Josephus problem: a(2*n) = 2*a(n)-1, a(2*n+1) = 2*a(n)+1.

Terms

a(0) =0a(1) =1a(2) =1a(3) =3a(4) =1a(5) =3a(6) =5a(7) =7a(8) =1a(9) =3a(10) =5a(11) =7a(12) =9a(13) =11a(14) =13a(15) =15a(16) =1a(17) =3a(18) =5a(19) =7a(20) =9a(21) =11a(22) =13a(23) =15a(24) =17a(25) =19a(26) =21a(27) =23a(28) =25a(29) =27