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

Author

R. J. Mathar, May 07 2007

Keywords

• nonn

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.

Examples

			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.
		

Links

• Index entries for sequences related to the Josephus Problem

Crossrefs

Cf. A007495, A128982.

Programs

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;