a(n)=(# 1's)-(# 0's) in first n terms of A031269.

Terms

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