0, 1, 10, 11, 12, 100, 101, 110, 111, 112, 120, 121, 122, 123, 1000, 1001, 1010, 1011, 1012, 1100, 1101, 1110, 1111, 1112, 1120, 1121, 1122, 1123, 1200, 1201, 1210, 1211, 1212, 1220, 1221, 1222, 1223, 1230, 1231, 1232, 1233, 1234, 10000, 10001, 10010, 10011
Offset: 0
Catalan's Triangle T(row,col) = A009766 begins with row n=0 and 0<=col<=n as:
Row 0: 1
Row 1: 1, 1
Row 2: 1, 2, 2
Row 3: 1, 3, 5, 5
Row 4: 1, 4, 9, 14, 14
Row 5: 1, 5, 14, 28, 42, 42
Row 6: 1, 6, 20, 48, 90, 132, 132
(the leftmost diagonal of 1s is "column 0").
...
For example, for n=38, we find that A081290(38)=14, which occurs on row A081288(n)-1 = 4, in columns A081288(n)-1 and A081288(n)-2, i.e., as T(4,4) and T(4,3). Thus we subtract 38-14 to get 24, and we see that the next term downward on the same diagonal, 28, is too large to accommodate into the same sum, so we go one diagonal up, starting now from T(3,2) = 5. This fits in, so we now have 24 - 5 = 19, and also the next term on the same diagonal, T(4,2) = 9, fits in, so we now have 19-9 = 10. The next term on the same diagonal, T(5,2) = 14, would not fit in anymore, so we rewind ourselves back to penultimate column, but one step up from where we started on this diagonal, so T(2,1) = 2, which fits in, 10 - 2 = 8, also the next one T(3,1) = 3, 8 - 3 = 5, and the next one T(4,1) = 4, 5 - 4 = 1, after which comes T(5,1) = 5 > 1, thus we jump to T(1,0) = 1, 1-1 = 0, and T(2,0)=1 would not fit anymore, thus next time the row would be zero, and the algorithm is ready with 1 (14), 2 (5+9), 3 (2+3+4) and 1 (1) terms collected, whose total sum 14+5+9+2+3+4+1 = 38, thus a(38) = 1231.
For n=20, the same algorithm results in 1 (14), 1 (5), 0 (not even the first tentative term T(2,1) = 2 from the column 1 would fit, so it is skipped), and from one row higher we get the needed 1 (1), so the total sum of these is 14+5+0+1 = 20, thus a(20) = 1101.
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