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Mar 16, 2017 at 15:42 vote accept T. Amdeberhan
Mar 16, 2017 at 12:18 history edited T. Amdeberhan CC BY-SA 3.0
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Mar 16, 2017 at 9:03 comment added Wolfgang Interesting to note that the determinant $\det B_n=\prod(b_1\pm\cdots\pm b_n)$ is symmetric in $b_2,...,b_n$, which is not straightforward from the recursions.
Mar 16, 2017 at 8:13 history edited Wolfgang CC BY-SA 3.0
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Mar 16, 2017 at 7:58 history answered Wolfgang CC BY-SA 3.0