a(n)=(# 1's)-(# 0's) in first n terms of A030353.

Terms

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