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Sep 24, 2021 at 12:08 comment added Peter Taylor This should follow straightforwardly from the characterisation that $a(2^k + m)$ with $1 \le m \le 2^k$ is given by $4^k + 2^k (m-1) + \operatorname{revinv}_k(m-1)$ where $\operatorname{revinv}_k$ reverses and inverts the bits of a $k$-bit number.
Sep 23, 2021 at 6:26 history edited Notamathematician CC BY-SA 4.0
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Sep 21, 2021 at 16:41 history asked Notamathematician CC BY-SA 4.0