a(n)=(# 2's)-(# 3's) in first n terms of A030567.

Terms

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