a(n)=(# 1's)-(# 2'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) =4a(10) =5a(11) =4a(12) =4a(13) =3a(14) =3a(15) =3a(16) =4a(17) =4a(18) =4a(19) =4a(20) =4a(21) =4a(22) =5a(23) =5a(24) =5a(25) =5a(26) =5a(27) =5a(28) =6a(29) =6