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Search: id:A124801
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| A124801 |
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Triangle, row sums = Fibonacci numbers in two ways. |
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+0
2
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| 1, 1, 0, 1, 0, 1, 1, 0, 3, -1, 1, 0, 6, -4, 1, 1, 0, 10, -10, 10, -3, 1, 0, 15, -20, 30, -18, 5, 1, 0, 21, -35, 70, -63, 35, -8, 1, 0, 28, -56, 140, -168, 140, -64, 13
(list; table; graph; listen)
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OFFSET
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1,9
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COMMENT
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n-th row sum of signed terms = Fn; n-th row sum of unsigned terms = F(2n-3).
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FORMULA
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Diagonalize the inverse binomial transform of the Fibonacci sequence as an infinite matrix, M; and P = Pascal's triangle as an infinite lower triangular matrix. The triangle A124802 = P*M, with the zeros deleted.
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EXAMPLE
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A039834, (1, 0, 1, -1, 2, -3, 5, -8, 13...) = the diagonal of M, then the first few rows of P*M =
1;
1, 0;
1, 0, 1;
1, 0, 3, -1;
1, 0, 6, -4, 2;
1, 0, 10, -10, 10, -3;
1, 0, 15, -20, 30, -18, 5;
1, 0, 21, -35, 70, -63, 35, -8;
...
Row 7 terms (signed) = F7 = 13 = (1 + 15 -20 + 30 - 18 + 5).
Row 7 terms (unsigned) = F11 = 28 = (1 + 15 + 20 + 30 + 18 + 5).
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CROSSREFS
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Cf. A039834, A124802.
Sequence in context: A130115 A130160 A162169 this_sequence A124926 A115378 A120060
Adjacent sequences: A124798 A124799 A124800 this_sequence A124802 A124803 A124804
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KEYWORD
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tabl,sign
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AUTHOR
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Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 08 2006
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