|
|
A165194
|
|
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.
|
|
3
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,6
|
|
COMMENTS
|
Row sums = A000110, the Bell sequence starting with offset 1; (1, 2, 5, 15,...).
|
|
LINKS
|
|
|
FORMULA
|
Given the Bell sequence, A000110: (1, 1, 2, 5, 15,...); row 1 = 1, row 2 =
(1, 1);...where left half of row (n+1) = row n. Right half of row (n+1)
= reversal of row n, replacing terms with the next Bell number.
|
|
EXAMPLE
|
First few rows of the triangle =
1;
1, 1;
1, 1, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1;
1, 1, 2, 1, 2, 5, 2, 1, 2, 5, 15, 5, 2, 5, 2, 1;
...
For example: row 4, left half = (1, 1, 2, 1); right half = (1, 2, 1, 1)
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).
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|