Additional comments from Gottfried Helms, Apr 16 2008: Following Richard Mathar's construction using Stirling2 numbers, we can say the following: In a matrix-multiplication scheme we have St2 = // Stirling numbers 2'nd kind | 1 . . . . . | | 1 1 . . . . | | 1 3 1 . . . | | 1 7 6 1 . . | | 1 15 25 10 1 . | | 1 31 90 65 15 1 | X1 = // Stirling numbers 2'nd kind, row shifted | 1 . . . . . | | . 1 1 . . . | | . . 1 3 1 . | | . . . 1 7 6 | | . . . . 1 15 | | . . . . . 1 | Then simply H3 = St2 * X1 In a matrix-multiplication-scheme X1= | 1 . . . . . | | . 1 1 . . . | | . . 1 3 1 . | | . . . 1 7 6 | | . . . . 1 15 | | . . . . . 1 | St2= H3 = | 1 . . . . . | | 1 . . . . . | | 1 1 . . . . | | 1 1 1 . . . | | 1 3 1 . . . | | 1 3 4 3 1 . | | 1 7 6 1 . . | | 1 7 13 19 13 6 | | 1 15 25 10 1 . | | 1 15 40 85 96 75 | | 1 31 90 65 15 1 | | 1 31 121 335 560 616 | It is nice that X1 is a simple rowshift of St2. A second triangle of interest, H4, can be computed by a similar simple scheme: H4 = | 1 . . . . . . . . . . . . | | 1 1 1 1 . . . . . . . . . | | 1 3 4 6 4 3 1 . . . . . . | | 1 7 13 26 31 31 25 13 6 1 . . . | | 1 15 40 100 171 220 255 215 156 85 35 10 1 | Row-shifting H3 in the obvious way gives X4 X4= | 1 . . . . . . . . . . . . | | . 1 1 1 . . . . . . . . . | | . . 1 3 4 3 1 . . . . . . | | . . . 1 7 13 19 13 6 1 . . . | | . . . . 1 15 40 85 96 75 35 10 1 | and then again simply H4 = St2 * X4 H4 = | 1 . . . . . . . . . . . . | | 1 1 1 1 . . . . . . . . . | | 1 3 4 6 4 3 1 . . . . . . | | 1 7 13 26 31 31 25 13 6 1 . . . | | 1 15 40 100 171 220 255 215 156 85 35 10 1 | Gottfried Helms