a(n)=(# 1's)-(# 0's) in first n terms of A031253.

Terms

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