remains.
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3
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1, 1, 2, 2, 1, 5, 2, 6, 1, 5, 9, 1, 5, 9, 13, 1, 5, 9, 13, 17, 21, 3, 7, 11, 15, 19, 23, 27, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| If one counts only one place (resp. two places) at each stage to determine the element to be deleted, we get A006257 (resp. A054995).
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REFERENCES
| See A054995 for references and links.
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FORMULA
| It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n-1)+4) mod n. The conjecture is false; counterexample: a(21)=21; a(20)=17; (a(20)+4)mod 21=0; corrected formula: a(n)=(a(n-1)+3) mod n +1;
The conjecture is true. After removing the 4th number, we are reduced to the n-1 case, but starting with 5 instead of 1. - David Wasserman, Aug 08 2005
a(n) = A032434(n,4) if n>=4. - R. J. Mathar, May 04 2007
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CROSSREFS
| Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.
Sequence in context: A123398 A188945 A102849 * A016538 A134226 A184050
Adjacent sequences: A088330 A088331 A088332 * A088334 A088335 A088336
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KEYWORD
| nonn,easy
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com), Nov 13 2003
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EXTENSIONS
| More terms from David Wasserman (wasserma(AT)spawar.navy.mil), Aug 08 2005
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