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Search: id:A124730
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| A124730 |
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Triangle, row sums = powers of 3. |
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3
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| 1, 1, 2, 1, 6, 2, 1, 14, 8, 4, 1, 30, 22, 24, 4, 1, 62, 52, 92, 28, 8, 1, 126, 114, 288, 120, 72, 8, 1, 254, 240, 804, 408, 384, 80, 16, 1, 510, 494, 2088, 1212, 1584, 46, 192, 16
(list; table; graph; listen)
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OFFSET
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0,3
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COMMENT
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In A124731, we switch the diagonals. In both triangles, row sums = powers of 3.
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FORMULA
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Let M = the infinite bidiagonal matrix with (1,2,1,2...) in the main diagonal and (2,1,2,1...) in the subdiagonal. The n-th row of the triangle (extracting the zeros) = M^n * [1,0,0,0...].
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EXAMPLE
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Row 2 = (1, 6, 2) since [1,0,0; 2,2,0; 0,1,1]^2 * [1,0,0] = [1,6,2].
First few rows of the triangle are:
1;
1, 2;
1, 6, 2;
1, 14, 8, 4;
1, 30, 22, 24, 4;
1, 62, 52, 92, 28, 8;
1, 126, 114, 288, 120, 72, 8;
...
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CROSSREFS
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Cf. A124731, A124732.
Sequence in context: A133200 A103881 A101024 this_sequence A114283 A106187 A110135
Adjacent sequences: A124727 A124728 A124729 this_sequence A124731 A124732 A124733
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KEYWORD
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nonn,tabl
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AUTHOR
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Gary W. Adamson & Roger L. Bagula (qntmpkt(AT)yahoo.com), Nov 05 2006
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