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A128529
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Survivor of the Josephus problem, counting direction reversed after each step.
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0
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1, 1, 1, 1, 3, 4, 1, 3, 5, 1, 9, 8, 3, 3, 11, 1, 15, 7, 7, 18, 19, 16, 3, 7, 15, 24, 25, 18, 9, 28, 19, 24, 7, 13, 21, 5, 31, 20, 11, 15, 21, 32, 3, 11, 31, 7, 39, 23, 25, 15, 35, 1, 47, 32, 15, 54, 55, 48, 9, 19, 39, 60, 59, 58, 63, 7, 49, 50, 11, 40, 27, 70, 63, 48, 23, 27, 47, 74, 67
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,5
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COMMENTS
| As in A007495, counting for elimination starts clockwise for the first elimination, then continues counter-clockwise from the eliminated place for the second, then toggles again to clockwise for the third elimination and changes direction in that manner after each elimination. Sequence shows original place of the survivor.
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EXAMPLE
| n=5 start with 1,2,3,4,5, count upwards to eliminate 5: 1,2,3,4. Count backwards
from 4 over 3 over 2 over 1 to eliminate 4: 1,2,3. Then count forwards from 1
(wrapping around and upwards of 4) over 2 etc. to eliminate 2: 1,3. Count backwards
starting at 1 (left of eliminated 2) to eliminate 1 and to leave a(5)=3.
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MAPLE
| a := proc(n) local l, dir, pos, i, c ; dir := 1 ; pos := 0 ; l := [seq(i, i=1..n)] ; for i from 1 to n-1 do pos := pos+n*dir ; pos := 1+((pos-1) mod nops(l)) ; l := subsop(pos=NULL, l) ; dir := -dir ; if dir > 0 then pos := pos-1 ; fi ; od ; RETURN(op(1, l)) ; end: for n from 1 to 85 do printf("%d, ", a(n)) ; od;
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CROSSREFS
| Cf. A007495, A128982.
Sequence in context: A117378 A088197 A087517 * A205547 A169782 A131228
Adjacent sequences: A128526 A128527 A128528 * A128530 A128531 A128532
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KEYWORD
| nonn
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AUTHOR
| R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2007
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