a(n)=(# 1's)-(# 0's) in first n terms of A031235.

Terms

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