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Mar 7, 2022 at 16:59 answer added Peter Taylor timeline score: 1
Mar 7, 2022 at 16:00 comment added T. Amdeberhan Thank you for this comment.
Mar 7, 2022 at 14:43 comment added Peter Taylor It's probably worth noting that you can substitute the Stirling numbers of the second kind for $U$, since the difference between the recurrence $U(n, k) = U(n-1, k-1) + k^2 U(n-1, k)$ and the recurrence $S_2(n, k) = S_2(n-1, k-1) + k S_2(n-1, k)$ disappears in characteristic 2.
Mar 5, 2022 at 20:22 history edited T. Amdeberhan CC BY-SA 4.0
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Mar 5, 2022 at 20:10 history edited YCor CC BY-SA 4.0
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Mar 5, 2022 at 19:45 history edited T. Amdeberhan CC BY-SA 4.0
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Mar 5, 2022 at 18:28 history asked T. Amdeberhan CC BY-SA 4.0