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Search: id:A128982
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| A128982 |
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If in a line of n persons every n-th person is eliminated until only one person is left, which position P should one assume in the original lineup to avoid being eliminated? |
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+0
2
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| 1, 1, 2, 2, 4, 2, 6, 2, 6, 6, 10, 2, 12, 2, 6, 8, 16, 2, 18, 2, 16, 18, 22, 2, 22, 12, 16, 8, 28, 2, 30, 2, 28, 18, 22, 12, 36, 2, 6, 8, 40, 2, 42, 2, 30, 42, 46, 2, 42, 14, 40, 30, 52, 2, 36, 24, 52, 54, 58, 2, 60, 2, 6, 30, 48, 24, 66, 2, 30, 18, 70, 2, 72, 2, 6, 20, 60, 18, 78, 2, 72, 78
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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The difference between this, A007495 and the diagonal of A032434 is that for each of the n-1 elimination processes, counting from 1 to n starts at the lowest position in the line that is still occupied, not right after the most recently eliminated position. Wrapping around when n exceeds the number of residual occupied positions still occurs in circular fashion as in the original Josephus problem. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2007
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FORMULA
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If n is prime then P = n - 1. If n is prime + 1 then P = 2.
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EXAMPLE
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Elimination at n=6: 1,2,3,4,5,6 -> 1,2,3,4,5 -> 2,3,4,5 -> 2,4,5 -> 2,4 -> 2. After the 3 is eliminated, counting does not start at 4 but again at 2.
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MAPLE
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A128982 := proc(n) local l ; l := [seq(i, i=1..n)] ; for i from 1 to n-1 do rm := ((n-1) mod nops(l))+1 ; l := subsop(rm=NULL, l) ; od ; RETURN(op(1, l)) ; end: for n from 1 to 85 do printf("%d, ", A128982(n)) ; od ; - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2007
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CROSSREFS
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Cf. A007495, A032434.
Adjacent sequences: A128979 A128980 A128981 this_sequence A128983 A128984 A128985
Sequence in context: A002322 A127835 A117004 this_sequence A096216 A121599 A080221
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KEYWORD
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nonn
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AUTHOR
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Harri Aaltonen (harri.aaltonen(AT)mail.vak.fi), Apr 30 2007
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EXTENSIONS
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This is a version of the Josephus problem. Several other versions are already in the OEIS. - njas, May 01 2007
Corrected and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2007
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