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May 18, 2020 at 1:52 comment added J. M.'s missing motivation Indeed, thank you @Greg. I thought about φ^((-1)^k) only quite a while after writing this answer.
May 18, 2020 at 1:51 history edited J. M.'s missing motivation CC BY-SA 4.0
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May 17, 2020 at 19:40 comment added Greg Martin φ^(1 - 2 Boole[Mod[k, 2] == 1]) is unnecessarily complicated; φ^(1 - 2 Mod[k, 2]) works just as well, as does φ^((-1)^k).
May 16, 2020 at 10:39 vote accept Shivam K
May 16, 2020 at 8:51 history answered J. M.'s missing motivation CC BY-SA 4.0