A128529 Survivor of the Josephus problem, counting direction reversed after each step.

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

Offset

1,5

Comments

As in A007495, counting for elimination starts clockwise for the first elimination, then continues counterclockwise 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.

Links

Table of n, a(n) for n=1..79.

Index entries for sequences related to the Josephus Problem

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.

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;

Crossrefs

Cf. A007495, A128982.

Sequence in context: A088197 A350605 A087517 * A205547 A169782 A131228

Adjacent sequences: A128526 A128527 A128528 * A128530 A128531 A128532

Keyword

nonn

Author

R. J. Mathar, May 07 2007

Status

approved