%I #4 Apr 08 2022 15:24:37

%S 1,1,1,1,1,2,1,1,1,2,1,2,5,2,1,1,1,2,1,2,5,2,1,1,2,5,15,5,2,5,2,1

%N Triangle of 2^n terms by rows, left half of (n+1)-th row = row n; right half = "reverse and increment" row n; using terms in A000110.

%C Row sums = A000110, the Bell sequence starting with offset 1; (1, 2, 5, 15,...).

%C Rows tend to A165195.

%F Given the Bell sequence, A000110: (1, 1, 2, 5, 15,...); row 1 = 1, row 2 =

%F (1, 1);...where left half of row (n+1) = row n. Right half of row (n+1)

%F = reversal of row n, replacing terms with the next Bell number.

%e First few rows of the triangle =

%e 1;

%e 1, 1;

%e 1, 1, 2, 1;

%e 1, 1, 2, 1, 2, 5, 2, 1;

%e 1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1;

%e ...

%e For example: row 4, left half = (1, 1, 2, 1); right half = (1, 2, 1, 1)

%e replaced with the next higher Bell numbers: (2, 5, 2, 1). Appending the two \kQ halves, we obtain row 4: (1, 1, 2, 1, 2, 5, 2, 1), sum = 15 = A000110(4).

%Y A000110, A165195, A165196

%K nonn,tabf

%O 1,6

%A _Gary W. Adamson_, Sep 06 2009