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A345226
Irregular triangle, row sums equal A000041; in the format of A233932.
1
1, 1, 1, 2, 1, 2, 1, 2, 4, 1, 2, 4, 5, 2, 8, 5, 2, 8, 5, 2, 7, 16, 5, 2, 7, 16, 17, 2, 7, 30, 17, 2, 7, 30, 17, 23, 7, 54, 17, 23, 7, 54, 51, 23, 7, 95, 51, 23, 7, 95, 51, 23, 7, 55, 161, 51, 23, 7, 55, 161, 139, 23, 7, 55, 266, 139, 23, 7, 55, 266, 139, 160, 7, 55, 431, 139, 160, 7, 55
OFFSET
1,4
COMMENTS
The format of A233932 has a Gray code property of one term change in the next row. Using the production matrix shown below, we can obtain an array with row sums of any target sequence.
FORMULA
Let P equal the infinite lower triangular matrix with 1's in every row: [(1), (1, 1), (1, 1, 1), ...]. Begin with the following matrix format such that M[n, A001511(k)] = 1, otherwise 0:
1
0, 1
1
0, 0, 1
1
0, 1
1
0, 0, 0, 1
...
Replace the 1's with A187219 (the first difference row of A000041), getting M:
1
0, 1
1
0, 0, 2
2
0, 4
4
0, 0, 0, 7
...
Perform the operation P * M, getting A345226 as an irregular matrix. The operation P * M is equivalent to taking partial sums of column terms from top to bottom.
EXAMPLE
The first few rows of the array equal A000041 with offset 1 as to sums:
1;
1, 1;
2, 1;
2, 1, 2;
4, 1, 2;
4, 5, 2;
8, 5, 2;
8, 5, 2, 7;
16, 5, 2, 7;
16, 17, 2, 7;
30, 17, 2, 7;
30, 17, 23, 7;
54, 17, 23, 7;
54, 51, 23, 7;
95, 51, 23, 7;
95, 51, 23, 7, 55;
161, 51, 23, 7, 55;
161, 139, 23, 7, 55;
...
The leftmost column is (1, 1, 2, 2, 4, 4, 8, 8, ...), being the partial sums of the first column in matrix M: (1, 0, 1, 0, 2, 0, 4, ...).
KEYWORD
nonn,tabf
AUTHOR
Gary W. Adamson, Jun 12 2021
STATUS
approved