a(n)=(# 1's)-(# 2's) in first n terms of A031269.

Terms

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