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COMMENTS
| Right border = A143805 (1, 1, 2, 7, 36, 250,...) = row sums shifted one place to the left, = (1, 2, 7, 36, 250,...). Sum of n-th row terms = rightmost term of next row.
A130534 = the Stirling cycle numbers: 1;
1, 1;
2, 3, 1;
6, 11, 6, 1;
...
First few rows of the triangle = 1;
1, 1;
2, 3, 2;
6, 11, 12, 7;
24, 50, 70, 70, 36;
120, 274, 450, 595, 540, 250;
720, 1764, 3248, 5145, 6300, 5250, 2229;
...
The triangle by rows, applies termwise products of the eigensequence terms of A130534: (1, 1, 2, 7, 36, 250,...) = A143805; to row terms of A130534. Thus row 3 = (6, 11, 12, 7) = (6, 11, 6, 1) and termwise product of the first 4 terms of A143805: (1, 1, 2, 7).
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