%I A128529
%S A128529 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,
%T A128529 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,
%U A128529 54,55,48,9,19,39,60,59,58,63,7,49,50,11,40,27,70,63,48,23,27,47,74,67
%N A128529 Survivor of the Josephus problem, counting direction reversed after each 
               step.
%C A128529 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.
%e A128529 n=5 start with 1,2,3,4,5, count upwards to eliminate 5: 1,2,3,4. Count 
               backwards
%e A128529 from 4 over 3 over 2 over 1 to eliminate 4: 1,2,3. Then count forwards 
               from 1
%e A128529 (wrapping around and upwards of 4) over 2 etc. to eliminate 2: 1,3. Count 
               backwards
%e A128529 starting at 1 (left of eliminated 2) to eliminate 1 and to leave a(5)=3.
%p A128529 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;
%Y A128529 Cf. A007495, A128982.
%Y A128529 Sequence in context: A117378 A088197 A087517 this_sequence A131228 A131129 
               A087694
%Y A128529 Adjacent sequences: A128526 A128527 A128528 this_sequence A128530 A128531 
               A128532
%K A128529 nonn
%O A128529 1,5
%A A128529 R. J. Mathar (mathar(AT)strw.leidenuniv.nl), May 07 2007