a(n)=(# 1's)-(# 0's) in first n terms of A030588.

Terms

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