A335552 Triangle T(n,k) read by rows: in the Josephus problem with n initial numbers on a line: eliminate each second and reverse left-right-direction of elimination. T(n,k) is the (n-k+1)st element removed, 1<=k<=n.

1, 3, 1, 3, 1, 4, 1, 5, 3, 4, 1, 5, 3, 6, 4, 3, 7, 1, 5, 6, 4, 3, 7, 1, 5, 8, 6, 4, 9, 1, 5, 3, 7, 8, 6, 4, 9, 1, 5, 3, 7, 10, 8, 6, 4, 11, 3, 7, 1, 5, 9, 10, 8, 6, 4, 11, 3, 7, 1, 5, 9, 12, 10, 8, 6, 4, 9, 1, 13, 5, 3, 7, 11, 12, 10, 8, 6, 4, 9, 1, 13, 5, 3, 7, 11, 14, 12, 10, 8, 6, 4, 11, 3, 15

Offset: 1

Author

R. J. Mathar, Jun 22 2020

Keywords

• nonn
• tabl

Examples

			The triangle starts
   1
   3   1
   3   1   4
   1   5   3   4
   1   5   3   6   4
   3   7   1   5   6   4
   3   7   1   5   8   6   4
   9   1   5   3   7   8   6   4
   9   1   5   3   7  10   8   6   4
  11   3   7   1   5   9  10   8   6   4
  11   3   7   1   5   9  12  10   8   6   4
   9   1  13   5   3   7  11  12  10   8   6   4
   9   1  13   5   3   7  11  14  12  10   8   6   4
  11   3  15   7   1   5   9  13  14  12  10   8   6   4
  11   3  15   7   1   5   9  13  16  14  12  10   8   6   4
   1  17   9  13   5   3   7  11  15  16  14  12  10   8   6   4
   1  17   9  13   5   3   7  11  15  18  16  14  12  10   8   6   4
   3  19  11  15   7   1   5   9  13  17  18  16  14  12  10   8   6   4
   3  19  11  15   7   1   5   9  13  17  20  18  16  14  12  10   8   6   4
		

Links

• Georg Fischer, Table of n, a(n) for n = 1..1000
• K. Matsumoto, T. Nakamigawa, M. Watanabe, On the switchback vertion of Josephus Problem, Yokohama Math. J. 53 (2007) 83, function f_k(n).
• Index to sequences related to the Josephus problem

Crossrefs

Cf. A090569 (column k=1).

Programs

Maple

sigr := proc(n,r)
    floor(n/2^r) ;
end proc:
# A063695
f := proc(n)
    local ndigs,fn,k ;
    ndigs := convert(n,base,2) ;
    fn := 0 ;
    for k from 2 to nops(ndigs) by 2 do
        fn := fn+op(k,ndigs)*2^(k-1)
    end do;
    fn ;
end proc:
g := proc(t,n)
    local r;
    if t =1 then
        0 ;
    elif t > 1 then
        r := ilog2( (n-1)/(t-1) ) ;
        (-2)^r*(f( sigr(2*n-1,r) )+f( sigr(n-1,r) )-2*t+3) ;
    end if;
end proc:
ft := proc(t,n)
    f(n-1)+1+g(t,n) ;
end proc:
for n from 1 to 20 do
    for t from 1 to n-1 do
        printf("%3d ", ft(t,n)) ;
    end do:
    printf("\n") ;
end do: